Episode 06.Pressure Relief Valve Sizing Method
Summary
A pressure relief valve is a device designed to protect a system from overpressure, commonly used in high-pressure systems. Selecting the appropriate valve size is crucial for effective pressure release. This episode covers how to size a pressure relief valve based on fluid type.
1. Introduction
(Figure 6.1)
A pressure relief valve is a type of valve that helps prevent excessive pressure within a system. When the pressure rises above a set limit, the valve opens to release fluid, thereby reducing the pressure and protecting the equipment or system.
When determining the size of the pressure relief valve, there are key factors to consider. The first step is to decide the valve’s set the pressure at which the valve will open. It is typically set to the design pressure or the Maximum Allowable Working Pressure (MAWP) to ensure the system’s safety.
The second factor is whether the valve can release sufficient fluid from the system when it opens. To achieve this, we need to calculate the orifice area of the pressure relief valve. This episode will focus on determining the area for various scenarios. The formula for calculating the orifice area typically follows the form:
\( {\small\displaystyle\begin{align} A \quad=\quad \frac{W}{BKG} \tag{Eq 6.1} \end{align} } \)
\({\small\displaystyle W}\) represents the mass flow rate of the discharged fluid, \({\small\displaystyle B}\) is a fixed value depending on the units used, \({\small\displaystyle K}\) represents various coefficients, and \({\small\displaystyle G}\) is the mass flux. While the method of determining the values for each term varies depending on the situation, the basic structure remains consistent.
2. Sizing pressure relief valve for gas service
When sizing the valve for gas relief, the calculation method differs depending on the gas flow condition – whether it is choked(=critical) flow or non-choked flow. How can we determine the type of gas flow during discharge? Assume no frictional losses and no difference in potential energy. Additionally, there is no work and no heat exchange with the surroundings. Then, the heat (or energy) balance equation simplifies to:
\( {\small\displaystyle\begin{align} d\left(\frac{v^2}{2}\right)\,+\,gdh\,+\,\frac{dP}{\rho}\,+\,dh_f \,=\, 0 \quad\longrightarrow\quad d\left(\frac{v^2}{2}\right) \,=\, -\frac{dP}{\rho} \tag*{} \end{align} } \)
The gas discharge process in a pressure relief valve follows an isentropic process. Then, \({\small\displaystyle P_0}\) is considered the stagnation state, the density becomes \({\small\displaystyle \rho_0}\), and the velocity is zero.
\( {\small\displaystyle\begin{gather} P\rho^{-\gamma} \,=\, \text{constant} \\[10pt] P\,=\,P_0, \qquad \rho\,=\,\rho_0, \qquad v\,=\,0 \tag*{} \end{gather} } \)
Then, we can derive (Eq 6.2) by integrating over the range from \({\small\displaystyle P_0}\) to \({\small\displaystyle P}\).
\( {\small\displaystyle\begin{align} \int_0^v\,d\left(\frac{v^2}{2}\right) \quad&=\quad \frac{P_0^{1/\gamma}}{\rho_0}\int_P^{P_0}\,P^{-1/\gamma}\,dP,\quad \frac{P_0}{\rho_0^{\gamma}}\,=\,\frac{P}{\rho^{\gamma}} \\[10pt] \frac{v^2}{2} \quad&=\quad \frac{\gamma}{\gamma\,-\,1}\cdot\frac{P_0}{\rho_0}\left[1\,-\,\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}\right] \\[15pt] v^2 \quad&=\quad \frac{2\gamma}{\gamma\,-\,1}\cdot\frac{P_0}{\rho_0}\left[1\,-\,\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}\right] \tag{Eq 6.2} \end{align} } \)
By expressing the velocity \({\small\displaystyle v}\), in terms of Mach number, we can rewrite (Eq 6.2) as follows.
\( {\small\displaystyle\begin{align} Ma^2\gamma\frac{P}{\rho} \quad&=\quad \frac{2\gamma}{\gamma\,-\,1}\cdot\frac{P_0}{\rho_0}\left[1\,-\,\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}\right],\quad Ma\,=\, \frac{v}{\sqrt{\gamma\dfrac{P}{\rho}}} \\[10pt] Ma^2 \quad&=\quad \frac{2}{\gamma\,-\,1}\left[\left(\frac{P}{P_0}\right)^{\frac{1\,-\,\gamma}{\gamma}}\,-\,1\right],\quad \frac{\rho}{\rho_0} \,=\, \left(\frac{P}{P_0}\right)^{1/\gamma} \tag*{} \end{align} } \)
When the equation is rearranged based on the pressure ratio, it becomes (Eq 6.3). Using (Eq 6.3), we can calculate the pressure ratio (=critical pressure ratio) at which the flow becomes choked flow, meaning the point where \({\small\displaystyle Ma\,=\,1}\).
