Handbook of Turbomachinery Second Edition Revised - Ventech!

clearance. Since tip clearance represents a loss at the tip of the blade, the rest
of the blade does more than the average work. Therefore, the vector
diagram is calculated using the zero-clearance efficiency. Since we do not
know the turbine geometry at this point, we must make another assumption:
we assume that the tip clearance loss is 5%, so that the overall zero-clearance
efficiency is 0.84. Note that the required flow rate is calculated using the
overall efficiency with clearance, since that represents the energy available at
the turbine shaft. Equation (2) is used to calculate the actual enthalpy drops:
and
DhOA ¼ð0:8Þ 49:14 Btu
lbm
DhOA ZC ¼ð0:84Þ 49:14 Btu
lbm
¼ 39:31 Btu
lbm
¼ 41:28 Btu
lbm
The required turbine flow is found using Eq. (7):
_m ¼ P ð100 hpÞð:7069 Btu=sec=hpÞ
¼ ¼ 1:798 lbm= sec
DhOA 39:31 Btu=lbm
The mass flow rate is needed to calculate turbine flow area and is also a
system requirement.
We specify the vector diagram by selecting values of the turbine work
and flow coefficients. We also select a turbine hub-to-tip radius ratio of 0.7,
restricting the choice of mean work coefficient to values less than 1.356 in
ordertoavoidnegativereactionatthehub.FromSmith’schart(Fig.12),we
initially choose a work coefficient of 1.3 and a flow coefficient of 0.6 to result
in a zero-clearance, stator inlet to rotor exit total-to-total efficiency of 0.94.
We apply these coefficients at the mean radius of the turbine. From Eq. (22)
we calculate the mean blade speed, Um:
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
DhOA ZC ð32:174 ft lbf=ðlbm sec
Um ¼
¼
l
2 r
ÞÞð778:16 ft lbf=BtuÞð41:28 Btu=lbmÞ
1:3
¼ 891:6ft= sec
The axial velocity is calculated from Eq. (51):
Vx2 ¼ð0:6Þð891:6ft= secÞ ¼535:0ft= sec
In order to construct the vector diagram, we make two more assumptions:
(1) there is zero swirl leaving the turbine stage in order to minimize the exit
kinetic energy loss, and (2) the axial velocity is constant through the stage.
Copyright © 2003 Marcel Dekker, Inc.