Transport Phenomena - Fluid Mechanics Theory Differential Shell Momentum Balance in Rectangular Cartesian Coordinates A shell momentum balance is used below to derive a general differential equation that can be then employed to solve several fluid flow problems in rectangular Cartesian coordinates. For this purpose, consider an incompressible fluid in laminar flow under the effects of both pressure and gravity in a system of length L and width W, which is at an angle β to the vertical. End effects are neglected assuming the dimension of the system in the x-direction is relatively very small compared to those in the y-direction (W) and the zdirection (L).
Figure. Differential rectangular slab (shell) of fluid of thickness Δx used in z-momentum balance for flow in rectangular Cartesian coordinates. The y-axis is pointing outward from the plane of the computer screen. Since the fluid flow is in the z-direction, vx = 0, vy = 0, and only vz exists. For small flow rates, the viscous forces prevent continual acceleration of the fluid. So, vz is independent of z and it is meaningful to postulate that velocity vz = vz(x) and pressure p = p(z). The only nonvanishing components of the stress tensor are τxz = τzx, which depend only on x. Consider now a thin rectangular slab (shell) perpendicular to the x-direction extending a distance W in the y-direction and a distance L in the z-direction. A 'rate of z-momentum' balance over this thin shell of thickness Δx in the fluid is of the form: Rate of z-momentum
In − Out + Generation = Accumulation
At steady-state, the accumulation term is zero. Momentum can go 'in' and 'out' of the shell
by both the convective and molecular mechanisms. Since vz(x) is the same at both ends of the system, the convective terms cancel out because (ρ vz vz W Δx)|z = 0 = (ρ vz vz W Δx)|z = L. Only the molecular term (L W τxz ) remains to be considered, whose 'in' and 'out' directions are taken in the positive direction of the x-axis. Generation of z-momentum occurs by the pressure force acting on the surface [p W Δx] and gravity force acting on the volume [(ρ g cos β) L W Δx]. The different contributions may be listed as follows: •
rate of z-momentum in by viscous transfer across surface at x is (L W τxz )| x
•
rate of z-momentum out by viscous transfer across surface at x + Δx is (L W τxz )| x + Δx
•
rate of z-momentum in by overall bulk fluid motion across surface at z = 0 is (ρ vz vz W Δx )| z = 0
•
rate of z-momentum out by overall bulk fluid motion across surface at z = L is (ρ vz vz W Δx )| z = L
•
pressure force acting on surface at z = 0 is p0 W Δx
•
pressure force acting on surface at z = L is − pL W Δx
•
gravity force acting in z-direction on volume of rectangular slab is (ρ g cos β) L W Δx
On substituting these contributions into the z-momentum balance, we get (L W τxz ) | x − (L W τxz ) | x+Δx+ ( p 0 − p L ) W Δx + (ρ g cos β) L W Δx = 0
(1)
Dividing the equation by L W Δx yields τxz | x+Δx − τxz | x p − p L + ρ g L cos β = 0 Δx L
(2)
On taking the limit as Δx → 0, the left-hand side of the above equation is exactly the definition of the derivative. The right-hand side may be written in a compact and convenient way by introducing the modified pressure P, which is the sum of the pressure and gravitational terms. The general definition of the modified pressure is P = p + ρ g h , where h is the distance upward (in the direction opposed to gravity) from a reference plane of choice. The advantages of using the modified pressure P are that (i) the components of the gravity vector g need not be calculated; (ii) the solution holds for any flow orientation; and (iii) the fluid may flow as a result of a pressure difference, gravity or both. Here, h is negative since the z-axis points downward, giving h = − z cos β and therefore P = p − ρ g z cos β. Thus, P0 = p0 at z = 0 and PL = pL − ρ g L cos β at z = L giving p0 − pL + ρ g L cos β = P0 − PL ≡ ΔP. Thus, equation (2) yields dτxz = ΔP dx L
(3)
The first-order differential equation may be simply integrated to give Δ τxz = P x + C1 L
(4)
Here, C1 is an integration constant, which is determined using an appropriate boundary condition based on the flow problem. Equation (4) shows that the momentum flux (or shear stress) distribution is linear in systems in rectangular Cartesian coordinates. Since equations (3) and (4) have been derived without making any assumption about the type of fluid, they are applicable to both Newtonian and non-Newtonian fluids. Some of the axial flow problems in rectangular Cartesian coordinates where these equations may be used as starting points are given below
Transport Phenomena - Fluid Mechanics Theory Differential Shell Momentum Balance in Cylindrical Coordinates A shell momentum balance is used below to derive a general differential equation that can be then employed to solve several fluid flow problems in cylindrical coordinates. For this purpose, consider an incompressible fluid in laminar flow under the effects of both pressure and gravity in a system of length L, which is at an angle β to the vertical. End effects are neglected assuming the dimension of the system in the radial direction is relatively very small compared to that in the axial direction (L).
Figure. Differential cylindrical shell of fluid of thickness Δr used in z-momentum balance for axial flow in cylindrical coordinates. Since the fluid flow is in the axial direction, vr = 0, vθ = 0, and only vz exists. For small flow rates, the viscous forces prevent continual acceleration of the fluid. So, vz is independent of z and it is meaningful to postulate that velocity vz = vz(r) and pressure p = p(z). The only nonvanishing components of the stress tensor are τrz = τzr, which depend only on r. Consider now a thin cylindrical shell perpendicular to the radial direction extending a distance L in the z-direction. A 'rate of z-momentum' balance over this thin shell of thickness Δr in the fluid is of the form: Rate of z-momentum
In − Out + Generation = Accumulation
At steady-state, the accumulation term is zero. Momentum can go 'in' and 'out' of the shell by both the convective and molecular mechanisms. Since vz(r) is the same at both ends of the system, the convective terms cancel out because (ρ vz vz 2πr Δr)|z = 0 = (ρ vz vz 2πr Δr)|z = L. Only the molecular term (2πr L τrz ) remains to be considered, whose 'in' and 'out' directions are taken in the positive direction of the r-axis. Generation of z-momentum occurs by the pressure force acting on the surface [p 2πr Δr] and gravity force acting on the volume [(ρ g cos β) 2πr Δr L]. The different contributions may be listed as follows: •
rate of z-momentum in by viscous transfer across cylindrical surface at r is (2πr L τrz )|r
•
rate of z-momentum out by viscous transfer across cylindrical surface at r + Δr is (2πr L τrz )|r + Δr
•
rate of z-momentum in by overall bulk fluid motion across annular surface at z = 0
is (ρ vz vz 2πr Δr)|z = 0 •
rate of z-momentum out by overall bulk fluid motion across annular surface at z = L is (ρ vz vz 2πr Δr)|z = L
•
pressure force acting on annular surface at z = 0 is p0 2πr Δr
•
pressure force acting on annular surface at z = L is − pL 2πr Δr
•
gravity force acting in z-direction on volume of cylindrical shell is (ρ g cos β) 2πr Δr L
On substituting these contributions into the z-momentum balance, we get (2πr L τrz ) | r − (2πr L τrz ) | r + Δr+ ( p 0 − p L ) 2πr Δr + (ρ g cos β) 2πr Δr L = 0
(1)
Dividing the equation by 2π L Δr yields p −p + 0 L (r τrz ) | r + Δr − (r τrz ) | r = ρ g L cos β Δr L
r
(2)
On taking the limit as Δr → 0, the left-hand side of the above equation is the definition of the first derivative. The right-hand side may be written in a compact and convenient way by introducing the modified pressure P, which is the sum of the pressure and gravitational terms. The general definition of the modified pressure is P = p + ρ g h , where h is the height (in the direction opposed to gravity) above some arbitrary preselected datum plane. The advantages of using the modified pressure P are that (i) the components of the gravity vector g need not be calculated in cylindrical coordinates; (ii) the solution holds for any flow orientation; and (iii) the fluid may flow as a result of a pressure difference, gravity or both. Here, h is negative since the z-axis points downward, giving h = − z cos β and therefore P = p − ρ g z cos β. Thus, P0 = p0 at z = 0 and PL = pL − ρ g L cos β at z = L giving p0 − pL + ρ g L cos β = P0 − PL ≡ ΔP. Thus, equation (2) yields Δ d (r τrz) = P r dr L
(3)
The above differential equation may be simply integrated to give the general expression for the momentum flux (or shear stress) distribution in cylindrical coordinate systems as Δ P r C1 τrz = 2+ r L
(4)
For Newtonian fluids, Newton's law of viscosity may be substituted for τrz in equation (4) to obtain Δ r P −μ z = + 2 dr L dv
C1 r
The above differential equation is simply integrated to obtain the following general expression for the velocity profile for Newtonian fluids: ΔP 2 C r vz = − 4 μ ln r + C2 − 1 L μ
(5)
(6)
Here, C1 and C2 are integration constants, which are determined using appropriate boundary conditions based on the flow problem. Since equations (3) and (4) have been derived without making any assumption about the type of fluid, they are applicable to both Newtonian and non-Newtonian fluids. On the other hand, equation (6) is applicable to only Newtonian fluids. Some of the axial flow problems in cylindrical coordinates where these equations may be used as starting points are given below.
Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid flow in a plane narrow slit
Problem. Consider a fluid (of density ρ) in incompressible, laminar flow in a plane narrow slit of length L and width W formed by two flat parallel walls that are a distance 2B apart. End effects may be neglected because B << W << L. The fluid flows under the influence of both a pressure difference Δp and gravity.
Figure. Fluid flow in plane narrow slit. a) Using a differential shell momentum balance, determine expressions for the steady-state shear stress distribution and the velocity profile for a Newtonian fluid (of viscosity μ). b) Obtain expressions for the maximum velocity, average velocity and the mass flow rate for slit flow.
