Instrumentation Process Design Solved Sample Problems

Sample Problem – control valve sizing

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 at the heater is 2.0 barg. Normal, minimum and maximum oil flowrates are 200m3/hr, 60m3/hr and 220 m3/hr respectively. Size a level control valve to determine the control valve flow coefficient or valve Cv.

Oil properties are –
Density at given conditions = 700 kg/m3
Viscosity at given conditions = 5 cP
Critical pressure = 60 bara
Vapor pressure = 1.5 bara

The details of the oil line are –
Line size 8”
Total length of the line = 50m
Heater inlet nozzle elevation – vessel outlet nozzle elevation = -4.0 m (heater is on the ground and vessel is elevated)
Fittings – 12 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 pressure drop 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 the pressure 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 using EnggCyclopedia’s pipe pressure drop calculator for single phase flow as 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)

As per EnggCyclopedia’s 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 = 0.0545 bar
For minimum flow case, pressure loss = 0.128 bar/km; frictional pressure drop in straight pipe = 0.05 X 0.128 = 0.0064 bar
For maximum flow case, pressure loss = 1.445 bar/km; frictional pressure drop in straight pipe = 0.05 X 1.445= 0.0723 bar

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/s
For 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 gravitational acceleration 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 is calculated using EnggCyclopedia’s 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 ρv2/2 for each case, where ‘v’ is the velocity in m/s. Velocity for each flow case is calculated in EnggCyclopedia’s pipe pressure drop calculator for single phase flow.

For normal flow case, fittings pressure drop = K X ρv2/2 = 5.64 X 700 X 1.712 / (2 X 105) = 0.0577
For minimum flow case, fittings pressure drop = K X ρv2/2 = 5.64 X 700 X 0.512 / (2 X 105) = 0.0051
For maximum flow case, fittings pressure drop = K X ρv2/2 = 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 + 0.0577 – 0. 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 allowable pressure drop across the control valve for each of the three cases. It is calculated as,
Pressure drop across control valve = Vessel outlet pressure – Heater inlet pressure – line pressure drop
For normal flow, ΔP = 3.5 – 2 – (-0.1623) = 1.6622 bar
For normal flow, ΔP = 1.7623 bar
For normal flow, ΔP = 1.6224 bar

Step 3

The final step of solving this sample problem is to determine the control valve flow coefficient or valve Cv using EnggCyclopedia’s control valve sizing calculator. The inlet pressure to the valve is 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 valve opening. 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.

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