(a) 10 g of water at 20 °C is converted into ice at –10 °C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant at 4·2 J/g-K and that of ice to be half of this value and taking the latent heat of fusion of i

Question

(a) 10 g of water at 20 °C is converted into ice at –10 °C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant at 4·2 J/g-K and that of ice to be half of this value and taking the latent heat of fusion of ice at 0 °C to be 335 J/g, calculate the total entropy change of the system. (20 marks)

(b) A shaft having diameter of 5 cm rotates in a bearing made of cast iron. The shaft rotates at 4500 r.p.m. The bearing is 15 cm long, 8 cm outer diameter and has thermal conductivity of 70 W/m-K. There is a uniform clearance between the shaft and the bearing of 0·6 mm. The clearance is filled with a lubricating oil having thermal conductivity of 0·14 W/m-K and dynamic viscosity of 0·03 N-s/m². The bearing is cooled externally by a liquid, and its outer surface is maintained at 40 °C. Disregarding the heat conduction through the shaft and assuming only one-dimensional heat transfer, determine (i) the rate of heat transfer to the coolant, (ii) the surface temperature of the shaft and (iii) the mechanical power wasted by the viscous dissipation in the lubricating oil. (20 marks)

(c) Air (C_p = 1·05 kJ/kg-K, γ = 1·38) at 3 bar pressure and T = 600 K is flowing with a velocity of 180 m/s inside a 20 cm diameter duct. Calculate the—

(i) mass flow rate;

(ii) stagnation temperature;

(iii) Mach number;

(iv) stagnation pressure assuming flow to be (1) compressible and (2) incompressible. (10 marks)

UPSC Answer Check · Accepted Answer

(a) Entropy is a state property, so replace the irreversible process by a reversible path. For the water–ice system:
Cooling liquid water from 20 °C to 0 °C:
ΔS₁ = m c_w ln(T₂/T₁) = 10 × 4.2 × ln(273/293)
= 42 × ln(0.931741) = −2.969 J/K.
Freezing at 0 °C:
ΔS₂ = −m L/T₂ = −10 × 335/273 = −12.271 J/K.
Cooling ice from 0 °C to −10 °C:
ΔS₃ = m c_ice ln(T₃/T₂) = 10 × 2.1 × ln(263/273)
= 21 × ln(0.963370) = −0.784 J/K.

Total entropy change of the system:
ΔS_total = −2.969 − 12.271 − 0.784 = −16.024 J/K.

The negative sign is expected because the system cools and freezes, rejecting heat.

(b) Using the thin-film (Couette) approximation, valid because clearance c = 0.6 mm is small compared with shaft radius r₁ = 25 mm.

Given:
r₁ = 0.025 m, c = 0.0006 m,
r₂ = bearing inner radius = r₁ + c = 0.0256 m,
r₃ = bearing outer radius = 0.04 m,
L = 0.15 m, N = 4500/60 = 75 rev/s.

Shaft surface velocity:
V = ω r₁ = 2πN r₁ = 2π × 75 × 0.025 = 3.75π = 11.781 m/s.

Shear stress in oil:
τ = μV/c = 0.03 × 11.781/0.0006 = 589.05 N/m².

Shaft area:
A = 2π r₁ L = 2π × 0.025 × 0.15 = 0.023562 m².

Viscous force:
F = τA = 589.05 × 0.023562 = 13.879 N.

Mechanical power wasted:
P = FV = 13.879 × 11.781 = 163.51 W.

(b)(i) In steady state, all viscous heat goes to the coolant:
Rate of heat transfer to coolant = 163.51 W.

(b)(ii) Since shaft conduction is neglected, the oil–shaft interface is adiabatic. For the oil film with internal generation q‴:

q‴ = Q/[π(r₂² − r₁²)L] = 163.51/[π(0.0256² − 0.025²) × 0.15]
= 1.143 × 10⁷ W/m³.

For radial conduction with internal generation and dT/dr = 0 at r₁:
T(r₁) − T(r₂) = q‴/(4k_oil)[r₂² − r₁² + 2r₁² ln(r₁/r₂)]
= 14.58 K.

Bearing conduction resistance:
R_b = ln(r₃/r₂)/(2π k_b L)
= ln(0.04/0.0256)/(2π × 70 × 0.15)
= 0.006765 K/W.

Temperature drop through bearing:
ΔT_b = Q R_b = 163.51 × 0.006765 = 1.106 K.

Thus bearing inner surface temperature = 40 + 1.106 = 41.106 °C.

Shaft surface temperature:
T_shaft = 41.106 + 14.58 = 55.69 °C.

(b)(iii) Mechanical power wasted by viscous dissipation is the same as the heat generated:
P_wasted = 163.51 W.

(c) For ideal gas:
R = C_p(γ − 1)/γ = 1050 × (1.38 − 1)/1.38 = 289.13 J/kg-K.

Density:
ρ = P/(RT) = 300000/(289.13 × 600) = 1.729 kg/m³.

Duct area:
A = πD²/4 = π × 0.2²/4 = 0.031416 m².

(c)(i) Mass flow rate:
m_dot = ρAV = 1.729 × 0.031416 × 180
= 9.779 kg/s.

(c)(ii) Stagnation temperature:
T₀ = T + V²/(2C_p)
= 600 + 180²/(2 × 1050)
= 600 + 15.429 = 615.43 K.

(c)(iii) Speed of sound:
a = sqrt(γRT) = sqrt(1.38 × 289.13 × 600)
= sqrt(239400) = 489.29 m/s.

Mach number:
M = V/a = 180/489.29 = 0.368.

(c)(iv)(1) Compressible stagnation pressure:
P₀ = P[1 + (γ − 1)M²/2]^γ/(γ−1)
= 3[1 + 0.19 × 0.368²]^3.6316
= 3 × 1.09659
= 3.290 bar.

(c)(iv)(2) Incompressible stagnation pressure:
P₀ = P + ρV²/2
= 3 bar + (1.729 × 180²)/(2 × 10⁵)
= 3 + 0.28015
= 3.280 bar.

(a) 10 g of water at 20 °C is converted into ice at –10 °C at constant atmospheric pressure. Assuming the specific heat of liquid…

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