\( {\small\displaystyle\begin{align} \frac{P}{P_0} \quad&=\quad \left(\frac{\gamma\,-\,1}{2}Ma^2\,+\,1\right)^{\frac{\gamma}{1\,-\,\gamma}} \tag{Eq 6.3} \\[15pt] \frac{P^*}{P_0} \quad&=\quad \left(\frac{2}{\gamma\,+\,1}\right)^{\frac{\gamma}{\gamma\,-\,1}} \tag{Eq 6.4} \end{align} } \)
Choked flow
Let’s look at the case where the discharged fluid is a choked flow. For choked flow to occur, the pressure ratio \({\small\displaystyle P_B/P_1}\) must be (Eq 6.4) or less. \({\small\displaystyle P_1}\) is the inlet pressure, defined as the sum of the set pressure and the allowable overpressure. \({\small\displaystyle P_B}\) represents the backpressure on the discharge side.
\( {\small\displaystyle\begin{align} \frac{P_B}{P_1} \quad \le\quad \left(\frac{2}{\gamma\,+\,1}\right)^{\frac{\gamma}{\gamma\,-\,1}} \tag{Eq 6.5} \end{align} } \)
Under these conditions, we can calculate the mass flux. Mass flux, \({\small\displaystyle G}\) is expressed as the product of velocity and density. By substituting \({\small\displaystyle v}\) using (Eq 6.2), we can simplify it as follows.
\( {\small\displaystyle\begin{align} \qquad G \quad&=\quad \rho\cdot v \\[10pt] &=\quad \rho\sqrt{\frac{2\gamma}{\gamma\,-\,1}\cdot\frac{P_1}{\rho_1}\cdot\left[1\,-\,\left(\frac{P_2}{P_1}\right)^{(\gamma\,-\,1)/\gamma}\right]} \,=\, \sqrt{\frac{2\gamma\rho_2 P_1}{\gamma\,-\,1}\cdot\frac{\rho_2}{\rho_1}\cdot\frac{\gamma\,-\,1}{\gamma\,+\,1}} \,=\, \sqrt{\frac{2\gamma\rho_2P_1}{\gamma\,+\,1}\cdot\left(\frac{P_2}{P_1}\right)^{1/\gamma}} \\[10pt] &=\quad \sqrt{\frac{2\gamma\rho_2P_1}{\gamma\,+\,1}\cdot\left(\frac{2}{\gamma\,+\,1}\right)^{\frac{1}{\gamma\,-\,1}}} \,=\, \sqrt{\gamma\cdot\left(\frac{2}{\gamma\,+\,1}\right)^{\gamma/(\gamma\,-\,1)}\cdot\rho_1\cdot\left(\frac{P_2}{P_1}\right)^{1/\gamma}\cdot P_1} \\[10pt] &=\quad \sqrt{\gamma\rho_1P_1\left(\frac{2}{\gamma\,+\,1}\right)^{(\gamma\,+\,1)/(\gamma\,-\,1)}} \end{align} } \)
\( {\small\displaystyle\begin{align} G \quad=\quad P_1\sqrt{\frac{M_w\gamma}{ZRT_1}\left(\frac{2}{\gamma\,+\,1}\right)^{(\gamma\,+\,1)/(\gamma\,-\,1)}} \tag{Eq 6.6} \end{align} } \)
Substituting the above equation into (Eq 6.1) gives the following result.
\( {\small\displaystyle\begin{align} A \quad&=\quad \frac{W}{BKP_1\sqrt{\dfrac{M_w\gamma}{ZRT_1}\cdot\left(\dfrac{2}{\gamma\,+\,1}\right)^{(\gamma\,+\,1)/(\gamma\,-\,1)}}} \\[10pt] &=\quad \frac{W}{BK_bK_cK_dP_1\sqrt{\dfrac{M_w\gamma}{ZRT_1}\cdot\left(\dfrac{2}{\gamma\,+\,1}\right)^{(\gamma\,+\,1)/(\gamma\,-\,1)}}} \tag{Eq 6.7} \end{align} } \)
| \( {\displaystyle A } \) | \( {\displaystyle W } \) | \( {\displaystyle B } \) | \( {\displaystyle P } \) | \( {\displaystyle R } \) |
| \( {\displaystyle \text{m}^2 } \) | \( {\displaystyle \text{kg/s} } \) | \( {\displaystyle 1 } \) | \( {\displaystyle \text{Pa} } \) | \( {\displaystyle 8,314\frac{\text{J}}{\text{kmol}\cdot\text{K} }} \) |
| \( {\displaystyle \text{in}^2 } \) | \( {\displaystyle \text{lb/hr} } \) | \( {\displaystyle 1702 } \) | \( {\displaystyle \text{psia} } \) | \( {\displaystyle 10.73\frac{\text{psia}\cdot\text{ft}^3}{\text{lbmol}\cdot ^\circ\text{R}} } \) |
The value of \({\small\displaystyle B}\) depends on the units used. For details, please refer to (Table 6.1). \({\small\displaystyle K_b}\) is the coefficient accounting for back pressure, \({\small\displaystyle K_c}\) is the combination coefficient, which depends on whether only a safety valve is installed or if it is installed in series with a rupture disk, and \({\small\displaystyle K_d}\) is the discharge coefficient. We can find the value for each \({\small\displaystyle K}\) in API STD 520 Part 1 or provided by the manufacturer.[1]
Non-choked flow
Now, let’s consider the case where the discharged fluid flow is a non-choked flow. Unlike choked flow, it occurs when the pressure ratio \({\small\displaystyle P_B/P_1}\) is greater than (Eq 6.4).
\( {\small\displaystyle\begin{align} \frac{P_B}{P_1} \quad >\quad \left(\frac{2}{\gamma\,+\,1}\right)^{\frac{\gamma}{\gamma\,-\,1}} \tag{Eq 6.8} \end{align} } \)
Then, mass flux \({\small\displaystyle G}\) is:
\( {\small\displaystyle\begin{align} \qquad G \quad&=\quad \sqrt{\frac{2\gamma\rho P_1}{\gamma\,-\,1}\cdot\frac{\rho}{\rho_0}\left[1\,-\,\left(\frac{P_2}{P_1}\right)^{(\gamma\,-\,1)/\gamma}\right]} \,=\, \sqrt{\frac{2\gamma\rho_1P_1}{\gamma\,-\,1}\cdot\left(\frac{P_2}{P_1}\right)^{2/\gamma}\left[1\,-\,\left(\frac{P_2}{P_1}\right)^{(\gamma\,-\,1)/\gamma}\right]} \\[10pt] &=\quad \sqrt{\frac{2\gamma\rho_1P_1}{\gamma\,-\,1}\left[\left(\frac{P_2}{P_1}\right)^{2/\gamma}\,-\,\left(\frac{P_2}{P_1}\right)^{(\gamma\,+\,1)/\gamma}\right]} \end{align} } \)
\( {\small\displaystyle\begin{align} G \quad=\quad P_1\sqrt{\frac{M_w}{ZRT_1}\cdot\frac{2\gamma}{\gamma\,-\,1}\left[\left(\frac{P_2}{P_1}\right)^{2/\gamma}\,-\,\left(\frac{P_2}{P_1}\right)^{(\gamma\,+\,1)/\gamma}\right]} \tag{Eq 6.9} \end{align} } \)
\({\small\displaystyle P_2}\) is the total backpressure, defined as the sum of the backpressure and the allowable overpressure. By substituting (Eq 6.9) into (Eq 6.1), we can calculate the valve’s required area under non-choked flow conditions.
\( {\small\displaystyle\begin{align} A \quad=\quad \frac{W} {BK_cK_dP_1\sqrt{\dfrac{M_w}{ZRT_1}\cdot\dfrac{2\gamma}{\gamma\,-\,1}\left[\left(\dfrac{P_2}{P_1}\right)^{2/\gamma}\,-\,\left(\dfrac{P_2}{P_1}\right)^{(\gamma\,+\,1)/\gamma}\right]}} \tag{Eq 6.10} \end{align} } \)
Sizing for steam
For steam in choked flow conditions, (Eq 6.11) is used instead of (Eq 6.7).
\( {\small\displaystyle\begin{align} A[\text{m}^2] \quad&=\quad \frac{686W}{P_1K_bK_cK_dK_nK_{sh}} \\ \tag{Eq 6.11} \\ A[\text{in}^2] \quad&=\quad \frac{W}{51.5P_1K_bK_cK_dK_nK_{sh}} \end{align} } \)
\({\small\displaystyle K_n}\) is the correction factor for the Napier equation. When \({\small\displaystyle P_1}\) is 10.3Mpa(1,500psig) or below, \({\small\displaystyle K_n}\) equals 1. For \({\small\displaystyle P_1}\) exceeding 10.3Mpa up to 22.1Mpa(3200psig), \({\small\displaystyle K_n}\) can be caculated using the following equation.
\( {\small\displaystyle\begin{align} K_n \quad&=\quad \frac{27.644P_1\,-\,1000}{33.242P_1\,-\,1061} \qquad P_1:[\text{mPa}] \\ \tag{Eq 6.12} \\ K_n \quad&=\quad \frac{0.1906P_1\,-\,1000}{0.2292P_1\,-\,1061} \qquad P_1:[\text{psia}] \end{align} } \)
\({\small\displaystyle K_{sh}}\) is a correction factor applied when steam is in a superheated state; its value can be found in API STD 520 Part 1.[1] For saturated steam, \({\small\displaystyle K_{sh}}\) is 1.
3. Sizing pressure relief valve for liquid service
How do we calculate the size of the pressure relief valve when the fluid is a liquid? The basic structure remains the same. The mass flux of the liquid phase can be derived through the following steps: Since the fluid is a liquid, we can assume that its density remains constant.
\( {\small\displaystyle\begin{align} \frac{dP}{\rho} \,=\, -d\left(\frac{v^2}{2}\right) \quad&\longrightarrow\quad \int_{P_1}^{P_B}\,\frac{1}{\rho}\,dP \,=\, -\int_0^{v}\,d\left(\frac{v^2}{2}\right) \\[15pt] v \quad&=\quad \sqrt{\frac{2\left(P_1\,-\,P_B\right)}{\rho}} \tag{Eq 6.13} \\[15pt] G \quad&=\quad \sqrt{2\rho\left(P_1\,-\,P_B\right)} \tag{Eq 6.14} \end{align} } \)
Therefore, the size of the valve is:
\( {\small\displaystyle\begin{align} A \quad&=\quad \frac{W}{BK\sqrt{2\rho\left(P_1\,-\,P_B\right)}} \\[10pt] &=\quad \frac{W}{BK_cK_dK_vK_w\sqrt{2\rho\left(P_1\,-\,P_B\right)}} \tag{Eq 6.15} \end{align} } \)
\({\small\displaystyle K_v}\) is the coefficient related to fluid viscosity, which can be determined using (Eq 6.16). \({\small\displaystyle K_w}\) is the correction factor for back pressure and should be applied when using a bellows-type valve.
\( {\small\displaystyle\begin{align} K_v \quad=\quad \left(0.9935\,+\,\frac{2.878}{Re^{0.5}}\,+\,\frac{342.75}{Re^{1.5}}\right)^{-1} \tag{Eq 6.16} \end{align} } \)
Nomenclature
\({\small\displaystyle A}\) : area
\({\small\displaystyle B}\) : fixed value depending on units used
\({\small\displaystyle G}\) : mass flux
\({\small\displaystyle g}\) : gravity acceleration
\({\small\displaystyle h}\) : height
\({\small\displaystyle h_f}\) : friction head
\({\small\displaystyle K}\) : coefficients
\({\small\displaystyle K_b}\) : correction factor due to backpressure
\({\small\displaystyle K_c}\) : combination coefficient
\({\small\displaystyle K_d}\) : discharge coefficient
\({\small\displaystyle K_n}\) : correction factor for Napier equation
\({\small\displaystyle K_sh}\) : correction factor for superheated steam
\({\small\displaystyle K_v}\) : correction factor due to viscosity
\({\small\displaystyle K_w}\) : correction factor due to backpressure when using a bellows-type valve
\({\small\displaystyle Ma}\) : mach number
\({\small\displaystyle P}\) : pressure
\({\small\displaystyle P_1}\) : inlet pressure (absolute)
\({\small\displaystyle P_2}\) : outlet pressure(absolute)
\({\small\displaystyle P_B}\) : backpressure
\({\small\displaystyle P^*}\) : choked flow pressure
\({\small\displaystyle R}\) : gas constant
\({\small\displaystyle Re}\) : Reynolds number
\({\small\displaystyle T}\) : temperature
\({\small\displaystyle v}\) : velocity
\({\small\displaystyle W}\) : mass flowrate
\({\small\displaystyle Z}\) : compressibility factor
\({\small\displaystyle \gamma}\) : isentropic coefficient or ratio of the specific heats
\({\small\displaystyle \rho}\) : density
Reference
[1] : American Petroleum Institute. API Standard 520, Sizing, Selection, and installation of Pressure-relieving Devices Part1-Sizing and Selection. API | API 520 Part I, 10th Ed.