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. The compression and expansion processes shown in the pressure-volume diagram are adiabatic (constant entropy) and isentropic (constant volume) processes, respectively.
b.
Process 1-2: Adiabatic compression
From the pressure-volume diagram, we can see that the compression ratio is 8. Therefore, the final volume at state 2 is 1/8 of the initial volume at state 1:
V2 = V1/8 = 0.02155 m3/mol
Using the ideal gas law:
P1V1 = nRT1
P2V2 = nRT2
Since the amount of air (n = 1.67 mol) is constant throughout the cycle, we can combine these equations to get:
P1V1/T1 = P2V2/T2
Since the compression is adiabatic (Q = 0) and the air is an ideal gas with a constant specific heat ratio, we can use the following relation:
P2/P1 = (V1/V2)γ
where γ = c_P/c_V = 1.4 is the molar specific heat ratio. Substituting the values and solving for P2, we get:
P2 = P1*(V1/V2)γ = 1296.4 kPa
The work done during the compression process is given by:
W12 = -nRT1*(V2 - V1)/(γ - 1)
Substituting the values:
W12 = -1.67*8.31*290*(0.02155 - 0.2155)/(1.4 - 1) = 2212.6 J
The change in entropy during the adiabatic compression is zero:
ΔS12 = 0
From the temperature-entropy diagram, we can see that the temperature at state 2 can be found by following the adiabatic curve that passes through state 1:
T2 = T1*(V1/V2)γ-1 = 773.3 K
Process 2-3: Isochoric heat addition
During this process, heat is transferred to the air at constant volume (isochoric process), so the work done is zero:
W23 = 0
The change in entropy during an isochoric process is given by:
ΔS23 = Q23/T2 = 2320/(1.67*8.31*773.3) = 0.405 J/K
From the ideal gas law:
P3 = P2
T3 = (P3V3)/(nR) = (P2V2)/(nR)
Since the volume is constant during this process, V3 = V2 = 0.02155 m3/mol. Substituting the values, we get:
T3 = (1296.4*0.02155)/(1.67*8.31) = 342.6 K
Process 3-4: Adiabatic expansion
Similar to the adiabatic compression process, we can use the adiabatic relation:
P4/P3 = (V3/V4)γ
where V4 is the final volume at state 4. Since the expansion ratio is the same as the compression ratio, V4 = 8V3 = 0.1724 m3/mol. Substituting the values, we get:
P4 = P3*(V3/V4)γ = 101.2 kPa
The work done during the expansion process is given by:
W34 = -nRT3*(V3 - V4)/(γ - 1)
Substituting the values:
W34 = -1.67*8.31*342.6*(0.1724 - 0.02155)/(1.4 - 1) = -2516.4 J
The change in entropy during the adiabatic expansion is zero:
ΔS34 = 0
From the temperature-entropy diagram, we can see that the temperature at state 4 can be found by following the adiabatic curve that passes through state 3:
T4 = T3*(V3/V4)γ-1 = 200.0 K
a. The compression process (1-2) is an isentropic process, as it occurs at constant entropy. The expansion process (3-4) is also an isentropic process, as it occurs at constant entropy.
b. At the end of the compression process (2), the temperature can be determined by using the formula:
T2 = T1 * (V1 / V2)^(γ-1)
where T1 is the initial temperature, V1 is the initial volume, V2 is the final volume, and γ is the molar specific heat ratio. Plugging in the given values:
T1 = 17°C + 273 = 290 K
V1 = 8 * V2
γ = 1.4
T2 = 290 K * (1/8)^(1.4 - 1)
T2 ≈ 163 K
The pressure at the end of the compression process (2) can be determined using the relationship between pressure and volume for an isentropic process:
P2 = P1 * (V1 / V2)^γ
where P1 is the initial pressure, V1 is the initial volume, V2 is the final volume, and γ is the molar specific heat ratio. Plugging in the given values:
P1 = 100 kPa
V1 = 8 * V2
γ = 1.4
P2 = 100 kPa * (1/8)^1.4
P2 ≈ 1.16 kPa
The work done during the compression process can be determined using the equation:
W = P1 * V1 * (1 - (V2 / V1)^γ / (γ - 1))
where P1 is the initial pressure, V1 is the initial volume, V2 is the final volume, and γ is the molar specific heat ratio. Plugging in the given values:
P1 = 100 kPa
V1 = 8 * V2
γ = 1.4
W = 100 kPa * 8 * V2 * (1 - (1/8)^1.4 / (1.4 - 1))
W ≈ 548 kJ
The change in entropy during the compression process is zero, as it is an isentropic process.
At the end of the expansion process (4), the temperature is the same as at the end of the compression process, which is approximately 163 K.
