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Sample Problem Statement

A level control valve is to be installed on an 8 oil line going from an oil water separator to an

oil heater. The oil water separator operates at 3.5 barg and 250C. Inlet pressure requirement atthe heater is 2.0 barg. Normal, minimum and maximum oil flowrates are 200m

3/hr, 60m

3/hr and

220 m3

/hr respectively. Size a level control valve to determine the control valve flow coefficientor valve Cv.

Oil propert ies areDensity at given conditions = 700 kg/m3

Viscosity at given conditions = 5 cP

Critical pressure = 60 baraVapor pressure = 1.5 bara

The detail s of the oil li ne areLine size 8

Total length of the line = 50mHeater inlet nozzle elevationvessel outlet nozzle elevation = -4.0 m (heater is on the ground

and vessel is elevated)Fittings12 nos. of 900 elbows and 2 gate valves

Step 1

First step of solving this control valve sizing sample problem is to determine the line pressuredrop resulting due to frictional losses from pipe and fittings plus elevational losses. Normally for

this case, the level control valve would be located close to the separator vessel. Hence thepressure drop between vessel and the control valve has been neglected and inlet pressure to the

control valve has been assumed to be the same as vessel outlet pressure.Frictional losses from straight pipe alone can be easily calculated usingEnggCyclopedias pipe

pressure drop calculator for single phase flowas follows.

Mass flow of oil = 200 X 700 = 140000 kg/hr (Normal flow case)Mass flow of oil = 60 X 700 = 42000 kg/hr (Minimum flow case)

Mass flow of oil = 220 X 700 = 164000 kg/hr (Maximum flow case)

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As perEnggCyclopedias calculator, pressure drop in bar/km of straight pipe is reported here

for the 3 cases,

For normal flow case,

pressure loss = 1.09 bar/km; frictional pressure drop in straight pipe = 0.05 X 1.09 = 0.0545bar

For minimum flow case,pressure loss = 0.128 bar/km; frictional pressure drop in straight pipe = 0.05 X 0.128 = 0.0064bar

For maximum flow case,

pressure loss = 1.445 bar/km; frictional pressure drop in straight pipe = 0.05 X 1.445= 0.0723bar

Fluid velocity expressed in m/s, for each flow case is also calculated at this time. This velocity

will be later used for determination of pressure drop due to fittings.

For normal flow, velocity = 1.71 m/sFor minimum flow, velocity = 0.51 m/s

For maximum flow, velocity = 2.01 m/s

Elevational pressure loss for all three cases is the same and is equal to (density X gravitationalacceleration X elevation change). Hence for all the three cases pressure loss is

= 700 X 9.8 X (-4.0) / 105 bar = -0.2744 bar.

Negative value indicates pressure gain instead of pressure loss due to drop in height.

To determine frictional pressure loss due to fittings, first the combined K-factor of fittings iscalculated usingEnggCyclopedias K-factor calculator.

For 12 nos. of 900 elbows and 2 gate valves, K factor = 5.64.

Pressure drop due to fittings is obtained by multiplying the K-factor by v2/2 for each case,where v is the velocity in m/s. Velocity for each flow case is calculated in EnggCyclopedias

pipe pressure drop calculator for single phase flow.

For normal flow case, fittings pressure drop = K Xv2/2 = 5.64 X 700 X 1.712 / (2 X 105) =

0.0577For minimum flow case, fittings pressure drop = K X v2/2 = 5.64 X 700 X 0.512 / (2 X 105) =

0.0051

For maximum flow case, fittings pressure drop = K X v2/2 = 5.64 X 700 X 2.012 / (2 X 105) =0.0798

Total pressure drop can be the calculated by adding the 3 components calculated independently

for each case,Line pressure drop for normal case = 0.0545 + 0.05770. 2744 = -0.1623 bar

Line pressure drop for normal case = -0.2629 bar

Line pressure drop for normal case = -0.1224 bar.

Note that, due to drop in height, the net pressure drop in the line has turned out to be negative.

Step 2

Next step for solving the control valve sizing sample problem is to determine the allowablepressure drop across the control valve for each of the three cases. It is calculated as,

Pressure drop across control valve = Vessel outlet pressureHeater inlet pressureline

pressure dropFor normal flow, P = 3.5 2(-0.1623) = 1.6622 bar

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For normal flow, P = 1.7623 bar

For normalflow, P = 1.6224 bar

Step 3

The final step of solving this sample problem is to determine the control valve flow coefficient orvalve Cv using EnggCyclopedias control valve sizing calculator. The inlet pressure to the valveis taken as vessel operating pressure since the valve is very close to the vessel. The outlet

pressure is taken by considering the allowable pressure drop across the control valve. Following

valve Cv values are calculated for the 3 cases.Normal flow Case, Cv = 150.44

Minimum flow case, Cv = 44.31

Maximum flow case, Cv = 178.94

The chosen valve Cv is always higher than the maximum Cv requirement with a margin for valveopening. These valve Cv values are given to the control valve manufacturer along with

corresponding flowrate values and subsequently a suitable valve with a higher valve Cv is

chosen to be installed.

Reference:

http://www.enggcyclopedia.com/2011/06/problem-solving-control-valve-sizing/