Solution. a) Differential equation from shell momentum balance For a plane narrow slit, the natural choice is rectangular Cartesian coordinates. Since the fluid flow is in the z-direction, vx = 0, vy = 0, and only vz exists. Further, vz is independent of z and it is meaningful to postulate that velocity vz = vz(x) and pressure p = p(z). The only nonvanishing components of the stress tensor are τxz = τzx, which depend only on x. Consider now a thin rectangular slab (shell) perpendicular to the x-direction extending a distance W in the y-direction and a distance L in the z-direction. A 'rate of z-momentum' balance over this thin shell of thickness Δx in the fluid is of the form: Rate of z-momentum
In − Out + Generation = Accumulation
At steady-state, the accumulation term is zero. Momentum can go 'in' and 'out' of the shell by both the convective and molecular mechanisms. Since vz(x) is the same at both ends of the slit, the convective terms cancel out because (ρ vz vz W Δx)|z = 0 = (ρ vz vz W Δx)|z = L. Only the molecular term (L W τxz ) remains to be considered. Generation of z-momentum occurs by the
pressure force (p W Δx) and gravity force (ρ g W L Δx). On substituting these contributions into the z-momentum balance, we get (L W τxz ) | x − (L W τxz ) | x+Δx+ ( p 0 − p L ) W Δx + ρ g W L Δx = 0
(1)
Dividing the above equation by L W Δx yields τxz| x+Δx − τxz| x
=
Δx
p0 − pL + ρ g L L
(2)
On taking the limit as Δx → 0, the left-hand side of the above equation is the definition of the derivative. The right-hand side may be compactly and conveniently written by introducing the modified pressure P, which is the sum of the pressure and gravitational terms. The general definition of the modified pressure is P = p + ρ g h , where h is the distance upward (in the direction opposed to gravity) from a reference plane of choice. Since the z-axis points downward in this problem, h = − z and therefore P = p − ρ g z . Thus, P0 = p0 at z = 0 and PL = pL − ρ g L at z = L giving p0 − pL + ρ g L = P0 − PL ≡ ΔP. Thus, equation (2) gives dτxz = ΔP dx
L
(3)
Equation (3) on integration leads to the following expression for the shear stress distribution: Δ τxz = P x + C1
(4)
L
The constant of integration C1 is determined later using boundary conditions. It is worth noting that equations (3) and (4) apply to both Newtonian and non-Newtonian fluids, and provide starting points for many fluid flow problems in rectangular Cartesian coordinates Shear stress distribution and velocity profile Substituting Newton's law of viscosity for τxz in equation (4) gives dv −μ
z
dx
Δ = P x + C1
(5)
L
The above first-order differential equation is simply integrated to obtain the following velocity profile:
ΔP
C x2 1 vz = − 2 μ − L μ
x + C2
The no-slip boundary conditions at the two fixed walls are BC 1: at x = B, vz = 0 BC 2: at x = −B,
vz = 0
(6)
(7) (8)
Using these, the integration constants may be evaluated as C1 = 0 and C2 = ΔP B2 / (2 μ L). On substituting C1 = 0 in equation (4), the final expression for the shear stress (or momentum flux) distribution is found to be linear as given by Δ P τxz = x (9) L Further, substitution of the integration constants into equation (6) gives the final expression for the velocity profile as 2
ΔP x 2 1 B vz = − 2μL B
(10)
It is observed that the velocity distribution for laminar, incompressible flow of a Newtonian fluid in a plane narrow slit is parabolic. b) Maximum velocity, average velocity and mass flow rate From the velocity profile, various useful quantities may be derived. (i) The maximum velocity occurs at x = 0 (where dvz/dx = 0). Therefore, vz,max = vz| x = 0 =
ΔP B2 2μL
(11)
(ii) The average velocity is obtained by dividing the volumetric flow rate by the cross-sectional area as shown below. (12) vz,avg = 1B Δ 2 vz,max B − B∫ vz W = vz = P = dx 2 ∫ d B2
∫
−B
B
W dx
− BB x
3 μ 3 L
Thus, the ratio of the average velocity to the maximum velocity for Newtonian fluid flow in a narrow slit is 2/3. (iii) The mass rate of flow is obtained by integrating the velocity profile over the cross section of the slit as follows. B ∫ w = ρ vz W dx = ρ W (2 B) vz,avg − B
(13)
Thus, the mass flow rate is the product of the density ρ, the cross-sectional area (2 B W) and the average velocity vz,avg. On substituting vz,avg from equation (12), the final expression for the mass rate of flow is w =
2 ΔP B3 W ρ 3μL
(14)
The flow rate vs. pressure drop (w vs. ΔP) expression above is the slit analog of the HagenPoiseuille equation (originally for circular tubes). It is a result worth noting because it provides the starting point for creeping flow in many systems (e.g., radial flow between two parallel circular disks; and flow between two stationary concentric spheres). Finally, it may be noted that the above analysis is valid when B << W. If the slit thickness B is of the same order of magnitude as the slit width W, then vz = vz (x, y), i.e., vz is not a function of only x. If W = 2B, then a solution can be obtained for flow in a square duct.
Transport Phenomena - Fluid mechanics Problem : Newtonian fluid flow in a falling film Problem. Consider a Newtonian liquid (of viscosity μ and density ρ) in laminar flow down an inclined flat plate of length L and width W. The liquid flows as a falling film with negligible rippling under the influence of gravity. End effects may be neglected because L and W are large compared to the film thickness δ.
Figure. Newtonian liquid flow in a falling film. a) Determine the steady-state velocity distribution. b) Obtain the mass rate of flow and average velocity in the falling film. c) What is the force exerted by the liquid on the plate in the flow direction? d) Derive the velocity distribution and average velocity for the case where x is replaced by a coordinate x' measured away from the plate (i.e., x' = 0 at the plate and x' = δ at the liquid-gas interface).
Solution. a) Shear stress distribution For axial flow in rectangular Cartesian coordinates, the differential equation for the momentum flux is dτxz = ΔP dx
L
(1)
where P is a modified pressure, which is the sum of both the pressure and gravity terms, that is, ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the z-axis with the vertical. Since the flow is solely under the influence of gravity, Δp = 0 and therefore ΔP/L = ρ g cos β. On integration, τxz = ρ g x cos β + C1
(2)
On using the boundary condition at the liquid-gas interface (τxz = 0 at x = 0), the constant of integration C1 is found to be zero. This gives the expression for the shear stress τxz for laminar flow in a falling film as τxz = ρ g x cos β (3)
It must be noted that the above momentum flux expression holds for both Newtonian and nonNewtonian fluids (and does not depend on the type of fluid).
The shear stress for a Newtonian fluid (as per Newton's law of viscosity) is given by τxz = − μ
dvz
(4)
dx
Velocity distribution To obtain the velocity distribution, there are two possible approaches. In the first approach, equations (3) and (4) are combined to eliminate τxz and get a first-order differential equation for the velocity as given below. dvz dx
ρg = − cos β
x
(5)
μ
On integrating, vz = − ρ g x2 cos β/(2μ) + C2. On using the no-slip boundary condition at the solid surface (vz = 0 at x = δ ), C2 = ρ g δ2 cos β/(2μ). Thus, the final expression for the velocity distribution is parabolic as given below. 2
vz =
ρ g δ2 x cos β 1 − 2μ δ
(6)
In the second approach, equations (1) and (4) are directly combined to eliminate τxz and get a second-order differential equation for the velocity as given below. d2 − μ vz = ρ g cos β (7) dx2 Integration gives dvz dx
ρg cos β =−
x + C1
(8)
μ
On integrating again, ρ g x2 vz = − cos β 2μ
+ C1 x + C2
(9)
On using the boundary condition at the liquid-gas interface (at x = 0, τxz = 0 ⇒ dvz/dx = 0), the constant of integration C1 is found to be zero from equation (8). On using the other boundary condition at the solid surface (at x = δ, vz = 0), equation (9) gives C2 = ρ g δ2 cos β/(2μ). On substituting C1 and C2 in equation (9), the parabolic velocity profile in equation (6) is again obtained. b) Mass flow rate The mass flow rate may be obtained by integrating the velocity profile given by equation (6) over the film thickness as shown below. 1 δ w =
∫ 0
2
ρ vz W dx
ρ gWδ = cos β 2μ
w =
3
2
x x 1 ∫ d − δ δ 0
ρ2 g W δ3 cos β 3μ
(10)
(11)
When the mass flow rate w in equation (11) is divided by the density ρ and the film crosssectional area (W δ), an expression for the average velocity is obtained. Also, the maximum velocity vz, max occurs at the interface x = 0 and is readily obtained from equation (6). Thus, equations (6) and (11) give =
ρ g δ2 2 cos β = 3μ
vz, max
(12)
3
c) Force on the plate The z-component of the force of the fluid on the solid surface is given by the shear stress integrated over the wetted surface area. Using equation (3) gives Fz = L W τxz|x = δ = ρ g δ L W cos β
(13)
As expected, the force exerted by the fluid on the plate is simply the z-component of the weight of the liquid film. d) Choice of coordinates Consider the coordinate x' starting from the plate (i.e., x' = 0 is the plate surface and x' = δ is the liquid-gas interface). The form of the differential equation for the momentum flux given by equation (1) will remain unchanged. Thus, equation (2) for the new choice of coordinates will be
τx'z = ρ g x' cos β + C1
(14)
On using the boundary condition at the liquid-gas interface (τx'z = 0 at x' = δ), the constant of integration is now found to be given by C1 = − ρ g δ cos β. This gives the shear stress as τx'z = ρ g (x' − δ) cos β (15) Note that the momentum flux is in the negative x'-direction. On substituting Newton's law of viscosity, ρg dvz = cos β (δ − x') (16) dx' μ On integrating, vz = (ρg/μ) (δ x' − ½ x'2) cos β + C2. On using the no-slip boundary condition at the solid surface (vz = 0 at x' = 0), it is found that C2 = 0. Thus, the final expression for the velocity distribution is 2
vz =
ρ g δ2 x 1 x cos β '− ' μ δ 2δ
(17)
The average velocity may be obtained by integrating the velocity profile given in equation (17) over the film thickness as shown below. 1 2 δ ρ g δ2 x 1 x x 1 = cos β ' ' w = ∫ vz (18) ∫ − d δ dx' μ δ 2δ δ 0 0 Integration of equation (18) yields the same expression for obtained earlier in equation (12). It must be noted that the two coordinates x and x' are related as follows: x + x' = δ. By substituting (x/δ)2 = [1 − (x'/δ)]2 = 1 − 2(x'/δ) + (x'/δ)2, equation (17) is directly obtained from equation (6).
Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid flow in a circular tube
Problem. Consider a fluid (of constant density ρ) in incompressible, laminar flow in a tube of circular cross section, inclined at an angle β to the vertical. End effects may be neglected because the tube length L is relatively very large compared to the tube radius R. The fluid flows under the influence of both a pressure difference Δp and gravity.
Figure. Fluid flow in circular tube. a) Using a differential shell momentum balance, determine expressions for the steady-state shear stress distribution and the velocity profile for a Newtonian fluid (of constant viscosity μ). b) Obtain expressions for the maximum velocity, average velocity and the mass flow rate for pipe flow. c) Find the force exerted by the fluid along the pipe wall.
Solution. a) Flow in pipes occurs in a large variety of situations in the real world and is studied in various engineering disciplines as well as in physics, chemistry, and biology. Differential equation from shell momentum balance For a circular tube, the natural choice is cylindrical coordinates. Since the fluid flow is in the zdirection, vr = 0, vθ = 0, and only vz exists. Further, vz is independent of z and it is meaningful to postulate that velocity vz = vz(r) and pressure p = p(z). The only nonvanishing components of the stress tensor are τrz = τzr, which depend only on r. Consider now a thin cylindrical shell perpendicular to the radial direction and of length L. A 'rate of z-momentum' balance over this thin shell of thickness Δr in the fluid is of the form: Rate of z-momentum
In − Out + Generation = Accumulation
At steady-state, the accumulation term is zero. Momentum can go 'in' and 'out' of the shell by both the convective and molecular mechanisms. Since vz(r) is the same at both ends of the tube, the convective terms cancel out because (ρ vz vz 2πr Δr)|z = 0 = (ρ vz vz 2πr Δr)|z = L. Only the molecular term (2πr L τrz ) remains to be considered, whose 'in' and 'out' directions are taken in the positive direction of the r-axis. Generation of z-momentum occurs by the pressure force
acting on the surface [p 2πr Δr] and gravity force acting on the volume [(ρ g cos β) 2πr Δr L]. On substituting these contributions into the z-momentum balance, we get (2πr L τrz ) | r − (2πr L τrz ) | r + Δr+ ( p 0 − p L ) 2πr Δr + (ρ g cos β) 2πr Δr L = 0
(1)
Dividing the above equation by 2π L Δr yields p0 − pL + = ρ g L cos β L
(r τrz ) | r + Δr − (r τrz ) | r Δr
r
(2)
On taking the limit as Δr → 0, the left-hand side of the above equation is the definition of the first derivative. The right-hand side may be written in a compact and convenient way by introducing the modified pressure P, which is the sum of the pressure and gravitational terms. The general definition of the modified pressure is P = p + ρ g h , where h is the height (in the direction opposed to gravity) above some arbitrary preselected datum plane. The advantages of using the modified pressure P are that (i) the components of the gravity vector g need not be calculated in cylindrical coordinates; (ii) the solution holds for any orientation of the tube axis; and (iii) the effects of both pressure and gravity are in general considered. Here, h is negative since the z-axis points downward, giving h = − z cos β and therefore P = p − ρ g z cos β. Thus, P0 = p0 at z = 0 and PL = pL − ρ g L cos β at z = L giving p0 − pL + ρ g L cos β = P0 − PL ≡ ΔP. Thus, equation (2) gives Δ d (r τ ) = P rz r (3) dr L
Equation (3) on integration leads to the following expression for the shear stress distribution: Δ P r C1 τrz = 2+ r L
(4)
The constant of integration C1 is determined later using boundary conditions. It is worth noting that equations (3) and (4) apply to both Newtonian and non-Newtonian fluids, and provide starting points for many fluid flow problems in cylindrical coordinates Shear stress distribution and velocity profile Substituting Newton's law of viscosity for τrz in equation (4) gives − μ dv z
Δr = P+
C1
(5)
dr
2 L
r
The above differential equation is simply integrated to obtain the following velocity profile: ΔP C r2 1 vz = − 4 μ ln r + C2 − L μ The integration constants C1 and C2 are evaluated from the following boundary conditions: BC 1: at r = 0, τrz and vz are finite BC 2: at r = R,
vz = 0
(6)
(7) (8)
From BC 1 (which states that the momentum flux and velocity at the tube axis cannot be infinite), C1 = 0. From BC 2 (which is the no-slip condition at the fixed tube wall), C2 = ΔP R2 / (4 μ L). On substituting C1 = 0 in equation (4), the final expression for the shear stress (or momentum flux) distribution is found to be linear as given by Δ P τrz =
r
(9)
2 L Further, substitution of the integration constants into equation (6) gives the final expression for the velocity profile as 2
ΔP r 2 1 vz = R − 4μL R
(10)
It is observed that the velocity distribution for laminar, incompressible flow of a Newtonian fluid in a long circular tube is parabolic. b) Maximum velocity, average velocity and mass flow rate From the velocity profile, various useful quantities may be derived. (i) The maximum velocity occurs at r = 0 (where dvz/dr = 0). Therefore,
vz,max = vz| r = 0 =
ΔP R2 4μL
(11)
(ii) The average velocity is obtained by dividing the volumetric flow rate by the cross-sectional area as shown below. Δ R P R vz 2 π r 0∫ 2 2 1 ∫ vz R dr vz,avg = =R = = vz,max (12) r 8 2 0 dr μ 2 R 2 π r dr 0∫ L Thus, the ratio of the average velocity to the maximum velocity for Newtonian fluid flow in a circular tube is ½. (iii) The mass rate of flow is obtained by integrating the velocity profile over the cross section of the circular tube as follows. R w =
∫ 0
ρ vz 2 π r dr = ρ π R2 vz,avg
(13)
Thus, the mass flow rate is the product of the density ρ, the cross-sectional area (π R2) and the average velocity vz,avg. On substituting vz,avg from equation (12), the final expression for the mass rate of flow is w =
π ΔP R4 ρ 8μL
(14)
The flow rate vs. pressure drop (w vs. ΔP) expression above is well-known as the HagenPoiseuille equation. It is a result worth noting because it provides the starting point for flow in many systems (e.g., flow in slightly tapered tubes). c) Force along tube wall The z-component of the force, Fz, exerted by the fluid on the tube wall is given by the shear stress integrated over the wetted surface area. Therefore, on using equation (9), Fz = (2 π R L) τrz| r = R = π R2 ΔP = π R2 Δp + π R2 ρ g L cos β
(15)
where the pressure difference Δp = p0 − pL. The above equation simply states that the viscous force is balanced by the net pressure force and the gravity force.
Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid flow in a slightly tapered tube Problem. A fluid (of constant density ρ) is in incompressible, laminar flow through a tube of length L. The radius of the tube of circular cross section changes linearly from R0 at the tube entrance (z = 0) to a slightly smaller value RL at the tube exit (z = L).
Figure. Fluid flow in a slightly tapered tube. Using the lubrication approximation, determine the mass flow rate vs. pressure drop (w vs. ΔP) relationship for a Newtonian fluid (of constant viscosity μ).
Solution Hagen-Poiseuille equation and lubrication approximation The mass flow rate vs. pressure drop (w vs. ΔP) relationship for a Newtonian fluid in a circular tube of constant radius R is w =
π ΔP R 4 ρ 8μL
The above equation, which is the famous Hagen-Poiseuille equation, may be re-arranged as
(1)
8 μ ΔP = w L ρ π
1
(2)
R4
For the tapered tube, note that the mass flow rate w does not change with axial distance z. If the above equation is assumed to be approximately valid for a differential length dz of the tube whose radius R is slowly changing with axial distance z, then it may be re-written as d P −
8μ = w
dz
ρπ
1 [R(z)] 4
(3)
The approximation used above where a flow between non-parallel surfaces is treated locally as a flow between parallel surfaces is commonly called the lubrication approximation because it is often employed in the theory of lubrication. The lubrication approximation, simply speaking, is a local application of a one-dimensional solution and therefore may be referred to as a quasi-onedimensional approach. Equation (3) may be integrated to obtain the pressure drop across the tube on substituting the taper function R(z), which is determined next. Taper function As the tube radius R varies linearly from R0 at the tube entrance (z = 0) to RL at the tube exit (z = L), the taper function may be expressed as R(z) = R0 + (RL − R0) z / L. On differentiating with respect to z, we get dR = RL − R0 dz
(4)
L
Pressure P as a function of radius R Equation (3) is readily integrated with respect to radius R rather than axial distance z. Using equation (4) to eliminate dz from equation (3) yields 8μ = w (− dP ) ρπ
L RL − R0
dR R4
Integrating the above equation between z = 0 and z = L, we get
(5)
R PL
∫ P0
(−dP )
8μ = w ρπ
L
L
dR
RL ∫ − R0 R
R4
(6)
0
P0 − PL L
1 1 8μ 1 = w R− 3 (RL − R 0 ρπ R0 ) 3 L
(7)
3
Equation (7) may be re-arranged into the following standard form in terms of mass flow rate: 3 (λ − 3 λ3 π ΔP R0 4 π ΔP 1) 4 ρ w = = R0 ρ 1 + − 1 − λ λ + 8μL 8μL 3 λ2
(8)
where the taper ratio λ ≡ RL / R0. The term in square brackets on the right-hand side of the above equation may be viewed as a taper correction to equation (1).
Transport Phenomena - Fluid Mechanics Problem : Annular drag flow with inner cylinder moving axially - Newtonian fluid in wire coating die Problem. A wire - coating die essentially consists of a cylindrical wire of radius κR moving horizontally at a constant velocity V along the axis of a cylindrical die of radius R. If the pressure in the die is uniform, then the Newtonian fluid (of constant viscosity μ and constant density ρ) flows through the narrow annular region solely by the drag due to the motion of the wire (and is referred to as 'axial annular Couette flow'). Neglect end effects and assume an isothermal system.
Figure. Fluid flow in wire coating die. a) Establish the expression for the steady-state velocity profile in the annular region of the die. b) Obtain the mass flow rate through the annular die region. c) Estimate the coating thickness δ some distance downstream of the die exit. d) Find the force that must be applied per unit length of the wire. e) Write the force expression in d) as a 'plane slit' formula with a 'curvature correction'.
Solution. a) Shear stress and velocity distribution The fluid flow is in the z-direction shearing constant-r surfaces. Therefore, the velocity component that exists is vz(r) and the shear stress component is τrz(r). A momentum balance in cylindrical coordinates gives Δ d (r τ ) = P rz dr
r
(1)
L
Since there is no pressure gradient and gravity does not play any role in the motion of the fluid, ΔP = 0. Equation (1) with the right-hand side set to zero leads to the following expression for the shear stress distribution on integration: τrz =
C1 r
(2)
The constant of integration C1 is determined later using boundary conditions. Substituting Newton's law of viscosity for τrz in equation (2) gives
−μ
d vz d r
=
C1
(3)
r
The above differential equation is simply integrated to obtain the following velocity profile: C vz = − 1 ln r + C2
(4)
μ The integration constants C1 and C2 are evaluated from the following no-slip boundary conditions: BC 1: at r = R, vz = 0
(5)
BC 2: at r = κR,
(6)
vz = V
From BC 1, C2 = (C1/μ) ln R. So, vz = (C1/μ) ln (R/r) and BC 2 yields C1 = − μV/ ln κ. On substituting the integration constants into equations (2) and (4), the final expressions for the shear stress and velocity distribution are
τrz =
μV r ln (1/κ)
vz = ln (r/R) V
ln κ
(7)
(8)
b) Mass flow rate The mass rate of flow is obtained by integrating the velocity profile over the cross section of the annular region as follows. w = 2 π ρ R rvz
2 π R2 1rl rdr
(9)
∫
κ dr R
= Vρ
∫ n
ln κ R R R κ
Integration by parts then gives 2
2 π R2 V 1r ρ w = ln κ 2R
2
l r 1r − n R 4R
1
(10) κ
On evaluation at the limits of integration, the final expression for the mass rate of flow is 1− π R2 κ2 − w = V ρ ln( 2 2 2 1/κ κ ) c) Coating thickness
(11)
Some distance downstream of the die exit, the Newtonian liquid is transported as a coating of thickness δ exposed to air. A momentum balance in the 'air region' yields the same shear stress expression [equation (2)]. The boundary condition at the liquid-gas interface is τrz = 0 at r = κR + δ, which now gives C1 = 0 and τrz = 0 throughout the coating. Furthermore, equations (4) and (6) are valid and give the velocity profile as vz = V through the entire coating. As the velocity becomes uniform and identical to that of the wire, the liquid is transported as a rigid coating. The mass flow rate in the air region is given by wair = 2πVρ
κR + δ
∫
r dr = πR2Vρ [(κ + δ/R)2 − κ2]
(12)
κR
Equating the above expression for the flow rate wair in the air region to the flow rate w in the die region gives δ R
w
1 / 2
= κ2 πR2 + Vρ
−κ
(13)
On substituting w from equation (11), the final expression for the coating thickness is 1
1 − κ2 / 2 δ = 2 ln(1/κ R )
−κ
(14)
d) Force applied to wire The force applied to the wire is given by the shear stress integrated over the wetted surface area. Therefore, on using equation (7), Fz = (2 π κR) τ | rz r = κR = L
2πμV ln (1/κ)
(15)
e) Plane slit formula for force A very narrow annular region with outer radius R and inner radius (1 − ε)R, where ε is very small, may be approximated by a plane slit with area = 2π(1 − ½ε)RL ≈ 2πRL and gap = εR. Then, the velocity profile in the plane slit will be linear and the force applied to the 'plane' wire is given by Fplane 2πRL
V
= μ ε R
Fp or
lan e
=
2πμV ε
(16)
L
On substituting kappa; = 1 − ε (since R − κR = εR for the thin annular gap) in equation (15), the force applied to the wire is given by Fz = L
2πμV − ln (1 − ε)
The Taylor series expansion given below is next used. 1 1 1 1 ε2 ε3 ε4 ln (1 − ε) = −ε − ε5 − ... − − − 2 3 4 5
(17)
(18)
1 1 1 1 1 1 1 9 1 ε ε2 ε ε4 1 = 1 1 ε ε2 ε3 ε4 + 3 = 7− − − 1 − 2 − ... − ln (1 − ε) ε + + + + ... 2 3 4 5 ε 2 2 2 4 0 − 1
1 1
(19)
Equation (19) may be obtained by using (1 + u)−1 = 1 − u + u2 − u3 + u4 − ... (which is an infinite geometric progression with |u| < 1). On substituting in equation (17), the following force expression is finally obtained as a 'plane slit' formula with a 'curvature correction'. 1 2πμ 1 1 1 9 1 ε ε2 ε3 ε4 Fz = V (20) 1 2 7 − . − − − − L .. ε 2 2 4 2 0
Transport Phenomena - Fluid Mechanics Problem : Tangential annular flow between two coaxial rotating cylinders Problem. Consider an incompressible isothermal fluid in laminar flow between two coaxial cylinders, whose inner and outer wetted surfaces have radii of κ R and R, respectively. The inner and outer cylinders are rotating at angular velocities Ω i and Ω o, respectively. End effects may be neglected.
Figure. Tangential annular flow between two slowly rotating cylinders. a) Determine the steady-state velocity distribution in the fluid (for small values of Ω i and Ω o). b) Find the torques acting on the two cylinders during the tangential annular flow of a Newtonian fluid.
Solution. a) Simplification of continuity equation In steady laminar flow, the fluid is expected to travel in a circular motion (for low values of Ω i and Ω o). Only the tangential component of velocity exists. The radial and axial components of velocity are zero; so, vr = 0 and vz = 0. For an incompressible fluid, the continuity equation gives
.v = 0.
In cylindrical coordinates, (1)
So, vθ = vθ (r, z). If end effects are neglected, then vθ does not depend on z. Thus, vθ = vθ (r). Simplification of equation of motion There is no pressure gradient in the θ -direction. The components of the equation of motion simplify to (2)
(3)
(4)
The r-component provides the radial pressure distribution due to centrifugal forces and the zcomponent gives the axial pressure distribution due to gravitational forces (the hydrostatic head effect). Velocity profile by solving differential equation The θ -component of the equation of motion is integrated to get the velocity profile: (5)
(6)
The no-slip boundary conditions at the two cylindrical surfaces are (7)
(8)
The integration constants are thus given by (9)
On substituting for C1 and C2, the velocity profile is obtained as (10)
The velocity distribution given by the above expression can be written in the following alternative form: (11)
In the above form, the first term on the right-hand-side corresponds to the velocity profile when the inner cylinder is held stationary and the outer cylinder is rotating with an angular velocity Ω o. This is essentially the configuration of the Couette - Hatschek viscometer for determining viscosity by measuring the rate of rotation of the outer cylinder under the application of a given torque (as discussed below). The second term on the right-hand-side of the above equation corresponds to the velocity profile when the outer cylinder is fixed and the inner cylinder is rotating with an angular velocity Ω i. This is essentially the configuration of the Stormer viscometer for determining viscosity by measuring the rate of rotation of the inner cylinder under the application of a given torque (as discussed next). b) Determination of momentum flux From the velocity profile, the momentum flux (shear stress) is determined as (12)
Determination of torque The torque is obtained as the product of the force and the lever arm. The force acting on the inner cylinder itself is the product of the inward momentum flux and the wetted surface area of the inner cylinder. Thus, (13)
Similarly, the torque acting on the outer cylinder is obtained as the product of the outward momentum flux, the wetted surface area of the outer cylinder, and the lever arm. Thus, (14)
The tangential annular flow system above provides the basic model for rotational viscometers (to determine fluid viscosities from measurements of torque and angular velocities) and for some kinds of friction bearings.
Transport Phenomena - Fluid Mechanics Problem : Free surface shape of liquid in tangential annular flow Problem. An incompressible isothermal liquid in laminar flow is open to the atmosphere at the top. Determine the shape of the free liquid surface z(r) at steady state (neglecting end effects, if any) for the following cases:
Case a)
Case b)
Case c)
Case d)
Figure. Free surface shape of liquid in four cases of tangential flow. a) when the liquid is in the annular space between two vertical coaxial cylinders, whose inner and outer wetted surfaces have radii of κ R and R, respectively. Here, the inner cylinder is rotating at a constant angular velocity Ω i and the outer cylinder is stationary. Let zR be the liquid height at the outer cylinder. b) when a single vertical cylindrical rod of radius Ri is rotating at a constant angular velocity Ω i in a large body of quiescent liquid. Let zRo be the liquid height far away from the rotating rod. Hint: Use the result in a). c) when the liquid is in the annular space between two vertical coaxial cylinders, whose inner and outer wetted surfaces have radii of κ R and R, respectively. Here, the outer cylinder is rotating at a constant angular velocity Ω o and the inner cylinder is stationary. Let zR be the liquid height at the outer cylinder. d) when the liquid is in a vertical cylindrical vessel of radius R, which is rotating about its own axis at a constant angular velocity Ω o. Let zR be the liquid height at the vessel wall. Hint: Use the result in c).
Solution. Consider the liquid in the annular space between two vertical coaxial cylinders, whose inner and outer wetted surfaces have radii of κ R and R, respectively. Let the inner and outer cylinders be rotating at constant angular velocities of Ω i and Ω o, respectively. The first task is to determine the velocity profile by simplifying the continuity equation and the equation of motion. The next task is to obtain the free surface shape of the liquid from the velocity profile. Simplification of continuity equation In steady laminar flow, the liquid is expected to travel in a circular motion with only the tangential component of velocity. The radial and axial components of velocity are zero; so, vr = 0 and vz = 0. For an incompressible fluid, the continuity equation gives
.v = 0. In cylindrical coordinates,
(1)
So, vθ = vθ (r, z). If end effects are neglected, then vθ does not depend on z. Thus, vθ = vθ (r).
Simplification of equation of motion There is no pressure gradient in the θ -direction. The components of the equation of motion simplify to (2)
(3)
(4)
As discussed later, the r-component provides the radial pressure distribution due to centrifugal forces and the z-component gives the axial pressure distribution due to gravitational forces (the hydrostatic head effect). Velocity profile by solving differential equation The θ -component of the equation of motion is integrated to get the velocity profile: (5)
(6)
The no-slip boundary conditions at the two cylindrical surfaces are (7)
(8)
The integration constants are thus given by (9)
On substituting for C1 and C2, the velocity profile is obtained as (10)
The velocity distribution given by the above expression can be written in the following alternative form: (11)
In the above form, the first term on the right-hand-side corresponds to the velocity profile when the inner cylinder is held stationary and the outer cylinder is rotating with an angular velocity Ω o. The second term on the right-hand-side of the above equation corresponds to the velocity profile when the outer cylinder is fixed and the inner cylinder is rotating with an angular velocity Ω i. a) Free surface shape for rotating inner cylinder and fixed outer cylinder Now, consider the case where the inner cylinder is rotating at a constant angular velocity Ω i and the outer cylinder is stationary. With Ω o = 0, the velocity profile in Eq. 11 is substituted into the r-component of the equation of motion to get (12)
Integration gives (13)
The above expression for the pressure p is next substituted into the z-component of the equation of motion to obtain (14)
On substituting the expression for f1 in Eq. 13, (15)
The constant C3 is determined from the following condition: (16)
where zR is the liquid height at the outer cylinder and patm denotes the atmospheric pressure. Subtraction of the above two equations allows for elimination of C3 and the determination of the pressure distribution in the liquid as (17)
Since p = patm at the liquid-air interface, the shape of the free liquid surface is obtained as (18) b) Free surface shape for rotating cylinder in an infinite expanse of liquid The result in a) may be used to derive the free surface shape when a single vertical cylindrical rod of radius Ri is rotating at a constant angular velocity Ω i in a large body of quiescent liquid. On substituting R = Ro and κ = Ri / Ro, Eq. 11 (with Ω o = 0) and Eq. 18 yield (19)
The above equations give the velocity profile and the shape of the free liquid surface for the case where the inner cylinder of radius Ri is rotating at a constant angular velocity Ω i and the outer cylinder of radius Ro is stationary. Here, zRo be the liquid height far away from the rotating rod.
In the limit as Ro tends to infinity, the velocity profile and free surface shape for a single vertical rod rotating in a large body of quiescent liquid are obtained as (20) c) Free surface shape for rotating outer cylinder and fixed inner cylinder Next, consider the case where the outer cylinder is rotating at a constant angular velocity Ω o and the inner cylinder is stationary. With Ω i = 0, the velocity profile in Eq. 11 is substituted into the r-component of the equation of motion to get (21)
Integration gives (22)
The above expression for the pressure p is now substituted into the z-component of the equation of motion to obtain (23)
On substituting the expression for f2 in Eq. 22, (24)
The constant C4 is found from the following condition: (25)
where zR is the liquid height at the outer cylinder and patm denotes the atmospheric pressure. Subtraction of the above two equations eliminates C4 and gives the pressure distribution in the liquid as (26)
Since p = patm at the liquid-air interface, the shape of the free liquid surface is finally obtained as (27) An alternative approach is to obtain the above equation from the result of a) by substituting r = κ R in Eq. 18 to get (28)
The above expression gives the difference in elevations of the liquid surface at the outer and inner cylinders. Subtracting Eq. 28 from Eq. 18 yields (29)
The cylinders must now be swapped, i.e., the rotating cylinder should be made the outer cylinder rather than the inner one. So, let κ R = Ro and R = β Ro. On substituting κ = 1/β and R = β Ro in the above equation, (30)
On recognizing that β (which is a dummy parameter) can be simply replaced by κ , Ro may be replaced by R, and Ω i by Ω o, the above equation is identical to Eq. 27. d) Free surface shape for cylindrical vessel rotating about its own axis The result in c) may be used to derive the free surface shape when a liquid is in a vertical cylindrical vessel of radius R, which is rotating about its own axis at a constant angular velocity Ω o. In the limit as κ tends to zero, Eq. 11 (with Ω i = 0) and Eq. 27 yield the velocity profile and free surface shape as (31)
Note that as κ tends to zero, the first and second terms in square brackets in Eq. 27 also tend to zero. The velocity profile suggests that each element of the liquid in a rotating cylindrical vessel moves like an element of a rigid body. The free surface of the rotating liquid in a cylindrical vessel is a paraboloid of revolution (since Eq. 31 corresponds to a parabola).
Transport Phenomena - Fluid Mechanics Problem : Parabolic mirror fabrication from free surface shape of rotating liquid Problem. Top of Form
A plastic resin is in a vertical cylindrical vessel of radius R, which is rotating about its own axis at a constant angular velocity Ω .
Figure. Parabolic mirror from free surface shape of rotating liquid.
a) Determine the shape of the free surface z(r) at steady state (neglecting end effects, if any). Let zR be the liquid height at the vessel wall. b) Show that the radius of curvature (of the free surface shape) is given by
c) The cylinder is rotated until the resin hardens in order to fabricate the backing for a parabolic mirror. Determine the angular velocity (in rpm) necessary for a mirror of focal length 173 cm. Given: The focal length f is one-half the radius of curvature rc at the axis.
Solution. a) Simplification of continuity equation In steady laminar flow, the liquid is expected to travel in a circular motion with only the tangential component of velocity. The radial and axial components of velocity are zero; so, vr = 0 and vz = 0. For an incompressible fluid, the continuity equation gives
.v = 0. In cylindrical coordinates, (1 )
So, vθ = vθ (r, z). If end effects are neglected, then vθ does not depend on z. Thus, vθ = vθ (r). Simplification of equation of motion There is no pressure gradient in the θ -direction. The components of the equation of motion simplify to (2 )
(3 )
(4 )
As discussed later, the r-component provides the radial pressure distribution due to centrifugal forces and the z-component gives the axial pressure distribution due to gravitational forces (the hydrostatic head effect). Velocity profile by solving differential equation The θ -component of the equation of motion is integrated to get the velocity profile:
(5 )
(6 )
Since the velocity vθ is finite at r = 0, the integration constant C2 must be zero. Because vθ = ΩR at r = R, C1 = 2Ω. Thus, vθ = Ω r
(7 )
The above velocity profile suggests that each element of the liquid in a rotating cylindrical vessel moves like an element of a rigid body.
Free surface shape for rotating liquid in cylinder Next, the velocity profile is substituted into the r-component of the equation of motion (Eq. 2) to get ρ vθ
∂p =
2
∂r
= ρΩ2r
(8 )
r
Integration gives p = ρΩ2r2/2 + f1(z)
(9 )
The above expression for the pressure p is now substituted into the z-component of the equation of motion (Eq. 4) to obtain
∂p ∂z
− = ρg
⇒
d f1 = d z
−ρg or f1 = −ρgz + C3
(10 )
On substituting the expression for f1 in Eq. 9, p = ρΩ2r2/2 − ρgz + C3
(11 )
The constant C3 is found from the following condition: at r = R, z = zR and p = patm
⇒
patm = ρΩ2R2/2 − ρgzR + C3
(12 )
where zR is the liquid height at the vessel wall and patm denotes the atmospheric pressure.
Subtraction of the above two equations eliminates C and gives the pressure distribution in the liquid as p − patm = ρΩ2(r2 − R2)/2 + ρg(zR − z)
(13 )
Since p = patm at the liquid-air interface, the shape of the free liquid surface is finally obtained as zR − z = Ω2(R2 − r2)/(2g)
(14 )
The free surface of the rotating liquid in a cylindrical vessel is a paraboloid of revolution (since Eq. 14 corresponds to a parabola).
b) General expression for radius of curvature The circle that is tangent to a given curve at point P, whose center lies on the concave side of the curve and which has the same curvature as the curve has at P, is referred to as the circle of curvature. It radius (CP in the figure) defines the radius of curvature at P. The circle of curvature has its first and second derivatives respectively equal to the first and second derivatives of the curve itself at P. If the coordinates of the center C and point P are (r1, z1) and (r, z) respectively, then the radius of curvature CP is given by rc = [(z1 − z)2 + (r − r1)2]1/2
(15 )
Now, the radius CP is perpendicular to the tangent to the curve at P. So, the slope of the radius CP is given by
−
d r = d z
z1 − z
(16 )
r1 − r
Thus, the first and second derivatives of the curve at point P are d z = d r
d
r− r1
2
and z1 −z
z
1
r −d r1 z
= z1 + = (z1 − d − d z 2 r z)2 r
[1 + (dz / dr)2] z1 − z
(17 )
On substituting the above derivatives in Eq. 15, the general expression for the radius of curvature is obtained as (18 )
c) Step. Angular velocity necessary for parabolic mirror fabrication Let z0 be the height of the liquid surface at the centerline of the cylindrical vessel. Substituting z = z0 at r = 0 in Eq. 14 gives zR − z0 = Ω2R2/(2g). Thus, an alternative expression (in terms of z0) for the shape of the free surface is z = z0 + Ω2r2/(2g)
(19 )
Differentiating the above expression twice gives d 2
d z = d r
Ω2 r g
z and
Ω2 =
d r
g
(20 )
2
On substituting these derivatives into Eq. 18, the radius of curvature of the parabola is given by rc = [1 + Ω4r2/g2]3/2 (g/Ω2)
(21 )
Since the focal length f of the mirror is one-half the radius of curvature rc at the axis, f = ½ rc|r = 0 = g/(2Ω2)
(22 )
Finally, the rotational speed Ω necessary to fabricate a mirror of focal length f is given by Ω = [g/(2f)]1/2
(23 )
Substitution of numerical values 173
On substituting the focal length f = cm and the gravitational acceleration g = 981 cm/s2 in the above equation, the rotational speed may be calculated as 1.68 rad/s = 1.68 x 60/(2π) rpm = 16.08 rpm. The numerical value above for the focal length f in cm may be changed and the rotational speed recalculated by clicking on the button below. true
Recalculate
Bottom of Form
Transport Phenomena - Fluid Mechanics Problem : Flow between two concentric rotating spheres Problem. Consider an incompressible isothermal fluid in laminar flow between two concentric spheres, whose inner and outer wetted surfaces have radii of κ R and R, respectively. The inner and outer spheres are rotating at constant angular velocities Ω i and Ω o, respectively. The spheres rotate slowly enough that the creeping flow assumption is valid.
Figure. Flow between two slowly rotating spheres.
a) Determine the steady-state velocity distribution in the fluid (for small values of Ω i and Ω o). b) Find the torques on the two spheres required to maintain the flow of a Newtonian fluid. c) Simplify the velocity and torque expressions for the case of a single solid sphere of radius Ri rotating slowly at a constant angular velocity Ω i in a very large body of quiescent fluid.
Solution. a) Simplification of continuity equation In steady laminar flow, the fluid is expected to travel in a circular motion (for low values of Ω i and Ω o). So, vr = 0 and vθ = 0. For an incompressible fluid, the continuity equation gives
.v = 0.
In spherical coordinates, (1 )
So, vφ = vφ (r, θ ). Simplification of equation of motion When the flow is very slow, the term ρ v. v in the equation of motion can be neglected because it is quadratic in the velocity. This is referred to as the creeping flow assumption. Thus, for steady creeping flow, the entire left-hand side of the equation of motion is zero and we get 0=− p+µ
v+ρ g=− P+µ
2
2
v
(2 )
where P is the modified pressure given by P = p + ρ gh and h is the elevation in the gravitational field (i.e., the distance upward in the direction opposite to gravity from some chosen reference plane). In this problem, h = z = r cos θ . The above equation is referred to as the Stokes flow equation or the creeping flow equation of motion. The components of the creeping flow equation in spherical coordinates simplify to (3 )
(4 )
(5 )
There is no dependence on the angle φ due to symmetry about the vertical axis. The above partial differential equation must be solved for the velocity distribution vφ (r, θ ). Velocity profile by solving differential equation The no-slip boundary conditions at the two spherical surfaces are (6 )
(7 )
From the boundary conditions, the form vφ (r, θ ) = f(r) sin θ appears to be an educated guess to solve the partial differential equation for the velocity profile. On substituting this form in the φ -component of the equation of motion and simplifying, we get (8 )
The above ordinary differential equation may be solved by substituting f(r) = rn. The resulting characteristic equation is n2 + n − 2 = 0, whose solution is n = 1, −2. Thus, (9 )
On imposing the boundary conditions, (10 )
The integration constants are thus given by (11 )
On substituting for C1 and C2, the velocity profile is obtained as (1 2)
The velocity distribution given by the above expression can be written in the following alternative form:
(13 )
In the above form, the first term on the right-hand-side corresponds to the velocity profile when the inner sphere is held stationary and the outer sphere is rotating with an angular velocity Ω o. The second term on the right-hand-side of the above equation corresponds to the velocity profile when the outer sphere is fixed and the inner sphere is rotating with an angular velocity Ω i.
b) Determination of momentum flux From the velocity profile, the momentum flux (shear stress) is determined as (14 )
Determination of torque The torque required to rotate the inner sphere is obtained as the integral over the sphere surface of the product of the force exerted on the fluid by the solid surface element and the lever arm κ R sin θ . The force required on the inner sphere is the product of the outward momentum flux and the wetted surface area of the inner sphere. Thus, the differential torque on the inner sphere surface element is (15 )
On integrating from 0 to π , the torque needed on the inner sphere is (16 )
Note that the integral above evaluates to 4/3. Similarly, the differential torque needed on the outer sphere surface element is obtained as the product of the inward momentum flux, the wetted surface area of the outer sphere, and the lever arm. Thus,
(17 )
On integrating from 0 to π , the torque needed on the outer sphere is (18 )
c) Limiting case of single rotating sphere For the case of a single solid sphere of radius Ri rotating slowly at a constant angular velocity Ω i in a very large body of quiescent fluid, let R = Ro and κ = Ri / Ro. Then, on letting Ω o = 0, Eqs. (13) and (16) give (19 )
The above equations hold when the outer sphere of radius Ro is stationary and the inner sphere of radius Ri is rotating at an angular velocity Ω i. In the limit as Ro tends to infinity, the results for a single rotating sphere in an infinite body of quiescent fluid are obtained as (20 )
Transport Phenomena - Fluid Mechanics Problem : Radial fluid flow between two porous concentric spheres Problem. An isothermal, incompressible fluid of density ρ flows radially outward owing to a pressure difference between two fixed porous, concentric spherical shells of radii κR and R. Note that the velocity is not zero at the solid surfaces. Assume negligible end effects and steady laminar flow in the region κR ≤ r ≤ R.
Figure. Radial flow between two porous concentric spheres.
a) Simplify the equation of continuity to show that r2 vr = constant. b) Simplify the equation of motion for a Newtonian fluid of viscosity μ. c) Obtain the pressure profile P(r) in terms of PR and vR, the pressure and velocity at the sphere of radius R. d) Determine the nonzero components of the viscous stress tensor for the Newtonian case.
Solution a) Simplification of continuity equation Since the steady laminar flow is directed radially outward, only the radial velocity component vr exists. The other two components of velocity are zero; so, vθ = 0 and vφ = 0. For incompressible flow, the continuity equation gives ∇.v = 0. In spherical coordinates, 1
∂ r r 2
1
∂ (r2 vr ) +
1
∂
r ∂ sin θ θ
(vθ sin θ)+
∂ v φ
r sin ∂ θ φ
∂ =0 ⇒
(r2 vr ) = 0
(1 )
∂ r
On integrating the simplified continuity equation, r2 vr = f(θ, φ). There is no dependence expected on the angles θ and φ from symmetry arguments. In other words, r2 vr = C, where C is a
constant. This is simply explained from the fact that mass (or volume, if density ρ is constant) is conserved; so, ρ (4 π r2 vr ) = w is constant and C = w/(4πρ). b) Simplification of equation of motion The equation of motion is D v = −∇p − ∇ . τ + ρg
ρ
(2 )
D t
The above equation is simply Newton's second law of motion for a fluid element. It states that, on a per unit volume basis, the mass multiplied by the acceleration is because of three forces, namely the pressure force, the viscous force, and the gravity force. For an incompressible, Newtonian fluid, the term −∇ . τ = μ ∇2v and the equation of motion yields the Navier - Stokes equation. On noting that r2 vr = constant from the continuity equation, the components of the equation of motion for steady flow in spherical coordinates may be simplified as given below.
rcomponent :
ρ
v
∂ p
∂ vr
= −
r ∂r
∂ p
0 = −
+ ρ g sin θ r
φ-
(3 )
∂ r
1 θcomponent :
− ρ g cos θ
(4 )
∂ θ
0 = ∂p
(5
component :
)
∂φ
Note that the modified pressure P may be defined as P = p + ρgh, where h is the elevation or the height above some arbitrary datum plane, to avoid calculating the components of the gravitational acceleration vector g in spherical coordinates. Here, P = p + ρg r cos θ giving ∂P/∂r = ∂p/∂r + ρg cos θ and ∂P/∂θ = ∂p/∂θ − ρg r sin θ. Then, the components of the equation of motion in terms of the modified pressure are
rcomponent :
ρ
∂ vr v
r
∂P =
(6 )
∂ − r
∂r
∂P θcomponent :
0 =
(7 ) ∂θ
∂P φcomponent :
0 =
(8 ) ∂φ
Note that the equations (6) - (8) could have been directly obtained from the following form of the equation of motion: ρ D = −∇P − v ∇.τ
(9 )
D t
in which the modified pressure P includes both the pressure and gravitational terms.
From equations (7) and (8), P is a function of only r. Substituting P = P(r) and vr = vr(r) in equation (6) then gives d P
= −ρ vr
dr
dvr (10 ) dr
c) Pressure profile On substituting vr = C / r2, equation (10) gives d P
C2 = 2ρ
(11 ) r5
dr
Integration gives ρ C 2
P= −
+ C1 2 r4
(12 )
Since the pressure is P = PR and the velocity is vr = vR at r = R, the integration constants may be evaluated as C1 = PR + ρ C2/(2R4) and C = R2 vR . Then, substitution in equation (12) yields the pressure profile as ρ C2
R
1 R4 = ρ 1 1 PR v − 2 + − R
P = PR + 2 R4
4
r
2
(13 )
r
d) Determination of nonzero components of viscous stress tensor For an incompressible, Newtonian fluid, the viscous stress tensor is given by τ = −μ [ ∇v + (∇v)t ]. Since the only velocity component that exists is vr(r) = C / r2, the nonzero components of the viscous stress tensor are d vr τrr = −2 μ
= d r
4 μ C
r
v r
τθθ μ
and = = τφφ = −2 −
3
2μC (14 ) r
3
r
Note that the shear stresses are all zero. The normal stresses in equation (10) are partly compressive and partly tensile. Not only is the φ-component of [∇ . τ] directly zero, but the r-component and the θ-components are zero too as evaluated below.
1 d [∇ . τ]r =
[∇ . τ]θ =
1
4 4 τθθ μ μ + C C (r 2 τφφ = τrr) − + = 0
r d 2 r
∂ (τθθ sin θ) −
r sin ∂ θ θ
− r
r
r
4
4
τφφ 2μ 2μ = 0 cot = C + C θ − cot cot θ θ
(15 )
(16 )
r
r
4
r
4
Since [∇ . τ] = 0, viscous forces can be neglected in this flow and the Euler equation for inviscid fluids may be directly used.
