Alternative Energy Systems and Applications Solutions Manual B. K. Hodge Mississippi State University
The solutio ns for the problems in the textbook w ere worked in Mathcad. Mathcad. In many many prob lems, specific locations and/or energy energy systems components are prescribed. Instructors may wish to particularize these to other locations and/or and/or co mponents. Ad di ti on all y, m any pr ob lem s c an b e mo di fi ed t o f oc us on sp ecif ec if ic concepts of interest or posed as parametric parametric studi es to illustrate alternative alternative system operating condition s.
Chapter 2 For the problem solutions in this chapter, Eqs. (2-11) and (2-17) are expressed so the results are positive if the device is a pump or compressor; thus Power fluid = mdot ⋅ ( U2 ⋅ Vu2
−
2
U1 ⋅ Vu1) and H = −0.5 ⋅ U1
−
2
U2
+
2
Vr 2
−
2
Vr 1
+
2
V1
−
2
V2
1. A centrifugal pump running at 3500 RPM pumps water at a flow rate of 0.01 m 3/s. The water enters axially and leaves the rotor at 5 m/s relative to the blades, that are radial at the exit. If the pump requires 5 kW and is 67 percent efficient, estimate the basic dimensions (rotor exit diameter and width). RPM :=
2⋅π
ω :=
min
Power req := 5.0 ⋅ kW
3500 ⋅ RPM
Power fluid :=
3
Q := 0.01 ⋅
m
ρ :=
sec
1000 ⋅
ω=
η ⋅ Power req
-1
η :=
366.519 s
0.67
Power fluid = 3.35 ⋅ kW
kg
mdot := Q ⋅
3
ρ
mdot
=
-1
10 kg ⋅ s
m
The velocity triangles are
Vr 1
V2
V1
Vr 2
U1
Vr 2 := 5 ⋅
U2
Power fluid = mdot ⋅ ( U2 ⋅ Vu2
−
m sec
U1 ⋅ Vu1)
Since V1 is radial, Vu1 = 0. Because the exit velocity triangle is a right triangle, Vu2 = U2 . Then 2
Power fluid = mdot ⋅ U2
U2 :=
Power fluid mdot
U2
=
-1
18.303m ⋅ s
r 2 :=
U2
r 2 = 0.05 m
ω
The exit diameter is thus D2 = 0.10 m. the rotor width at the exit must pass the mass flow with a radial velocity of 5 m/sec. width :=
mdot
π ⋅ 2 ⋅ r 2 ⋅ ρ ⋅ Vr 2
width
=
−3
6.374 × 10
m
width
=
0.637 ⋅ cm
2. The dimensions of a centrifugal pump are provided in the table. Dimension Inlet Outlet Radius, mm Blade width, mm Blade angle (degrees)
175 50 65
500 30 70
The pump handles water and is driven at 750 RPM. Calculate the increase in head and the power input if the flow rate is 0.75 m 3/s. RPM :=
2⋅π
ω :=
min
750 ⋅ RPM
ω=
D1 := 2 ⋅ 1 75 ⋅ mm
width1 := 50 ⋅ m m
U1 :=
U1
ω ⋅ 175 ⋅ mm
=
-1
13.744 m ⋅ s
3
-1
Q := 0.75 ⋅
78.54 s
Vm 1
V1
s
D2 := 2 ⋅ 5 00 ⋅ mm
width 2 := 30 ⋅ m m
U2 :=
U2
ω ⋅ 500 ⋅ mm
=
-1
39.27 m ⋅ s
Vr 2
V2
Vr 1
m
Vm 2
65 o
70o
The volume flow rate can be used to find the radial velocity (Vm) components. Q=
π ⋅ D ⋅ width ⋅ Vm
Vm1 :=
Q
Vm1
π D1 ⋅ width1
=
-1
13.642m ⋅ s
Vm2 :=
Q
π ⋅ D 2 ⋅ width2
Vm2
The power to the fluid is Power fluid = mdot ⋅ ( U2 ⋅ Vu2
−
U1 ⋅ Vu1)
ρ :=
kg
1000 ⋅
3
m Vu1 := U1 Vu1
=
−
Vm1 ⋅ cot( 65 ⋅ deg) -1
7.383 m ⋅ s
Power fluid :=
ρ ⋅ Q ⋅ ( U2 ⋅ Vu2 − U1 ⋅ Vu1)
Vu2 := U2 Vu2
=
−
Vm2 ⋅ cot( 70 ⋅ deg) -1
36.374m ⋅ s
Power fluid = 995.18 ⋅ kW
=
-1
7.958 m ⋅ s
3. Velocity components are given for various turbomachines in the sketches. The following information is the same for all of the sketches: outer radius = 300 mm, inner radius = 150 mm, Q = 2 m 3/sec, ρ = 1000 kg/m3, and ω = 25 rad/sec. Determine the torque, power, change in head, and reaction for each set of conditions. r o := 300 ⋅ mm
