This research focuses primarily on the development of a numerical model of a feed line with a single control valve in operation. The control valve is modeled using its installed characteristics followed by its validation with Test data. Water is used as a medium to carry out the analysis owing to its ease of availability, nontoxicity, stability at room temperature, and less susceptibility to cavitation.
The current analysis aims at minimizing the variation in flowrates and pressure at the upstream of test article to ensure required flow parameters within permissible range of variation. A mathematical model of a feed line is developed and presented in this paper which includes flow components like constant pressure tank, feed line, Control valve, elbows and fittings. This model, upon successful validation, will be used to simulate some predetermined process test conditions like pressure and flow rate at the inlet of turbopumps of rocket engine during its testing. Based on the agreement of test results and numerical model, as discussed in this paper, this model shall be used further to model the pressure and flow characteristics during flow transition from one tank to other using simultaneous operation of two valves. It is expected that the changeover of flow from one from one feed circuit to another will cause pressure and flow transients at the engine inlet which are detrimental to the performance of the engine.
Modeling principles and classical assumptions
The analysis is carried out by solving the 1D equations obtained by applying the conservation of mass and momentum principles to a control volume. The solution involves following steps:

1.
Derivation of the partial differentiation equation.

2.
Parameters affecting the flow transients in a pipeline.

3.
Solving the equation through partial differential equation using “method of characteristics”.

4.
Characterization of the boundary conditions.

5.
Modeling the feed line circuit with control valve and assigning proper boundary conditions.

6.
Analysis of the results derived from above steps.
The existing fundamental theory for the transient flows in pipelines is drawn from classical assumptions: [13, 14]

1.
Onedimensional flow, with pipe full of incompressible fluid at all times.

2.
Dynamic liquidpipe interactions are neglected and a quasisteady interaction [15] between pipe and liquid is assumed and friction factors obtained under steady conditions applied to unsteady flows.

3.
Liquid velocity is much less than wave velocity.

4.
Rigid feed lines, which is merely an approximation as the feed line has expansion joints and flexible hoses.
Governing equations and numerical solution [16]
Newton second law of motion can be applied to the case of unsteady flow of a compressible fluid to an infinitesimal mass of fluid in an elastic pipe, as illustrated in Fig. 1, to get momentum equation, as follows:
$$ \rho Adx\frac{dV}{dt}= pA\left( pA+\frac{\partial p}{\partial x} Adx\right)+\rho gAdx\ sin\theta \tau \pi Ddx $$
(1)
As wall shear stress τ can be written as \( \boldsymbol{\tau} =\frac{\boldsymbol{\rho} \boldsymbol{fV}\left\boldsymbol{V}\right}{\mathbf{8}} \) ,
Thus, Eq. (1) can be rewritten as:
$$ \boldsymbol{L}\mathbf{1}:\frac{\mathbf{\partial}\boldsymbol{V}}{\mathbf{\partial}\boldsymbol{t}}+\frac{\mathbf{1}}{\boldsymbol{\rho}}\frac{\mathbf{\partial}\boldsymbol{p}}{\mathbf{\partial}\boldsymbol{x}}+\boldsymbol{g}\frac{\boldsymbol{dz}}{\boldsymbol{dx}}+\frac{\boldsymbol{fV}\mid \boldsymbol{V}\mid }{\mathbf{2}\boldsymbol{D}}=\mathbf{0} $$
(2)
Considering conservation of mass, the continuity equation can be obtained from the flow element and can be written as:
$$ \boldsymbol{L}\mathbf{2}:\frac{\boldsymbol{dp}}{\boldsymbol{dt}}+{\uprho \mathrm{a}}^2\frac{\boldsymbol{\partial V}}{\boldsymbol{\partial x}}+\boldsymbol{v}\ \frac{\partial p}{\partial x}=0 $$
(3)
The method of characteristics proceeds by making a linear combination of Eqs. (2) and (3) using a Lagrangian multiplier λ which, by appropriate substitution, results in two characteristic equations [17]
C^{+} equation:
$$ \frac{\boldsymbol{dV}}{\boldsymbol{dt}}+\frac{\mathbf{1}}{\boldsymbol{\rho} \boldsymbol{a}}\frac{\boldsymbol{dp}}{\boldsymbol{dt}}+\boldsymbol{g}\frac{\boldsymbol{dz}}{\boldsymbol{dx}}+\frac{\boldsymbol{fV}\left\boldsymbol{V}\right}{\mathbf{2}\boldsymbol{D}}=\mathbf{0} $$
(4)
Equation (4) is applicable if \( \frac{dx}{dt} \) = v + a
C^{−} equation:
$$ \frac{dV}{dt}\frac{1}{\rho a}\frac{dp}{dt}+g\frac{dz}{dx}+\frac{fV\leftV\right}{2D}=0 $$
(5)
Equation (5) is applicable if \( \frac{dx}{dt}=\mathrm{v}\mathrm{a} \)
The above equation forms the basis for the finite difference approach to compute the numerical solution of the transient characteristics in feed lines of fluids.
\( \frac{\boldsymbol{dx}}{\boldsymbol{dt}} \) = v ± a represents a pair of straight lines between t and x on which C+ and C− equations are valid, as shown in given Fig. 2.
Equations (4) and (5) are nonlinear coupled equation in flow velocity and pressure; therefore, first order firstorder finite difference method is used to solve the system of equations and compute the numerical solution.
A pipe of length L is discretized into N number of elements, giving N + 1 number of nodes. For every time step Δt, the pressure and flow velocity is computed at each node. The time step is determined by the pipe length and the wave speed according to Δt = Δx/a.
The MOC requires viscous flow in pipes to satisfy the Courant condition in Eq. 6. The Courant number (Co), is defined as the ratio of the actual wave speed and the numerical wave speed (Δx/Δt).
$$ \mathrm{Co}=\frac{a}{\raisebox{1ex}{$\Delta x$}\!\left/ \!\raisebox{1ex}{$\Delta t$}\right.} $$
(6)
