Fluid_Flow_Rules_of_Thumb_for_Chemical

14

Rulesof Thumb for Chemical Engineers

M~ Graph

fL/D - (1/M22)(p1/P2)2 [1 - (P2/p1)2 ] - ln(P1/P2) 2

(4)

100

- , , - M-0.1 - , , - M- 0 . 2 I o M-0.3 I -~- M- 0 . 4 I - * - M- 0 . 5 I - ~ M-O.6 I - ~ M-O.7 I ...,.,.. M- 0 . 8 I [] M-O.9 I - - - Cr i t i ca l I

\ \

Critical pressure at the pipe outlet in psia

10

~ ~ ~ i i l l i ~ m m m ~ m m m m m ~ ' ~ m m m m m ~ ~ ~ ~ ~ ~ m m ~ ~ m m m m m ~ ~ m m ~ ~ ~ ~ ~ mmmmmi .Emmmm i l i ~

(5)

- (W/408d 2)(ZT/Mw)~

P c r i t

For comparison the author has generated an Excel | plot (Figure 3) using the data from Figure 2. This is for those readers who work with this popular spreadsheet.

m m m m m m m m m m m m ~

0.1

m m m m m m m m m m m m m k ~ m m m m m m m m m m m m m m

0.01

Example

0.1

0.2

0 . 3

0 . 4

0 . 5

0 . 6

0 . 7

0 . 8

0 . 9

P2/P1

Given (see sketch following):

Figure 3. Excel| version of M1 chart.

Calculate how much gas will flow to the vessel through the 1 in line with the normally closed hand valve fully opened. Use the psv full open pressure of 136 psia as the vessel pressure. The equivalent length of 200 ft includes the fully opened hand valve. The 1 in pipe's inside diameter is 1.049 in. Assume Z = 1.0.

The Mak Isothermal flow chart is such a useful tool that the author has used it for cases where P1 is known instead of P2 with a trial and error approach. The author has now generated a graph (Figure 2) based upon M~ using Equa- tion 2. The Isothermal flow chart (Figure 1) based on M2 uses Equation 4. Figure 2 facilitates the following case.

Given: Find: Steps:

P1 and W

P2 1. Get f from GPSA graph (Figure 4). 2. Determine fL/D. 3. Obtain Z. Figures 5, 6, and 7 are provided for convenience. 4. Calculate M1. See Equation 1. 5. Get Pz/P1 from Figure 2. The critical curve indicates where M1 -- P2/Pl. When this happens M2 = 1 since M2 = Ml(P1/P2). The design pipe diameter might have to be changed to provide a possible set of conditions. 6. Calculate P2.

Calculations:

Note that if the AP was across a restriction orifice, sonic velocity would occur since the AP is greater than 2 : 1 (315/136 = 2.31). However, the AP is along a length of pipe, so we will use Mak' s method. For commercial steel pipe:

f - 0.023 id - 1.049in - 0.0874 ft

P2/P1 - 136/135 - 0.43 fL/D - 0.023(200)/0.0874 - 52.6 M2 - 0 . 2 8 ( f r om Figure 1)

Some calculations require knowing the critical pres- sure at which sonic velocity occurs. This is calculated with Equation 5. The applicable equations are

Note that if the flow were critical, M2 would be 1.

M2 - 1.702 x 10-5[W/(PzDZ)][ZT/Mw] ~ P2 - 136 psia D 2 - 0 . 0 8 7 4 2 - 0.00764 T - 460 + 60 - 520 ~ Mw - 1 6 Z - 1.0 (given)

Based on M1

(1)

M1 - 1.702 • 10-5[W/(P1D 2)](ZT/Mw)~

(2)

fL/D = (l/M12)[1 - (P2/p1)2] _ ln(P~/P2):

0 . 28 - 1.702 x 10-5[W/(136 x 0.00764)][1.0(520)/16] 0.5

Based on M2

(3)

W - 0.28 x 1.039 x 105/(5.70 x 1 . 702) - 30001b/hr

M2 - 1.702 • 10-5[W/(P2D 2)](ZT/Mw)~