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Handbook of Turbomachinery Second Edition Revised - Ventech!

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The density can now be determined:<br />

r 2 ¼<br />

44:1lbf =in 2<br />

3<br />

53:34<br />

144 in2<br />

ft 2<br />

ft lbf<br />

lbm R ð564:2RÞ<br />

and the required flow area:<br />

A2 ¼<br />

1:798 lbm=sec<br />

¼ 0:0703 lbm<br />

ft 3<br />

ð0:0703 lbm=ft 3 Þð535:0ft=secÞ<br />

144 in:2<br />

ft 2<br />

¼ 6:882 in:2<br />

The rotor exit hub and tip radii cannot be uniquely determined until<br />

either shaft speed, blade height, or hub-to-tip radius ratio is specified. Once<br />

one parameter is specified, the others are determined. For this example, we<br />

choose a hub-to-tip ratio <strong>of</strong> 0.7 as a compromise between performance and<br />

manufacturability. If the turbine shaft speed were restricted to a certain<br />

value or range <strong>of</strong> values, it would make more sense to specify the shaft<br />

speed. The turbine tip radius is determined from<br />

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi<br />

A2<br />

rt2 ¼<br />

p½1 ðrh=rtÞ 2 s<br />

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi<br />

6:882 in:<br />

¼<br />

Š<br />

2<br />

p½1 ð0:7Þ 2 s<br />

¼ 2:073 in:<br />

Š<br />

This results in a hub radius <strong>of</strong> 1.451 in., a mean radius <strong>of</strong> 1.762 in. and<br />

a blade height <strong>of</strong> 0.622 in. The shaft speed is found from Eq. (8):<br />

o ¼ Um=rm ¼<br />

891:6ft=sec<br />

¼ 6073 rad=sec<br />

ð1:762 inÞð1ft=12 inÞ<br />

or 57,600 rpm. The tip speed <strong>of</strong> the turbine is 1,049 ft/sec, well within our<br />

guidelines.<br />

The next step is to calculate the overall efficiency. From Smith’s chart,<br />

a stator inlet to rotor exit total-to-total efficiency at zero clearance is<br />

available. We must correct this for tip clearance effects, the inlet loss, and<br />

the exit kinetic energy loss. At l ¼ 1:3 and f ¼ 0:6, Smith’s chart predicts<br />

Z 0 0 2 0 ZC ¼ 0:94<br />

Assuming a tip clearance <strong>of</strong> 0.015 in., the total-to-total efficiency including<br />

the tip clearance loss is calculated from Eq. (55) using a value <strong>of</strong> 2 for Kc:<br />

Z 0 0 2 0 ¼ðZ 0 0 2 0 ZCÞ 1 2 rt<br />

rm<br />

d<br />

h<br />

2:073 0:015<br />

¼ 0:94 1 2<br />

1:762 0:622<br />

¼ 0:8867<br />

Equation (56) is used to determine the overall efficiency including inlet and<br />

exit losses. From the problem statement, we know that the overall pressure<br />

Copyright © 2003 Marcel Dekker, Inc.

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