Question

This is problem 6-12 from El-Wakil’s Powerplant Technology book -- a condensing only feedwater heater uses 7/8-in-OD 90-10 copper-nickel tubes. It receives 84,000 lb_m/h of 95 percent quality bled steam at 20 psia, and 160,000 lb_m/h of drain from the next higher pressure heater at 240°F. 3.9x10⁶ lb_m/h of feedwater goes through the heater at 7 ft/s, 2000 psia, and 195°F. The terminal temperature difference is 5°F. Determine the size, length, and number of tubes based on a U-tube design. Take a maximum allowable stress in the tubes of 15,000 psi.

Answer 1

Calculation of tube thickness

The tube membrane stress is given by

P= tube inside pressure=2000 psi

d= internal diameter=D-2t

t= thickness

D=OD=7/8 in=0.875 in

We assume allowable stress is 15000 psi

Hence

t=0.055 in

We select 16 BWG tubes with thickness 0.065 in

Calculation of heat transfer surface area

The heat load is given by

= steam flow=84000 lbm/h

= drain flow= 160000 lbm/h

steam inlet enthalpy at 20 psia, 96% quality= 1108.19 Btu/lbm

drain inlet enthalpy at 240⁰ F saturated= 208.47 Btu/lbm

Drain outlet enthalpy at 20 psia saturated water=196.24 Btu /lbm

Hence

Specific heat of water is c_p= 1.005 Btu/lbm-⁰F

Outlet temperature of water is

T_wi= water inlet temperature= 195 ⁰F

= water flow= 3.9x10⁶ lbm/h

Hence

Steam inlet temperature is

T_si= 227.92 ⁰F

Water inlet temperature T_wi= 195 ⁰F

Hence

The heat transfer surface area is given by

F= correction factor= 0.8 app for shell and tube configuration

U= overall heat transfer coefficient= 283 Btu/h-ft². ⁰F

Hence

We consider effective length of tube 20 ft in one pass

Hence overall length of U tube=2x20= 40 ft app

Heat transfer surface of 1 tube

Hence no of tubes is

Please note that if maximum allowable pressure drop in water is indicated we can check the adequacy of length of tube.