Question

An internally reversible Rankine cycle with both reheat and regeneration operates between 400OC and 3.8 MPa and 50OC. The first stage turbine exhausts 700 kPa where part part of the steam is extracted for use in an open feed-water heater. The remainder is heated to 500 OC before entering the second stage. Saturated steam leaves the feedwater heater. Compute the cycle efficiency. USE THE T-s CHART TO FIND THE ENTHALPIES. Neglect pump work.

Answer 1

$\textbf{Given data}\rightarrow$

P1 3.8 MPa

$T_{1}=400\: \: \: ^{o}\textup{C}$

$P_{2}=700\textup{ kPa}$

P3 = 700 kPa

$T_{3}=500\: \: \: ^{o}\textup{C}$

$P_{4}=700\textup{ kPa}$

$T_{5}=50\: \: \: ^{o}\textup{C}$

T-s diagram →

T 3 1 kg 11 у 9,10 2,4 7,8 1-Y 5 S

$\textbf{Solution}\rightarrow$

$\textup{from T-s diagram enthalpy calculated from various points are }$

$h_{1}=3217.86\textup{ kJ/kg}$

$1 = 6.799 kJ/kgK

$s_{1}=s_{2}$

$h_{2}=2803.06\textup{ kJ/kg}$

$h_{3}=3481.81\textup{ kJ/kg}$

$s_{3}=7.932\textup{ kJ/kgK}$

$h_{4}=2803.06\textup{ kJ/kg}$

$s_{3}=s_{5}$

$h_{5}=2545.061\textup{ kJ/kg}$

$h_{7}=h_{f}\textup{ at 50 }^{o}\textup{C}=209.34\textup{ kJ/kg}$

$h_{11}=h_{f}\textup{ at 700 kPa }=697\textup{ kJ/kg}$

$\textbf{Solution}\rightarrow$

$\textup{energy balance for open feed water heater}$

$\left ( 1-y \right )h_{7}+y*\left ( h_{4} \right )=1*h_{11}$

$\left ( 1-y \right )*209.31+y*2803.06=1*697$

$209.34-209.34y+y*2803.06=1*697$

$y=\frac{697-209.34}{2803.06-209.34}$

$y=0.188$

Turbine work (WT) →

wt = (hi – h2) + (1 - y) * (h2 – h4) + (1 - y) * (h3 - 95)

$w_{T}=\left ( 3217.86-2803.06 \right )+\left ( 1-0.188 \right )*\left ( 2803.06-2803.06 \right )+\left ( 1-0.188 \right )*\left ( 3481.81-2545.061 \right )$

$w_{T}=1175.44\textup{ kJ/kg}$

Heat supplied in boiler (3) ►

$q_{s}=\left ( h_{1}-h_{11} \right )+\left ( 1-y \right )*\left ( h_{3}-h_{2} \right )$

$q_{s}=\left ( 3217.86-697 \right )+\left ( 1-0.188 \right )*\left ( 3481.81-2803.06 \right )$

$q_{s}=3072.005\textup{ kJ/kg}$

$\eta =\frac{w_{T}}{q_{s}}$

$\eta =\frac{1175.44}{3072.005}$

$\eta =0.3826$