Question

A typical coal-fired power plant generates 1000 MW of usable power at an overall thermal efficiency of 40%.

PART A:

What is the rate of heat input to the plant? Answer in MW

PART B:

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? Answer in kg

PART C:

At what rate is heat ejected into the cool reservoir, which is the nearby river? Answer in MW

PART D:

The river's temperature is 18.0 ∘C before it reaches the power plant and 18.5 ∘C after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. Answer in m^3/s

PART E:

By how much does the river's entropy increase each second? Answer in J/K

Answer 1

heat of combustion L= 2.65x10⁷ J/kg
Power generated W = 1000 MW = 10⁹ W
Efficiency η = 0.40
Intial Tempature Ti = 18 deg. C = 291 K
Final Tempature Tf = 18.5 deg. C = 291.5 K

PART A.

Efficiency η = (heat output)/(heat input)

since heat output = power generated(W)

heat input Q_in = (W) /η

heat input Q_in = 2500 MW

PART B. Let Coal per day consumed is m kg/day.
η = W / m*L
m= W / η*L
m= 10⁹ / (0.4* 2.65x10⁷)
m = 94.3396226415 kg/s

m = 8150943.396225601 kg/day

PART C.

heat ejected rate Q_loss= Q_in - W = 2500 -1000 = 1500 MW.

PART D.

let river's flow rate is m_w than
Q_loss = Q_w
m_w*C*ΔT = 1500 * 10³ kW

m_w = (1500x 103)/(4.18x0.5 )
m_w = 717703.34 kg/s = 717.703 m3/sec

PART E.

increase in river's entropy

ΔS_w= m_w*C*ln(Tf/Ti)
ΔS_w = 717703.34*4.18* ln(291.5/291)
ΔS_w = 5150.2183 W/K
ΔS_w = 5150.2183 J/sec-K