In an engine working on Dual cycle, the temperature and pressure at the beginning of the cycle are 90°C and 1 bar respectively. The compression ratio is 9. The maximum pressure is limited to 68 bar and total heat supplied per kg of air is 1750 kJ.
Determine:
(i) Pressure and temperatures at all salient points
(ii) Air standard efficiency
(iii) Mean effective pressure.
To solve this problem, we will use the air standard dual cycle equations:
(i) Pressure and temperatures at all salient points:
1. Beginning of the cycle (Point 0):
Temperature (T0) = 90°C + 273.15 = 363.15 K
Pressure (P0) = 1 bar
2. Point 1 (End of the isentropic compression):
The compression ratio (r) is given as 9.
From the isentropic process equation for compression, we have:
P1 / P0 = (V0 / V1)^(γ)
where P1 and P0 are the pressures at points 1 and 0 respectively, and V0 and V1 are the specific volumes at points 0 and 1 respectively.
Since the process is isentropic, γ is the heat capacity ratio and is given as 1.4 for air.
Given that V1 / V0 = 1 / r = 1 / 9, we can substitute this value into the equation to solve for P1:
P1 / 1 bar = (1 / 9)^(1.4)
P1 = 1 bar * (1 / 9)^(1.4)
P1 = 0.4461 bar
To calculate T1, we use the equation:
P0 * V0^γ-1 = P1 * V1^γ-1
Since V1 / V0 = 1 / r = 1 / 9, we can rewrite the equation as:
P0 * V0^γ-1 = P1 * (1 / r)^(γ-1) * V0^γ-1
Substituting the given values, we get:
1 bar * (V0)^1.4 = 0.4461 bar * (1 / 9)^(1.4) * (V0)^1.4
Simplifying the equation, we find:
V0 = (0.4461 / 1)^(1.4 / 0.4)
V0 = 0.3699 m^3/kg
Using the ideal gas equation P0 * V0 = m * R * T0, where m is the mass of air per kg and R is the gas constant for air, we can solve for T0:
T0 = P0 * V0 / (m * R)
T0 = 1 bar * 0.3699 m^3/kg / (1 kg * 287.1 J/(kg*K))
T0 = 1.2857 K
Now, we can calculate T1 using the equation:
T1 = T0 * (P1 / P0)^((γ - 1) / γ)
T1 = 1.2857 K * (0.4461 bar / 1 bar)^((1.4 - 1) / 1.4)
T1 = 0.7967 K
3. Point 2 (End of the constant volume heat addition):
Using the First Law of Thermodynamics, we have:
Qin - Wc = ΔU
Qin = Wc + ΔU
Qin = cp * (T3 - T2) + cv * (T4 - T3)
Given that ΔU = cv * (T4 - T3) and cp = γ * R / (γ - 1), we can substitute the values into the equation:
Qin = γ * R / (γ - 1) * (T3 - T2) + cv * (T4 - T3)
Qin = γ * 287.1 J/(kg*K) / (1.4 - 1) * (T3 - T2) + 0.718 * (T4 - T3)
1750 kJ = γ * 287.1 J/(kg*K) / 0.4 * (T3 - 0.7967 K) + 0.718 * (T4 - T3)
Using the constant specific heat model, we have:
Qin = cp * m * (T3 - T2) + cp * m * (T4 - T3)
Given that cp = γ * R / (γ - 1), m = 1 kg, and cp = cv + R, we can substitute the values into the equation:
Qin = γ * R / (γ - 1) * (T3 - 0.7967) + (γ * R / (γ - 1) - R) * (T4 - T3)
1750 kJ = γ * 287.1 J/(kg*K) / (1.4 - 1) * (T3 - 0.7967 K) + (γ * 287.1 J/(kg*K) / (1.4 - 1) - 287.1 J/(kg*K)) * (T4 - T3)
Now, we have two equations with two unknowns: T3 and T4. We can solve these equations simultaneously to find the values of T3 and T4.
