(a) Consider the system shown in Fig. 1(a). The two chambers initially have equal volumes of 28 litres and contain air (C_p = 1·005 kJ/kg-K and C_v = 0·717 kJ/kg-K) and hydrogen (C_p = 14·32 kJ/kg-K and C_v = 10·17 kJ/kg-K), respectively. The chamber

Question

(a) Consider the system shown in Fig. 1(a). The two chambers initially have equal volumes of 28 litres and contain air (C_p = 1·005 kJ/kg-K and C_v = 0·717 kJ/kg-K) and hydrogen (C_p = 14·32 kJ/kg-K and C_v = 10·17 kJ/kg-K), respectively. The chambers are separated by a frictionless piston which is non-heat-conducting. Both the gases are initially at 140 kPa and 40 °C. Heat is added to the air side until the pressure of both the gases reaches 280 kPa. All outside walls of the chambers are insulated except for the surface where heat is added to air. Calculate the final temperature of the air. (10 marks)

(b) What is 'choked flow' in a convergent-divergent nozzle? Explain, with diagram, the effect of pressure ratio on exit velocity of compressible gas in a convergent-divergent nozzle. (10 marks)

(c) An axial flow compressor with inlet and outlet angles of 40° and 15°, respectively has been designed for 50% reaction. The compressor has a pressure ratio of 6 : 1 and overall isentropic efficiency of 0·80, when inlet static temperature is 41 °C. The blade speed and axial velocity are constant throughout. Assuming a value of 210 m/s for blade speed, find the number of stages required if the work done factor is 0·88 for all the stages. Take Cp = 1·005 kJ/kg-K and γ = 1·4 for air. (10 marks)

(d) Hot water is flowing through a pipe made of cast iron having thermal conductivity of 52 W/m-°C, with an average velocity of 1·5 m/s. The inner and outer diameters of the pipe are 3 cm and 3·5 cm, respectively. The pipe passes through a 15 m long section of a basement whose temperature is 15 °C. The temperature of the water drops from 70 °C to 67 °C as it passes through the basement. The heat transfer coefficient on the inner surface of the pipe is 400 W/m²-°C. Determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe. (10 marks)

(e) (i) Define the total and spectral black body emissive powers. How are they related to each other? (5 marks)

(ii) Consider two identical bodies, one at 1000 K and the other at 1500 K. Which body emits more radiation in the shorter wavelength region? Which body emits more radiation at a wavelength of 20 μm? (5 marks)

UPSC Answer Check · Accepted Answer

(a)
For the hydrogen side, the walls are insulated and the piston is frictionless and non-heat-conducting, so hydrogen undergoes a reversible adiabatic compression. Let hydrogen be denoted by H.

Initial state: P₁ = 140 kPa, T₁ = 40 °C = 313.15 K, V₁ = 28 L = 0.028 m³ for each gas.
Final pressure: P₂ = 280 kPa.

For hydrogen,
γ_H = C_p/C_v = 14.32/10.17 = 1.40806.

Using the reversible adiabatic relation P V^γ = constant:
V_H2 = V₁ (P₁/P₂)^(1/γ_H)
= 28 × (140/280)^(1/1.40806)
= 28 × 0.5^0.71020
= 17.11 L.

Total volume = 28 + 28 = 56 L.
Final air volume:
V_air2 = 56 − 17.11 = 38.89 L.

For air, using the ideal gas relation between initial and final states:
T_air2 = T₁ (P₂/P₁)(V_air2/V₁)
= 313.15 × 2 × (38.89/28)
= 869.8 K approximately.

Final temperature of air ≈ 869.8 K ≈ 596.7 °C.

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(b)
Choked flow: In a convergent-divergent nozzle, choked flow occurs when the throat reaches sonic velocity, M = 1. At this condition, the mass flow rate reaches its maximum value for the given upstream stagnation pressure and temperature. Once choked, further reduction of downstream pressure cannot increase the mass flow rate; it only changes the pressure distribution in the diverging section.

For air, the critical pressure ratio is
p*/p₀ = (2/(γ+1))^(γ/(γ−1)) ≈ 0.528 for γ = 1.4.

Schematic diagram:

p₀, T₀ → converging section → throat A_t, M = 1 when choked → diverging section → exit A_e

Effect of pressure ratio p_b/p₀ on exit velocity:
For p_b/p₀ = 1, there is no flow.
As p_b/p₀ decreases but remains above the critical value, the flow is subsonic throughout. The exit velocity increases as p_b/p₀ decreases.
At the critical pressure ratio, the throat becomes sonic. Choking starts, and mass flow rate becomes maximum.
For p_b/p₀ below critical, the diverging section can accelerate the flow supersonically. A normal shock may appear in the diverging section depending on back pressure.
At the design pressure ratio, the nozzle is correctly expanded and the exit velocity reaches its maximum supersonic value for the given area ratio.
If p_b/p₀ is reduced below the design value, the nozzle becomes underexpanded. The exit velocity at the nozzle exit remains approximately constant, and further expansion occurs outside the nozzle.

Thus, the exit velocity generally increases as p_b/p₀ decreases, but after choking the mass flow becomes constant and the diverging section controls supersonic expansion.

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(c)
For a 50% reaction axial compressor stage with constant axial velocity C_a and blade speed U, the velocity triangles are symmetrical:
α₁ = β₂ and α₂ = β₁.

Given β₁ = 40°, β₂ = 15°, U = 210 m/s.

Using the velocity triangle relation:
C_a = U/(tan β₁ + tan β₂).

tan 40° = 0.83910, tan 15° = 0.26795.

