Page 198 - Standard Handbook Of Petroleum & Natural Gas Engineering

Fluid Mechanics 183

If the pressure ratio is less than or equal to that specified by Equation 2-67, the flow
will be sonic at the choke throat and the temperature at the throat can be found from

T2 = TI( &) (2-68)

The flow rate can be found from

(2-69)

For critical flow the discharge coefficient is dependent upon the geometry of the
choke and its diameter or the ratio p of its diameter to that of the upstream pipe (see
Figure 2-24).

Example 2-1 7
A 0.6 gravity hydrocarbon gas flows from a 2-in. ID pipe through a 1-in. ID orifice
plate. The upstream temperature and pressure are 75°F and 800 psia, respectively.
The downstream pressure is 200 psia. Does heating need to be applied to assure that
frost does not clog the orifice? What will be the flow rate?
Check for Critical Flow. From Table 2-8 it is determined that k = 1.299. Checking
the pressure ratio, Equation 2-67, gives
-- - 0.25 < (-) 2 k'(k-l) = 0.546
"
PI k+l

Therefore critical flow conditions exist and the fluid velocity in the choke is sonic.

Temperature in the Choke. From Equation 2-68 the temperature in the throat is
calculated as

T, = TI(-&-) = 535( ) = 465.4"R =5.4"F
1 + 1.299
Therefore, if the gas contains water, icing or hydrate formation may occur causing
the throat to clog. A heating system may be needed.
Flowrate. Assume that the discharge coefficient is C = 1.0. The choke area is A =
R( 1/2), = 0.785 in?. Thus, from Equation 2-69,

(1.299t1)/(1.29%1) 0.5
800 3
Q = 610(1.0)(0.785)
[1.2gg( 1.299+1)
[( 0. 6)(535)]0'5
= 14.021 scfm