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

The log-mean-temperature-difference is 24.03⁰C

Step-by-step 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;

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

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

Step-by-step explanation:

Below is an attachment containing the solution.

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation: