Consider a 25-mm-diameter and 15-m-long smooth tube that is used for heating fluids. The wall is heated electrically to provide a constant surface heat flux along the entire tube. Fluids enter the tube at 508C and exit at 1508C. If the mass flow rate is maintained at 0.01 kg/s, determine the convection heat transfer coefficients at the tube outlet for water, engine oil, and liquid mercury.

Water: h = 35.53 W/m².k

Engine oil: h = 18.84 W/m².k

Mercury: h = 1.19 W/m².k

Explanation:

Assuming the steady state, one-dimensional heat flow, it is clear that the added to the fluid by tube heat will be equal to the heat transfer through convection outside the tube.

Therefore,

mCΔT = hAΔT

mC = hA

h = mC/A

where,

h = convection coefficient

m = mass flow rate =  0.01 kg/s

C = specific heat capacity of fluid

A = surface area of tube = 2πrL = 2π(0.0125 m)(15 m) = 1.178 m²

FOR WATER:

C = 4186 J/Kg.k

Therefore,

h = (0.01 kg/s)(4186 J/Kg.k)/(1.178 m²)

h = 35.53 W/m².k

FOR ENGINE OIL:

C = 2220 J/Kg.k

Therefore,

h = (0.01 kg/s)(2220 J/Kg.k)/(1.178 m²)

h = 18.84 W/m².k

FOR LIQUID MERCURY:

C = 140 J/Kg.k

Therefore,

h = (0.01 kg/s)(140 J/Kg.k)/(1.178 m²)

h = 1.19 W/m².k

answer:

rawr

explanation: