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

(a)(i) Effect of pressure ratio on work and efficiency

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) Efficiency at maximum work

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) Net heat transfer to the air

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) Temperature rise of the air

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) Drag on the windshield by the Chilton–Colburn analogy

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.