Question

An oil engine working on dual engine cycle takes air at 1.0 bar and ${16}^{\circ }C$. The maximum cycle pressure is 60 bar. The compression ratio is 14:1. The heat addition at constant pressures equal to heat addition at constant volume. What is the air standard efficiency of the cycle?

• A. $82.5\%$
• B. $64.4\%$
• C. $78.1\%$
• D. $74.6\%$

Answer 1

The correct option is B $64.4\%$

$\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$

$=(14{)}^{1.4-1}=2.873$

$T_1=16+273=289K$

$T_2=2.873\times 289=830.5K$

In 2 to 3 (V = C)

$\frac{P_3}{P_2}=\frac{T_3}{T_2}$

$\Rightarrow T_3=\frac{P_3}{P_2}\times T_2=\frac{60}{P_2}\times 830.5$

$and\frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }=(14{)}^{1.4}=40.23$

$P_2=40.23\times 1=40.23bar$

$\Rightarrow T_3=\frac{60\times 830.5}{40.23}=1238.54K$

Given

$(Q_s{)}_{V=C}=(Q_s{)}_{F=C}$

$\Rightarrow c_v(T_5-T_2)=c_p(T_4-T_3)$

$\Rightarrow 0.718(1238.54-830.5)=1.005(T_4-1238.54)$

$\Rightarrow T_4=\frac{0.718\times 408.044}{1.005}+1238.54=1530.058K$

Now $\frac{V_4}{V_3}=\frac{T_4}{T_5}=\frac{1530.058}{1238.54}=1.2354$

$\therefore \frac{V_5}{V_4}=\frac{V_1}{V_4}=\frac{V_1}{V_2}\times \frac{V_3}{V_4}=\frac{14\times 1}{1.2354}=11.332$

$\frac{T_4}{T_5}={\left(\frac{V_5}{V_4}\right)}^{\gamma -1}=(11.332{)}^{0.4}=2.640755$

$T_5=\frac{1530.058}{2.640755}=579.402K$

$\therefore \text{Heat supplied}=c_p(T_3-T_2)+c_p(T_4-T_3)$

$=2c_v(T_3-T_2)$

$=2\times 0.718(1238.54-830.5)$

$=585.94kJ/kg$

$\text{and heat rejected}=c_v(T_5-T_1)=0.718(579.402-289)=208.508kJ/kg$

$\therefore \eta =1-\frac{Q_{rej}}{Q_s}=1-\frac{208.508}{585.94}=0.6442=64.42\%$