Transport Phenomena - Fluid Mechanics Problem : Radial fluid flow between two porous coaxial cylinders Problem. An isothermal, incompressible fluid of density ρ flows radially outward owing to a pressure difference between two fixed porous, coaxial cylindrical shells of radii κR and R. Note that the velocity is not zero at the solid surfaces. Assume long cylinders with negligible end effects and steady laminar flow in the region κR ≤ r ≤ R.
Figure. Radial flow between two porous coaxial cylinders (top view).
a) Simplify the equation of continuity to show that r vr = constant. b) Simplify the equation of motion for a Newtonian fluid of viscosity μ. c) Obtain the pressure profile P(r) in terms of PR and vR, the pressure and velocity at the cylinder of radius R. d) Determine the nonzero components of the viscous stress tensor for the Newtonian case.
Solution. a) Simplification of continuity equation
Since the steady laminar flow is directed radially outward, only the radial velocity component vr exists. The tangential and axial components of velocity are zero; so, vθ = 0 and vz = 0. For incompressible flow, the continuity equation gives ∇.v = 0. In cylindrical coordinates, 1
∂
(r vr ) ∂+ r r
1
∂ v
∂ vθ
z
+ r
∂ θ
∂ =0 ⇒
(r vr ) = 0
(1 )
∂ r
∂ z
On integrating the simplified continuity equation, r vr = f(θ, z). Since the solution is expected to be symmetric about the z-axis, there is no dependence on the angle θ. Furthermore, the cylinders are long with negligible end effects (considered unbounded in z-direction); so, there is no zdependence. In other words, r vr = C, where C is a constant. This is simply explained from the fact that mass (or volume, if density ρ is constant) is conserved; so, ρ (2 π r vr L) = w is constant and C = w/(2πLρ). b) Simplification of equation of motion The equation of motion is D v = −∇P − ∇.τ
ρ
(2 )
D t
in which P includes both the pressure and gravitational terms. Note that the modified pressure P is defined as P = p + ρgh, where h is the elevation or the height above some arbitrary datum plane. Then, the formulation is valid for any orientation of the cylinder axis. For an incompressible, Newtonian fluid, the term −∇ . τ = μ ∇2v and the equation of motion yields the Navier - Stokes equation. On noting that r vr = constant from the continuity equation, the components of the equation of motion for steady flow in cylindrical coordinates may be simplified as given below. r-
ρv ∂ r
∂P =
(3
vr
component :
∂ r
−
∂r
)
∂P θcomponent :
0 =
(4 ) ∂θ
∂P zcomponent :
0 =
(5 ) ∂z
From equations (4) and (5), P is a function of only r. Substituting P = P(r) and vr = vr(r) in equation (3) then gives d P
= −ρ vr
dr
dvr (6 ) dr
c) Pressure profile On substituting vr = C / r, equation (6) gives d P
C2 = ρ
dr
Integration gives
(7 ) r3
ρ C 2
P= −
+ C1
(8 )
2 r2
Since the pressure is P = PR and the velocity is vr = vR at r = R, the integration constants may be evaluated as C1 = PR + ρ C2/(2R2) and C = R vR . Then, substitution in equation (8) yields the pressure profile as ρ C2 P = PR + 2 R2
R
1 R2 = ρ 1 1 PR v 2 − + − R 2
r
2
(9 )
r
d) Determination of nonzero components of viscous stress tensor For an incompressible, Newtonian fluid, the viscous stress tensor is given by τ = −μ [ ∇v + (∇v)t ]. Since the only velocity component that exists is vr(r) = C / r, the nonzero components of the viscous stress tensor are d vr τrr = −2 μ
= d r
2 μ C
v τθθ μ
r
and = −2
r
= −
2
2μC (10 ) r
2
r
Note that the shear stresses are all zero. The normal stresses in equation (10) are equal in magnitude and opposite in sign (partly compressive and partly tensile). Not only are the θ-component and z-component of [∇ . τ] zero, but the r-component is zero too because [∇ . τ]r 1 d (r
τ
2 2 = 0
(11
θ
=
τrr) d− r r
θ
r
μ μ C C = + − r r 3
)
3
Since [∇ . τ] = 0, viscous forces can be neglected in this flow and the Euler equation for inviscid fluids may be directly used.
Transport Phenomena - Fluid Mechanics Problem : Radial flow of a Newtonian fluid between parallel disks Problem. Steady, laminar flow occurs in the space between two fixed parallel, circular disks separated by a small gap 2b. The fluid flows radially outward owing to a pressure difference (P1 − P2) between the inner and outer radii r1 and r2, respectively. Neglect end effects and consider the region r1 ≤ r ≤ r2 only. Such a flow occurs when a lubricant flows in certain lubrication systems.
Figure. Radial flow between two parallel disks.
a) Simplify the equation of continuity to show that r vr = f, where f is a function of only z. b) Simplify the equation of motion for incompressible flow of a Newtonian fluid of viscosity μ and density ρ. c) Obtain the velocity profile assuming creeping flow. d) Sketch the velocity profile vr (r, z) and the pressure profile P(r). e) Determine an expression for the mass flow rate by integrating the velocity profile. f) Derive the mass flow rate expression in e) using an alternative short-cut method by adapting the plane narrow slit solution.
Solution.
a) Simplification of continuity equation Since the steady laminar flow is directed radially outward, only the radial velocity component vr exists. The tangential and axial components of velocity are zero; so, vθ = 0 and vz = 0. For incompressible flow, the continuity equation gives ∇.v = 0. In cylindrical coordinates, 1
∂
(r vr ) ∂+ r r
∂ v
∂ 1 vθ
z
+ r
∂ θ
∂ =0 ⇒
(r vr ) = 0
(1 )
∂ r
∂ z
On integrating the simplified continuity equation, r vr = f(θ, z). Since the solution is expected to be symmetric about the z-axis, there is no dependence on the angle θ. Thus, f is a function of z only and not of r or θ. In other words, r vr = f(z). This is simply explained from the fact that mass (or volume, if density ρ is constant) is conserved; so, ρ (2 π r vr dz) = dw is constant (at a given z) and is independent of r. b) Simplification of equation of motion For a Newtonian fluid, the Navier - Stokes equation is D v = −∇P + μ∇2v
ρ
(2 )
D t
in which P includes both the pressure and gravitational terms. On noting that vr = vr(r, z), its components for steady flow in cylindrical coordinates may be simplified as given below. rcomponent :
∂ 1∂ (r ∂ ρv ∂ = ∂ − P+ vr +2 r vr μ ) v ∂r
∂ r
∂ r r∂ r
r
(3 )
∂ z 2
∂P θcomponent :
0 =
(4 ) ∂θ
∂P zcomponent :
0 =
(5 ) ∂z
Recall that r vr = f(z) from the continuity equation. Substituting vr = f / r and P = P(r) in equation (3) then gives
f
d P
μ
2
− ρ
d2f
= + − r3
d r
(6 ) r
dz2
c) Velocity profile Equation (6) has no solution unless the nonlinear term (that is, the f 2 term on the left-hand side) is neglected. Under this 'creeping flow' assumption, equation (6) may be written as r d d2f P = μ dz2
(7 )
d r
The left-hand side of equation (7) is a function of r only, whereas the right-hand side is a function of z only. This is only possible if each side equals a constant (say, C0). Integration with respect to r from the inner radius r1 to the outer radius r2 then gives P2 − P1 = C0 ln (r2/r1). On replacing C0 in terms of f, r 2
μ l n r
0 = (P1 − P2) +
d2f (8 ) dz2
1
The above equation may be integrated twice with respect to z as follows. df
− ΔP =
d z
z z2 + + = 2μr C ln 1 r (r2/r1) vr
2
2 μ ln (r2/r1)
μ ln (r2/r1)
− ΔP
− ΔP f =
(9 )
z + C1
z + C1 z + C2 ⇒
C2 (10 ) r
Here, ΔP ≡ P1 − P2. Equation (10) is valid in the region r1 ≤ r ≤ r2 and −b ≤ z ≤ b.
Imposing the no-slip boundary conditions at the two stationary disk surfaces (vz = 0 at z = +b and any r) gives C1 = 0 and C2 = ΔP b2 / [2 μ ln (r2/r1)]. On substituting the integration constants in equation (10), the velocity profile is ultimately obtained as ΔP b2 vr =
z2 1
2 μ r ln (r2/r1)
− b
(11 )
d) Sketch of velocity profile and pressure profile The velocity profile from equation (11) is observed to be parabolic for each value of r with vr,max = ΔP b2 / [2 μ r ln (r2/r1)]. The maximum velocity at z = 0 is thus inversely proportional to r. In general, it is observed from equation (11) that vr itself is inversely proportional to r. Sketches of vr(z) for different values of r and vr(r) for different values of |z| may be plotted. The pressure profile obtained by integrating the left-hand side of equation (7) is (P − P2) / (P1 − P2) = [ln(r/r2)] / [ln(r1/r2)]. A sketch of P(r) may be plotted which holds for all z. e) Mass flow rate by integrating velocity profile The mass flow rate w is rigorously obtained by integrating the velocity profile using w = ∫ n . ρv dS, where n is the unit normal to the element of surface area dS and v is the fluid velocity vector. For the radial flow between parallel disks, n = δr , v = vrδr , and dS = 2πr dz. Then, substituting the velocity profile from equation (11) and integrating gives
w =
b
∫ − b
π ΔP b2 ρ ρ vr (2 π r) dz = μ ln (r2/r1)
z 3
z
b
4π ΔP b3 ρ (12 )
=
−3 − b b
3μ ln (r2/r1)
2
f) Mass flow rate using short-cut method by adapting narrow slit solution The plane narrow slit solution may be applied locally by recognizing that at all points between the disks the flow resembles the flow between parallel plates provided vr is small (that is, the creeping flow is valid).
The mass flow rate for a Newtonian fluid in a plane narrow slit of width W, length L and thickness 2B is given by w = 2 ΔP B3 W ρ / (3 μ L). In this expression, (ΔP/L) is replaced by (− dP/dr), B by b, and W by 2πr. Note that mass is conserved; so, w is constant. Then, integrating from r1 to r2 gives the same mass flow rate expression [equation (12)] as shown below. r d
4π P r b3 ρ 2 2 (−dP) w ∫ = ∫ ⇒ w = r 1
P
r
3μ
1
4π (P1 − P2) b3 ρ (13 ) 3μ ln (r2/r1)
This alternative short-cut method for determining the mass flow rate starting from the narrow slit solution is very powerful because the approach may be used for nonNewtonian fluids where analytical solutions are difficult to obtain.
Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid in a parallel - disk viscometer Problem. A parallel - disk viscometer consists of two circular disks of radius R separated by a small gap B (with R >> B). A fluid of constant density ρ, whose viscosity μ is to be measured, is placed in the gap between the disks. The lower disk at z = 0 is fixed. The torque Tz necessary to rotate the upper disk (at z = B) with a constant angular velocity Ω is measured. The task here is to deduce a working equation for the viscosity when the angular velocity Ω is small (creeping flow).
Figure. Parallel - disk viscometer.
a) Simplify the equations of continuity and motion to describe the flow in the parallel - disk viscometer. b) Obtain the tangential velocity profile after writing down appropriate boundary conditions. c) Derive the formula for determining the viscosity μ of a Newtonian fluid from measurements of the torque Tz and angular velocity Ω in a parallel - disk viscometer. Neglect the pressure term.
