Heat Transfer Miscellaneous
 Steam in the condenser of a thermal power plant is to be condensed at a temperature of 30°C with cooling water which enters the tubes of the condenser at 14°C and exits at 22°C. The total surface area of the tubes is 50 m^{2}, and the overall heat transfer coefficient is 2000 W/ m^{2}. The heat transfer (in MW) to the condenser is ______(correct to two decimal places).

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Condenser is given.
LMTD = θ_{1}  θ_{2} In θ_{1} θ_{2} = 16  8 = 11.54°C In 16 8
Q = UA (LMTD)
= 2000 × 50 × 11.54 = 1.154 MWCorrect Option: A
Condenser is given.
LMTD = θ_{1}  θ_{2} In θ_{1} θ_{2} = 16  8 = 11.54°C In 16 8
Q = UA (LMTD)
= 2000 × 50 × 11.54 = 1.154 MW
 Air enters a counter flow HE at 70°C and leaves at 40°C. Water enters at 30°C and leaves at 50°C, the LMTD in degree C is

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∴ ∆T_{1} = 70 – 50 = 20°C, ∆T_{2} = 40 – 30 = 10°C
LMTD = ∆T_{1} ∆T_{2} log ∆T_{1} ∆T_{2} = 20  10 = 14.43 log 20 10 Correct Option: B
∴ ∆T_{1} = 70 – 50 = 20°C, ∆T_{2} = 40 – 30 = 10°C
LMTD = ∆T_{1} ∆T_{2} log ∆T_{1} ∆T_{2} = 20  10 = 14.43 log 20 10
 The value of Biot number is very small (less than 0.01),when

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When conductive resistance of solid is negligible, then biot number is very small.
Correct Option: C
When conductive resistance of solid is negligible, then biot number is very small.
 A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer coefficient on bead surface is 400 W/m^{2}K, Thermophysical properties of thermocouple material are k = 20 W/mK, c = 400 J/kgK and ρ = 8500 kg/m^{3}. If the thermocouple initially at 30°C is placed in a hot stream of 300°C, the time taken by the bead to reach 298°C, is

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Given h = 400 W/m^{2}K, k = 20 W/mK
c = 400 J/kg.K, ρ = 8500 kg/m^{3} T(t) = 298 °C, T_{i} = 30 °C, T_{∞} = 300 °CL = V = (π / 6)D^{3} = D A πD^{2} 6 = 1 × 0.706 × 10^{3} = 1.76 × 10^{4} m 6 Biot Number = hL k = 400 × 1.176 × 10^{4} = 0.0023 20
Since B_{i} < 0.1, lumped system analysis can be usedT(t) = e hAT T_{i}  T_{∞} ρcv 298  300 = e hAT 30  300 ρcV 2 = e hAT 270 ρcV ln 270 = hAT 2 ρcV ⇒ T = ρcV ln 270 = ρc V ln 135 cA 2 h A = 8500 × 400 × 1.176 × 10^{4} × 4.90 ≈ 4.9 sec. 400 Correct Option: B
Given h = 400 W/m^{2}K, k = 20 W/mK
c = 400 J/kg.K, ρ = 8500 kg/m^{3} T(t) = 298 °C, T_{i} = 30 °C, T_{∞} = 300 °CL = V = (π / 6)D^{3} = D A πD^{2} 6 = 1 × 0.706 × 10^{3} = 1.76 × 10^{4} m 6 Biot Number = hL k = 400 × 1.176 × 10^{4} = 0.0023 20
Since B_{i} < 0.1, lumped system analysis can be usedT(t) = e hAT T_{i}  T_{∞} ρcv 298  300 = e hAT 30  300 ρcV 2 = e hAT 270 ρcV ln 270 = hAT 2 ρcV ⇒ T = ρcV ln 270 = ρc V ln 135 cA 2 h A = 8500 × 400 × 1.176 × 10^{4} × 4.90 ≈ 4.9 sec. 400
 A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/mK, its density 9000 kg/m^{3} and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m^{2}K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s,

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Given data: K = 400 W/mK; T_{i} = 500 K ; ρ = 9000 Kg/m^{3}.
T_{∞} = 300 K; C_{p} = 385 J/KG K; h = 250 W /m^{2}KL_{c} = V = D = (0.005) = 8.33 × 10^{4} A 6 6 T = T_{∞} + (T_{i}  T_{∞}) e hAT ρcV dT = hAT (T_{i}  T_{∞}) e hAT dt ρcV ρcV dT = hA (T_{i}  T_{∞}) dt _{t = 0} ρcV = 250 × 200 = 17.31 K / S 9000 × 385 × 8.33 × 10^{4} Correct Option: C
Given data: K = 400 W/mK; T_{i} = 500 K ; ρ = 9000 Kg/m^{3}.
T_{∞} = 300 K; C_{p} = 385 J/KG K; h = 250 W /m^{2}KL_{c} = V = D = (0.005) = 8.33 × 10^{4} A 6 6 T = T_{∞} + (T_{i}  T_{∞}) e hAT ρcV dT = hAT (T_{i}  T_{∞}) e hAT dt ρcV ρcV dT = hA (T_{i}  T_{∞}) dt _{t = 0} ρcV = 250 × 200 = 17.31 K / S 9000 × 385 × 8.33 × 10^{4}