Answer to Question #132889 in Mechanical Engineering for Mohammed

Question

0.5 kg of air is taken through a constant pressure cycle. At the beginning of
adiabatic compression are 96.5 kN/m 2 and 15 0 C. The pressure ratio of
compression is 6. Constant pressure heat addition occurs after adiabatic
compression until the volume is doubled. Take γ = 1.4 and R= 0.289 kJ/kg K.
Determine for the cycle:
(a) The pressure, volume and temperature at cycle state points. (b) The heat received.
(c) The thermal efficiency. (d) The work ratio.
(e) The heat rejected.

Answer 1

mass of air =0.5 kg, P₁₌96.5 kN/m²,T₁= 15 ⁰C,compression ratio=6, Heat addition takes place at constant pressure to volume become doubled.

(a)

$\frac{P_2}{P_1}= r_p$ ,

$P_2= r_p\times P_1=6\times 96.5= 579 \frac{kN}{m^2}$

$\frac{T_1}{T_2}=(\frac{P_1}{P_2})^{\frac{\gamma -1}{\gamma}}$

$\frac{288}{T_2}=(\frac{96.5}{579})^{\frac{1.4 -1}{1.4}}$

$T_2=480.8 K$

$\frac{P_1}{P_2}^{\frac{1}{\gamma}}= \frac{V_2}{V_1}$

for finding v1 we use ideal gas equation

$P_1V_1=nRT_1$

$95.5\times V_1=1\times 0.289\times 288$

$V_1=0.8625 m^3$

$\frac{96.5}{579}^{\frac{1}{1.4}}= \frac{V_2}{0.8625}$

$V_2=0.2398 m^3$

(b)$H_1= m C_p \Delta T=0.5\times 1.0115\times 192.8= 97.508 kJ$

(c) Thermal efficiency

$\eta_t= 1- (\frac{V_2}{V_1})^{\gamma -1}$

$\eta = 0.40$

(d)

Work ratio=$1- (\frac{T_1}{T_2})\times r^{\frac{\gamma -1}{\gamma}}$

Work ratio= 0.01

for heat rejected

$\eta = 1- \frac{Q_2}{Q_1}$

$Q_2= 58.5 kJ$ it is heat rejected