Solution

A detailed flow diagram of the process is provided in Figure 9.

Assuming the process to be steady state and noting that there is no heat loss or shaft work, the energy balance on a rate basis is

This equation indicates that the sum of the enthalpy changes for the two streams must equal zero.

The change in enthalpy for the vaporization of the water stream is

¿¿vaporization = (1000 lb/h)(l 151 - 28.1 Btu/lb)

The change of enthalpy for the cooling of the superheated steam may now be determined. Since the mass flowrate of one stream is unknown, its AH must be expressed in terms of this mass flowrate, which is represented in Figure 9 asm lb/h.

Superheated steam

H=1572 Btu/lb m lb/h ^

600 F, 20 atm H=1316 Btu/lb

Water ^ 60°F, 1 atm

Saturated steam 212°F, 1 atm

1000 lb/h

1000 lb/h

Since the overall AH is zero, the enthalpy changes of the two streams must total zero. Thus,

^^vaporization ^^cooling ^

Tables of enthalpies and other state properties of many substances may be found in R. H. Perry and D. W. Green, Ed., Perry's Chemical Engineers' Handbook, 7th ed., McGraw-Hill, New York, 1996.

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