Hydrocarbon Processing - July 2022 - 36

Valves, Pumps and Turbomachinery
A sample problem demonstrates the usefulness of this flowrate
calculation method. Light oil (ρ = 960 kg/m3
, μ = 1.2 Cp)
is flowing through a valve station. This piping system contains
20 m of horizontal new commercial 3 in.-diameter Schedule 40
pipe, three gate valves, four 90° welded elbows (r/d = 1.5), one
conventional globe valve and one thin orifice plate with a loss
coefficient of k = 3.6. With all valves wide open, the pressure
drop in this system is 11,844 Pa. Calculate the flowrate of oil in
l/min (liters per minute) as shown below.
1. The 3 in.-diameter Schedule 40 pipe data is: d = 77.9 mm;
D = 0.0779 m; ft
= 0.018; and ε = 0.046 mm.
2. Kinematic viscosity (Eq. 11):
ν = [1.2/(1,000 × 960)] = 1.25 × 10-6
m2/sec
(11)
3. Summarizing K for valves and fittings (Eq. 12):
∑K = (3 × 8 + 4 × 20 + 1 × 340)0.018 + 3.6 = 11.592 (12)
4. Total head loss applying Bernoulli equations (Eq. 13):
hf
= [11,844 / (960 × 9.806)] = 1.2581 m
5. Equivalent length of pipe estimated for valves and
fittings (Eq. 14):
Le = 33 × 11.592 ×
0.07791.36 ×
⎛
⎝
= 42 + 20 ≈ 62m
Λ=
960
1.2
f =
⎛
⎜⎝
1.2
960
1.2581
20
⎞
⎟⎠
⎞
⎠
−0.24
×
(14)
0.12
≈ 42m
6. Estimated for adjusted pipe (Eq. 15):
La
7. Conventional number Λ (Eq. 16):
× 1.2581 ×
8. Friction factor (Eq. 17):
0.25
⎡
⎢⎣
log
⎛
⎜⎝
0.000576
2.478
+ 0.27
0.046
77.9
⎞
⎟⎠
⎤
⎥⎦
0.07793
62
1
= 2.478
(16)
2
3
(13)
NOMENCLATURE
d = Internal diameter of pipe, mm
D = Internal diameter of pipe, m
f = Darcy friction factor
ft
hf
K = Fitting loss coefficient
K0
= Friction factor in zone of complete turbulence
= Total head loss, m
= Fitting loss coefficient for a specific type of fitting
Le = Equivalent length of pipe, m
Ls = Straight length of pipe, m
La = Adjusted length of pipe, m
(L/D)e
q = Flowrate, m3/sec
(15)
Q = Flowrate, l/min
Re = Reynolds number
μ = Dynamic viscosity of fluid, cP
ν = Kinematic viscosity of fluid, m2
/sec
LITERATURE CITED
Churchill, S. W., " Friction factor equations spans all fluid flow regime, " Chemical
Engineering, November 1977.
Crane Valve Group (CVG), " Flow of fluids through valves, fittings & pipe, " Crane
Technical Paper No. 410 (TP-410), 1988.
Darby, R., " Correlate pressure drops through fittings, " Chemical Engineering, July
1999.
2 = 0.0215
9. Actual equivalent length of pipe (Eq. 18):
Le
= (11.592/0.0215) × 0.0779 = 42 m
10. Flowrate (Eq. 19):
q = 3.47835 × 0.07792.5 ×
0.00572347 m3 / sec
1.2581
=
0.0215 × 62
Q = q × 60,000 = 343.4 l/min
11. Checking results (Eq. 20):
36 JULY 2022 | HydrocarbonProcessing.com
(19)
ALEJANDRO GARCIA is a mechanical engineer that graduated
from the University of Cienfuegos Cuba. He received his MS
degree in mechanical engineering with a specialty in materials
from the Autonomous University of Nuevo Leon, Mexico. He
gained several years of experience in power plants and the
automotive industry as a static and dynamic equipment specialist.
He now works in a steel and wire plant in Houston, Texas as a
mechanical engineer. The author can be reached alessandromilan88@gmail.com.
4 Mott, R. L., Applied fluid mechanics, 4th Ed., Prentice Hall PTR, 1996.
5
(17)
Verma, C. P., " Solve pipe flow problems directly, " Hydrocarbon Processing, August
1979.
(18)
ISRAEL GARCIA graduated from the University of
Cienfuegos Cuba with an MS degree in mechanical engineering.
Mr. Rodriguez has been attached to the mechanical engineering
faculty of that university since 1985 as a Professor in fluid
mechanics, heat transfer and science materials. Mr. Garcia has
more than 30 yr of industrial experience in chemical plants
and power stations and has presented several papers that
deal with the design of heat exchangers, pressure vessels and piping systems.
The author can be reached at isgaro47@gmail.com.
υ = [(4 × 0.00572347) / (3.1416 × 0.07792
1.2 m/s
Re = [(1.2 × 0.0779 × 106
ε/D = (0.046 / 77.9) = 0.0005905
f = 0.0216
Λ = 0.0002258 × 74,784 × √0.0216 = 2.48
The friction factor calculated in Step 8 (Eq. 17) is slightly
less than the calculation above, but the difference is small
enough to forego any correction of the equivalent length of pipe
and flowrate. Therefore, the calculated flowrate is correct and
the condition given in Eq. 8 is fulfilled.
Takeaway. The proposed calculation methods here determine
flowrate trough valves, fittings and pipe for most liquid and vapors
without tedious trial and error methods. The equations
to determinate unknown parameters are derived from classical
equations published in literature.1,4,5
These methods are also applicable
in open tanks draining through piping systems and in
piping systems with elevation above the reference level containing
numerous fittings and valves.
)] =
(20)
) / 1.25] = 74,784
= Equivalent length of pipe in numbers of pipe diameters
http://www.HydrocarbonProcessing.com
Hydrocarbon Processing - July 2022

Table of Contents for the Digital Edition of Hydrocarbon Processing - July 2022
