Question

Common Data Question:
In a simple Bryaton cycle, the pressure ratio is $8$ and temperature at the entrance of compressor and turbine are $300K$ and $1400K$ respectively. Both compressor and gas turbine have isentropic efficiencies equal to $80$%. or the gas, assume a constant value of $c_p$ (specific heat at constant pressure) equal to $1kJ/kgK$ and ratio of specific heats as $14$. Neglect changes in kinetic and potential energies

The power required by the compressor in $kW/kg$ of gas flow are is

• A. $194.7$
• B. $243.4$
• C. $304.3$
• D. $378.5$

Answer 1

The correct option is C $304.3$

Given data:
Pressure ratio,
$r_p=8$
$T_1=300K;T_3=1400K$
${\eta }_c={\eta }_T=0.8$
$C_p=1kJ/kgK$
$\gamma =1.4$

For adiabatic isentropic process $1-2s$
$\frac{T_{2s}}{T_1}={\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}$
$\frac{T_{2s}}{300}=(r_p{)}^{\frac{\gamma -1}{\gamma }}=(8{)}^{\frac{1.4-1}{1.4}}=(8{)}^{0.2857}$
$\frac{T_{2s}}{300}=1.811$
or $T_{2s}=300\times 1.811=543.3K$
${\eta }_C=\frac{\text{Isentropic rise in temperature}}{\text{Actual rise in temperature}}$
$=\frac{T_{2s}-T_1}{T_2-T_1}$
$\therefore ,0.8=\frac{543.3-300}{T_2-300}$
$T_2-300=\frac{243.3}{0.8}=304.12$
or $T_2=604.12K$
Power requried by compressor,
$w_C=c_p(T_2-T_1)$
$=1(604.12-300)$
$304.12kJ/kg$