C_a = 210/(0.83910 + 0.26795)
= 210/1.10705
= 189.69 m/s.

Ideal work per stage from Euler work equation:
w_stage = U C_a (tan β₁ − tan β₂)
= 210 × 189.69 × (0.83910 − 0.26795)
= 210 × 189.69 × 0.57115
= 22 752 J/kg
= 22.75 kJ/kg.

With work-done factor 0.88:
w_actual = 0.88 × 22.75 = 20.02 kJ/kg per stage.

At inlet, for 50% reaction, α₁ = β₂ = 15°.
C₁ = C_a/cos 15° = 189.69/0.96593 = 196.39 m/s.

T₁ = 41 °C = 314.15 K.

Inlet stagnation temperature:
T₀₁ = T₁ + C₁²/(2C_p)
= 314.15 + 196.39²/(2 × 1005)
= 314.15 + 19.19
= 333.34 K.

For pressure ratio 6, isentropic outlet stagnation temperature:
T₀₂s = T₀₁ × 6^((γ−1)/γ)
= 333.34 × 6^0.2857
= 333.34 × 1.66846
= 556.16 K.

Overall isentropic efficiency η_c = 0.80:
T₀₂ − T₀₁ = (T₀₂s − T₀₁)/η_c
= (556.16 − 333.34)/0.80
= 278.53 K.

Total work required:
w_total = C_p (T₀₂ − T₀₁)
= 1.005 × 278.53
= 279.92 kJ/kg.

Number of stages:
N = w_total/w_actual
= 279.92/20.02
= 13.98.

Therefore, number of stages required = 14.

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(d)
Assume water properties at average temperature 68.5 °C:
ρ = 1000 kg/m³, C_p = 4.18 kJ/kg·°C.

Pipe inner diameter D_i = 0.03 m, outer diameter D_o = 0.035 m, length L = 15 m.
Water velocity V = 1.5 m/s.

Flow area:
A_c = πD_i²/4 = π(0.03)²/4 = 7.0686 × 10⁻⁴ m².

Mass flow rate:
ṁ = ρ A_c V
= 1000 × 7.0686 × 10⁻⁴ × 1.5
= 1.0603 kg/s.

Heat lost by water:
Q = ṁ C_p ΔT
= 1.0603 × 4.18 × (70 − 67)
= 13.296 kW
= 13 296 W.

Using LMTD for water 70 °C to 67 °C and surroundings at 15 °C:
ΔT₁ = 70 − 15 = 55 °C,
ΔT₂ = 67 − 15 = 52 °C.

ΔT_lm = (55 − 52)/ln(55/52)
= 3/ln(1.05769)
= 53.49 °C.

Total thermal resistance:
R_total = ΔT_lm/Q
= 53.49/13 296
= 4.023 × 10⁻³ K/W.

Inner convection resistance:
A_i = πD_iL = π × 0.03 × 15 = 1.4137 m².
R_i = 1/(h_i A_i)
= 1/(400 × 1.4137)
= 1.768 × 10⁻³ K/W.

Pipe wall resistance:
R_wall = ln(D_o/D_i)/(2πkL)
= ln(0.035/0.03)/(2π × 52 × 15)
= 3.145 × 10⁻⁵ K/W.

Outer resistance:
R_o = R_total − R_i − R_wall
= 4.023 × 10⁻³ − 1.768 × 10⁻³ − 0.03145 × 10⁻³
= 2.223 × 10⁻³ K/W.

Outer surface area:
A_o = πD_oL = π × 0.035 × 15 = 1.6493 m².

Combined outer heat transfer coefficient:
h_o = 1/(R_o A_o)
= 1/(2.223 × 10⁻³ × 1.6493)
= 272.8 W/m²·°C.

Combined convection and radiation heat transfer coefficient at outer surface ≈ 273 W/m²·°C.

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(e)(i)
Total black body emissive power, E_b, is the total radiant energy emitted per unit time per unit surface area of a black body over all wavelengths. Its unit is W/m².

Spectral black body emissive power, E_bλ, is the radiant energy emitted per unit time per unit surface area per unit wavelength interval at wavelength λ. Its unit is W/m²·μm.

They are related by:
E_b = ∫ from 0 to ∞ E_bλ dλ.

Using Planck’s distribution:
E_bλ = C₁/[λ⁵(exp(C₂/(λT)) − 1)]
where C₁ = 2πhc₀² and C₂ = hc₀/k.

Integration over all wavelengths gives the Stefan-Boltzmann law:
E_b = σT⁴.

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(e)(ii)
By Wien’s displacement law:
λ_max T = 2898 μm·K.

For T = 1000 K:
λ_max = 2898/1000 = 2.90 μm.

For T = 1500 K:
λ_max = 2898/1500 = 1.93 μm.

Thus, the 1500 K body has its peak radiation at a shorter wavelength, so it emits more radiation in the shorter wavelength region.

At 20 μm, using Planck’s law:
E_λ ∝ 1/[λ⁵(exp(C₂/(λT)) − 1)].

Ratio at 20 μm:
E_λ(1500)/E_λ(1000)
= [exp(C₂/(20 × 1000)) − 1]/[exp(C₂/(20 × 1500)) − 1]
= [exp(0.7194) − 1]/[exp(0.4796) − 1]
= (2.053 − 1)/(1.615 − 1)
= 1.053/0.615
≈ 1.71.

So at a wavelength of 20 μm, the 1500 K body emits more radiation than the 1000 K body, by a factor of about 1.71.

(a) Consider the system shown in Fig. 1(a). The two chambers initially have equal volumes of 28 litres and contain air (C_p =…

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