Modeling of low Mach number unsteady turbulent pipe flows

Abstract

Under adiabatic conditions, and neglecting temperature variations due to entropy production, we present a set of Reynolds Averaged Navier–Stokes (RANS) equations for fluids of low compressibility, i.e., fluids in the liquid state. In the low Mach number limit, we specialize the RANS equations to the one-dimensional unsteady pipe flow, and we deduce the dimensionless number that plays a predominant role in the flow behavior. We reduce the system of equations to a linear damped wave equation, and use its analytical solution to investigate the propagation of large amplitude pressure waves in liquid-filled pipes (water hammer phenomenon). We test the model reliability by comparing the analytical solution of the proposed model against experimental data available in the literature.

1 Introduction

We consider the Reynolds Averaged Navier–Stokes (RANS) equations for fluids with low compressibility, i.e., fluids in the liquid state. Based on the Reynolds decomposition, the RANS equations provide a statistical description of turbulent flows [1]. For incompressible fluids, the Reynolds decomposition is coupled with the time-averaging. When the ergodicity is violated the ensemble averaging is used instead [2]. For compressible fluids, the Reynolds decomposition is typically coupled with the Favre averaging [3]. The RANS equations involve the turbulence modeling of the terms that appear in the system of equations to describe the effects of turbulent fluctuations. Indeed, these additional terms need closure equations. These topics represent crucial aspects of fluid mechanics research. Following the suggestions proposed in [4], under adiabatic conditions, and neglecting the temperature variations due to entropy production, in Sect. 2 we deduce a set of RANS equations without invoking the Favre averaging. After discussing the physical insights of the additional terms in the system of equations, we provide a few remarks about the Boussinesq turbulence model [5] to close the momentum and entropy equations. In Sect. 3 we specialize the RANS equations to 1D unsteady pipe flow, and, in the low Mach number limit, we reduce the system of equations to a linear damped wave equation. In formulating the dimensionless equations, we introduce a dimensionless number that plays a predominant role in the unsteady flow. We use the linear damped wave equation to investigate the propagation of large amplitude pressure waves in liquid-filled pipes (water hammer phenomenon [6]). To test the model reliability, we provide a comparison of experimental data available in the literature [7] against the analytical solution of the proposed equation (the analytical solution of the linear damped wave equation is provided in “Appendix”). Experimental data concern the thermodynamic pressure time series collected during a water hammer phenomenon in a horizontal steel pipe. Due to the strong energy loss that accompanies the wave propagation, we report deeper insights into the physical aspect related to the entropy production and the kinetic energy equation. In Sect. 4, we summarize the main findings.

2 RANS equations for fluids with low compressibility

Under adiabatic conditions, the balance equations for (first order) viscous fluids read as (se, e.g., [8]):

$$\begin{aligned}{} & {} \frac{\partial \rho }{\partial t}+\nabla \cdot \left( \rho \varvec{v}\right) =0,\end{aligned}$$

(1)

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho \varvec{v}\right) +\nabla \cdot \left( \rho \varvec{v}\otimes \varvec{v}\right) =\rho \frac{\partial \varvec{v}}{\partial t}+\rho \nabla \varvec{v}\cdot \varvec{v}=-\rho g{\hat{\varvec{e}}}_z \nonumber \\{} & {} \quad -\nabla p-\nabla \cdot {\underline{\varvec{T}}}_{vis}, \end{aligned}$$

(2)

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho e_m\right) +\nabla \cdot \left( \rho e_m\varvec{v}+p\varvec{v}+{\underline{\varvec{T}}}_{vis}\cdot \varvec{v}\right) =p\nabla \cdot \varvec{v}\nonumber \\{} & {} \quad +{\underline{\varvec{T}}}_{vis}:\nabla \varvec{v}, \end{aligned}$$

(3)

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho e_t\right) +\nabla \cdot \left( \rho e_t\varvec{v}+p\varvec{v}+{\underline{\varvec{T}}}_{vis}\cdot \varvec{v}\right) =0, \end{aligned}$$

(4)

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho u_i\right) +\nabla \cdot \left( \rho u_i\varvec{v}\right) =-p\nabla \cdot \varvec{v}-{\underline{\varvec{T}}}_{vis}:\nabla \varvec{v}, \end{aligned}$$

(5)

$$\begin{aligned}{} & {} T\frac{\partial }{\partial t}\left( \rho e_s\right) +T\nabla \cdot \left( \rho e_s\varvec{v}\right) =-{\underline{\varvec{T}}}_{vis}:\nabla \varvec{v}, \end{aligned}$$

(6)

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho e_p\right) +\nabla \cdot \left( \rho e_p\varvec{v}\right) =-p\nabla \cdot \varvec{v}, \end{aligned}$$

