100 lbmh of oil at 130 degree F is needed for a process modi

100 lbm/h of oil at 130 degree F is needed for a process modification. The oil will be available at 70 degree F, so a plant engineer designed a heater consisting of a 1-in. ID tube, 10 ft long, the wall of which will be maintained at 215 degree F by condensing steam. Will the heater work as required? Properties of the oil may be assumed constant at the following values: C_P = 0.49 Btu/lbm middot degree F mu = 1.42 lbm/ft middot h k = 0.0825 Btu/h middot ft middot degree F rho = 55 lbm/ft^3

Solution

Area A = 3.14/4 *1^2 = 0.785 in^2

Perimeter P = 3.14*(1/12) = 0.2617 ft

Velocity V = m/(rho*A) = 100/(55*(0.785/12^2)) = 29.98 ft/hr

Revnolds number Re = rho*VD/meu = 55*29.98*(1/12)/1.42 = 96.8

Since Re < 2300, flow is laminar and hence Nusselt number Nu = 3.66

Nu = hD/k

Heat transfer coeff h = Nu*k/D = 3.66*0.0825 / (1/12) = 3.62 Btu/h .ft^2 - F

(Tw - Tout) / (Tw - Tin) = e^ (-PLh / mCp)

(215 - Tout) / (215 - 70) = e^ (-0.2617*10*3.62 / (100*0.49))

Tout = 95.5 deg F

Since 95.5 < 130 deg F, heater will not work as required.