Control valves are an essential component in many industrial processes, used to regulate the flow of fluids such as gases and liquids. Proper sizing of control valves is crucial for maintaining precise control over the process parameters. Inaccurate valve sizing can result in issues such as cavitation, flashing, and reduced performance.

Control valve sizing calculation is crucial for maintaining efficient and reliable industrial processes. By following the proper sizing process and selecting the appropriate valve type, engineers and operators can ensure the optimal performance of their industrial processes.

**Table of content:
**Control valve sizing

Control valve sizing calculation

## Control valve sizing

Control valve sizing is the process of selecting the appropriate size of a control valve for a specific industrial process application. Control valves are used to regulate the flow of fluids such as gases and liquids, and proper sizing is essential for maintaining precise control over process parameters such as flow rate, pressure, and temperature.

The sizing process involves determining the process parameters, selecting the appropriate valve type, calculating the required valve coefficient (C_{v }), determining the valve size, and verifying valve performance. Proper sizing ensures that the valve can handle the maximum expected flow rate while maintaining the desired level of control, and can prevent issues such as cavitation, flashing, and reduced performance.

Selecting the right valve type is also important in control valve sizing. Common types of control valves include globe valves, ball valves, and butterfly valves, each with their unique advantages and disadvantages. The selection of the valve type depends on factors such as the process fluid, the operating conditions, and the required level of control.

Factors to consider in ensuring proper valve performance include proper installation, valve position, and control system settings. It is important to verify valve performance after installation through testing and adjustment of the valve's position and control system settings.

## Control valve sizing calculation

#### 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 200m^{3}/hr, 60m^{3}/hr and 220 m^{3}/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 C_{v }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 C_{v } values are calculated for the 3 cases.

Normal flow Case, C_{v } = 150.44

Minimum flow case, C_{v }= 44.31

Maximum flow case, C_{v }= 178.94

The chosen valve C_{v }is always higher than the maximum C_{v }requirement with a margin for valve opening. These valve C_{v }values are given to the control valve manufacturer along with corresponding flowrate values and subsequently a suitable valve with a higher valve C_{v }is chosen to be installed.