Solution a)Simplification of continuity equation In steady laminar flow, the fluid is expected to travel in a circular motion (for small values of Ω). Only the tangential component of velocity exists. The radial and axial components of velocity are zero; so, vr = 0 and vz = 0. For incompressible flow, the continuity equation gives ∇.v = 0. In cylindrical coordinates, 1
∂
(r vr ) ∂+ r r
∂ 1v
∂ v
θ
z
+ r ∂ θ
∂ v =0 ⇒
∂ z
θ
(1 )
=0 ∂ θ
So, vθ = vθ (r, z). Simplification of equation of motion For a Newtonian fluid, the components of the Navier - Stokes equation may be simplified as given below. ρv rcomponent :
∂p
2 θ
−
= − r
(2 ) ∂r
∂ 1∂ θcomponent :
zcomponent :
(r 0 = vθ + ∂ ∂) r rr
0 = ∂ −ρ − pg
∂2vθ (3 ) ∂z2
(4 )
∂ z
From symmetry arguments, there is no pressure gradient in the θ-direction; so, p = p(r, z).
b) Velocity profile The no-slip boundary conditions at the two disk surfaces are BC 1 :
at z = 0, any r, vθ = 0
(5 )
BC 2 :
at z = B, any r, vθ = Ω r
(6 )
It is possible to make an educated guess for the form of the tangential velocity taking a clue from BC 2. Thus, it is reasonable to postulate vθ (r, z) = r f(z). Substituting in equation (3) then gives d2 f
d f = 0 ⇒
dz 2
= C1 z + C2
or
f = C1
(7 )
d z
Equation (5) gives f = 0 at z = 0 or C2 = 0. On the other hand, equation (6) gives f = Ω at z = B or C1 = Ω/B. On substituting f = Ω z/B, the tangential velocity profile is ultimately obtained as vθ = Ω r z
(8 )
B
The above result is expected and could have been guessed by considering a fluid contained between two parallel plates (with the lower plate at z = 0 held stationary and the upper plate at z = B moving at a constant velocity V). Then, the linear velocity distribution is given by v = V z/B, which is the velocity profile at any r in the parallel - disk viscometer. On noting that V = Ω r at the upper plate, equation (8) is obtained.
c) Determination of torque From the velocity profile in equation (8), the momentum flux (shear stress) is determined as ∂v Ω θ θ = r and − τrθ = − μ = 0 μ r ∂ ∂z B rr ∂v
τzθ = −μ
(9 )
The differential torque dT is rigorously obtained as the cross product of the lever arm r and the differential force dF exerted on the fluid by a solid surface element. Thus, dT = r x dF where dF = n.(p δ + τ) dS, r is the vector drawn from the axis of rotation to the element of surface area dS, n is the unit normal to the surface, p is the pressure, δ is the identity (unit) tensor, and τ is the viscous stress tensor. For the parallel - disk viscometer, dT = r x n.τ dS = r δr x (−δz).τzθ δzδθ (2 π r dr)
(10 )
The pressure term is neglected in equation (10). Since δz.δz = 1 and δr x δθ = δz, we obtain dT = r (−τzθ) (2 π r dr) δz. Thus, the z-component of the differential torque is dTz = r (−τzθ) (2 π r dr), which is simply the product of the lever arm, the momentum flux and the differential surface area. Integrating over the area of the upper disk after substituting for τzθ from equation (9) yields Tz |z = B =
R[ (−τzθ) (2 π r2 dr)]z = B = ∫
2 Rr3 π dr μ ∫=
π μ Ω R4
(11 )
Ω 0
0
2B
B
Equation (11) provides the formula for determining the viscosity μ of a Newtonian fluid from measurements of the torque Tz and angular velocity Ω in a parallel - disk viscometer as μ = 2BTz/(πΩR4).
Note that the shear stress is not uniform [as given by equation 9)] in a parallel - disk viscometer, whereas it is uniform throughout the gap in a cone - and - plate viscometer.
Power law fluid flow in a plane narrow slit Problem. Consider a fluid (of density ρ) in incompressible, laminar flow in a plane narrow slit of length L and width W formed by two flat parallel walls that are a distance 2B apart. End effects may be neglected because B << W << L. The fluid flows under the influence of a pressure difference Δp, gravity or both.
Figure. Fluid flow in plane narrow slit.
a) Determine the steady-state velocity distribution for a non-Newtonian fluid that obeys the power law model (e.g., a polymer liquid).
b) Obtain the mass flow rate for a power law fluid in slit flow.
Solution. a) Shear stress distribution For axial flow in rectangular Cartesian coordinates, the differential equation for the momentum flux is dτ
ΔP
xz
(1 )
= L
dx
where P is a modified pressure, which is the sum of both the pressure and gravity terms, i.e., ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the z-axis with the vertical. On integration, this gives the expression for the shear stress τxz for laminar flow in a plane narrow slit as Δ P τxz =
(2 )
x L
It must be noted that the above momentum flux expression holds for both Newtonian and nonNewtonian fluids (and does not depend on the type of fluid). Further, τxz = 0 at x = 0 in slit flow based on symmetry arguments, i.e., the velocity profile is symmetric about the midplane x = 0. For the region 0 ≤ x ≤ B, the velocity decreases with increasing x and therefore dvz / dx ≤ 0. On the other hand, for the region −B ≤ x ≤ 0, the velocity decreases with decreasing x and therefore dvz / dx ≥ 0. Thus, the power law model is given as τxz = −d n m v z
d
for 0 ≤ x ≤B
(3 )
x
d v
z n
τxz = − m
for −B ≤ x≤0
(4 )
d x Velocity distribution
To obtain the velocity distribution for 0 ≤ x ≤ B, equations (2) and (3) are combined to eliminate τxz and get d vz −
ΔP 1 x / =
d x
n
for 0 ≤ x ≤B
(5 )
m L
On integrating and using the no-slip boundary condition (vz = 0 at x = B ), we get
vz =
ΔP B 1/
B
n
(1/ n) +1
m L
x (1/ n)
1 +1 −
for 0 ≤ x ≤ B
(6 )
B
Similarly, for −B ≤ x ≤ 0, equations (2) and (4) may be combined to eliminate τxz and obtain dv z
−ΔP 1/ = x n
for −B ≤ x ≤0
(7 )
m L
dx
Again, integrating and using the no-slip boundary condition (vz = 0 at x = −B) gives ΔP B 1/
vz =
B
n
m L
(1/ n) +1
x (1/ n) 1 − +1 −
for −B ≤ x ≤ 0
(8 )
B
Equations (6) and (8) may be combined and represented as a single equation for the velocity profile as follows: ΔP B 1/
vz =
B
n
m L
(1/ n) +1
| (1/ x n) 1 | +1 −
for | x | ≤ B
(9 )
B
Note that the maximum velocity in the slit occurs at the midplane x = 0. b) Mass flow rate Since the velocity profile is symmetric about the midplane x = 0, the mass flow rate may be obtained by integrating the velocity profile over half the cross section of the slit as shown below. B w = ∫ ρ vz W 2 dx 0
Δ B2 P 1 W ρ 1 x (1/ x / n) B n 1 + = d 1 2 (1/ − m n) + 0 B B L 1
∫
(10 )
2 B2 Wρ w =
Δ P B
1/n
(1/ n) m +2 L
(11 )
When the mass flow rate w in equation (11) is divided by the density ρ and the cross-sectional area (2 B W), an expression for the average velocity is obtained. For n = 1 and m = μ, equation (11) reduces to the Newtonian result, i.e., w = 2 ΔP B3 W ρ / (3 μ L).
Transport Phenomena - Fluid mechanics Problem : Power law fluid flow in a falling film Problem. Consider a liquid (of density ρ) in laminar flow down an inclined flat plate of length L and width W. The fluid flows as a falling film with negligible rippling under the influence of gravity. End effects may be neglected because L and W are large compared to the film thickness δ.
Figure. Fluid flow in a falling film.
a) Determine the steady-state velocity distribution for a non-Newtonian fluid that obeys the power law model (e.g., a polymer liquid). Reduce the result to the Newtonian case. b) Obtain the mass flow rate for a power law fluid. Simplify for a Newtonian fluid. c) What is the force exerted by the fluid on the plate in the flow direction?
Solution. a) Shear stress distribution For axial flow in rectangular Cartesian coordinates, the differential equation for the momentum flux is dτ
ΔP
xz
(1 )
= dx
L
where P is a modified pressure, which is the sum of both the pressure and gravity terms, that is, ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the z-axis with the vertical. Since the flow is solely under the influence of gravity, Δp = 0 and therefore ΔP/L = ρ g cos β. On integration, τxz = ρ g x cos β + C1
(2 )
On using the boundary condition at the gas-liquid interface (τxz = 0 at x = 0), the constant of integration C1 is found to be zero. This gives the expression for the shear stress τxz for laminar flow in a falling film as τxz = ρ g x cos β
(3 )
It must be noted that the above momentum flux expression holds for both Newtonian and nonNewtonian fluids (and does not depend on the type of fluid). In the falling film, the velocity decreases with increasing x and therefore dvz / dx ≤ 0. Thus, the power law model is given as τxz = −d n m v
(4 )
z
d x
Velocity distribution To obtain the velocity distribution, equations (3) and (4) are combined to eliminate τxz and get d vz − d x
ρg cos β =
1 / n
x1/n
(5 )
m
On integrating, vz = − (ρ g cos β/m)1/n x1/n + 1 /(1/n + 1) + C2. On using the no-slip boundary condition at the solid surface (vz = 0 at x = δ ), C2 = (ρ g cos β/m)1/n δ1/n + 1 /(1/n + 1). Thus, the final expression for the velocity distribution is
vz =
ρ g δ 1/ cos β n
m
δ
x (1/ n)
1 + (1/ − n) + δ 1
1
(6a )
Equation (6a) when simplified for a Newtonian fluid (n = 1 and m = μ) gives the following parabolic velocity profile.
vz =
b) Mass flow rate
ρ g δ2 x2 cos β 1 − 2μ δ
(6b )
The mass flow rate may be obtained by integrating the velocity profile over the film thickness as shown below. δ2 1 1 Wρ
ρgδ / cos β n
δ w =
∫ ρ vz W = dx 0
x (1/
x
n) 1 + d 1
∫
m
(1/ − n) + δ 1 0
w =
ρg δ2 W δ ρ cos β (1/ n) +2 m
(7 )
δ
1/n
(8a )
Equation (8a) for a Newtonian fluid (n = 1 and m = μ) gives
w =
ρ2 g W δ 3 cos β
(8b )
3μ
When the mass flow rate w in equation (8) is divided by the density ρ and the film crosssectional area (W δ), an expression for the average velocity is obtained. Also, the maximum velocity vz, max occurs at the interface x = 0 and is readily obtained from equation (6). Thus, equations (6a) and (8a) give =
δ
(1/ n) +2
ρgδ cos β
1
(1/
/ = n) n +
1 m (1/ n)
vz, max
(9a )
+ 2 For n = 1 and m = μ, equation (9a) reduces to the Newtonian result given below.
=
ρ g δ2 cos β
2 =
3μ
vz, max
(9b )
3
c) Force on the plate The z-component of the force of the fluid on the solid surface is given by the shear stress integrated over the wetted surface area. Using equation (3) gives Fz = L W τxz|x = δ = ρ g δ L W cos β
(10 )
As expected, the force exerted by the fluid on the plate is simply the z-component of the weight of the liquid film.
Transport Phenomena - Fluid mechanics Problem : Power law fluid flow in a circular tube Problem. Consider a fluid (of constant density ρ) in incompressible, laminar flow in a long circular tube of radius R and length L. End effects may be neglected because the tube length L is relatively very large compared to the tube radius R. The fluid flows under the influence of a pressure difference Δp, gravity or both.
Figure. Fluid flow in circular tube.
a) Determine the steady-state velocity distribution for a non-Newtonian fluid that obeys the power law model (e.g., a polymeric liquid). b) Obtain the mass flow rate for a power law fluid in tube flow.
