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
<h3>
Answer: True</h3>
Often we replace "units" with something like "feet" or "meters"
examples: square feet, square meters, square yards, etc
The "square" refers to the idea that any surface area can be unwrapped and lay flat, then you can break the area of that flat figure into smaller squares. A good example of this is a dice has 6 faces and each face is a square itself, so the total surface area is the sum of all the 6 square areas.
Answer:
alright
Step-by-step explanation:
Answer:
6
Step-by-step explanation:
Hayden uses 3 cups of water to make a broth for soup
He uses 3 times as many cups of milk
Therefore the number of broth ounces made by Hayden can be calculated as follows
= 3+3
= 6
Answer:
- 40
Step-by-step explanation:
