In an ideal Otto cycle, the pressure and temperature at the beginning of compression are 97 kN/m2 and 50oC respectively. The volume ratio of compression is 5:1, and the heat input into the working fluid during the cycle is 930 kJ/kg.
Show this on PV and TS diagrams, and determine:
(a) The maximum temperature attained in the cycle;
(b) The thermal efficiency of the cycle; and
(c) The work done during the cycle per kilogram of the working fluid.
Assume Cv = 0.717 kJ/kg K, 𝛶 = 1.4 and R = 0.2837 kJ/kg
To solve this problem, we need to go through the different processes in the Otto cycle and use the given information to calculate the required values.
First, let's start by plotting the process on a PV (pressure-volume) diagram:
1. Process 1-2: Isentropic Compression
In this process, the volume reduces while the pressure increases. The volume ratio of compression is given as 5:1, so the final volume (V2) can be calculated by dividing the initial volume (V1) by 5. However, since we don't have values for V1 and V2, we will use a different approach. We know that the working fluid undergoes isentropic compression in this process, so we can use the isentropic relation:
P2/P1 = (V1/V2)^(ζ)
Where P2 and P1 are the final and initial pressures, and ζ is the ratio of specific heats. Rearranging this equation, we can solve for V2:
V2 = V1 * (P1/P2)^(1/ζ)
Given:
P1 = 97 kN/m2
P2 = ?
ζ = 1.4
Substituting the values into the equation, we can calculate V2:
V2 = V1 * (97 kN/m2 / P2)^(1/1.4)
Next, let's plot this process on the PV diagram. Draw a line between points 1 and 2, indicating the compression.
2. Process 2-3: Constant Volume Heat Addition
In this process, the volume remains constant, but the pressure and temperature increase due to the heat input. We know the heat input per kilogram of working fluid is 930 kJ/kg. Let's make this process vertical on the PV diagram from point 2 to point 3.
3. Process 3-4: Isentropic Expansion
In this process, the volume increases while the pressure decreases. We know the volume ratio of compression is 5:1, so the final volume (V4) is 5 times the final volume (V3). However, since we don't have values for V3 and V4, we will use the same approach as the compression process. We will use the isentropic relation:
P3/P4 = (V4/V3)^(ζ)
Where P3 and P4 are the initial and final pressures. Rearranging this equation, we can solve for V4:
V4 = V3 * (P3/P4)^(1/ζ)
Given:
P3 = ?
P4 = 97 kN/m2
ζ = 1.4
Substituting the values into the equation, we can calculate V4:
V4 = V3 * (P3 / 97 kN/m2)^(1/1.4)
Next, let's plot this process on the PV diagram. Draw a line between points 3 and 4, indicating the expansion.
4. Process 4-1: Constant Volume Heat Rejection
In this process, the volume remains constant, but the pressure and temperature decrease due to the heat rejection. We will make this process vertical on the PV diagram from point 4 to point 1.
Now, let's move on to the TS (temperature-entropy) diagram:
1. Process 1-2: Isentropic Compression
Since this process is isentropic, it will be represented as a vertical line on the TS diagram.
2. Process 2-3: Constant Volume Heat Addition
This process will be represented as a horizontal line on the TS diagram.
3. Process 3-4: Isentropic Expansion
Since this process is isentropic, it will be represented as a vertical line on the TS diagram.
4. Process 4-1: Constant Volume Heat Rejection
This process will be represented as a horizontal line on the TS diagram.
Now, let's calculate the required values:
(a) The Maximum Temperature Attained in the Cycle:
The maximum temperature in the cycle is the temperature at point 3. From the given information, we know the temperature at the beginning of compression (point 1) is 50°C. Since the compression and expansion processes are isentropic, the temperature change during these processes is reversible adiabatic. Using the formula for the temperature change in an isentropic process:
T2 = T1 * (P2/P1)^((ζ-1)/ζ)
Substituting the given values, we can solve for T2:
T2 = 50°C * (P2 / 97 kN/m2)^((1.4-1)/1.4)
(b) The Thermal Efficiency of the Cycle:
The thermal efficiency of the cycle is given by the formula:
η = 1 - (1 / (Vr)^(ζ-1))
Where Vr is the volume ratio of compression. Substituting the given value, we can calculate η.
(c) The Work Done During the Cycle per kilogram of the Working Fluid:
The work done during the cycle per kilogram of the working fluid is given by the formula:
W = Cv * (T3 - T2)
Where Cv is the specific heat at constant volume. Substituting the given values, we can calculate W.
0.5 kg of air (ideal gas) executes a Carnot power cycle having a thermal efficiency of 50 per cent. The heat transfer to the air during the isothermal expansion is 40 kJ. At the beginning of the isothermal expansion the pressure is 7 bar and the volume is 0.12 m3.
Determine:
(i) The maximum and minimum temperatures for the cycle in K;
(ii) The volume at the end of isothermal expansion in m3; and
(iii) The heat transfer for each of the four processes in kJ.
For air Cv = 0.721 kJ/kg K, and Cp = 1.008 kJ/kg K.
To solve this problem, let's go through the different processes in the Carnot power cycle and use the given information to calculate the required values.
Given:
m = 0.5 kg (mass of air)
η = 50% (thermal efficiency)
Q3 = 40 kJ (heat transfer during isothermal expansion)
P2 = 7 bar (pressure at the beginning of isothermal expansion)
V2 = 0.12 m^3 (volume at the beginning of isothermal expansion)
Cv = 0.721 kJ/kg K (specific heat at constant volume)
Cp = 1.008 kJ/kg K (specific heat at constant pressure)
(i) The Maximum and Minimum Temperatures:
The thermal efficiency is given by the formula:
η = 1 - (T2 / T1)
Rearranging the equation, we get:
T1 = T2 / (1 - η)
Substituting the given values, we can calculate T1.
(ii) The Volume at the End of Isothermal Expansion:
In a Carnot cycle, the expansion and compression processes are isothermal. Therefore, the temperature remains constant during these processes. We can use the ideal gas equation:
P1 * V1 = P2 * V2
Substituting the given values, we can calculate V1.
(iii) The Heat Transfer for Each of the Four Processes:
In a Carnot cycle, the heat transfer during isothermal expansion is equal to the heat transfer during isothermal compression. And the heat transfer during the adiabatic processes is zero. We can use the first law of thermodynamics:
Q = W + ΔU
For the isothermal expansion and compression processes, the work done is given by:
W = P * V * ln(V2/V1)
For the adiabatic processes, the work done is given by:
W = Cp * (T3 - T2)
Using these formulas, we can calculate the heat transfer for each process.
Once we have these calculations, we will have the answers to all of the questions. Let's plug in the given values and do the calculations.
To determine the maximum temperature, thermal efficiency, and work done during the cycle in an ideal Otto cycle, we can follow these steps:
Step 1: Calculate the intermediate pressures:
Since the compression volume ratio is 5:1, we can use the formula:
Compression ratio (rc) = V1/V2 = (V1/V2)^(1/𝛶)
where V1 and V2 are the initial and final volumes, respectively.
Given:
Volume ratio (V1/V2) = 5
rc = (5)^(1/𝛶)
rc = (5)^(1/1.4)
rc ≈ 2.808
The intermediate pressures can be calculated using the equation:
p2/p1 = (rc)^(𝛶)
Given:
p1 = 97 kN/m^2
p2/p1 = (2.808)^(1.4)
p2/p1 ≈ 4.097
p2 = p1 * (2.808)^(1.4)
p2 ≈ 398.37 kN/m^2
Step 2: Calculate the maximum temperature:
The maximum temperature (T3) can be determined using the formula:
T3 = T2 * rc^(𝛶 - 1)
where T2 is the temperature at the end of combustion.
Given:
T2 = 50°C = 50 + 273.15 = 323.15 K
T3 = 323.15 K * (2.808)^(1.4 - 1)
T3 ≈ 731.16 K
Step 3: Sketch the PV and TS diagrams:
On the PV (Pressure-Volume) diagram, plot the following points:
Point 1: (V1, p1)
Point 2: (V2, p2)
Point 3: (V2, p3) (same volume as point 2)
On the TS (Temperature-Entropy) diagram, plot the following points:
Point 1: (s1, T1)
Point 2: (s2, T2)
Point 3: (s3, T3)
Step 4: Calculate the thermal efficiency:
The thermal efficiency (η) can be determined using the equation:
η = 1 - (1/rc)^(𝛶 - 1)
Given:
rc = 2.808
η = 1 - (1/2.808)^(1.4 - 1)
η ≈ 0.582 = 58.2%
Step 5: Calculate the work done per kilogram of working fluid:
The work done per kilogram of working fluid can be calculated using the equation:
w = cp * (T3 - T2) - cv * (T1 - T4)
Given:
cp = 𝛶 * R = 1.4 * 0.2837 = 0.39658 kJ/kg K
cv = Cv = 0.717 kJ/kg K
T4 = T1 = 50°C = 323.15 K
w = 0.39658 * (731.16 - 323.15) - 0.717 * (323.15 - 323.15)
w ≈ 188.36 kJ/kg
Therefore, in an ideal Otto cycle:
(a) The maximum temperature attained in the cycle is approximately 731.16 K.
(b) The thermal efficiency of the cycle is approximately 58.2%.
(c) The work done during the cycle per kilogram of the working fluid is approximately 188.36 kJ.