Figure 1 below are the pressure-volume and temperature-entropy diagrams for a four
stroke Otto cycle which uses 1.67 moles of air as the working fluid (the gas that undergoes
the processes shown). The air has a molar specific heat ratio of 1.4 with a constant volume
molar specific heat capacity of 208 J.K-1
.mol-1
. The air is at 100 kPa and 17°C at the
beginning of the compression (1 – 2), which decreases its volume by a factor of 8. After
compression, 2320 J heat is isochorically transferred to the gas. Both compression and
expansion processes occur at constant entropy.
a. Using the information provided in the diagram, specify type(s) of thermodynamic
processes associated with the compression and expansion of the air.
b. Determine the temperature, pressure, work, and change in entropy at the end of
each thermodynamic process.
a. Both the compression (process 1-2) and expansion (process 3-4) processes are isentropic (constant entropy) processes.
b.
- At the beginning of the compression (state 1), P1 = 100 kPa and T1 = 17°C = 290 K.
- At the end of the compression (state 2), the volume is decreased by a factor of 8, so V2 = V1/8. From the PV diagram, we can see that P2 = 800 kPa. Since the compression is isentropic, we can use the equation for adiabatic processes to find T2:
T2 = T1*(P2/P1)^((γ-1)/γ) = 290*(800/100)^((1.4-1)/1.4) = 1047 K
- The heat transfer during the process 2-3 is isochoric, so ΔV = 0 and hence no work is done. Therefore, from the first law of thermodynamics:
Q23 = ΔU23 = n*Cv*(T3 - T2) = 1.67*208*(T3 - 1047)
- At the end of the isochoric process (state 3), the heat transfer increases the temperature to T3. Since the process is isentropic, we can use the equation for adiabatic processes to find P3:
P3 = P2*(T3/T2)^(γ/(γ-1)) = 800*(T3/1047)^(1.4/(1.4-1))
- During the expansion process 3-4, the gas expands while the entropy remains constant, so it is again an isentropic process. Therefore, we can use the adiabatic equation to find the final state (state 4):
T4 = T3*(P4/P3)^((γ-1)/γ)
Since the expansion ratio is also 8, we can calculate V4 = 8*V3 = 8*V2 = 8*V1/8 = V1. Using the PV equation:
P3*V3^γ = P4*V4^γ
we can solve for P4:
P4 = P3*(V3/V4)^γ = P3*(1/8)^1.4 = 36.4 kPa
- Finally, from the first law of thermodynamics applied to process 4-1, we can find the heat transfer during this isochoric process:
Q41 = ΔU41 = n*Cv*(T1 - T4) = 1.67*208*(290 - T4)
Putting all the values together:
T2 = 1047 K
P2 = 800 kPa
Q23 = 1.67*208*(T3 - 1047)
P3 = 800*(T3/1047)^(1.4/(1.4-1))
T4 = T3*(36.4/800)^((1.4-1)/1.4)
P4 = 36.4 kPa
Q41 = 1.67*208*(290 - T4)
where T3 is the temperature at the end of the isochoric process (state 3).
a. From the given information, it is stated that both the compression and expansion processes occur at constant entropy. Therefore, the compression process is an isentropic process (constant entropy), and the expansion process is also an isentropic process.
b. Let's analyze each step of the process:
1. Beginning of Compression (state 1):
- Temperature: 17°C (Convert to Kelvin: T1 = 17 + 273 = 290 K)
- Pressure: 100 kPa
- Process: Isentropic (Compression)
- Specific heat ratio (γ): 1.4
2. End of Compression (state 2):
- Volume: Decreased by a factor of 8 (V2 = V1/8)
- Process: Isentropic (Compression)
To determine the state 2 properties such as temperature, pressure, work, and change in entropy, we need additional information or data such as the polytropic efficiency of the compressor (η).
3. Isochoric Heat Transfer (state 3):
- Heat transfer: 2320 J
- Process: Isochoric (Constant volume)
4. Beginning of Expansion (state 4):
- Process: Isentropic (Expansion)
- Specific heat ratio (γ): 1.4
To determine the state 4 properties such as temperature, pressure, work, and change in entropy, we need additional information or data.
Without additional data, it is not possible to fully determine the temperature, pressure, work, and change in entropy at the end of each thermodynamic process.