Question

An air-standard dual cycle has a compression ratio of 9.1 and displacement of Vd = 2.2 L. At the beginning of compression, p1 = 95 kPa, and T1 = 290 K. The heat addition is 4.25 kJ, with one quarter added at constant volume and the rest added at constant pressure. Determine: a) each of the unknown temperatures at the various states, in K. b) the net work of the cycle, in kJ. c) the power developed at 3000 cycles per minute, in kW. d) the thermal efficiency. e) the mean effective pressure, in kPa.

Answer 1

a) T₂ is 701.479 K

T₃ is 1226.05 K

T₄ is 2350.34 K

T₅ is 1260.56 K

b) The net work of the cycle in kJ is 2.28 kJ

c) The power developed is 114.2 kW

d) The thermal efficiency,
`\eta _(dual)` is 53.78%

e) The mean effective pressure is 1038.25 kPa

Step-by-step explanation:

a) Here we have;

`(T_(2))/(T_(1))=\left ((v_(1))/(v_(2)) \right )^(\gamma -1) = \left (r \right )^(\gamma -1) = \left ((p_(2))/(p_(1)) \right )^{(\gamma -1)/(\gamma )}`

Where:

p₁ = Initial pressure = 95 kPa

p₂ = Final pressure =

T₁ = Initial temperature = 290 K

T₂ = Final temperature

v₁ = Initial volume

v₂ = Final volume

`v_d` = Displacement volume =

γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air

r = Compression ratio = 9.1

Total heat added = 4.25 kJ

1/4 × Total heat added =
`c_v * (T_3 - T_2)`

3/4 × Total heat added =
`c_p * (T_4 - T_3)`

`c_v` = Specific heat at constant volume = 0.718×2.821× 10⁻³

`c_p` = Specific heat at constant pressure = 1.005×2.821× 10⁻³

v₁ - v₂ = 2.2 L

`\left (v_(1))/(v_(2)) \right =r \right = 9.1`

v₁ = v₂·9.1

∴ 9.1·v₂ - v₂ = 2.2 L = 2.2 × 10⁻³ m³

8.1·v₂ = 2.2 × 10⁻³ m³

v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³

v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³

Plugging in the values, we have;

`{T_(2)}= T_(1) * \left (r \right )^(\gamma -1) = 290 * 9.1^(1.4 - 1) = 701.479 \, K`

From;

`\left ((p_(2))/(p_(1)) \right )^{(\gamma -1)/(\gamma )}= \left (r \right )^(\gamma -1)` we have;

`p_(2) = p_(1)} * \left (r \right )^(\gamma ) = 95 * \left (9.1 \right )^(1.4) = 2091.13 \ kPa`

1/4×4.25 =
`0.718 * 2.821 * 10^(-3)* (T_3 - 701.479)`

∴ T₃ = 1226.05 K

Also;

3/4 × Total heat added =
`c_p * (T_4 - T_3)` gives;

3/4 × 4.25 =
`1.005 * 2.821 * 10^(-3) * (T_4 - 1226.05)` gives;

T₄ = 2350.34 K

`(T_(4))/(T_(5))=\left ((v_(5))/(v_(4)) \right )^(\gamma -1) = \left ((r)/(\rho ) \right )^(\gamma -1)`

`\rho = (T_4)/(T_3) = (2350.34)/(1226.04) = 1.92`

`T_(5) = (T_(4))/(\left ((r)/(\rho ) \right )^(\gamma -1))= (2350.34 )/(\left ((9.1)/(1.92 ) \right )^(1.4-1)) =1260.56 \ K`

b) Heat rejected =
`c_v * (T_5 - T_1)`

`Therefore \ heat \ rejected = 0.718 * 2.821 * 10^(-3)* (1260.56 - 290) = 1.966 kJ`

The net work done = Heat added - Heat rejected

∴ The net work done = 4.25 - 1.966 = 2.28 kJ

The net work of the cycle in kJ = 2.28 kJ

c) Power = Work done per each cycle × Number of cycles completed each second

Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second

Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW

d)

`Thermal \ efficiency, \, \eta _(dual) = (Work \ done)/(Heat \ supplied) = (2.28)/(4.25) * 100 = 53.74 \%`

The thermal efficiency,
`\eta _(dual)` = 53.78%

e) The mean effective pressure,
`p_m`, is found as follows;

`p_m = (W)/(v_1 - v_2) =(2.28)/(2.2 * 10^(-3)) = 1038.25 \ kPa`

The mean effective pressure = 1038.25 kPa.