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:
Yes.
Step-by-step explanation:
We are using a line regression tool to solve the parameters asked in the problem. We can use online tools or that of Excel. According to the tool, the best fit values are
<span><span> </span><span>Slope0.3848 ± 0.03956
</span><span>Y-intercept0.6053 ± 0.6370
</span><span>X-intercept-1.573
</span><span>1/Slope<span>2.598
Best fit lines make sure that the standard deviation at each point is minimum from the best fit line.</span></span></span>
So,
1. Graph each inequality separately.
2. Choose a test point to determine which side of the line needs to be shaded.
3. The solution to the system will be the area where the shadings from each inequality overlap one another (purple area)
As for the system of inequalities we say it's unbounded.