Based on the analytical studies and procedures proposed by O’brien [18] and considering linearized equation, Perkins [19] has proved that for the process to be stable, the time step Δt should be calculated such that the Courant number should be less than or equal to 1 for better stability and faster convergence of the model. This shows that characteristics through the point C, as shown in Fig. 2 shall not fall outside segment AB [20].
The C^{+}, hereafter mentioned as CP is valid upstream, i.e., when using information from the previous node in the previous time step, represented by the characteristic line with positive slope in Fig. 2. The C^{−}, hereafter mentioned as CM is valid downstream, i.e., when using information from the next node in the previous time step, represented by the characteristic line with negative slope in Fig. 2. Integration of equations followed by first order approximation, results in following equations.
$$ \left({V}_{\mathrm{C}}{V}_{\mathrm{A}}\right)+\left({P}_{\mathrm{C}}{P}_{\mathrm{A}}\right)/\uprho \mathrm{a}+\mathrm{g}\frac{dz}{dx}\ \left({t}_{\mathrm{C}}{t}_{\mathrm{A}}\right)+{\mathrm{fV}}_{\mathrm{A}}\mid {V}_{\mathrm{A}}\mid \left({t}_{\mathrm{C}}{t}_{\mathrm{A}}\right)/2\mathrm{D}=0 $$
(7)
$$ {\mathrm{x}}_{\mathrm{C}}{\mathrm{x}}_{\mathrm{A}}=\left({V}_{\mathrm{A}}+\mathrm{a}\right)\ \left({t}_{\mathrm{C}}{t}_{\mathrm{A}}\right) $$
(8)
$$ \left({V}_{\mathrm{C}}{V}_{\mathrm{B}}\right)\left({P}_{\mathrm{C}}{P}_{\mathrm{B}}\right)/\uprho \mathrm{a}+\mathrm{g}\frac{dz}{dx}\ \left({t}_{\mathrm{C}}{\mathrm{t}}_{\mathrm{B}}\right)+{\mathrm{fV}}_{\mathrm{B}}\mid {V}_{\mathrm{B}}\mid \left({t}_{\mathrm{C}}{t}_{\mathrm{B}}\right)/2\mathrm{D}=0 $$
(9)
$$ {\mathrm{x}}_{\mathrm{C}}{\mathrm{x}}_{\mathrm{B}}=\left({V}_{\mathrm{B}}+\mathrm{a}\right)\ \left({t}_{\mathrm{C}}{\mathrm{t}}_{\mathrm{B}}\right) $$
(10)
Rewriting above equations in terms of flow rate Q and solving Eqs. (7) and (9) for QC and PC,
$$ Q\mathrm{c}=\mathrm{CP}\mathrm{BPc} $$
(11)
$$ Q\mathrm{c}=\mathrm{CM}+\mathrm{BPc} $$
(12)
where
$$ {\left.\mathrm{CP}={Q}_{\mathrm{A}}+{\mathrm{BP}}_{\mathrm{A}}{R}_{\mathrm{A}}{\mathrm{FFQ}}_{\mathrm{A}}\right}_{\mathrm{A}}\mid $$
(13)
$$ \mathrm{CM}={Q}_{\mathrm{B}}{\mathrm{B}\mathrm{P}}_{\mathrm{B}}{R}_{\mathrm{B}}{\mathrm{FFQ}}_{\mathrm{B}\mid \mathrm{B}\mid } $$
(14)
$$ B=A/\uprho \mathrm{a} $$
(15)
$$ R= gA\Delta \mathrm{t}\frac{dz}{dx} $$
(16)
$$ \mathrm{FF}=\mathrm{f}\Delta \mathrm{t}/2\mathrm{DA} $$
(17)
Solution to Eqs. (11) and (12) gives
$$ Q\mathrm{c}=\left(\mathrm{CM}+\mathrm{CP}\right)/2 $$
(18)
Equation (18) can be substituted in Eqs. (11) or (12) to get pressure Pc at that node.
Flowchart for computation of solution [11]
This section describes the steps involved in calculating the Pressure and flow rate for the pipeline transient problem. Corresponding steps involved in modeling process in the form of flowchart, as illustrated in Fig. 3.
Boundary conditions [20]
The finite difference equations, as defined in above equations, are applicable when the end boundary equations remain same, which is usually the case with steadystate flows. As soon as the boundary conditions start varying, it start influencing the interior points in the subsequent time step. Thus, in order to compute the pressure and flow at every time step, proper boundary conditions become necessary to be defined.
At the inlet and outlet of pipe, only one equation is available at either end. Thus, with respect to the previous time step, Eq. (13) is used to compute positive characteristic line at the upstream end of pipe while Eq. (14) can be used to get negative characteristic equation on the downstream side of the pipe, as shown in Fig. 4. With known pressure or flow conditions at the inlet or outlet of pipe, Eqs. (11) and (12) are used to calculate the other flow parameters on either end.
