(a) (i) How do the specific work output and efficiency vary with pressure ratio in a gas turbine?

(ii) Prove that the efficiency of a gas turbine corresponding to the maximum work done in a Brayton cycle is given by the relation

η_w max = 1 - 1/(√t

Question

(a) (i) How do the specific work output and efficiency vary with pressure ratio in a gas turbine?

(ii) Prove that the efficiency of a gas turbine corresponding to the maximum work done in a Brayton cycle is given by the relation

η_w max = 1 - 1/(√t)

where t is the ratio of the maximum and minimum temperatures.

(20 marks)

(b)

A solar collector, as shown in Fig. 3(b) below, having dimensions as 1 m wide and 5 m long, has constant spacing of 3 cm between the glass cover and the collector plate. Air enters the collector at 30 °C and at a rate of 0·15 m³/s through the 1 m wide edge and flows along the 5 m long passageway. If the average temperatures of the glass cover and the collector plate are 20 °C and 60 °C, respectively, determine (i) the net rate of heat transfer to the air in the collector and (ii) the temperature rise of air as it flows through the collector.

Fig. 3(b)

The properties of air at 1 atm and an estimated average temperature of 35 °C may be taken as :

ρ = 1·145 kg/m³, k = 0·02625 W/m-°C, ν = 1·655×10⁻⁵ m²/s, Cₚ = 1007 J/kg-°C, Pr = 0·7268

20 marks

Fig. 3(b)

The properties of air at 1 atm and an estimated average temperature of 35 °C may be taken as :

ρ = 1·145 kg/m³, k = 0·02625 W/m-°C, ν = 1·655×10⁻⁵ m²/s,
Cₚ = 1007 J/kg-°C, Pr = 0·7268

(20 marks)

(c)

A windshield of a car, having dimensions as 0·6 m high and 1·8 m long, is electrically heated and is subjected to parallel winds at 1 atm, 0 °C and 80 km/hr. The electrical power consumption is observed to be 50 W, when the exposed surface temperature of the windshield is 4 °C. Disregarding the radiation and heat transfer from the inner surface and using the momentum heat transfer analogy, determine the drag force the wind exerts on the windshield. The properties of air at 0 °C and 1 atm may be taken as :

ρ = 1·292 kg/m³, Cₚ = 1·006 kJ/kg-K, Pr = 0·7362

10 marks

ρ = 1·292 kg/m³, Cₚ = 1·006 kJ/kg-K, Pr = 0·7362

(10 marks)

UPSC Answer Check · Accepted Answer

(a)(i) For an ideal Brayton gas-turbine cycle with fixed minimum temperature T₁ and maximum temperature T₃, let pressure ratio r_p = p₂/p₁, a = (γ−1)/γ, and x = r_p^a = T₂/T₁. Then
Compressor work: w_C = Cp T₁(x−1)
Turbine work: w_T = Cp T₃(1−1/x)
Net specific work: w_net = Cp[T₃(1−1/x) − T₁(x−1)]

As r_p increases from 1, w_net rises from zero, reaches a maximum at x = √(T₃/T₁), then decreases; beyond a high pressure ratio the compressor work becomes dominant and net work may become zero or negative. The ideal thermal efficiency is

η = 1 − T₁/T₂ = 1 − 1/x = 1 − 1/r_p^a

so η increases monotonically with pressure ratio and approaches 1. Thus, higher r_p always improves ideal efficiency, but specific work is maximum only at an optimum pressure ratio depending on T₃/T₁.

(a)(ii) For maximum work, differentiate w_net/Cp = T₃ + T₁ − T₁x − T₃/x with respect to x:

d/dx [T₃ + T₁ − T₁x − T₃/x] = −T₁ + T₃/x² = 0

Hence T₃/T₁ = x² = t, so x = √t. At this condition,

η_wmax = 1 − 1/x = 1 − 1/√t

Therefore, η_wmax = 1 − 1/√t, valid for an ideal Brayton cycle with constant Cp, no pressure losses, and no regeneration.

(b)(i) From Fig. 3(b), width W = 1 m, length L = 5 m, spacing H = 0.03 m.
Cross-sectional area:

A_c = WH = 1 × 0.03 = 0.03 m²

Velocity:

u = V_dot/A_c = 0.15/0.03 = 5 m/s

Hydraulic diameter:

D_h = 4A_c/P = 4(0.03)/[2(1+0.03)] = 0.05825 m

Reynolds number:

Re = uD_h/ν = 5 × 0.05825 / (1.655×10⁻⁵) = 1.76×10⁴

Flow is turbulent. Using Dittus–Boelter for heating, n = 0.4:

Nu = 0.023 Re^0.8 Pr^0.4
Nu = 0.023 × (1.76×10⁴)^0.8 × 0.7268^0.4 = 50.43

Heat-transfer coefficient:

h = Nu k/D_h = 50.43 × 0.02625 / 0.05825 = 22.7 W/m²°C

Each heat-transfer surface area:

A_s = WL = 1 × 5 = 5 m²

Using estimated air average temperature 35 °C, net heat to air is heat gained from collector plate minus heat lost to glass cover:

q_net = hA_s[(60−35) − (35−20)]
q_net = 22.7 × 5 × 10 = 1135 W

Answer (b)(i): q_net ≈ 1.14 kW.

(b)(ii) Mass flow rate:

m_dot = ρV_dot = 1.145 × 0.15 = 0.17175 kg/s

Energy balance:

q_net = m_dot Cp ΔT

ΔT = q_net/(m_dot Cp) = 1135/(0.17175 × 1007) = 6.57 °C

Answer (b)(ii): temperature rise ≈ 6.57 °C; outlet air temperature ≈ 36.57 °C.

(c) Wind speed:

U = 80 km/hr = 80 × 1000/3600 = 22.22 m/s

Temperature difference:

ΔT = T_s − T_∞ = 4 − 0 = 4 K

Using the Chilton–Colburn momentum–heat transfer analogy:

St Pr^(2/3) = C_f/2

Thus drag force:

F_D = C_f(1/2 ρU²A) = St Pr^(2/3) ρU²A

But heat transfer is:

q = hAΔT = St ρU Cp A ΔT

Eliminating St:

F_D = q Pr^(2/3) U/(Cp ΔT)

Now:

Pr^(2/3) = 0.7362^(2/3) = 0.8153

Cp = 1.006 kJ/kg-K = 1006 J/kg-K

F_D = 50 × 0.8153 × 22.22 / (1006 × 4) = 0.225 N

Answer (c): drag force ≈ 0.225 N.

(a) (i) How do the specific work output and efficiency vary with pressure ratio in a gas turbine? (ii) Prove that the efficiency…

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