ME 1065 - LMTD and Effectiveness-NTU Derivations L. Schaefer September 12, 2005
Brief Brief Deriv Derivation of the LMTD To design or predict the performance of a heat exchanger, the LMTD and the effectiveness-NTU methods are both useful. I’ll first touch on the LMTD method, to give you an overview of its derivation and meaning. One circumstance in designing or predicting the performance of a hxgr is the need to relate the heat transfer rate to quantities like the inlet and outlet temperatures, the U and the A. Examine an exchange of heat between two streams, separated by a thin sheet of area A, as seen in the next Figure. Let’s write the expressions for mass conservation and energy conservation, and see what assumptions we need to mak make. e. Also need to determin determinee the control control volume. volume. Assume: • Uniform • Steady flow • All the heat that comes from the hot stream goes into the cold stream (the hxgr is insulated from its
surroundings) • No phase change • Constant specific heats • Negligible Kinetic and potential energy • U is constant (or nearly so)
Mass conservation: ∂ ∂t
ρd— V +
CV
· ˆ ρV · V ndA ndA = 0
CS
· ˆ ρV · V ndA ndA = 0
CS
m ˙ c = ρ c,oV o Ao = ρ c,i V i Ai
Energy conservation: conservation: ∂ ∂t
CV
ρed— V +
· ˆ ˙ ρeV · V ndA ndA = Q˙ − W
∗
CS
· ˆ ρ(e + P v )V · V ndA ndA = 0
CS
Q˙ = ρ c,ohc,oV o Ao − ρc,i hc,i V i Ai Q˙ = m ˙ c (hc,o − hc,i )
1
Q˙ = m ˙ h (hh,i − hh,o)
So with the specific heat assumption that we made (and treating the fluids as incompressible), we can write Q˙ = mc ˙ p ∆T
We’ll also define C i = m ˙ i c p,i as the heat capacity rate. We want to relate the inlet and outlet temperatures, the U, and the A, to the rate of heat transferred, and we can do this as: Q˙ = U A∆T mean
. We now need to find what this mean temperature is. To do this, we’ll look at a differential area of the heat exchanger where a differential amount of heat is transferred. Using this equation to examine the hot and cold streams separately yields: δ Q˙ = −C h dT h ; C h = (m ˙ c p )h
(1)
δ Q˙ = ±C c dT c ; C c = (m ˙ c p )c
(2)
where ± or ∓ appears, the top sign designates parallel-flow, and the bottom sign designates counterflow. A heat exchanger relationship for expressing the heat transfer between fluids over a differential area is: δ Q˙ = U (T h − T c ) dA , or
T h − T c =
(3)
δ Q˙ U dA
(4)
Rearranging equations 1 and 2 yields: dT h = −
δ Q˙ δ Q˙ and dT c = ± ; so C h C c
dT h − dT c = d (T h − T c ) = δ Q˙ ( −
1 C h
∓
1 C c
)
(5)
Dividing the above equation by equation 4 results in: d(T h − T c ) 1 1 ∓ = U (− ) dA T h − T c C h C c
(6)
Equation 6 can then be integrated over the heat exchanger: Parallel-flow: T h − T c 1 1 ln[ ] = U A ( − ) − 2
2
T h − T c 1
Counterflow: ln[
C h
1
C c
T h − T c 1 1 ] = U A ( − + ) T h − T c C h C c 2
1
1
2
(7)
Replacing the specific heats of equation 7 with those found before yields Parallel-flow: T h − T c UA ln[ ]= [(T h − T h ) + ( T c − T c )] T h − T c Q˙ 2
2
1
1
2
Counterflow: ln[
1
1
2
T h − T c UA ]= [(T h − T h ) + ( T c − T c )] T h − T c Q˙ 2
1
1
2
2
2
1
2
1
Q˙ = U A
∆T 1 − ∆T 2 T ln[ ∆ ] ∆T 1 2
∆T LogMean =
∆T 1 − ∆T 2 T ln[ ∆ ] ∆T 1
(8)
2
This equation is valid only if C c = C h ; otherwise, ∆T 1 = ∆T 2 and a denominator of zero results. Example 11.1 from Incrop. and DeW. is a good example that incorporates the convection coefficient and the LMTD.
Effectiveness-NTU method If only the inlet temperatures (and not the outlet temperatures) are known, the LMTD method requires iteration. In these cases (and in some others), the effectiveness-NTU method should be used instead ( ε − N T U ). REMEMBER: Q˙ = q . C We want to show that ε = ε(NTU, C ) We’ll first show that this is true for parallel-flow, where C h = C min. So, what is q equal to, in terms of the temperatures and heat capacity rate of the hot fluid? min
max
q = C h (T hi − T ho ) ε =
C h (T hi − T ho ) T hi − T ho = C min(T hi − T ci ) T hi − T ci
C r =
C min C h T co − T ci = = C max C c T hi − T ho
Now, returning to the derivation of the LMTD, we know that for a parallel-flow hxgr, ln(
UA C h UA ∆T 2 1 1 ) = −U A( + ) = − (1 + ) = − (1 + C r ) C h C c C h C c C min ∆T 1
ln(
T ho − T co ) = −N T U (1 + C r ) T hi − T ci
and T ho − T co = exp(−N T U (1 + C r )) T hi − T ci
Now we have three expressions for temperature that relate to ε, N T U , and C r , and we can eliminate temperature altogether to find ε only in terms of N T U and C r . The derivation is as follows: T co = C r (T hi − T ho ) + T ci T ho − T co (T ho − T hi ) + (T hi − T co ) = T hi − T ci T hi − T ci T ho − T co T ho − T hi T hi − T ci T hi − T ho = + − C r T hi − T ci T hi − T ci T hi − T ci T hi − T ci T ho − T co = ( −ε) + (1) − C r (ε) = exp(−N T U (1 + C r )) T hi − T ci
and 1 − exp(−N T U (1 + C r )) 1 + C r We could also go through and do the same derivation for C c = C min. You’ll find more expressions for ε and N T U in any heat transfer textbook. ε =
3
