A steam-electric power plant delivers 900 MW of electric power. The surplus heat

Question

A steam-electric power plant delivers 900 MW of electric power. The surplus heat is exhausted into a river with a flow of 1.22×105 kg/s, causing a change in temperature of 3.71 oC.

Part a.) What is the efficiency of the power plant (in % Don't enter a unit.)?

Part b.) What is the rate of the thermal source?

My online homework says the correct answer for Part a.) is 3.22×101 and the correct answer for Part b.) is 2.80×109 W.

When I worked the problem my answers did not match my online homework correct answers . I would really appreciate it if someone could please work this problem. Thank you in advance!

Explanation / Answer

Pout = 900*10^6 W

surplus power = 1.22*10^5*4190*3.71 = 1896477800

Pin = 900*10^6 + 1896477800 = 2796477800

efficiency = (pin - Pout)/Pin = (2796477800-1896477800)/2796477800 = 3.21*10^1

+++++++++++++++++++++++

rate of thermal source = Pin = 2.8*10^9 W

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