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}]}](https://tex.z-dn.net/?f=L.M.T.D%20%3D%5Cfrac%7B%28T_h_f_1%20-T_c_f_2%29-%28T_h_f_2%20-T_c_f_1%29%7D%7B2.3log%5B%5Cfrac%7BT_h_f_1%20-T_c_f_2%7D%7BT_h_f_2%20-T_c_f_1%7D%5D%7D)
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](https://tex.z-dn.net/?f=L.M.T.D%20%3D%20%5Cfrac%7B28.5%20-20%7D%7B2.3Log%5B%5Cfrac%7B28.5%7D%7B20%7D%5D%7D%20%5C%5C%5C%5CL.M.T.D%20%3D%20%5Cfrac%7B8.5%7D%7B0.3538%7D%20%3D24.03%5EoC)
Therefore, the log-mean-temperature-difference is 24.03⁰C
Answer:
This means that you should do what is possible within parentheses first, then exponents, then multiplication and division (from left to right), and then addition and subtraction (from left to right).
Step-by-step explanation:
Answer:
The missing exponent is 5
Step-by-step explanation:
Answer:
i told you in the comment it equals 7
Step-by-step explanation: