Need 2 and 4 stat A system undergoes a cycle consisting of four processes. All u
ID: 2995357 • Letter: N
Question
Need 2 and 4 stat
A system undergoes a cycle consisting of four processes. All units are KJ. On a separate sheet of paper, copy the following table and fill in the missing values. A system consisting of 7 kg of air is initially at 200 kPa and 30 degree C. Determine the heat transfer necessary to increase the pressure by a factor of tow, at constant volume. (Note: Air under these conditions behave as a ideal gas with constant specific heat). (Note: Assume constant volume) Find: Q = ? [in kJ] The average person emits approximately 400 Btu of heat per one hour. There are 36 in an unventilated room measuring 30 ft Times 25 ft 10 ft. Estimate the increase in of the air after 10 minutes, assuming constant pressure conditions. (Note: 10 minutes = (1/6) hour) (Note: Assume an initial temperature of 70 degree F and an initial pressure of 14.7 psia.)(Note: Air under these conditions behaves as a ideal gas with constant specific head.) Find: the temperature increase Delta T = ? [in degree F] 4. If 0.05 kg of air undergoes the cycle shown in the figure, calculate the work output. Find: W = ? [in kJ] HOMEWORK: Hand in solutions, showing work and calculations, to:Explanation / Answer
2)
In Const Vol
W = 0
P1 = 200 kPa
T1 = 30 deg = 303 K
P2 = 400 kPa
In a const vol
P1/T1 = P2/T2
T2 = (P2/P1)*T1
T2 = 606 K
Heat input = m*Cv*dT = 7*0.717*(606-303) = 1520.76 kJ
4)
V1 = 0.003 m^3
V2 = V3 = 0.03 m^3
P3 = 100 kPa
In 3-1
P*V^1.4 = C
P3*V3^1.4 = P1*V1^1.4
P1 = P3*(V3/V1)^1.4
P1 = 100*(10)^1.4
P1 = 2511.89 kPa
P1*V1 = n*R*T1
T1 = 2511.89*10^3*0.003*28.9/(0.05*10^3*8.314)
T1 = 524 K
P3*V3 = n*R*T3
T3 = 100*10^3*0.03*28.9/(0.05*8.314)
T3 = 208.56 K
In 2-3
T2 = T1 = 524 K
P2/T2 = P3/T3
P2 = P3*(T2/T3)
P2 = 100*(524/208.56)
P2 = 251.25 kPa
Work done in 1 to 2
W12 = n*R*T2*ln(P1/P2) = (0.05*10^3/28.9)*8.314*524*ln(2511.89/251.25)
W12 = 17.354 kJ
Work done in 2 to 3
W23 = 0
Work done in 3 to 1
W31 = -(P3*V3-P1*V1)/(1-1.4) = -(100*0.03-2511.89*0.003)/(1-1.4)
W31 = -11.339 kJ
W = W12 + W23 + W31 = 6.015 kJ
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