Consider a 25-mm-diameter and 15-m-long smooth tube that is maintained at a constant surface temperature. Fluids enter the tube at 50°C with a mass flow rate of 0.01 kg/s. Determine the tube surface temperatures necessary to heat water, engine oil, and liquid mercury to the desired outlet temperature of 150°C

⚠this is not my answer it from hamzaahmeds⚠

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