Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency
= 3%
where;
![\zeta = \dfrac{W_{out}}{Q_{supplied }}](https://tex.z-dn.net/?f=%5Czeta%20%3D%20%5Cdfrac%7BW_%7Bout%7D%7D%7BQ_%7Bsupplied%20%7D%7D)
![Q_{supplied } = \dfrac{2}{0.03} \ MW](https://tex.z-dn.net/?f=Q_%7Bsupplied%20%7D%20%3D%20%5Cdfrac%7B2%7D%7B0.03%7D%20%5C%20MW)
![Q_{supplied } = 66.66 \ MW](https://tex.z-dn.net/?f=Q_%7Bsupplied%20%7D%20%3D%2066.66%20%5C%20MW)
However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;
![LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}](https://tex.z-dn.net/?f=LMTD%20%3D%20%5Cdfrac%7B%5CDelta%20T_1%20-%20%5CDelta%20T_2%7D%7BIn%20%28%5Cdfrac%7B%5CDelta%20T_1%7D%7B%5CDelta%20T_2%7D%20%29%7D)
Also;
![\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K](https://tex.z-dn.net/?f=%5CDelta%20T_1%20%3D%20T_%7Bh_%7Bin%7D%7D-%20T_%7Bc_%7Bout%7D%7D%20%5C%5C%20%5C%5C%20%5CDelta%20T_1%20%3D%20300%20-290%20%5C%5C%20%5C%5C%20%5CDelta%20T_1%20%3D%2010%20%5C%20K)
![\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K](https://tex.z-dn.net/?f=%5CDelta%20T_2%20%3D%20T_%7Bh_%7Bin%7D%7D-%20T_%7Bc_%7Bout%7D%7D%20%5C%5C%20%5C%5C%20%5CDelta%20T_2%20%3D%20292%20-290%20%5C%5C%20%5C%5C%20%5CDelta%20T_2%20%3D%202%5C%20K)
![LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}](https://tex.z-dn.net/?f=LMTD%20%3D%20%5Cdfrac%7B10%20-2%7D%7BIn%20%28%5Cdfrac%7B10%7D%7B2%7D%20%29%7D)
![LMTD = \dfrac{8}{In (5)}](https://tex.z-dn.net/?f=LMTD%20%3D%20%5Cdfrac%7B8%7D%7BIn%20%285%29%7D)
LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:
![Q_H = UA (LMTD)](https://tex.z-dn.net/?f=Q_H%20%3D%20UA%20%28LMTD%29)
where;
U = overall heat coefficient given as 1200 W/m².K
![66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}](https://tex.z-dn.net/?f=66.667%20%5Ctimes%2010%5E6%20%3D%201200%20%5Ctimes%20A%20%5Ctimes%204.97%20%5C%5C%20%5C%5C%20%20A%3D%20%5Cdfrac%7B66.667%20%5Ctimes%2010%5E6%7D%7B1200%20%5Ctimes%204.97%7D%20%5C%5C%20%5C%5C%20%20%5Cmathbf%7BA%20%3D%2011178.236%20%5C%20m%5E2%7D)
The mass flow rate:
![Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}](https://tex.z-dn.net/?f=Q_%7BH%7D%20%3D%20mC_p%28T_%7Bin%7D%20-T_%7Bout%7D%20%29%20%20%5C%5C%20%5C%5C%20%2066.667%20%5Ctimes%2010%5E6%3D%20m%20%5Ctimes%204.18%20%28300%20-292%29%20%5C%5C%20%5C%5C%20m%20%3D%20%5Cdfrac%7B%20%2066.667%20%5Ctimes%2010%5E6%7D%7B4.18%20%5Ctimes%208%7D%20%5C%5C%20%5C%5C%20%20%5Cmathbf%7Bm%20%3D%201993630.383%20%5C%20kg%2Fs%7D)
Answer:
20 m/s westward
Explanation:
Taking eastward as positive direction, we have:
is the velocity of Bill with respect to Amy (which is stationary)
is the velocity of Carlos with respect to Amy
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
![v_B = +5 m/s](https://tex.z-dn.net/?f=v_B%20%3D%20%2B5%20m%2Fs)
Therefore, Carlos velocity in Bill's reference frame will be
![v_C' = v_C - v_B = -15 m/s - (+5 m/s) = -20 m/s](https://tex.z-dn.net/?f=v_C%27%20%3D%20v_C%20-%20v_B%20%3D%20-15%20m%2Fs%20-%20%28%2B5%20m%2Fs%29%20%3D%20-20%20m%2Fs)
and the direction will be westward (negative sign).
Answer:
353225
Explanation:8uhhhhhhhhhlkgg
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:
![F=G*\frac{m_1m_2}{r^2}](https://tex.z-dn.net/?f=F%3DG%2A%5Cfrac%7Bm_1m_2%7D%7Br%5E2%7D)
By writing a dimensional analysis with the proper algebra handling, we obtain:
![N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}](https://tex.z-dn.net/?f=N%5B%3D%5DG%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%20%5C%5Ckg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%5B%3D%5DG%20%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%5C%5CG%5B%3D%5D%5Cfrac%7Bkg%2Am%2Am%5E2%7D%7Bkg%5E2%2As%5E2%7D%5C%5C%20%5C%5CG%5B%3D%5D%5Cfrac%7Bm%5E3%7D%7Bkg%2As%5E2%7D)
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards