Question 11.A.1: Calculate the LMTD correction factor for the following cases...

(a) Equation (11.A.5) is applicable for this case. Setting P = 0.75 gives the following equation for Φ :

P  =  1  – \left(\frac{1  +  \Phi^{-1}}{2  +  \ln \Phi}\right)  \quad    (R = 0.5)                 (11.A.5)

0.75=1-\left(\frac{1+\Phi^{-1}}{2+\ln \Phi}\right)

Using the nonlinear equation solver on a TI-80 series calculator, the solution is found to be Φ = 10.728, and ln Φ = 2.37286. Next, ψ is calculated using Equation (11.A.1):

\Psi = \frac{R  –  1}{\ln [(1  –  P)/(1  –  PR)]}   \quad   (R \neq 1)                           (11.A.1)

\Psi=\frac{0.5  –  1.0}{\ln [(1  –  0.75) /(1  –  0.75  \times  0.5)]}=0.54568

Finally, Equation (11.A.3) is solved for F:

\phi = \text{exp}  (1/ F \Psi)                (11.A.3)

F=(\Psi \ln \Phi)^{-1}=(0.54568 \times 2.37286)^{-1}=0.7723

(b) Equation (11.A.6) is applicable for this case. Setting R = 1.0 and P = 0.5 gives the following equation for Φ:

P = \left(\frac{R \phi^{R}}{\phi^{R}  –  1}  +  \frac{\phi}{\phi  –  1}  –  \frac{1}{\ln \phi}\right)^{-1}             (11.A.6)

0.5=\left(\frac{\Phi}{\Phi  –  1}+\frac{\Phi}{\Phi  –  1}-\frac{1}{\ln \Phi}\right)^{-1}

The solution is found using a TI calculator:  Φ = 3.51286, from which ln Φ = 1.25643. Since R = 1.0, the value of ψ is obtained from Equation (11.A.2):

\Psi=(1-P) / P \quad \quad  (R  = 1)                      (11.A.2)

\Psi=(1-P) / P=(1-0.5) / 0.5=1.0

The LMTD correction factor is found by solving Equation (11.A.3) as before:

F=(\Psi \ln \Phi)^{-1}=(1.0  \times 1.25643)^{-1}=0.7959

The reader can verify that the calculated values of F are in agreement with the graphs.