An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in 1987 was designed to operate between the temperature li
mits of 86°F at the ocean surface and 41°F at a depth of 2100 ft. About 13,300 gpm of cold seawater was to be pumped from deep ocean through a 40-in-diameter pipe to serve as the cooling medium or heat sink. If the cooling water experiences a temperature rise of 6°F and the thermal efficiency is 2.5 percent, determine the amount of power generated. Take the density of seawater to be 64 lbm/ft3. Also, take the specific heat of water to be c = 1.0 Btu/lbm·°F.
2 answers:
Answer:
power generated = 307.84 Kw
Explanation:
The Given data
Tk = 86 f = 303.15k
To = 41 f = 278.15k
thermal efficiency ( n ) = 0.025
Q = 13300 gpm = 0.84 m^3/sec
C = 1.0 Btu/Ibmf
density of seawater ( d ) = 64 Ibm/ft3 = 1025.18 kg/m^3
rise in temp = 6 f
we have to make assumptions
- steady state
- fluid is in compressible and constant
first we calculate
mass flow rate of water ( Mw )
Mw = density of seawater * Q
= 64 * 13300 * 1/7.4804
= 113790 Ibm/min = 1896.51 Ibm/sec
therefore the rejection of the cooling water will be
= mass flow rate * C * rise in temperature
= 1896.51 * 1.0 * 6
= 11379.06 BTU/s (Qout)
Determine the amount of power generated using thermal efficiency formula
n =
W = power generated
n = thermal efficiency
Qout = rejection of cooling water
0.025 =
W = 0.025 W + 284.48
W ( 1 - 0.025 ) = 284.48
therefore W = 284.48 / 0.975
W = 291.77 BTU/s
note : 1 kw = 0.9478 BTU/s
x = 291.77 BTU/s
W in kw ( x ) = 291.77 / 0.9478 = 307.84 Kw
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