Independent Research Project
Aim: To calculate the estimated force acting on the valve during the different operating conditions.
Given Problem Statement:
This estimated force should be above the target cracking force (opening force) of the valve which is 300N. The solid geometry will be provided, which can be simplified to extract fluid volume (spring can be neglected). The boundary conditions on which the simulation has to be performed are given as
Operating Conditions:
Case 1. At 0 Deg C (Fluid Viscosity: 416.73 centipoise, Density – 886kg/m3)
- Flow Rate 160 l/min, Inlet Pressure 20 bar
- Flow Rate 160 l/min, Inlet Pressure 10 bar
- Flow Rate 125 l/min, Inlet Pressure 20 bar
- Flow Rate 125 l/min, Inlet Pressure 20 bar
Case 2. At 40 Deg C (Fluid Viscosity: 27.52 centipoise, Density – 859.6 kg/m3)
- Flow Rate 160 l/min, Inlet Pressure 20 bar
- Flow Rate 160 l/min, Inlet Pressure 10 bar
- Flow Rate 125 l/min, Inlet Pressure 20 bar
- Flow Rate 125 l/min, Inlet Pressure 20 bar
Software Used: ANSYS Fluent
Problem Solution:
How to calculate the force near the valve?
For calculation of the force we have to calculate the pressure and area of the valve.
Pressure = Force / Area
Force = Pressure * Area
- Pressure calculation: We have to run the CFD Analysis of the valve for different given conditions.
- Area calculation: Using Space Claim we can calculate the Area of the valve.
Figure 1: Surface area of the Valve
We know the area of the valve, now we have to calculate the Pressure for different operating conditions.
Theory:
Valves are devices that work to control, regulate or direct flow within a system or process. Valves are looking to control flow, provide safety in a system that is piping liquids, solids, gasses or anything in between them.
Valves provide several functions, including:
- Starting or stopping flow based on the valve state
- Regulating flow and pressure within a piping system
- Controlling the direction of flow within a piping system
- Throttling flow rates within a piping system
- Improving safety through relieving pressure or vacuum in a piping system
Pressure relief valve:
A pressure relief valve or relief valve is a special type of safety valve system used to control or limit the pressure in a system. It can be manually or automatically controlled from a pressurized vessel or piping system. The pressurized fluid or gas is discharged to a reservoir or atmosphere to relieve pressure in excess of the maximum allowable working pressure.
The primary purpose of a pressure relief valve is to protect pressure vessels or system from catastrophic failure. Catastrophic failure could be disastrous during an overpressure event, could either be liquid or gaseous.
This device is widely used in petrochemical, petroleum refining, chemical manufacturing industries. Industries where natural gas processing occurs and power generation, as well as water supply industries, also make good use of pressure relief valve. Though it’s generally known as relief valve depending on its field of application it can be called pressure relief valve (PRV), pressure safety valve (PSV), or safety valve. You should note the design and operation of these valves are slightly different.
Figure 2: Parts of Pressure relief valve
Things to Consider Before Selecting a Pressure Relief Valve:
Below are things an engineer or professional in this field consider while selecting a relief valve:
- Design Considerations which include the size, material, product, complexity, etc.
- The individual relief load should be determined.
- Causes of overpressure.
- Various types of relief devices and their characteristics should consider seeing the best option.
- Selecting appropriate relief devices to handle the imposed loads. Several components must be put into consideration. Parts like set pressure, backpressure, dual relief valve, and multiple relief valves must suit the capacity of the relieving device.
- Installation considerations
- Inlet piping and discharge piping
- Reactive forces
- Tailpipe considerations
- Rapid Cycling
- Resonant chatter which occurs when the inlet piping produces excessive loss at the valve inlet.
- Liquid discharge considerations.
