(a) A lump of ice with a mass of 1·5 kg at an initial temperature of 260 K melts at the pressure of 1 bar as a result of heat transfer from the environment. After some time, the resulting water attains the temperature of environment, 293 K. Calculate

Question

(a) A lump of ice with a mass of 1·5 kg at an initial temperature of 260 K melts at the pressure of 1 bar as a result of heat transfer from the environment. After some time, the resulting water attains the temperature of environment, 293 K. Calculate the entropy production associated with this process. The latent heat of fusion of ice is 334 kJ/kg, the specific heat of ice and water are 2·07 kJ/kg-K and 4·2 kJ/kg-K, respectively. Assume that ice melts at 273·15 K. (10 marks)

(b) Downstream of a normal shock wave, the characteristic Mach number M*_y = 0·5 and the stagnation pressure is 2 bar. Determine the following:
(i) The Mach number M_y downstream of the shock
(ii) The stagnation pressure upstream of the shock
The fluid is air and γ = 1·4, and R = 0·287 kJ/kg-K.
Normal shock table and Isentropic flow table provided at the end, may be used. (10 marks)

(c) Show, in the form of a table, how the following parameters change (increase/decrease/remain constant) across a normal shock wave:
Static pressure, Static temperature, Stagnation pressure, Stagnation temperature, Stagnation density, Mach number, Entropy and Stagnation enthalpy (10 marks)

(d) At a certain instant of time, the temperature across a large wall of thickness 50 cm is given as T(x) = 90 - 80x + 16x² + 32x³ - 25x⁴ where x is in metres and measured from the left face of the wall as shown in the figure below and T is in °C. If the area of the wall is 10 m² and there is no generation of heat in the wall, compute the following:
(i) The rate of heat entering and leaving the wall
(ii) The rate of heat energy stored in the wall
(iii) The rate of temperature change with time at x = 0 and x = 0·5 m (10 marks)

(e) A 120 W electric bulb has a filament temperature of 3005 °C. Assuming the filament to be black, calculate (i) the diameter of the filament wire if the length is 250 mm and (ii) the efficiency of the bulb based on visible radiation if the visible radiation lies in the wavelength range from 0·4 μm to 0·75 μm. Assume Stefan-Boltzmann constant (σ) as 5.67×10⁻⁸ W/m²-K⁴. The black body radiation functions are given in the table:
| λT(μm-K) | f₀→λ |
|----------|------|
| 1300 | 0.004317 |
| 1400 | 0.007791 |
| 2400 | 0.140266 |
| 2500 | 0.161366 | (10 marks)

UPSC Answer Check · Accepted Answer

(a) Take the environment as a reservoir at T₀ = 293 K. The ice warms from 260 K to 273.15 K, melts at 273.15 K, then the resulting water warms from 273.15 K to 293 K.
ΔS_ice = m c_i ln(T_m/T_i) = 1.5 × 2.07 × ln(273.15/260) = 0.1532 kJ/K
ΔS_melt = m L/T_m = 1.5 × 334/273.15 = 1.8342 kJ/K
ΔS_water = m c_w ln(293/273.15) = 1.5 × 4.2 × ln(293/273.15) = 0.4420 kJ/K

Thus,
ΔS_system = 0.1532 + 1.8342 + 0.4420 = 2.4293 kJ/K.

Heat taken from the environment:
Q = 1.5 × 2.07 × (273.15 − 260) + 1.5 × 334 + 1.5 × 4.2 × (293 − 273.15)
Q = 40.8308 + 501 + 125.055 = 666.8858 kJ

Entropy change of the environment:
ΔS_env = −Q/T₀ = −666.8858/293 = −2.2761 kJ/K

Entropy production:
S_gen = ΔS_system + ΔS_env = 2.4293 − 2.2761 = 0.1532 kJ/K

Final: S_gen ≈ 0.153 kJ/K, or 0.1533 kJ/K to four decimal places.

(b)(i) For a normal shock, the characteristic Mach numbers satisfy M*ₓ M*ᵧ = 1.
Given M*ᵧ = 0.5, therefore M*ₓ = 2.0.

The relation between M and M* is:
M*² = 2.4 M²/(2 + 0.4 M²)

For downstream:
0.25 = 2.4 Mᵧ²/(2 + 0.4 Mᵧ²)
0.5 + 0.1 Mᵧ² = 2.4 Mᵧ²
Mᵧ² = 0.5/2.3 = 5/23 = 0.21739
Mᵧ = √(5/23) = 0.4663

Final: Mᵧ = 0.4663.

