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Sindrei [870]
2 years ago
10

Pls answer ASAP I’ll give brainliest

Mathematics
2 answers:
Vikki [24]2 years ago
3 0

Answer:

I believe it would be either the 30° or the 60° the image is a little blurry .

andre [41]2 years ago
3 0
It’s constructing a 30 degree angle.
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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 co
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Answer:

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;

L.M.T.D =\frac{(T_h_f_1 -T_c_f_2)-(T_h_f_2 -T_c_f_1)}{2.3log[\frac{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 = \frac{28.5 -20}{2.3Log[\frac{28.5}{20}]} \\\\L.M.T.D = \frac{8.5}{0.3538} =24.03^oC

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

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