Geometry:
Figure 3: Given Geometry
Extract Volume Region:
Figure 4: selected the edges of the volumetric region
Figure 5: volumetric region
Meshing:
Figure 6: Meshing
Mesh Properties:
Mesh properties for given model are:
Element type: Triangle
Statistics:
Case Setup:
Case 1: Temp = 0 degree C, Flow Rate 160 l/min and Inlet Pressure 20 bar
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 886 kg/m3
Viscosity: 4.1673 poise
Boundary Conditions:
Inlet:
Pressure inlet = 20 bar,
Mass flow rate = 160 l/min = 2.666 kg/sec
Temperature = 273 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Initialization:
Residual Plot:
Figure 7: Case1-Residual plot
Pressure plot:
Figure 8: Case1-Pressure Plot
Contours:
Pressure Contour:
Figure 9: Case1-Pressure contour
Stream lines:
Figure 10: Case1-Stream Lines
Temperature:
Figure 11: Case1-Temperature Contour
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 127667.61 * 0.0028050
= 358.107 N
Force = 358.1 N
Case 2: Temp = 0 degree C, Flow Rate 160 l/min and Inlet Pressure 10 bar
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 886 kg/m3
Viscosity: 4.1673 poise
Boundary Conditions:
Inlet:
Pressure inlet = 10 bar,
Mass flow rate = 160 l/min = 2.666 kg/sec
Temperature = 273 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Residual Plot:
Figure 12: Case2-Residual Plot
Pressure:
Figure 13: Case2 – Pressure plot
Contours:
Pressure:
Figure 14: Case2 – Pressure contour
Temperature:
Figure 15: Case2 – Temperature contour
Streamlines:
Figure 16: Case2 – Stream lines
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 127667.61 * 0.0028050
= 358.107 N
Force = 358.1 N
Case 3: Temp = 0 degree, Flow Rate 125 l/min and Inlet Pressure 20 bar:
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 886 kg/m3
Viscosity: 4.1673 poise
Boundary Conditions:
Inlet:
Pressure inlet = 20 bar,
Mass flow rate = 125 l/min = 2.0833 kg/sec
Temperature = 273 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Residual Plot:
Figure 17: Case3 – Residual Plot
Pressure:
Figure 18: Case3 – Pressure Plot
Contours:
Pressure:
Figure 19: Case3 – Pressure Contour
Temperature:
Figure 20: Case3 – Temperature Contour
Stream lines:
Figure 21: Case3 – Stream Lines
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 83214.431 * 0.0028050
= 233.4164 N
Force = 233.4164 N
Case 4: Temp = 0 degree, Flow Rate 125 l/min and Inlet Pressure 10 bar:
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 886 kg/m3
Viscosity: 4.1673 poise
Boundary Conditions:
Inlet:
Pressure inlet = 10 bar,
Mass flow rate = 125 l/min = 2.0833 kg/sec
Temperature = 273 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Residual Plot:
Figure 22: Case4 – Residual Plot
Pressure:
Figure 23: Case4 – Pressure Plot
Contours:
Pressure:
Figure 24: Case4 – Pressure contour
Temperature:
Figure 25: Case4 – Temperature Contour
Stream lines:
Figure 26: Case4 – Stream Lines
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 83214.431 * 0.0028050
= 233.4164 N
Force = 233.4164 N
Case 5: Temp = 40 degree, Flow Rate 160 l/min and Inlet Pressure 20 bar:
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 859.6 kg/m3
Viscosity: 0.2752 poise
Boundary Conditions:
Inlet:
Pressure inlet = 20 bar,
Mass flow rate = 125 l/min = 2.6667 kg/sec
Temperature = 313 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Residual Plot:
Figure 27: Case5 – Residual Plot
Pressure:
Figure 28: Case5 – Pressure Plot
Contours:
Pressure:
Figure 29: Case5 – Pressure Contour
Temperature:
Figure 30: Case5 – Temperature Contour
Stream Lines:
Figure 31: Case5 – Stream lines
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 91113.836 * 0.0028050
= 255.574 N
Force = 255.574 N
Case 6: Temp = 40 degree, Flow Rate 160 l/min and Inlet Pressure 10 bar:
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 859.6 kg/m3
Viscosity: 0.2752 poise
Boundary Conditions:
Inlet:
Pressure inlet = 10 bar,
Mass flow rate = 125 l/min = 2.6667 kg/sec
Temperature = 313 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Residual Plot:
Figure 32: Case 6 – Residual Plot
Pressure:
Figure 33: Case 6 – Pressure Plot
Contours:
Pressure:
Figure 34: Case 6 – Pressure Contour
Temperature:
Figure 35: Case 6 – Temperature Contour
Stream Lines:
Figure 36: Case 6 – Steam lines
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 93092.75 * 0.0028050
= 261.1251 N
Force = 261.1251 N
Case 7 Temp = 40 degree, Flow Rate 125 l/min and Inlet Pressure 20 bar:
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 859.6 kg/m3
Viscosity: 0.2752 poise
Boundary Conditions:
Inlet:
Pressure inlet = 20 bar,
Mass flow rate = 125 l/min = 2.0833 kg/sec
Temperature = 313 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Residual Plot:
Figure 37: Case 7 – Residual Plot
Pressure:
Figure 38: Case 7 – Pressure Plot
Contour:
Pressure:
Figure 39: Case 7 – Pressure contour
Temperature:
Figure 40: Case 7 – Temperature contour
Stream Lines:
Figure 41: Case 7 – Stream lines
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 56586.513 * 0.0028050
= 158.7251N
Force = 158.7251 N
Case 8 Temp = 40 degree, Flow Rate 125 l/min and Inlet Pressure 10 bar:
Solver: Pressure based
Analysis: Steady state simulation
Models:
K-Epsilon, Standard model, Energy equation
Materials: Air
Density: 859.6 kg/m3
Viscosity: 0.2752 poise
Boundary Conditions:
Inlet:
Pressure inlet = 10 bar,
Mass flow rate = 125 l/min = 2.0833 kg/sec
Temperature = 313 K
Outlet: Pressure outlet, Gauge pressure-0 pa
Residual Plot:
Figure 42: Case 8 – Residual Plot
Pressure:
Figure 43: Case 8 – Pressure Plot
Contour:
Pressure:
Figure 44: Case 8 – Pressure contour
Temperature:
Figure 45: Case 8 – Temperature contour
Stream Lines:
Figure 46: Case 8 – Stream lines
Sample calculation:
Pressure = Force / Area
Force = Pressure * Area
Force = 56586.513 * 0.0028050
= 158.7251N
Force = 158.7251 N
Observations:
Figure 47: observation table
Graphs:
Condition 1
- Flow Velocity vs Pressure at Valve
Graph 1:Condition1- Flow Velocity vs Pr at Valve
- Flow Velocity vs Force
Graph 2: Condition 1- Flow velocity Vs Force
Condition2:
- Flow Velocity vs Pressure at Valve
Graph 3: Condition2- Flow Velocity vs Pr at Valve
- Flow Velocity vs Force
Graph 4: Condition 1- Flow velocity Vs Force
Observations:
- As the Flow velocity decreases, Pressure at the valve also decreases.
- Estimated Force at the valve decreases because of Pressure is directly proportionally to the Force.
- There is no much difference in estimated force, where pressure at the inlet at constant Flow rate.
- Temperature is also one of the parameter which depends on the force, if increases the temperature, estimated Force also decreases.
Conclusions:
- As per given different input conditions, we calculated the Pressure and estimated force at the valve.
- First we extract the fluid volume in Space claim for given geometry.
- Next we calculated the estimated force at the valve.
- For calculate the estimate force, there is a relation between the force and pressure.
- Force is directly proportionally to pressure, if we know the pressure and Area of the valve then we can calculate estimated Force (N).
- Pressure is calculated for running the different cases of the valve simulation using Fluent and area of the Valve is calculated using Space claim.
- Case1 and Case2 will gives more than 300 N of the estimated force, As per given problem Case1 and case2 satisfies the conditions of the estimated force.
- As per simulation,
- Mass flow rate decreases, estimated force also decreases at constant inlet pressure.
- Temperature increases, estimated force also decreases at constant mass flow rate.
- Estimated Force is depends on Mass flow rate and Temperature.
- Hence estimated force and Pressure at the values are calculated as per given conditions. Case1 and Case2 gives the estimated force greater than the cracking force.