Page 259 - Chemical process engineering design and economics

Compressors, Pumps, and Turbines 239

For an isentropic process (Equation 5.11.3), B! = s. Therefore, from Equa-
2S
tion5.ll.ll,
s 2S = 6.8301 = 0.6493 x s + 8.1502 (1 -x s)

The mass fraction of water in the exit stream for an isentropic process, x s, is equal
to 0.1760.
From Equation 5.11.12 for an isentropic process, the exit enthalpy for the
part-water, part-vapor stream,

h 2S= 0.1760 (191.83) + 0.824 (2584.7) = 2164.0 kJ/kg (930 Btu/lb)

The compressor power is within the range of single-stage turbines. If it is as-
sumed that the compressor will be directly coupled to the steam turbine, the com-
pressor shaft power must be matched by the steam-turbine shaft power. Allowing
for a 10% safety factor, the power delivered to the compressor will be 110 hp
(82.0 kW). From Equation 5.11.5, the turbine efficiency, T| B, at 36,000 rpm, 110
hp (82.0 kW), and 1.30 MPa (188.5 psi), is 36%.
The correction factor for superheated steam, c, is obtained from Equation
s
5.11.6 (Figure 5.20). The correction factor depends on the degrees of superheat at
the turbine inlet, and it is defined as the difference between the steam temperature
and the saturation temperature.

superheat = (260.0 - 191.6) °C (9 °F / 5 °C) = 123.1 °F

From Figure 5.20, c s » 0.87.
After solving Equation 5.11.2, 5.11.4, and 5.11.13 for x by eliminating h 2
and T| T, we obtain

hiv -h,+ (TiB/c s )(h 1 -h 2 s)
x = —————————————————
h 2V - h 2L + ( T| B / 2 c s) hi - h 2S)
2584.7 - 2954.0 + (0.36 / 0.87) (2954.0 - 2164.0)
x = ———————————————————————————— = - 0.0762
584.7 - 191.83 + [0.36 / 2 (0.87)] (2954.0 - 2164.0)
A negative sign means that no condensation occurs. This can also be shown
by calculating the actual enthalpy of the exit steam from Equation 5.11.2. If x = 0,
TIT = T)B/CS, as can be seen from Equation 5.11.4. Thus, from Equation 5.11.2, the
actual enthalpy,

h 2 = h, - (TIB/ c s) (h! - h 2S) = 2954.0 - (0.36 / 0.87) (2954.0 - 2164.0)

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