Question

Steam is to be condensed on the shell side of a heat exchanger at 75°F. Cooling water enters the tubes at 50°F at a rate of 46 lbm/s and leaves at 65°F. Assuming the heat exchanger to be well insulated, determine the rate of heat transfer in the heat exchanger and the rate of condensation of the steam. The specific heat of water is 1.0 Btu/lbm·°F and the enthalpy of vaporization of water at 75°F is 1050.9 Btu/lbm. The rate of heat transfer in the heat exchanger is Btu/s. The rate of condensation of the steam is lbm/s.

Answer 1

`m_(steam) = 0.656 lbm/s`

Step-by-step explanation:

given data:

flow rate = 46 lbm/s

`T_1 = 50` degree F

`T_2= 65` degree F

from energy balanced equation we have

`\Delta E = \Delta E_(IN)  - \Delta E_(OUT)`

AS CONDITION IS STEADY STATE HENCE

`\Delta E = 0` So

`\Delta E_(IN)  = \Delta E_(OUT)`

`Q_(in) + mh = mh_2`

`Q_(in) + mC_pT_1 = mC_pT_2`

`Q_(in) = mC_p(T_2 -T_1)`

`Q_(in) = 46 * 1(65 - 50)`

`Q_(in) = 690 Btu/s`

Rate of condensation
`m_(steam) = (Q_(in))/(h_(fg))`

`m_(steam) = (690)/(1050.9)`

`m_(steam) = 0.656 lbm/s`