An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in 1987 was designed to operate between the temperature limits 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.

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

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

answer:

explanation:

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answer:

a

explanation: