The plant burns anthracite coal, which has a heat of combustion of 2.65*107J/kg. How much coal does the plant use per day, if it operates continuously? A. The river's temperature is 18.0 ∘C before it reaches the power plant and 18.6 ∘C after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second.

B. By how much does the river's entropy increase each second?

C. The rate is heat ejected into the cool reservoir, which is the nearby river is 1330 MW

D. the rate of heat input to the plant is 2330 MW

Question

Kirby

Physics

10 December, 18:08

The plant burns anthracite coal, which has a heat of combustion of 2.65*107J/kg. How much coal does the plant use per day, if it operates continuously? A. The river's temperature is 18.0 ∘C before it reaches the power plant and 18.6 ∘C after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second.

B. By how much does the river's entropy increase each second?

C. The rate is heat ejected into the cool reservoir, which is the nearby river is 1330 MW

D. the rate of heat input to the plant is 2330 MW

+5

Answers (1)

Know the Answer?

Hill · Answer 1

a. m=7596679.24kg b. Vf=527.77777m3/s c. ds=6064755J

Explanation:

The precursor to this question would have been

a. how much coal is needed by the plant

fm the relation Qh/t * 1day (86400)

Qh=2330MW*86400

Qh=MJ

201.3*10^6MJ/2.65*10^7J/kg

m=7596679.24kg

b. firstly heat from the river is given as 1330MW

Recall that Q=mcΔ∅

Q=Quantity of required

m=mass flow rate of the river

c = specific eat capacity of water

Δ∅=change in temperature

divide bot sides bt time t

Qc/t = (m/t) cΔ∅

1330 = (m/t) * 4200 * (18.6-18)

(m/t) = 1330*10^6/2520

(m/t) = 527777.77kg/s

recall = density=mass rate/volumetric rate

1000kg/m3=527777.77kg/s/Vf

Vf=527.77777m3/s

3.ΔS=mcIn (t2/t1)

ΔS=527777.77*4200In (219.6/219)

=6064755J

ΔS = Entropy change