Solution. a) Shear stress distribution For axial flow in cylindrical coordinates, the differential equation for the momentum flux is Δ P
d (r τrz) d = r
r
(1 )
L
where P is a modified pressure, which is the sum of both the pressure and gravity terms, i.e., ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the tube axis with the vertical. On integration (with the condition that τrz must be finite at r = 0), this gives the expression for the shear stress τrz for steady laminar flow in a circular tube as Δ P τrz =
r 2 L
(2 )
It must be noted that the above momentum flux expression holds for both Newtonian and nonNewtonian fluids (and does not depend on the type of fluid). The velocity vz decreases with increasing r and therefore dvz / dr ≤ 0. Thus, the power law model is given as d v
z n τrz = − m
(3 )
d r Velocity distribution
Equations (2) and (3) are combined to eliminate τrz and get
−
Δ P r
d vz
= dr
1/n
(4 )
2 m L
On integrating and using the no-slip boundary condition (vz = 0 at r = R ), we get the velocity profile as ΔP R
vz =
R
r (1/ 1/ n) 1 + n − 1 (1/ 2 n) + m R 1 L
Note that the maximum velocity occurs at the tube centerline (r = 0).
(5 )
b)Mass flow rate : Hagen-Poiseuille equation analog The mass flow rate may be obtained by integrating the velocity distribution over the tube cross section as shown below. R w =
∫ 0
w = 2π
ΔP R
2 m L
ρ vz 2 π r dr
(6 )
R3 ρ 1
r (1/ r r n) 1 +1 − d
1/ n
∫
(1/ n) + 0 1
w =
R
(7 )
R R
Δ π R3 P ρ R
1/n
(8 )
(1/ n) 2 +3 m L
When the mass flow rate w in equation (8) is divided by the density ρ and the cross-sectional area (π R2), an expression for the average velocity is obtained. For n = 1 and m = μ, equation (8) reduces to the Hagen-Poiseuille equation for Newtonian fluids, i.e., w = π ΔP R4 ρ / (8 μ L).
Transport Phenomena - Fluid Mechanics Problem : Power law fluid flow in a slightly tapered tube Problem. A fluid (of constant density ρ) is in incompressible, laminar flow through a tube of length L. The radius of the tube of circular cross section changes linearly from R0 at the tube entrance (z = 0) to a slightly smaller value RL at the tube exit (z = L).
Figure. Fluid flow in a slightly tapered tube.
Using the lubrication approximation, determine the mass flow rate vs. pressure drop (w vs. ΔP) relationship for a fluid that can be described by the power law model (e.g., a polymeric liquid).
Solution. Click here for stepwise solution Step. Hagen-Poiseuille equation analog and lubrication approximation The mass flow rate vs. pressure drop (w vs. ΔP) relationship for a power law fluid in a circular tube of constant radius R is (click here for derivation) π R3 ρ w = (1/ n) +3
Δ P R
2 m L
1/n
(1 )
The above equation, which is the power law fluid analog of the Hagen-Poiseuille equation (originally for Newtonian fluids), may be re-arranged as Δ = 3n wn P 2m +
1
(2
1
ρ
L
)
R3n + 1
n π
For the tapered tube, note that the mass flow rate w does not change with axial distance z. If the above equation is assumed to be approximately valid for a differential length dz of the tube whose radius R is slowly changing with axial distance z, then it may be re-written as d P − d z
3n + 1 = 2m n
w n
1
ρ
(3 )
[R(z)]3n + 1
π
The approximation used above where a flow between non-parallel surfaces is treated locally as a flow between parallel surfaces is commonly called the lubrication approximation because it is often employed in the theory of lubrication. The lubrication approximation, simply speaking, is a local application of a one-dimensional solution and therefore may be referred to as a quasi-onedimensional approach. Equation (3) may be integrated to obtain the pressure drop across the tube on substituting the taper function R(z), which is determined next. Step. Taper function As the tube radius R varies linearly from R0 at the tube entrance (z = 0) to RL at the tube exit (z = L), the taper function may be expressed as R(z) = R0 + (RL − R0) z / L. On differentiating with respect to z, we get d R
dz
=
RL − R0
L
(4 )
Step. Pressure P as a function of radius R Equation (3) is readily integrated with respect to radius R rather than axial distance z. Using equation (4) to eliminate dz from equation (3) yields 3n w L n + 1 (− = RL dP ) 2m ρ − n π R0
dR (5 ) R3n + 1
Integrating the above equation between z = 0 and z = L, we get R L
3n w L n PL + 1 (−d = RL P ) 2m ρ − P0 n π R0 R
dR (6 )
∫
∫
R3n + 1
0
P0 − PL
3n + 1
w
= 2m
ρ n π
L
n
1
1
1
3n (RL − R0 )
− R R 0 3
n
(7 )
L 3 n
Equation (7) may be re-arranged into the following standard form in terms of mass flow rate: w =
π R03 ρ
ΔP 1 3n R0 / (λ n −
1/n
(8 )
1)
(1/ n) +3
2 m L
1− λ− 3n
where the taper ratio λ ≡ RL / R0. The last term on the right-hand side of the above equation may be viewed as a taper correction to equation (1).
Transport Phenomena - Fluid Mechanics Problem : Annular drag flow with inner cylinder moving axially - Power law fluid in wire coating die Problem. A wire - coating die essentially consists of a cylindrical wire of radius κR moving horizontally at a constant velocity V along the axis of a cylindrical die of radius R. If the pressure in the die is uniform, then the polymer melt (which may be considered a non-Newtonian fluid described by the power law model and of constant density ρ) flows through the narrow annular region solely by the drag due to the axial motion of the wire (which is referred to as 'axial annular Couette flow'). Neglect end effects and assume an isothermal system.
Figure. Fluid flow in wire coating die.
a) Establish the expression for the steady-state velocity profile in the annular region of the die. Simplify the expression for n = 1 (Newtonian fluid). b) Obtain the expression for the mass flow rate through the annular die region. Simplify the expression for n = 1 and n = 1/3. c) Estimate the coating thickness δ some distance downstream of the die exit. d) Find the force that must be applied per unit length of the wire.
Solution. Click here for stepwise solution a) Step. Shear stress and velocity distribution The fluid flows in the z-direction shearing constant-r surfaces. Therefore, the velocity component that exists is vz(r) and the shear stress component is τrz(r). A momentum balance in cylindrical coordinates (click here for detailed discussion) gives d (r τrz) d = r
Δ P r
(1 )
L
Since there is no pressure gradient and gravity does not play any role in the motion of the fluid, ΔP = 0. Equation (1) with the right-hand side set to zero leads to the following expression for the shear stress distribution on integration: C τrz =
1
(2 )
r
The constant of integration C1 is determined later using boundary conditions. Since the velocity vz decreases with increasing radial distance r, the velocity gradient dvz/dr is negative in the annular die region and the power law model may be substituted in the following form for τrz in equation (2). d v n C1 z m − = r d r
(3 )
Here, m and n are the consistency index and exponent in the power law model, respectively. The above differential equation is simply integrated to obtain the following velocity profile (for n ≠ 1): r C 1q /
1 n
vz = − m
+ C2
(4 )
q
where q = 1 − (1/n). The integration constants C1 and C2 are evaluated from the following no-slip boundary conditions: BC 1: at r = R,
vz = 0
(5 )
BC 2: at r = κR,
vz = V
(6 )
From BC 1, C2 = (C1/m)1/n Rq/q. So, vz = (C1/m)1/n Rq/q [1 − (r/R)q] and BC 2 yields V = (C1/m)1/n Rq/q [1 − κq]. On substituting the integration constants into equations (2) and (4), the final expressions for the shear stress and velocity distribution (for n ≠ 1) are m
qV
τrz =
n
q
R (1 − r κq )
v
(r/R)q −
(7a )
(7b )
1
z
= κq − 1
V
When n = 1, the right-hand sides of equations (7a) and (7b) evaluate to 0/0; therefore, the evaluation may be done using l'Hopital's rule by differentiating the numerator and denominator with respect to n (or q) and then evaluating the limit as n tends to 1 (or q → 0). On differentiating equation (7a), limq → 0 q/(1 − κq) = limq → 0 1/(− κq ln κ) = 1/(− ln κ)
Similarly, on differentiating equation (7b), limq → 0 [(r/R)q − 1]/(κq − 1) = limq → 0 [(r/R)q ln (r/R)]/(κq ln κ) = [ln (r/R)]/(ln κ)
Thus, on simplifying equations (7a) and (7b), the shear stress and velocity for a Newtonian fluid (n = 1 and m = μ) are μV τrz =
v
r ln (1/κ)
ln (r/R)
z
= V
(8a )
(8b )
ln κ
b) Step. Mass flow rate The mass rate of flow is obtained by integrating the velocity profile over the cross section of the annular region as follows. w = 2 R rvz 2 π 1rrq−dr πρ dr = R2 V
(9 )
ρ
∫
∫
1 R κq − κR R 1
κ R
Integration then gives
w =
2 π R2 Vρ
1
r 2 +
2 + κq − 1 R q
1r 2
q −
1
κ
(10 )
2R
On evaluation at the limits of integration, the final expression for the mass rate of flow (for n ≠ 1 and n ≠ 1/3) is
w =
1− 1 2 π R2 κ2 + − Vρ q κ 2 − κq − 1 2 + q
(11a )
2
When n = 1 (or q = 0), the right-hand side of the above equation gives 0/0. Using l'Hopital's rule gives (2 + q) (− κ2 +q lim ) ln κ − (1 lim − κ2 + q) q − = = q→ κq q → κ 0 2+ 0 ln −1 q 2 κ (2 + q)2 1
1− κ2 + q
1 − κ2
1
−2 κ2 ln κ − (1 − κ2)
4 ln κ
Substituting in equation (11a) gives the mass flow rate for a Newtonian fluid (n = 1) as
w =
1 − π R2 2 κ − Vρ 2 κ ln( 2 2 1/ κ)
(11b )
When n = 1/3 (or q = −2), the term (1 − κ2 + q)/(2 + q) becomes 0/0. Using l'Hopital's rule, limq → −2 (1 − κ2 + q)/(2 + q) = limq → −2 (− κ2 + q) ln κ = − ln κ
Substituting in equation (11a) gives the mass flow rate for a fluid with n = 1/3 (q = −2) as 2 π R2 Vρ w = κ−2 − 1
1 1 − l κ 2 n −
(11c )
κ 2
c) Step. Coating thickness Some distance downstream of the die exit, the liquid (polymer melt) is transported as a coating of thickness δ exposed to air. A momentum balance in the 'air region' yields the same shear stress expression [equation (2)]. The boundary condition at the liquid-gas interface is τrz = 0 at r = κR + δ, which now gives C1 = 0 and τrz = 0 throughout the coating. Furthermore, equations (4) and (6) are valid and give the velocity profile as vz = V through the entire coating. As the velocity becomes uniform and identical to that of the wire, the liquid is transported as a rigid coating. The mass flow rate in the air region is given by wair = 2πVρ
κR r dr = πR2Vρ [(κ + δ/R)2 + δ − κ2 ]
(12 )
∫ κR
Equating the above expression for the flow rate wair in the air region to the flow rate w in the die region gives w δ = R
1 / 2
κ2 + πR 2 V ρ
(13 )
−κ
On substituting the appropriate expression for w from equation (11a) or (11b) or (11c), the final expression for the coating thickness is obtained. It may be noted that the coating thickness δ depends only on R, κ and n, but not on V and m. d) Step. Force applied to wire The force applied to the wire is given by the shear stress integrated over the wetted surface area. Therefore, on using equation (7a) or equation (8a), qV n
Fz/L = (2 π κR) τrz| ⇒
r = κR
(Fz/L)|n ≠ q 1 = R 2πm (1 − κq)
or (Fz/L)| n=1 =
2πμV (14 ) ln (1/κ)
The above problem solution considers only drag flow in a wire-coating die. The problem is solved considering both drag and pressure by Malik, R. and Shenoy, U. V. (1991) 'Generalized annular Couette flow of a power-law fluid', Ind. Eng. Chem. Res. 30(8): 1950 - 1954. Related Problems in Transport Phenomena - Fluid Mechanics :
Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid in wire coating die
- Determination of shear stress, velocity profile, mass flow rate and coating thickness for Newtonian fluid rather than power law fluid in case of annular drag flow with inner cylinder moving axially
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