Question

In a steam power plant operating on the Rankine cycle, steam enters the turbine at $4$ MPa, ${350}^{o}C$ and exists at a pressure of $15$ kPa. Then it enters the condenser and exists as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is $90$%. The thermodynamic states of water and steam are given in the table.

h is specific enthalpy, s is specific entropy and v the specific volume; subscript f and g denote saturated liquid state and saturated vapour state.


The net work output $kJ{k}^{-}1$ of the cycle is

• A. $498$
• B. $775$
• C. $860$
• D. $957$

Answer 1

The correct option is C $860$

Given data:
$p_1=p_4=4MPa=4000kPa$
$T_1={350}^{o}C$
$p_2=p_3=15kPa$

From given saturated steam table at
$p_1=4MPa,T_1={350}^{o}C$
we get, $h_1=3092.5kJ/kg$
$s_1=6.5821kJ/kgK$
From given saturated steam table (pressure based)
$h_f=h_3=226.95kJ/kg$
$h_g=2599.1kJ/kg$
$s_f=0.7549kJ/kgK$
$s_g=8.0085kJ/kgK$
$v_f=v_3=0.001014m^{3}kg$
$s_1=s_{2s}=s_f+x_{2s}(s_g-s_f)$
$6.5821=0.7549+x_{2s}$
$(8.0085-0.7549)$
$65.821=0.7549+7.2536x_{2s}$
or $72536x_{2s}=5.8272$
or $x_{2s}=5.8272$
$h_{2s}=h_f+x_{2s}(h_g-g_f)$
$=226.95+0.8033$
$(2599.1-226.95)$
$226.95+1905.54$
$2132.49kJ/kg$
${\eta }_T=\frac{(\Delta h{)}_{act}}{(\Delta h{)}_{isen}}=\frac{h_1-h_2}{h_1-h)2s}$
$0.90=\frac{3092.5-h_2}{3092.5-2132.49}$
$0.90=\frac{3092.5-h_2}{960.01}$
or $960.01\times 0.90=3092.5-h_2$
or $864=3092.5-h_2$
or $h_2=2228.5kJ/kg$
Turbine work,
$w_T=h_1-h_2=3092.5-2228.5$
$=864kJ/kg$
Pump work,
$w_P=v_3(p_1-p_2)kJ/kg$
wherer $v_3$ in $m^3/kg$ and $p_1$ and $p_2$ in $kPa$
$\therefore w_P=0.001014(4000-15)$
$=4.04kJ/kg$
Net work output of the cycle,
$w_{net}=w_T-w_P=864-4.04$
$=859.96kJ/kg$
$=860kJ/kg$