A dual cycle which operates on air with a compression ratio of 16 has alow pressure of 200 kpa and a low temperature of 200 c . If tha cutoff ratio is 2 and the pressure ratio is 1.3 calculate tha thermal efficiency the heat input amd tha work output

1. The thermal efficiency of the cycle is 57.913%

2. The heat input is 32165.68 kJ/kg

3. The work output is -18.628.11 kJ/kg

Explanation:

1. T₁ = 200C = 473.15 K

The thermal efficiency of the cycle is given by the relation;

Where:

k = Pressure ratio = p₃/p₂ = 1.3

γ = Specific heat ratio = 1.4

β = v₄/v₃

= v₁/v₂

Which gives;

η = (1.3*(16/2)^1.4 - 1)/(((1.3 - 1) + 1.4*1.3*((16/2)-1))*16^0.4) = 0.57913

η = 57.913%

The thermal efficiency of the cycle = 57.913%

2. The heat input is given by the following relation;

Q = cv(T₂ - T₁) + cp(T₄ - T₃)

T₂/T₁ = (v₁/v₂)^(γ - 1) = 16^0.4 = 3.03

Therefore, T₂ = T₁×3.03= 473.15*3.03= 1434.323 K

T₃/T₂ = p₃/p₂ = 1.3

T₃ = T₂ ×1.3 = 1434.323 *1.3 = 1864.62 K

v₄/v₃ = T₄/T₃ = 16/2 = 8

T₄ = T₃×8 = 1864.62 × 8 = 14916.959 K

At 473.15 K, cv = 0.7386  cp = 1.026 kJ/(kg*K)

At 1434.323 K, cv = 0.9158 cp = 1.203

At 1864.62 K, cv = 0.9535 cp = 1.239

At 1864.62 K, cv = 0.874 cp = 1.24

Taking average values, we have

(0.7386   + 0.9158)/2= 0.8272 kJ/(kg*K)

cp = 1.24

The heat input is then found as follows;

Q = 0.874*(1434.323 - 473.15) + 2.4*(14916.96 - 1864.62) = 32165.68 kJ/kg

The heat input = 32165.68 kJ/kg

3. The work output is found as follows;

Given that η = -W/Q, we have;

-W = Q × η = 0.57913*32165.68 = -18.628.11 kJ/kg.

The work output = -18.628.11 kJ/kg.

answer:

a

explanation: