Question

A reheat Rankine cycle operates with water as the working fluid. Steam enters the first turbine at 8 MPa and 450°C and exits at 0.8 MPa. It is then reheated to 400°C before entering the second turbine, where it exits at 10 kPa. If the amount of work into the pump is 8.04 kJ/kg and the net work per cycle produced is 1410.25 kJ/kg, determine the thermal efficiency of the cycle

Answer 1

Answer:

The thermal efficiency, $\eta _{reheat}$, of the Rankine cycle with reheat is 36.81%

Explanation:

p₁ = 8 MPa = 80 Bars

T₁ = 450°C = 723.15 K

From steam tables, we have;

v₁ = 0.0381970 m³/kg

h₁ = 3273.23 kJ/kg

s₁ = 6.5577 kJ/(kg·K) = s₂

The p₂ = 0.8 MPa

T₂ = Saturation temperature at 0.8 MPa = 170.414°C = 443.564 K

h₂ = 2768.30 kJ/kg

$T_{2'}$ = 400°C = 673.15 K

$h_{2'}$ = at 400°C and 0.8 MPa = 3480.6 kJ/kg

p₃ = 10 kPa = 0.1 Bar

T₃ = Saturation temperature at 10 kPa = 45.805 °C = 318.955 K

h₃ = 2583.89 kJ/kg

h₄ = $h_{3f}$ = 191.812 kJ/kg

The thermal efficiency, $\eta _{reheat}$, of a Rankine cycle with reheat is given as follows;

$\eta _{reheat} = \dfrac{\left (h_{1}-h_{2} \right )+\left (h_{2'}-h_{3} \right )-W_{p}}{h_{1}-\left (h_{4}+W_{p} \right )+\left (h_{2'}-h_{2} \right )}$

Therefore, we have;

$\eta _{reheat} = \dfrac{(3273.23 -2768.30 ) + (3480.6 -2583.89 ) - 8.04)}{(3273.23 -(191.812 + 8.04) + (3480.6 -2768.30 ) } = 0.3681$

Which in percentage is 36.81%.