Question

A pipe, 4.5 cm in diameter and 1×104 cm in length, transports superheated vapor at a rate of 1.08× 106 grams/h. The pipe, which is located in a power plant at 300 K, has a uniform surface temperature of 370 K. If the temperature drop between the inlet and exit of the pipe is 35 K, determine the heat transfer coefficient as a result of convection between the pipe surface and the surroundings. Assume, the specific heat of the vapor is 2190 J/kg.K. (20 points)

Answer 1

Answer:

h = 23.237 W/m2 K

Explanation:

given data:

flow rate = 1.08*10^6 gm/h = 0.3 kg/s

D = 4.5 cm = 0.045 m

L = 10^4 cm = 100 m

surface temperature = 370 K

$\Delta T = 35K$

Surface heat of vapor = 2190 J/kg.k

From energy conservation principle we have

heat transfer btwn surface and air  = heat loss due to flow and temp. drop

where

heat transfer btwn surface and air is due to convection

$Q _{convection} = hA_s (T_S - T_∞)$

WHERE

$T_S = 370 K$

$T_∞ = 300 K$

$Heat\ loss = Q_{loss} = \dot m Cp \Delta T$

$\dot m = 0.3 kg/s$

from both above equation we have

$Q_{convection} = Q_{loss}$

$hA_s (T_S - T_∞) = \dot m Cp \Delta T$

putting all value to get heat transefer coefficient

$h = \frac{\dot m Cp \Delta T}{A_S((T_S - T_∞)}$

$h = \frac{0.3*2190*35}{14.137*(370-300)}$

h = 23.237 W/m2 K