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Answer to Question #132889 in Mechanical Engineering for Mohammed

Question #132889
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 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.
1
2020-09-16T05:13:50-0400

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

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