The simplest boundary conditions are specified values of the relevant variables, the pressure or flow in this case. At the inlet,
P_{c} = P_{reservoir} = constant.
Thus, using Eq. (14), negative characteristic equation is derived.
With pressure known at the inlet, Eq. (12) is used to compute flow at inlet.
Similarly, Eqs. (11) and (13) can be used to derive pressure/flow boundary condition at the outlet, provided either of one property out of flow or pressure is known.
It is to be noted that Eq. (18) and its associated equations are valid for interior points of the solution space, 2 ≤ i ≤ N.
Boundary conditions, valid for all values of t ≥ 0, must be imposed on the two points at either end. Initial conditions must be specified at t = 0.
Modeling of control valve [11]
The process of valve modeling defined in the present and subsequent sections is just a variation of the work by Berrier, Jr. [11] wherein the valve characteristics are defined as a function of valve response time and percentage lift of the valve. Berrier, Jr. [11] in his work presented valve pressuredischarge characteristic which follows exponential law and it is a function of time only and not a function of percentage lift of the valve. Moreover, the present work considers installed characteristics of the valve for analysis which evaluates flow and pressure transients across the valve based on variation of flow coefficient with percentage lift which is again a function of time.
Any flow component like a valveinline, as shown in Fig. 5, can be considered to be composed of two pipes connected by a valve. In this case, Eqs. (11) and (12) are valid for inlet and outlet of two pipes. Additional variables at valve inlet calls for additional auxiliary equations to obtain the complete solution. The continuity equation provides one such equation
$$ {\boldsymbol{Q}}_{\mathbf{P}}{\left{}_{\boldsymbol{J},\boldsymbol{N}+\mathbf{1}}={\mathbf{Q}}_{\mathbf{P}}\right}_{\mathbf{J}+\mathbf{1},\mathbf{1}} $$
(19)
while the fourth equation is deduced from the valve characteristic curve representing its flow coefficient vs. % lift of the valve.
Writing Eqs. (11) and (12) across the valve at nodes (J,N + 1) and (J + 1,1),
$$ {\left.{Q}_{\mathrm{P}}\right}_{\left(J,N+1\right)}={\mathrm{C}}_{\mathrm{p}}{\left{}_{\left(J,N+1\right)}{B}_1{\mathrm{P}}_{\mathrm{P}}\right}_{\left(J,N+1\right)} $$
(20)
$$ {\left.{Q}_{\mathrm{P}}\right}_{\left(J+1,1\right)}={\mathrm{C}}_{\mathrm{M}}{\left{}_{\left(J+1,1\right)}+{\mathrm{B}}_2{\mathrm{P}}_{\mathrm{P}}\right}_{\left(\mathrm{J}+1,1\right)} $$
(21)
where
$$ {\left.{C}_{\mathrm{p}}\right}_{\left(J,N+1\right)}={Q}_{\mathrm{a}}+{\mathrm{B}}_1{\mathrm{P}}_{\mathrm{a}}\mathrm{FF}\ {\mathrm{Q}}_{\mathrm{a}}\mid {\mathrm{Q}}_{\mathrm{a}}\mid $$
(22)
$$ {\left.{C}_{\mathrm{M}}\right}_{\left(J+1,1\right)=}\ {Q}_{\mathrm{b}}{\mathrm{B}}_2{\mathrm{P}}_{\mathrm{b}}\mathrm{FF}\ {\mathrm{Q}}_{\mathrm{b}}\mid {\mathrm{Q}}_{\mathrm{b}}\mid $$
(23)
When the valve is discharging into the downstream of the valve at some pressure, then for steadystate conditions at fully open condition, the flow rate through the valve can be written as:
$$ {Q}_0={C}_{\mathrm{VN}\ast}\kern0.5em \ast \surd {\Delta \mathrm{P}}_{\mathrm{o}} $$
(24)
