Thermodynamics Miscellaneous Easy Questions and Answers

Thermodynamics Miscellaneous

In an air-standard Otto cycle, the compression ratio is 10. The condition at the beginning of the compression process is 100 kPa and 27°C. Heat added at constant volume is 1500 kJ/kg, while 700 kJ/kg of heat is rejected during the other constant volume process in the cycle. Specific gas constant for air = 0.287 kJ/kgK. The mean effective pressure (in kPa) of the cycle is

1. 103
1. 310
1. 515
1. 1032

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Here, P₁ = 100 kPa
T₁ = 27°C
V₄ = V₁ = 10
V₃ V₂

C_v = 0.287 kJ/kgK
P_m × (V₁ – V₂ )= work done
= h₁ – h₂
= 1500 – 700 = 800 kJ
Now P₁V₁ = mRT₁
or 100 × 10³ × V₁ = 1× 0.287 × 10³ × 300
or V₁ = 0.861
∴ V₂ = 0.0861
Now P_m (0.861 – 0.0861) = 800
∴ P_m = 1032 kPa

Correct Option: D

Here, P₁ = 100 kPa
T₁ = 27°C
V₄ = V₁ = 10
V₃ V₂

C_v = 0.287 kJ/kgK
P_m × (V₁ – V₂ )= work done
= h₁ – h₂
= 1500 – 700 = 800 kJ
Now P₁V₁ = mRT₁
or 100 × 10³ × V₁ = 1× 0.287 × 10³ × 300
or V₁ = 0.861
∴ V₂ = 0.0861
Now P_m (0.861 – 0.0861) = 800
∴ P_m = 1032 kPa

Which one of the following is NOT a necessary assumption for the air-standard Otto cycle?

1. All processes are internally reversible.
1. Intake and exhaust processes are constant volume heat rejection processes.
1. The combustion process is a constant volume heat addition process.
1. The working fluid is an ideal gas with constant specific heats.

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Intake process is the constant volume heat addition process
Hence (b) is not the right assumption

Correct Option: B

Intake process is the constant volume heat addition process
Hence (b) is not the right assumption

Direction: In a steam power plant operating on the Rankine cycle, steam enters the turbine at 4 MPa, 350°C and exists at a pressure of 15 kPa. Then it enters the condenser and exists as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is 90%. The thermodynamic states of water and steam are given in the table.

h is specific enthalpy, s is specific entropy and v the specific volume; subscript f and g denote saturated liquid state and saturated vapour state.

Heat supplied (kJ kg^–1) to the cycle is

1. 2372
1. 2576
1. 2863
1. 3092

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Given: Rankine cycle on T-S diagram
P₃ = 4 MPa
T₃ = 350°C
η_t = 0.9
h₃ =3092.5 kJ/kg
s₃ = s₄ = 6.5821 kJ/kgk
h₁ = 225.94 kJ/kg = h_f
h_g = 2599.1 kJ/kg

h_fg = h_g – h_f = 2373.16 kJ/kg
s₁ = s_f = 0.7549 kJ/kgK
s_g = 8.0085 kJ/kgK
s_fg = 8.0085 – 0.7949 = 7.2536 kJ/kg K
Also, s₃ = s₄ = 6.5821
∴ s₄ = s₁ + x s_fg
= 0.7549 + x (7.2536) = 6.5821
⇒ x = 0.8
h₄ = h₁ + xh_fg
= 225.94 + (0.8 × 2375.16) = 2132.4 kJ/kg
Net work output of the cycle
= η_t (h₃ – h₄)
= 0.9 (3092.5 – 2132.4)
= 864 kJ/kg
≈ 860 kJ/kg
Heat supplied to the cycle = h₃ – h₂
Since we do not have information regarding h₂, therefore neglect compressor work and take
h₂ = h₁
∴ Q_s = 3092.5 – 2866 kJ/kg = 2863 kJ/kg

Correct Option: D

Given: Rankine cycle on T-S diagram
P₃ = 4 MPa
T₃ = 350°C
η_t = 0.9
h₃ =3092.5 kJ/kg
s₃ = s₄ = 6.5821 kJ/kgk
h₁ = 225.94 kJ/kg = h_f
h_g = 2599.1 kJ/kg

h_fg = h_g – h_f = 2373.16 kJ/kg
s₁ = s_f = 0.7549 kJ/kgK
s_g = 8.0085 kJ/kgK
s_fg = 8.0085 – 0.7949 = 7.2536 kJ/kg K
Also, s₃ = s₄ = 6.5821
∴ s₄ = s₁ + x s_fg
= 0.7549 + x (7.2536) = 6.5821
⇒ x = 0.8
h₄ = h₁ + xh_fg
= 225.94 + (0.8 × 2375.16) = 2132.4 kJ/kg
Net work output of the cycle
= η_t (h₃ – h₄)
= 0.9 (3092.5 – 2132.4)
= 864 kJ/kg
≈ 860 kJ/kg
Heat supplied to the cycle = h₃ – h₂
Since we do not have information regarding h₂, therefore neglect compressor work and take
h₂ = h₁
∴ Q_s = 3092.5 – 2866 kJ/kg = 2863 kJ/kg

The net work output (kJ kg^–1) of the cycle is

1. 498
1. 775
1. 860
1. 957

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Given: Rankine cycle on T-S diagram
P₃ = 4 MPa
T₃ = 350°C
η_t = 0.9
h₃ =3092.5 kJ/kg
s₃ = s₄ = 6.5821 kJ/kgk
h₁ = 225.94 kJ/kg = h_f
h_g = 2599.1 kJ/kg

h_fg = h_g – h_f = 2373.16 kJ/kg
s₁ = s_f = 0.7549 kJ/kgK
s_g = 8.0085 kJ/kgK
s_fg = 8.0085 – 0.7949 = 7.2536 kJ/kg K
Also, s₃ = s₄ = 6.5821
∴ s₄ = s₁ + x s_fg
= 0.7549 + x (7.2536) = 6.5821
⇒ x = 0.8
h₄ = h₁ + xh_fg
= 225.94 + (0.8 × 2375.16) = 2132.4 kJ/kg
Net work output of the cycle
= η_t (h₃ – h₄)
= 0.9 (3092.5 – 2132.4)
= 864 kJ/kg
≈ 860 kJ/kg
Heat supplied to the cycle = h₃ – h₂
Since we do not have information regarding h₂, therefore neglect compressor work and take
h₂ = h₁
∴ Q_s = 3092.5 – 2866 kJ/kg = 2863 kJ/kg

Correct Option: A

Given: Rankine cycle on T-S diagram
P₃ = 4 MPa
T₃ = 350°C
η_t = 0.9
h₃ =3092.5 kJ/kg
s₃ = s₄ = 6.5821 kJ/kgk
h₁ = 225.94 kJ/kg = h_f
h_g = 2599.1 kJ/kg

h_fg = h_g – h_f = 2373.16 kJ/kg
s₁ = s_f = 0.7549 kJ/kgK
s_g = 8.0085 kJ/kgK
s_fg = 8.0085 – 0.7949 = 7.2536 kJ/kg K
Also, s₃ = s₄ = 6.5821
∴ s₄ = s₁ + x s_fg
= 0.7549 + x (7.2536) = 6.5821
⇒ x = 0.8
h₄ = h₁ + xh_fg
= 225.94 + (0.8 × 2375.16) = 2132.4 kJ/kg
Net work output of the cycle
= η_t (h₃ – h₄)
= 0.9 (3092.5 – 2132.4)
= 864 kJ/kg
≈ 860 kJ/kg
Heat supplied to the cycle = h₃ – h₂
Since we do not have information regarding h₂, therefore neglect compressor work and take
h₂ = h₁
∴ Q_s = 3092.5 – 2866 kJ/kg = 2863 kJ/kg

A thermal power plant operates on a regenerative cycle with a single open feedwater heater, as shown in the figure. For the state points shown, the specific enthalpies are: h₁ = 2800 kJ/kg and h₂ = 200 kJ/kg. The bleed to the feedwater heater is 20% of the boiler steam generation rate. The specific enthalpy at state 3 is

1. 720 kJ/kg
1. 2280 kJ/kg
1. 1500 kJ/kg
1. 3000 kJ/kg

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h₃ = 0.2h₁ + h₂ × 0.8
= 560 + 160 = 720 kJ/kg

Correct Option: A

h₃ = 0.2h₁ + h₂ × 0.8
= 560 + 160 = 720 kJ/kg

	V₄	=	V₁	= 10
	V₃		V₂

	V₄	=	V₁	= 10
	V₃		V₂

Thermodynamics Miscellaneous

Thermodynamics Miscellaneous

Thermodynamics

Correct Option: D

Correct Option: B

Correct Option: D

Correct Option: A

Correct Option: A