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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 420 K. Heat is released into the river, and the water in the river flows by at a temperature of 24°C.

(a) What is the highest efficiency that this plant can have?

(b) How much heat is released into the river every second?

GW

(c) How much heat must be released by the core to supply 1.00 GW of electrical power?

GW

(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 0.60°C. What is the minimum flow rate that the water in the Hobbes River must have?

L/s

Edit
Added Thu, 12 Nov '15
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T2 = Temperature of water in river = 24 c = 24 + 273 = 297 K

T1 = Temperature of steam = 420 k

a) efficeincy is given as

\eta = 1 - (T2/T1 )

\eta = 1 - (297/420)

\eta = 0.293

b)

effiecieny is given as ::

\eta = Pout/Pin

Pout = 1 x 109 W

inserting the values

(0.293) = (1 x 109 )/Pin

Pin = 3.41 x 109 watt

heat realesed = heat input - heat output

heat relased = 3.41 x 109- 1 x 109

heat released = 2.41 x 109 watt

C)

Pin = 3.41 x 109 watt

d)

change in temperature = \DeltaT = 0.60 C = 0.60 K (since change remains same in c and K)

mass of water heated = m

c = specific heat of water = 4186 J/kgK

the heat required can be given as

Q = m c \DeltaT

mass if given as , m = density x volume = \rhoV

so Q = \rhoV c \DeltaT

dividing both side by ''t'

(Q/t) = \rho(V/t) c \DeltaT

inserting the values

(Q/t) = heat released = 2.41 x 109watt

2.41 x 109= (1000)(V/t) (4186) (0.6)

V/t = 959.55 m3/sec

V/t = 959.55 x 103 L/sec

Edit
Added Sat, 12 Dec '15
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