Thermodynamics Miscellaneous
- Consider a simple gas turbine (Brayton) cycle and a gas turbine cycle with perfect regeneration. In both the cycles, the pressure ratio is 6 and the ratio of the specific heats of the working medium is 1.4. The ratio of minimum to maximum temperatures is 0.3 (with temperatures expressed in K) in the regenerative cycle. The ratio of the thermal efficiency of the simple cycle to that of the regenerative cycle is._____.
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Brayton cycle
ηBrayton = 1 – 1 (rρ)γ-1/γ
given rρ = 6, γ = 1.4∴ ηBrayton = 1 – 1 = 0.4006 (6)1.4-1/1.4 
effectiveness ε = Cρ(T5 - T2) = 1 Cρ(T3 - T2)
T3 = T5Given Tmin = 0.3 = T1 Tmax T4
Wnet = WT – WC
= Cρ (T4 – T5) – Cρ (T2 – T1)
= Cρ (T4 – T3) – Cρ (T2 – T2 )[∵ T4 = T3]ηRegenerative = Wnet Heat-supplied = Cρ[(T4 – T3) - (T2 - T1)] Cρ(T4 – T3) = 1 - 
T2 - T1 
T4 - T3 = 1 - T1 T2 - 1 T1 1 - T4 1 - T3 T4 ηRegenerative = 1 - T2 - 1 T1 T1 T4 1 - T5 T4 ηRegenerative = 1 - - 1 T1 r(ρ)γ-1/γ T4 1 - 1 r(ρ)γ-1/γ = 1 - T1 r(ρ)γ-1/γ T4
= 1 - 0.3(6)1.4-1/1.4 = 0.4994ηBrayton = 0.4006 = 0.8021 ηRegenerative 0.4994 Correct Option: D
Brayton cycle
ηBrayton = 1 – 1 (rρ)γ-1/γ
given rρ = 6, γ = 1.4∴ ηBrayton = 1 – 1 = 0.4006 (6)1.4-1/1.4 effectiveness ε = Cρ(T5 - T2) = 1 Cρ(T3 - T2)
T3 = T5Given Tmin = 0.3 = T1 Tmax T4
Wnet = WT – WC
= Cρ (T4 – T5) – Cρ (T2 – T1)
= Cρ (T4 – T3) – Cρ (T2 – T2 )[∵ T4 = T3]ηRegenerative = Wnet Heat-supplied = Cρ[(T4 – T3) - (T2 - T1)] Cρ(T4 – T3) = 1 - T2 - T1 T4 - T3 = 1 - T1 T2 - 1 T1 1 - T4 1 - T3 T4 ηRegenerative = 1 - T2 - 1 T1 T1 T4 1 - T5 T4 ηRegenerative = 1 - - 1 T1 r(ρ)γ-1/γ T4 1 - 1 r(ρ)γ-1/γ = 1 - T1 r(ρ)γ-1/γ T4
= 1 - 0.3(6)1.4-1/1.4 = 0.4994ηBrayton = 0.4006 = 0.8021 ηRegenerative 0.4994
- In a Rankine cycle, the enthalpies at turbine entry and outlet are 3159 kJ/kg and 2187 kJ/ kg, respectively. If the specific pump work is 2 kJ/kg, the specific steam consumption (in kg/ kWh) of the cycle based on net output is.
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Specific steam consumption = 3600 WT - WP
Now, WT = h2 – h1
= 3159 – 2187 kJ/kg
∴ WT = 972 kJ/kg
and WP = 20 kJ/kgThus specific steam consumption = 3600 kg/kW-h 972 – 20 = 3.7815 kg/kW.h 972 - 20 Correct Option: C
Specific steam consumption = 3600 WT - WP
Now, WT = h2 – h1
= 3159 – 2187 kJ/kg
∴ WT = 972 kJ/kg
and WP = 20 kJ/kgThus specific steam consumption = 3600 kg/kW-h 972 – 20 = 3.7815 kg/kW.h 972 - 20
- Steam with specific enthalpy (h) 3214 kJ/kg enters an adiabatic turbine operating at steady state with a flow rate 10 kg/s. As it expands, at a point where his 2920 kJ/kg, 1.5 kg/s is extracted for heating purposes. The remaining 8.5 kg/s further expands to the turbine exit, where h = 2374 kJ/kg. Neglecting changes in kinetic and potential energies, the net power output (in kW) of the turbine is_____.
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ẇ = ṁtot(h1 - h2) + (ṁtot - ṁout)(h2 - h3)
ẇ = 10(3214 – 2920) + 8.5(2920 – 2374)
ẇ = 7581 kWCorrect Option: C
ẇ = ṁtot(h1 - h2) + (ṁtot - ṁout)(h2 - h3)
ẇ = 10(3214 – 2920) + 8.5(2920 – 2374)
ẇ = 7581 kW
- For a gas turbine power plant, identify the correct pair of statements. P. Smaller in size compared to steam power plant for same power output Q. Starts quickly compared to steam power plant R. Works on the principle of Rankine cycle S. Good compatibility with solid fuel
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Steam power plants are bulky due to presence of boiler and condenser. Gas turbines are compact, as compressor s and turbines are coupled on a commonshaft. In steam power plants, boiler takes lot of time to get started, as compared to Gas Turbines.
Correct Option: A
Steam power plants are bulky due to presence of boiler and condenser. Gas turbines are compact, as compressor s and turbines are coupled on a commonshaft. In steam power plants, boiler takes lot of time to get started, as compared to Gas Turbines.
- In an ideal Brayton cycle, atmospheric air (ratio of specific heats, cρ /cv = 1.4, specific heat at constant pressure = 1.005 kJ/kgK) at 1 bar and 300 K is compressed to 8 bar. The maximum temperature in the cycle is limited to 1280 K. If the heat is supplied at the rate of 80 MW, the mass flow rate (in kg/s) of air required in the cycle is _________.
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Given:
γ = 1.4,
cρ = 1.005 kJ/kg.k
P1 = 1 bar,
P2 = 8 bar
T1 = 300 K,
T3 = 1280 Kr = P2 = 8 P1
Qin = 80000 kW.
(1-2) isentropic processP2 = (rρ)γ-(1/γ) P1
T2 = 300 × (8)1.4 - 1/1.4
= 543.43 k
Qin = mcp × (T3 - T2)
80,000 = m × 1.005 (1280 – 300)
m = 108.07 kg/s.Correct Option: B
Given:
γ = 1.4,
cρ = 1.005 kJ/kg.k
P1 = 1 bar,
P2 = 8 bar
T1 = 300 K,
T3 = 1280 Kr = P2 = 8 P1
Qin = 80000 kW.
(1-2) isentropic processP2 = (rρ)γ-(1/γ) P1
T2 = 300 × (8)1.4 - 1/1.4
= 543.43 k
Qin = mcp × (T3 - T2)
80,000 = m × 1.005 (1280 – 300)
m = 108.07 kg/s.