The flow rate and pressure drop across the valve during the transient state conditions can be written as
$$ {Q}_{\mathrm{P}}=\boldsymbol{\uptau} .{C}_{\mathrm{VN}\ast}\kern0.5em \ast \surd \left({P}_{\mathrm{P}1}{P}_{\mathrm{P}2}\ \right) $$
(25)
Dividing Eqs. (24) by (25), we get
$$ \uptau =\frac{{\boldsymbol{Q}}_{\boldsymbol{p}}}{{\boldsymbol{Q}}_{\mathbf{0}}}\ast \sqrt{\frac{\Delta {\boldsymbol{P}}_{\mathbf{0}}}{\Big({\boldsymbol{P}}_{\boldsymbol{p}\mathbf{1}}{\boldsymbol{P}}_{\boldsymbol{p}\mathbf{2}\Big)}}}=\frac{{\boldsymbol{C}}_{\boldsymbol{V}}\ast }{{\boldsymbol{C}}_{\boldsymbol{V}\boldsymbol{N}\ast }}=\surd \frac{{\boldsymbol{C}}_{\boldsymbol{V}}}{{\boldsymbol{C}}_{\boldsymbol{V}\boldsymbol{N}}} $$
(26)
Rearranging the terms in Eq. (26),
$$ {Q}_{\mathrm{P}\mid J,N+1}=\uptau .Q0\ \sqrt{\frac{\Big({\boldsymbol{P}}_{\boldsymbol{p}\mathbf{1}}{\boldsymbol{P}}_{\boldsymbol{p}\mathbf{2}\Big)}}{\Big(\Delta {\boldsymbol{P}}_{\boldsymbol{o}\Big)}}} $$
(27)
Taking square on both sides and writing C_{V} as a function of C_{VN}
$$ {C}_{\mathrm{V}}={\boldsymbol{\uptau}}^{\mathbf{2}}\ {C}_{\mathrm{V}\mathrm{N}} $$
(28)
where,
$$ {\displaystyle \begin{array}{c}{C}_{\mathrm{VN}}=\frac{{\boldsymbol{Q}}_{\boldsymbol{o}}^{\mathbf{2}}}{2\mathbf{\Delta }{\mathbf{P}}_o}\\ {}\ {Q_{\mathrm{P}\mid J,N+1}}^2=2\ast \frac{{\boldsymbol{\uptau}}^{\mathbf{2}}.{\boldsymbol{Q}}_{\boldsymbol{o}}^{\mathbf{2}}}{2\mathbf{\Delta }{\mathbf{P}}_o}\ \left\{\frac{C_{P\left(J,N+1\right)}+{Q}_{P\left(J,N+1\right)}}{B_1}\right.\frac{C_{M\left(J+1,1\right)}{Q}_{P\left(J+1,N\right)}}{B_2}\ \Big\}\end{array}} $$
(29)
Rewriting,
$$ {Q_{\mathrm{P}\mid J,N+1}}^2\kern0.5em 2{C}_{\mathrm{v}}\ {Q}_{\mathrm{P}\mid J,N+1}\left(\frac{1}{B_1}+\frac{1}{B_2}\right)2{C}_{\mathrm{v}}\left\{\frac{{\boldsymbol{C}}_{\boldsymbol{P}}}{{\boldsymbol{B}}_{\mathbf{1}}}\frac{{\boldsymbol{C}}_{\boldsymbol{M}}}{{\boldsymbol{B}}_{\mathbf{2}}}\right\}=\mathbf{0} $$
(30)
The solution of above quadratic equation is based on the direction of flow which is decided by the value of \( \frac{C_P}{B_1}\frac{C_M}{B_2} \) if it is greater or less than 0.
Making substitutions and solving above equation for Q_{P}_{J,N + 1,} gives
$$ {Q}_{\mathrm{P}\mid J,N+1}={C}_{\mathrm{V}}\ \left(\frac{1}{B_1}+\frac{1}{B_2}\right)+\sqrt{\left({C_{\mathrm{V}}}^2{\left(\frac{1}{B_1}+\frac{1}{B_2}\right)}^2+2{C}_V\ \left(\frac{C_P}{B_1}\frac{C_M}{B_2}\right)\right)} $$
(31)
Above equation is applicable for positive flow where Eq. (26) is applicable and \( \frac{{\boldsymbol{C}}_{\boldsymbol{P}}}{{\boldsymbol{B}}_{\mathbf{1}}}\frac{{\boldsymbol{C}}_{\boldsymbol{M}}}{{\boldsymbol{B}}_{\mathbf{2}}} \) >0
For negative flow, Eq. (27) can be written as
$$ \uptau =\frac{{\boldsymbol{Q}}_{\boldsymbol{p}}}{{\boldsymbol{Q}}_{\mathbf{0}}}\ast \sqrt{\frac{{\boldsymbol{P}}_{\mathbf{0}}}{\Big({\boldsymbol{P}}_{\boldsymbol{p}\mathbf{2}}{\boldsymbol{P}}_{\boldsymbol{p}\mathbf{1}\Big)}}}\ \mathrm{and}\ \frac{{\boldsymbol{C}}_{\boldsymbol{P}}}{{\boldsymbol{B}}_{\mathbf{1}}}\frac{{\boldsymbol{C}}_{\boldsymbol{M}}}{{\boldsymbol{B}}_{\mathbf{2}}}<0 $$
In this case, Eq. (27) can be written as
$$ {Q}_{\mathrm{P}\mid J,N+1}={C}_{\mathrm{V}}\ \left(\frac{1}{B_1}+\frac{1}{B_2}\right)\sqrt{\left({C_{\mathrm{V}}}^2\left(\frac{1}{B_1}+\frac{1}{B_2}\right)2{C}_V\ \left(\frac{C_P}{B_1}\frac{C_M}{B_2}\right)\right)} $$
(32)
With Q_{P}_{J,N + 1} known, it can be substituted in continuity Eqs. (19), (20), (21) to compute Q_{P}_{J + 1,1}, P_{P1} and P_{P2.}