(b)(ii) For upstream, M*ₓ = 2, so:
4 = 2.4 Mₓ²/(2 + 0.4 Mₓ²)
8 + 1.6 Mₓ² = 2.4 Mₓ²
Mₓ² = 10
Mₓ = √10 = 3.1623

Using the normal-shock stagnation-pressure ratio:
p₀ᵧ/p₀ₓ = [2.4 Mₓ²/(2 + 0.4 Mₓ²)]^3.5 × [2.4/(2γMₓ² − 0.4)]^2.5

For Mₓ² = 10:
p₀ᵧ/p₀ₓ = [24/6]^3.5 × [2.4/27.6]^2.5
= 4^3.5 × (2/23)^2.5
= 0.28541

Given p₀ᵧ = 2 bar:
p₀ₓ = 2/0.28541 = 7.0075 bar

Final: p₀ₓ ≈ 7.008 bar.

(c) Across a normal shock wave:
Static pressure: increases
Static temperature: increases
Stagnation pressure: decreases
Stagnation temperature: remains constant
Stagnation density: decreases
Mach number: decreases
Entropy: increases
Stagnation enthalpy: remains constant

(d) The temperature profile is:
T(x) = 90 − 80x + 16x² + 32x³ − 25x⁴ °C

Differentiating:
dT/dx = −80 + 32x + 96x² − 100x³ °C/m
d²T/dx² = 32 + 192x − 300x² °C/m²

At x = 0:
dT/dx = −80 K/m
d²T/dx² = 32 K/m²

At x = 0.5 m:
dT/dx = −80 + 16 + 24 − 12.5 = −52.5 K/m
d²T/dx² = 32 + 96 − 75 = 53 K/m²

Using Fourier’s law, Q = −kA dT/dx.

(d)(i) Rate of heat entering at left face:
Q_in = −k(10)(−80) = 800k W

Rate of heat leaving at right face:
Q_out = −k(10)(−52.5) = 525k W

Here k is the thermal conductivity in W/m-K. Since k is not supplied in the question, the numerical W values cannot be fixed without it.

(d)(ii) Rate of heat energy stored:
Q_stored = Q_in − Q_out = 800k − 525k = 275k W

(d)(iii) For no heat generation:
∂T/∂t = α d²T/dx²

At x = 0:
∂T/∂t = 32α °C/s

At x = 0.5 m:
∂T/∂t = 53α °C/s

Here α = k/(ρc) is the thermal diffusivity in m²/s. Since α is not supplied, numerical °C/s values cannot be computed.

Final: Q_in = 800k W, Q_out = 525k W, Q_stored = 275k W; ∂T/∂t at x = 0 is 32α °C/s and at x = 0.5 m is 53α °C/s.

(e)(i) Filament temperature:
T = 3005 + 273.15 = 3278.15 K

For a black filament:
P = σA_s T⁴ = σπdL T⁴

T⁴ = 1.15482 × 10¹⁴ K⁴
σT⁴ = 5.67 × 10⁻⁸ × 1.15482 × 10¹⁴ = 6.5478 × 10⁶ W/m²

A_s = 120/(6.5478 × 10⁶) = 1.8327 × 10⁻⁵ m²

L = 250 mm = 0.25 m

d = A_s/(πL) = 1.8327 × 10⁻⁵/(π × 0.25)
d = 2.333 × 10⁻⁵ m = 0.02333 mm = 23.33 μm

Final: d ≈ 2.33 × 10⁻⁵ m.

(e)(ii) Visible range: 0.4 μm to 0.75 μm.

λ₁T = 0.4 × 3278.15 = 1311.26 μm-K
λ₂T = 0.75 × 3278.15 = 2458.61 μm-K

Interpolating from the given blackbody radiation functions:
f₁ ≈ 0.004317 + (11.26/100)(0.007791 − 0.004317) = 0.004708
f₂ ≈ 0.140266 + (58.61/100)(0.161366 − 0.140266) = 0.152633

Fraction of radiation in visible range:
f_visible = f₂ − f₁ = 0.152633 − 0.004708 = 0.147925

Efficiency = 0.147925 × 100 = 14.79%

Visible power = 0.147925 × 120 = 17.75 W

Final: visible-radiation efficiency ≈ 14.79%.

(a) A lump of ice with a mass of 1·5 kg at an initial temperature of 260 K melts at the pressure of 1 bar as a result of heat…

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