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232 Chapter 5

(5.36)
mi = m 2L + m 2V
If Equation 5.36 is divided by m t we find that

m 2L m 2V
——— ——=1 (5.37)
mi nil

but XL = m 2L /ni! the mass fraction of condensate and x v = m 2V/mi mass friction of
steam. Thus,

XL + x v = 1 (5.38)

This obvious relationship is used in Table 5.11 to obtain mass-fraction averages of
thermodynamic properties of steam-condensate mixtures. The macroscopic energy
balance, is used to obtain the steam flow rate. Like compressors, the kinetic and
potential energy terms are not significant, and the expansion is assumed to be
adiabatic.

Table 5.11 Summary of Equations for Sizing Steam Turbines_______

Subscripts: Isentropic process, s
First subscript: Entering steam, 1 — Exit steam, 2
Second subscript: Condensate, L — Steam, V
Energy Balance

P c' = r| T m(hi-h 2S ) (5.11.1)

h,-h 2
r, T = ——— __ definition (5.11.2)
h, - h 2S
— isentropic process (5.11.3)
Si = s 2S
Single Stage

TIT =(l-x/2)OiB/Cs) (5.11.4)

T| B = f((o',Pi',Pc') — Figure 5.19 (5.11.5)

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