## Heat Transfer Miscellaneous

#### Heat-Transfer

1. 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 m2, and the overall heat transfer coefficient is 2000 W/ m2. The heat transfer (in MW) to the condenser is ______(correct to two decimal places).

1. Condenser is given.

 LMTD = θ1 - θ2 In θ1 θ2

 = 16 - 8 = 11.54°C In 16 8

Q = UA (LMTD)
= 2000 × 50 × 11.54 = 1.154 MW

##### Correct Option: A

Condenser is given.

 LMTD = θ1 - θ2 In θ1 θ2

 = 16 - 8 = 11.54°C In 16 8

Q = UA (LMTD)
= 2000 × 50 × 11.54 = 1.154 MW

1. 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

1. ∴ ∆T1 = 70 – 50 = 20°C, ∆T2 = 40 – 30 = 10°C

 LMTD = ∆T1 ∆T2 log ∆T1 ∆T2

 = 20 - 10 = 14.43 log 20 10

##### Correct Option: B

∴ ∆T1 = 70 – 50 = 20°C, ∆T2 = 40 – 30 = 10°C

 LMTD = ∆T1 ∆T2 log ∆T1 ∆T2

 = 20 - 10 = 14.43 log 20 10

1. The value of Biot number is very small (less than 0.01),when

1. 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.

1. 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/m2K, Thermophysical properties of thermocouple material are k = 20 W/mK, c = 400 J/kgK and ρ = 8500 kg/m3. 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

1. Given h = 400 W/m2K, k = 20 W/mK
c = 400 J/kg.K, ρ = 8500 kg/m3 T(t) = 298 °C, Ti = 30 °C, T = 300 °C

 L = V = (π / 6)D3 = D A πD2 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 Bi < 0.1, lumped system analysis can be used

 T(t) = e -hAT Ti - 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/m2K, k = 20 W/mK
c = 400 J/kg.K, ρ = 8500 kg/m3 T(t) = 298 °C, Ti = 30 °C, T = 300 °C

 L = V = (π / 6)D3 = D A πD2 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 Bi < 0.1, lumped system analysis can be used

 T(t) = e -hAT Ti - 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

1. 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/m3 and its specific heat 385 J/kgK. If the heat transfer coefficient is 250 W/m2K 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,

1. Given data: K = 400 W/mK; Ti = 500 K ; ρ = 9000 Kg/m3.
T = 300 K; Cp = 385 J/KG K; h = 250 W /m2K

 Lc = V = D = (0.005) = 8.33 × 10-4 A 6 6

 T = T∞ + (Ti - T∞) e -hAT ρcV

 dT = -hAT (Ti - T∞) e -hAT dt ρcV ρcV

 dT = hA (Ti - 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; Ti = 500 K ; ρ = 9000 Kg/m3.
T = 300 K; Cp = 385 J/KG K; h = 250 W /m2K

 Lc = V = D = (0.005) = 8.33 × 10-4 A 6 6

 T = T∞ + (Ti - T∞) e -hAT ρcV

 dT = -hAT (Ti - T∞) e -hAT dt ρcV ρcV

 dT = hA (Ti - T∞) dt t = 0 ρcV

 = 250 × 200 = 17.31 K / S 9000 × 385 × 8.33 × 10-4