(ii) Air standard efficiency:
The air standard efficiency (η) is the ratio of the net work done to the heat supplied:
η = 1 - (1 / r)^(γ - 1)
η = 1 - (1 / 9)^(1.4 - 1)
(iii) Mean effective pressure:
The mean effective pressure (MEP) is defined as the average pressure over the entire cycle, given by the equation:
MEP = (p3 - p2) - (p4 - p1) / (V3 - V2)
MEP = (P3 - P2) - (P4 - P1) * Vc/v
where Vc is the clearance volume and v = Vc/V1.
Substituting the known values, we can calculate the MEP.
To solve this problem, we need to use the equations and formulas for the Dual cycle.
(i) Pressure and temperatures at all salient points:
Step 1: Determine the variables:
- Initial temperature (T1) = 90°C
- Initial pressure (P1) = 1 bar
- Compression Ratio (r) = 9
- Maximum Pressure (P3) = 68 bar
- Total heat supplied per kg of air (qin) = 1750 kJ
Step 2: Calculate the temperatures:
a) At the end of the isentropic compression (T2):
From the isentropic compression process:
P1 * (V1/V2)^(γ-1) = P2
Where γ is the specific heat ratio (Cp/Cv) and can be taken as 1.4 for air.
Substituting the values:
1 bar * (9)^(1.4-1) = P2
P2 ≈ 10.52 bar
b) At the end of the constant volume heat addition process (T3):
From the Dual cycle, the heat added is given as:
qin = Cp * (T3 - T2)
Rearranging the equation:
T3 = T2 + (qin / Cp)
Substituting the values:
T3 = T2 + (1750 kJ / Cp)
Since Cp = γ * R, where R is the specific gas constant for air, approximately 0.287 kJ/kg·K, we can rewrite the equation as:
T3 = T2 + (1750 kJ / (γ * R))
Substituting the values and solving for T3:
T3 = T2 + (1750 kJ) / (1.4 * 0.287 kJ/kg·K)
T3 ≈ T2 + 5024.648 K
c) At the end of the isentropic expansion (T4):
From the isentropic expansion process:
P3 * (V3/V4)^(γ-1) = P4
Substituting the values:
68 bar * (1/9)^(1.4-1) = P4
P4 ≈ 7.56 bar
d) At the end of the constant volume heat rejection process and the end of the cycle (T5):
Using the definition of the compression ratio:
r = V1/V2 = V4/V5
Substituting and rearranging the equation:
V4 = r * V5
From the isentropic expansion process:
P3 * (V3/V4)^(γ-1) = P4
Substituting the values and solving for V3 and V4:
68 bar * (1/9)^(1.4-1) = P4
V3 = r * V4 * (P3/P4)^((γ-1)/γ)
V3 ≈ 9 * V4 * (1/7.56)^((1.4-1)/1.4)
Since V4 = V5, we can solve for V5:
V5 = V3 / (r * (P3/P4)^((γ-1)/γ))
V5 ≈ V3 / (9 * (1/7.56)^((1.4-1)/1.4))
Using the ideal gas equation:
P5 * V5 = m * R * T5
Solving for T5:
T5 = P5 * V5 / (m * R)
Substituting the values and solving for T5:
T5 ≈ P5 * (V3 / (9 * (1/7.56)^((1.4-1)/1.4))) / (m * R)
(ii) Air standard efficiency:
The air standard efficiency (η) is given by the formula:
η = 1 - (1 / r)^(γ-1)
Substituting the value of r and γ:
η = 1 - (1 / 9)^(1.4-1)
(iii) Mean effective pressure:
Mean effective pressure (MEP) is given by the formula:
MEP = (qin - qout) / Vd
Where Vd is the displacement volume.
Substituting the value of qin:
MEP = (1750 kJ - qout) / Vd
To determine qout, we need to calculate the heat rejected per kg of air.
Using the formula for heat rejected in the constant volume process:
qout = Cv * (T5 - T4)
Substituting the values:
qout = Cv * (T5 - T4)
Finally, substitute the value of qout in the MEP equation to calculate the mean effective pressure.
Note: Cv is the specific heat at constant volume for air, approximately 0.718 kJ/kg·K.