Question

A concentric-tube heat exchanger cools hot water owing at 1 kg/s from 80C to 30C. It uses air owing at 5 kg/s to perform this cooling operation. Air enters at 10C and exits at 51.5C. (a) What is the log-mean-temperature-di erence

Answer 1

Explanation:

Below is an attachment containing the solution.

Answer 2

The log-mean-temperature-difference is 24.03⁰C

Explanation:

First we need to know if the heat exchanger is in parallel flow or counter-flow. However, counter flow arrangement is best used to recover heat.

L.M.T.D for counter flow is given as;

`L.M.T.D =((T_h_f_1 -T_c_f_2)-(T_h_f_2 -T_c_f_1))/(2.3log[(T_h_f_1 -T_c_f_2)/(T_h_f_2 -T_c_f_1)])`

where;

Thf₁ is the initial temperature of the hot fluid = 80°C

Tcf₂ is the final temperature of the cold fluid = 51.5°C

Thf₁ - Tcf₂ = 80 - 51.5 = 28.5⁰C

Thf₂ is the final temperature of the hot fluid = 30°C

Tcf₁ is the initial temperature of the cold fluid = 10°C

Thf₂ - Tcf₁ = 30 - 10 = 20⁰C

`L.M.T.D = (28.5 -20)/(2.3Log[(28.5)/(20)]) \\\\L.M.T.D = (8.5)/(0.3538) =24.03^oC`

Therefore, the log-mean-temperature-difference is 24.03⁰C