Question

An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are 30 ∘C and 6 ∘C, respectively. The surface and deep-water temperatures are 30 ∘C and 6 ∘C, respectively.

Part A What is the maximum theoretical efficiency of this power plant? Express your answer using two significant figures.

Part B

If the power plant is to produce a power of 190 kW , at what rate must heat be extracted from the warm water? Assume the maximum theoretical efficiency.

Express your answer using two significant figures.

Part C

At what rate must heat be absorbed by the cold water? Assume the maximum theoretical efficiency.

Express your answer using two significant figures.

Part D

The cold water that enters the plant leaves it at a temperature of 12 ∘C . What must be the flow rate of cold water through the system? Give your answer in kg/s and L/s .

Express your answer using two significant figures

Part E    find dm/dt =

Express your answer using two significant figures.

Answer 1

Theoretical efficiency of a heat engine

%eff =100( Th - Tc)/Th

Where Th = source temperature referenced to absolute zero (degrees K)
Tc = sink temperature referenced to absolute zero (degrees K)

degrees K = degrees C + 273
30C = 303 K
6C = 279K

%eff = 100 (303-279)/303

%eff= 7.9%

If the power plant is to produce a power of 190 kW , at what rate must heat be extracted from the warm water? Assume the maximum theoretical efficiency
heat required for 190kw
heat required = ouput/ eff

190/0.079 = 2405 kw or 2405 kJ/s

heat into cold water = input heat - kw output
= 2405 -190
= 2215 kw or 2215 kJ/s

The cold water that enters the plant leaves it at a temperature of 12 degrees C. What must be the flow rate of cold water through the system? Give your answer in kg/s and L/s.

specific heat of water = 4186J/Kg-C degree
water comes in at 6 degrees and leaves at 12 for a 6 degree difference

(2215 x10^3 watts / 4186 watt-sec/ kg-C) / 6 degrees= 88.19 kg/sec

1kg water = I liter
88.19 Liters/sec

Part E)

dm/dt = *A*v