The pressure at the end of the expansion process (4) can be determined using the relationship between pressure and volume for an isentropic process, similar to the compression process:
P4 = P3 * (V3 / V4)^γ
where P3 is the initial pressure, V3 is the initial volume, V4 is the final volume, and γ is the molar specific heat ratio. Plugging in the given values:
P3 = P2 ≈ 1.16 kPa
V3 = V2
γ = 1.4
P4 = 1.16 kPa * (8/1)^1.4
P4 ≈ 18.54 kPa
The work done during the expansion process can also be determined using the equation mentioned earlier:
W = P3 * V3 * (1 - (V4 / V3)^γ / (γ - 1))
where P3 is the initial pressure, V3 is the initial volume, V4 is the final volume, and γ is the molar specific heat ratio. Plugging in the given values:
P3 = 1.16 kPa
V3 = V2
γ = 1.4
W = 1.16 kPa * V2 * (1 - (1/8)^1.4 / (1.4 - 1))
W ≈ 180 kJ
The change in entropy during the expansion process is also zero, as it is an isentropic process.
a. From the information provided, we can determine the type(s) of thermodynamic processes associated with the compression and expansion of the air.
1. Compression Process (1-2): This process is represented by a vertical line (constant volume) on the pressure-volume diagram. The process is isochoric (constant volume) because the volume decreases by a factor of 8 while the pressure increases.
2. Heat Addition Process (2-3): This process is represented by a horizontal line (constant pressure) on the pressure-volume diagram. The process is isobaric (constant pressure) because the pressure remains the same while the volume increases due to heat transfer.
3. Expansion Process (3-4): This process is represented by a vertical line (constant volume) on the pressure-volume diagram. The process is isochoric (constant volume) because the volume remains the same while the pressure decreases.
4. Heat Rejection Process (4-1): This process is represented by a horizontal line (constant pressure) on the pressure-volume diagram. The process is isobaric (constant pressure) because the pressure remains the same while the volume decreases due to heat transfer.
b. Let's determine the temperature, pressure, work, and change in entropy at the end of each process:
1. Compression Process (1-2):
- Temperature (T2): End of compression process (point 2) = 17°C (given)
- Pressure (P2): End of compression process (point 2) = 8 times the initial pressure = 8 * 100 kPa = 800 kPa
- Work (W12): Area under the curve between points 1-2 on the PV diagram (requires numerical integration)
- Change in Entropy (S2): Since the process is isochoric, the change in entropy is zero.
2. Heat Addition Process (2-3):
- Temperature (T3): End of heat addition process (point 3) = Temperature at point 2 + Heat added/constant volume molar specific heat capacity
- Pressure (P3): End of heat addition process (point 3) = same as pressure at the end of compression (P2)
- Work (W23): Since the process is isobaric, work can be calculated as P * ΔV = P * (initial volume - final volume)
- Change in Entropy (S3): Since the process is isobaric, ΔS = Q / T = heat added / Temperature
3. Expansion Process (3-4):
- Temperature (T4): End of expansion process (point 4) = Temperature at point 3
- Pressure (P4): End of expansion process (point 4) = 1/8 times the pressure at the end of compression (P2) = 1/8 * 800 kPa = 100 kPa
- Work (W34): Area under the curve between points 3-4 on the PV diagram (requires numerical integration)
- Change in Entropy (S4): Since the process is isochoric, the change in entropy is zero.
4. Heat Rejection Process (4-1):
- Temperature (T1): End of heat rejection process (point 1) = Temperature at point 4
- Pressure (P1): End of heat rejection process (point 1) = same as initial pressure (P2)
- Work (W41): Since the process is isobaric, work can be calculated as P * ΔV = P * (initial volume - final volume) = P * (8 times the initial volume)
- Change in Entropy (S1): Since the process is isobaric, ΔS = Q / T = heat rejected / Temperature
Please note that the exact calculations for work and change in entropy in each process require numerical integration on the diagrams provided.
To answer this question, we need to analyze the information provided in the pressure-volume and temperature-entropy diagrams for the four-stroke Otto cycle. Let's break down the steps to find the required information:
a. Determining the type(s) of thermodynamic processes:
In the given information, it is mentioned that both the compression and expansion processes occur at constant entropy. A constant entropy process is represented as a vertical line in the temperature-entropy diagram. Therefore, the compression and expansion processes are both isentropic processes.
b. Determining the temperature, pressure, work, and change in entropy at the end of each process:
To determine the required values at the end of each process, we need to refer to the given diagrams. Let's go through each process step by step:
1. Compression (Process 1-2):
- From the pressure-volume (P-V) diagram, we can see that the volume decreases by a factor of 8 during compression.
- As the compression occurs at constant entropy, we can also determine the temperature change by referring to the temperature-entropy (T-S) diagram. The temperature decreases during compression.
2. Isochoric heat transfer (Process 2-3):
- The heat transfer occurs at constant volume (isochorically), so there is no change in volume during this process.
- The amount of heat transferred is given as 2320 J.
3. Expansion (Process 3-4):
- The expansion occurs at constant entropy. From the T-S diagram, we can determine the temperature change during expansion.
- The volume increases during expansion.
As the exact diagrams are not included in the question, we cannot determine the specific values without visual aid. I suggest referring to the provided diagrams to determine the temperature, pressure, work, and change in entropy at the end of each process.