Page 85 - Applied Process Design for Chemical and Petrochemical Plants Volume I

72 Applied Process Design for Chemical and Petrochemical Plants

in a piping system under all conditions of flow, includ- and thereby converting to a common base K, they are
ing laminar. then additive, when all referenced to the same size pipe. Flow
From the Darcy equation [3]: then can be determined for a fixed head system by
K = (f L/D) (2-25) GPM liquid = 19.65 d2 (hL/K)1/2 (2-29)

Head loss through a pipe, hL = (f) (L/D) (v2/2g) (2-26)
Of course, by selecting the proper equation, flows for
vapors and gases can be determined in the same way, as
Head loss through a valve (for instance), the K value is for the fitting or valve and not for the
hL = K(v2/2g) (2-27)
fluid.
where L/D is the equivalent length in pipe diameters of The head loss has been correlated as a function of the
straight pipe that will cause or develop the same pressure velocity head equation using K as the resistance coeffi-
drop as the fitting, component, or other obstruction cient in the equation.
under the same flow conditions. K is a constant for all flow
conditions through a given system component; thus, the
value of L/D for the specific component must vary hL = Kv2/2g = Kv2/64.4, ft of fluid (2-27)
inversely with the change in friction factor for varying
flow conditions [3]. For a system of multiple components of valves, pipe,
For various components’ K values, see Figures 2-1 2A, 2- and fittings, Equation 2-25 can be used to establish a com-
12B, 2-13A, 2-13B through 2-16 and Tables 2-2 and 2-3. ponent size to which each separate resistance can be
expressed as a “common denominator,” or common pipe
size. Under these conditions, the “corrected” K values are
Common Denominator for Use of “K” Factors in a additive and can be used as one number in Equation 2-27.
System of Varying Sizes of Internal Dimensions These types of corrections should be made to improve
and more accurately represent the pressure drop calcula-
K value can be adjusted to a common reference pipe tions.
size:
An example procedure connecting 3-in. and 6-in. pipe
and fittings, using Gin. as the final reference, is as follows:

where subscript 1 is the known resistance from standard K 1. From Table 2-2, read for 3 in. Sch. 40 pipe, fT = 0.018.
factor tables or charts (these are based on standard ANSI 2. Calculate R, for each pipe size.
pipe/fitting dimensions), and subscript 2 is the corrected
resistance coefficient used to identify the inside diameter 3. Read friction factor, f, from Figure 2-3, using Figure
of the actual pipe into which the valve or fitting is con- 2-1 1.
nected or installed. 4. Calculate K for 6-in. pipe:
The K values determined for various valves, fittings, K = 0.018 (L/d) (12), Lg” = ft Gin. pipe.
etc., are specific to the system, particularly valves. For
example, most reliable data* have been developed 5. Calculate Kfor 3-in. pipe, using Lr = ft of 3-in. pipe.
with valves and fittings installed in pipe of specific 6. Then, referencing to the 6-in. pipe throughout the
dimensions, then, if a larger or smaller inside diame- system:
ter valve or fitting is to be installed in a pipe of dif- K2 = (Ky) (d~/d~-)~, representing entire pipe part
ferent inside diameter, a correction of the K value of system.
should be made.
7. Add K values for individual fittings and valves from
Pressure drop through line systems containing more Figures 2-12A through 2-16 and Tables 2-2 and 2-3.
than one pipe size can be determined by (a) calculating
the pressure drop separately for each section at assumed 8. Using sum of Kvalues for 6-in. pipe, 3-in. pipe equiv-
flows, or (b) determining the K totals for each pipe size sep- alent calculated above in step 6, and all items in step
arately, and then converting to one selected size and using 7 above [3] :
that for pressure drop calculations. For example, using
h, = (0.00259 KQ‘)/(~v)~ (2-30)
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