Question

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.

Answers

GPT 3.5
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.

Related Questions