# A simple steam power cycle receives 100,000 kJ/min as heat transfer from hot combustion gases and rejects 66,000 kJ/min as heat transfer to the enviro

76 views
in General
closed
A simple steam power cycle receives 100,000 kJ/min as heat transfer from hot combustion gases and rejects 66,000 kJ/min as heat transfer to the environment. If the pump power required is 1400 kJ/min, the thermal efficiency of the cycle and turbine power output is
1. 51.5% and 590 kW
2. 34% and 590 kW
3. 51.5% and 566.6 kW
4. 34% and 566.6 kW

by (41.0k points)
selected

Correct Answer - Option 2 : 34% and 590 kW

Concept:

The net work done by the power plant is the difference between turbine work and pump work i.e. Wnet = WT - Wp

And in other words, net work done by the power plant is the difference between heat absorbed and heat rejected, i.e. Wnet = Q- QR

The thermal efficiency of the plant is given as, $\eta_{th}=1-\frac{Q_R}{Q_S}$

Calculation:

Given:

QS = 100000 kJ/min, QR = 66000 kJ/min, Wp = 1400 kJ/min

So, net work done is,

Wnet = 100000 - 66000 = 34000 kJ/min

And Turbine work,  WT = Wnet + Wp

WT = 34000 + 1400 = 35400 kJ/min ⇒ $\frac{35400}{60}=590~kW$

Thermal efficiency,

$\eta_{th}=1-\frac{66000}{100000}=0.34=34$%