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.

Question

Determine:
(i) Pressure and temperatures at all salient points
(ii) Air standard efficiency
(iii) Mean effective pressure.

Answer 1

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.