Question

A mass of 5 kilograms of air in a piston-cylinder assembly undergoes two processes in series from an initial state where p1 = 4 MPa, T1 = 264°C: Process 1–2: Constant-temperature expansion until the volume is twice the initial volume. Process 2–3: Constant-volume heating until the pressure is again 4 MPa. Sketch the two processes in series on a p-v diagram. Assuming ideal gas behavior, determine (a) the pressure at state 2, in MPa, (b) the temperature at state 3, in oC, and for each of the processes (c) the work and heat transfer, each in kJ.

Answer 1

a)

`p_2= 2MPa`

b)

`T_3 =1074^oC`

c)

`W_(12)=534.14kJ`

`W_(23)=0`

`Q_(12)=534.14kJ`

`Q_(23)=2192.7kJ`

Step-by-step explanation:

Given that:

P₁ = 4 MPa

T₁ = 264°C

mass (m) = 5kg

R = 0.287 kJ/kg.K

a) To calculate the pressure at state 2
`p_2`, we use ideal gas equation and since final volume is twice the initial volume,
`(V_2)/(V_1) = 1`

Therefore:
`p_2=p_1(V_1)/(V_2)=4MPa *0.5 = 2MPa`

b) To calculate the temperature at state 3
`T_3`, we use ideal gas equation and since
`T_2=T_1=264^oC=537K` and
`p_3=p_1`

Therefore:
`T_3=T_2(p_3)/(p_2)=537(4MPa)/(2MPa) =1074^oC`

c) To calculate work (
`W_(12)`), we use the isothermal equation

Therefore:
`W_(12)=mRT_1ln((V_2)/(V_1))=5*0.287*537*ln((4MPa)/(2MPa) ) =534.14kJ`

Since no volume change between process 2 and 3, work done (
`W_(23)=0`)

Between process 1 and 2, no temperature change,

Therefore:
`Q_(12)=W_(12)=534.14kJ`

To calculate the heat transfer between process 2 and 3, we use the

`u_3=827.88(kJ)/(kg) \\u_2=389.34(kJ)/(kg)`

`Q_(23)=m(u_3-u_2)+W_(23)=5(827.88-389.34)+0=2192.7kJ`