The chief engineer at a university that is constructing large number of new student dormitories decides install a counterflow co
ncentric tube heat exchanger on each of the dormitory shower drains. The thin walled copper drains are of diameter Di 50 m. Wastewater from the shower enters the heat exchanger at Th,i 38 C while fresh water enters the dormitory Tc,i 10 C. The wastewater flows down the vertical wall of the drain in a thin, falling film, providing hh 10,000 W/m2. K.
A) If the annular gap is d = 10 mm, the heat exchanger length is L = 1 m, and the water flow rate is mdot = 10 kg/min, determine the heat transfer rate and the outlet temperature of the warmed fresh water.
B) If a helical spring is installed in the annular gap so the fresh water is forced to follow a spiral path from the inlet to the fresh water outlet, resulting in h_c = 9050 W/m2·K, determine the heat transfer rate and the outlet temperature of the fresh water.
C) Based on the result for part (b), calculate the daily savings if 15,000 students each take a 10-minute shower per day and the cost of water heating is $0.07/kW·h.
A) If the annular gap is d = 10 mm, the heat exchanger length is L = 1 m, and the water flow rate is mdot = 10 kg/min, determine the heat transfer rate and the outlet temperature of the warmed fresh water.
B) If a helical spring is installed in the annular gap so the fresh water is forced to follow a spiral path from the inlet to the fresh water outlet, resulting in h_c = 9050 W/m2·K, determine the heat transfer rate and the outlet temperature of the fresh water.
C) Based on the result for part (b), calculate the daily savings if 15,000 students each take a 10-minute shower per day and the cost of water heating is $0.07/kW·h.
1 answer:
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The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh
However, the average power P, required for a year, t = ? Kw
Expressing their relationship, we will have
U = P x t
Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)
t = 8760 hours
P =
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Hence, the average power required during the year is 7,135Kw
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Recall, Efficiency, η = Power Output/Power Input (100)
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0.45 =
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the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules
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1 quads of energy = 10¹⁵ Joules
The total energy used is 0.000000139 quads
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Answer:
% reduction in area==PR=0.734=73.4%
% elongation=EL=0.42=42%
Explanation:
given do=12.8 mm
df=6.60
Lf=72.4 mm
Lo=50.8 mm
% reduction in area=((*(do/2)^2)-(
*(df/2)^2)))/
*(do/2)^2
substitute values
% reduction in area=73.4%
% elongation=EL=((Lf-Lo)/Lo))*100
% elongation=((72.4-50. 8)/50.8)*100=42%