A Nuclear Power Plant, located on a river, generates 1.00 GW of power. In this p

Question

A Nuclear Power Plant, located on a river, generates 1.00 GW of power. In this plant, liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine. Heat is absorbed by the liquid sodium in the core, and released by the liquid sodium (and into the superheated steam) in the heat exchanger. The temperature of the superheated steam is 420 K. Heat is released into the river, and the water in the river flows by at a temperature of 28°C.

a.) What is the highest efficiency that this plant can have?
b.) How much heat is released into the river every second?
c.) How much heat must be released by the core to supply 1.00 GW of electrical power?
d) Assume that new environmental laws have been passed so that the plant is not allowed to heat the river by more than 0.50°C. What is the minimum flow rate that the water in the river must have?

Explanation / Answer

a similar problem :

Q)

Calvin Cliffs Nuclear Power Plant, located on the Hobbes River, generates 1.00 GW of power. In this plant, liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine. Heat is absorbed by the liquid sodium in the core, and released by the liquid sodium (and into the superheated steam) in the heat exchanger. The temperature of the superheated steam is 500 K. Heat is released into the river, and the water in the river flows by at a temperature of 25ºC. (a) What is the highest efficiency that this plant can have? (b) How much heat is released into the river every second? (c) How much heat must be released by the core to supply 1.00 GW of electrical power? (d) Assume that new environmental laws have been passed to preserve the unique wildlife of the river. Because of these laws, the plant is not allowed to heat the river by more than 0.50ºC. What is the minimum flow rate that the water in the Hobbes River must have?

answer :

m We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1.00 GW of power and, from this value, the power that is wasted. The rate at which heat is being released to the river is related to the requisite flow rate of the river by dQ dt = cT dV dt.(a) The Carnot efficiency of a plant operating between temperatures Tcand Th is given by: hcmax C1TT = = Substitute numerical values and evaluate C: 0.404500K298K max= 1 =(c) The power that must be supplied, at 40.4% efficiency, to produce an output of 1.00 GW is given by: 2.48GW0.4041.00GWmaxoutputsupplied== =PP(b) Relate the wasted power to the power generated and the power supplied: Pwasted= Psupplied PgeneratedSubstitute numerical values and evaluate Pwasted: 1.48GWwasted2.48GW 1.00GW=P = (d) Express the rate at which heat is being dumped into the river: ( )dtdVc TVdtdc Tdtdmc TdtdQ= = = Solve for the flow rate dV/dt of the river: c TdQ dtdtdV

We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1.00 GW of power and, from this value, the power that is wasted. The rate at which heat is being released to the river is related to the requisite flow rate of the river by dQ dt = cT dV dt

(a) The Carnot efficiency of a plant operating between temperatures Tcand Th is given by:

(max) = (C) = 1 - Tc/Th

Substitute numerical values and evaluate C: 1 - 298/500 = 0.404

) The power that must be supplied, at 40.4% efficiency, to produce an output of 1.00 GW is given by:

Psupplied = Poutput/max = 1 GW/0.404 = 2.48GW

Relate the wasted power to the power generated and the power supplied:

Pwasted= Psupplied Pgenerated

= 2.48 - 1 = 1.48 GW

Express the rate at which heat is being dumped into the river:

dQ/dt = cdt dm/dt = cdt dv/dt

Solve for the flow rate dV/dt of the river:

and putting the values

we get, dv/dt = 7.1 * 10^5 L/s

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