(7)

where $\rho$ is the fluid density; $\varvec{v}$ the velocity; g the modulus of the gravitational acceleration; ${\hat{\varvec{e}}}_z$ the unit vector directed along the vertical direction; p the thermodynamic pressure [9]; ${\underline{\varvec{T}}}_{vis}$ is the viscous stress tensor given by:

$$\begin{aligned} {\underline{\varvec{T}}}_{vis}=\underline{\varvec{T}}-p\underline{\textbf{1}}, \end{aligned}$$

(8)

with $\underline{\varvec{T}}$ denoting the Cauchy stress tensor, and $\underline{\textbf{1}}$ the unit tensor; $e_m=e_k+gz$ the specific mechanical energy, with $e_k=\frac{\varvec{v}\cdot \varvec{v}}{2}=\frac{v^2}{2}$ being the specific kinetic energy, gz the specific gravitational potential energy, z the elevation in the gravitational field; $e_t=e_m+u_i$ the specific total energy, with $u_i$ indicating the specific internal energy; T the absolute temperature; $e_s$ the specific entropy; $e_p$ the specific elastic potential energy. We stress that Eq. (3) is derived from the scalar product of the momentum equation (2) and the velocity $\varvec{v}$; Eq. (4) by the application of the first principle of thermodynamics; Eq. (5) from the comparison between Eqs. (3) and (4); Eqs. (6) and (7) by using the Gibbs relationship:

$$\begin{aligned} \textrm{d}u_i=T\textrm{d}e_s+\frac{p}{\rho ^2}\textrm{d}\rho =T\textrm{d}e_s+\textrm{d}e_p. \end{aligned}$$

(9)

The dissipation function $\phi$, that describes the dissipation of the kinetic energy due to viscous effects, is given as:

$$\begin{aligned} \phi =-{\underline{\varvec{T}}}_{vis}:\nabla \varvec{v}. \end{aligned}$$

(10)

For linear isotropic viscous fluids, the constitutive mechanical equation reads as [8]:

$$\begin{aligned} {\underline{\varvec{T}}}_{vis}=-2\mu _1{\underline{\varvec{D}}}_{dev}-\eta \left( \nabla \cdot \varvec{v}\right) \underline{\textbf{1}}, \end{aligned}$$

(11)

where ${\underline{\varvec{D}}}_{dev}$ is the deviatoric part of the strain rate tensor $\underline{\varvec{D}}=\frac{1}{2}\left( \nabla \varvec{v}+{\nabla \varvec{v}}^\textrm{T}\right)$; $\mu _1$ the first viscous coefficient; $\eta$ the bulk viscosity. For Navier–Stokes fluids, the viscous coefficients can be regarded as constant [10]. Neglecting the temperature increase due to entropy production, the relationship between the density and the thermodynamic pressure is given by the barotropic equation [11]:

$$\begin{aligned} \frac{\textrm{d}\rho }{\rho }=\frac{\textrm{d}p}{\varepsilon }, \end{aligned}$$

(12)

where $\varepsilon$ is the fluid bulk modulus. According to Eq. (12), the wave celerity c is given as:

$$\begin{aligned} c=\sqrt{\frac{\varepsilon }{\rho }}. \end{aligned}$$

(13)

Setting $T\approx constant$, and assuming that the bulk modulus does not change significantly with pressure, Eq. (12) leads to:

$$\begin{aligned} \rho =\rho _0\textrm{e}^\frac{p-p_0}{\varepsilon }, \end{aligned}$$

(14)

where $\rho _0$ is the reference value of density at the pressure $p_0$. We express the Reynolds decomposition as:

$$\begin{aligned} b=\left[ b\right] +b^\prime , \end{aligned}$$

(15)

where $\left[ b\right]$ is the average value of a generic field b, and $b^\prime$ the turbulent fluctuation. We conjecture that for fluids with low compressibility, i.e., fluids in the liquid state for which $\varepsilon \gg 1$, the turbulent fluctuations of density are evanescent [4]. Accordingly:

$$\begin{aligned} \rho ^\prime =0, \end{aligned}$$

(16)

$$\begin{aligned} \rho =\left[ \rho \right] =\rho _0\textrm{e}^\frac{p-p_0}{\varepsilon }. \end{aligned}$$

(17)

Indeed, as $\varepsilon \gg 1$, $\frac{p^\prime }{\varepsilon }\ll 1$, $\textrm{e}^\frac{p^\prime }{\varepsilon }\approx 1$, and $\rho =\rho _0\textrm{e}^\frac{\left[ p\right] -p_0+p^\prime }{\varepsilon }\approx \rho _0\textrm{e}^\frac{\left[ p\right] -p_0}{\varepsilon }$. As the fluctuations of density are assumed as negligible and the fluid is assumed to be weakly compressible, the Favre averaging is no longer needed and the conventional RANS procedure [5] can be applied to obtain the following equations:

$$\begin{aligned}&\frac{\partial \rho }{\partial t}+\nabla \cdot \left( \rho \left[ \varvec{v}\right] \right) =0, \end{aligned}$$

(18)

$$\begin{aligned}&\frac{\partial }{\partial t}\left( \rho \left[ \varvec{v}\right] \right) +\nabla \cdot \left( \rho \left[ \varvec{v}\right] \otimes \left[ \varvec{v}\right] \right) =\rho \frac{\partial }{\partial t}\left[ \varvec{v}\right] +\rho \nabla \left[ \varvec{v}\right] \cdot \left[ \varvec{v}\right] \nonumber \\&\quad =-\rho g{\hat{\varvec{e}}}_z-\nabla \left[ p\right] -\nabla \cdot \left[ {\underline{\varvec{T}}}_{vis}\right] -\nabla \cdot {\underline{\varvec{T}}}_{Re}, \end{aligned}$$

(19)

$$\begin{aligned}&\frac{\partial }{\partial t}\left( \rho \left[ e_m\right] \right) +\nabla \cdot \left( \rho \left[ e_m\right] \left[ \varvec{v}\right] +\left[ p\right] \left[ \varvec{v}\right] +\left[ {\underline{\varvec{T}}}_{vis}\right] \cdot \left[ \varvec{v}\right] +\rho \left[ e_m^\prime \varvec{v}^\prime \right] \right. \nonumber \\&\qquad \left. +\left[ p^\prime \varvec{v}^\prime \right] +\left[ {\underline{\varvec{T}}}_{vis}^\prime \cdot \varvec{v}^\prime \right] \right) \nonumber \\&\quad = \left[ p\right] \nabla \cdot \left[ \varvec{v}\right] +\left[ {\underline{\varvec{T}}}_{vis}\right] \nonumber \\&\quad :\nabla \left[ \varvec{v}\right] +\left[ p^\prime \nabla \cdot \varvec{v}^\prime \right] +\left[ {\underline{\varvec{T}}}_{vis}^\prime :\nabla \varvec{v}^\prime \right] , \end{aligned}$$

(20)

$$\begin{aligned}&\frac{\partial }{\partial t}\left( \rho \left[ e_t\right] \right) +\nabla \cdot \left( \rho \left[ e_t\right] \right. \nonumber \\&\qquad \left. \left[ \varvec{v}\right] +\left[ p\right] \left[ \varvec{v}\right] +\left[ {\underline{\varvec{T}}}_{vis}\right] \right. \nonumber \\&\qquad \left. \cdot \left[ \varvec{v}\right] +\rho \left[ e_t^\prime \varvec{v}^\prime \right] +\left[ p^\prime \varvec{v}^\prime \right] +\left[ {\underline{\varvec{T}}}_{vis}^\prime \cdot \varvec{v}^\prime \right] \right) =0, \end{aligned}$$

(21)

$$\begin{aligned}&\frac{\partial }{\partial t}\left( \rho \left[ u_i\right] \right) +\nabla \cdot \left( \rho \left[ u_i\right] \left[ \varvec{v}\right] \right. \nonumber \\&\qquad \left. +\rho \left[ u_i^\prime \varvec{v}^\prime \right] \right) =\nonumber \\&\qquad -\left[ p\right] \nabla \cdot \left[ \varvec{v}\right] -\left[ {\underline{\varvec{T}}}_{vis}\right] \nonumber \\&\qquad :\nabla \left[ \varvec{v}\right] -\left[ p^\prime \nabla \cdot \varvec{v}^\prime \right] \nonumber \\&\qquad -\left[ {\underline{\varvec{T}}}_{vis}^\prime :\nabla \varvec{v}^\prime \right] , \end{aligned}$$

(22)

$$\begin{aligned}&\frac{\partial }{\partial t}\left( \rho T\left[ e_s\right] \right) +\nabla \cdot \left( \rho T\left[ e_s\right] \left[ \varvec{v}\right] \right. \nonumber \\&\qquad \left. +\rho T\left[ e_s^\prime \varvec{v}^\prime \right] \right) =-\left[ {\underline{\varvec{T}}}_{vis}\right] \nonumber \\&\qquad :\nabla \left[ \varvec{v}\right] -\left[ {\underline{\varvec{T}}}_{vis}^\prime :\nabla \varvec{v}^\prime \right] , \end{aligned}$$

(23)

$$\begin{aligned}&\frac{\partial }{\partial t}\left( \rho \left[ e_p\right] \right) +\nabla \cdot \left( \rho \left[ e_p\right] \left[ \varvec{v}\right] \right. \nonumber \\&\quad \left. +\rho \left[ e_p^\prime \varvec{v}^\prime \right] \right) =-\left[ p\right] \nabla \cdot \left[ \varvec{v}\right] \nonumber \\&\quad -\left[ p^\prime \nabla \cdot \varvec{v}^\prime \right] , \end{aligned}$$

(24)

where ${\underline{\varvec{T}}}_{Re}=\rho \left[ \varvec{v}^\prime \otimes \varvec{v}^\prime \right]$ is the Reynolds stress tensor. In analogy with Eq. (8), we decompose the Reynolds stress tensor in the form:

$$\begin{aligned} {\underline{\varvec{T}}}_{Re}=\mathcal {P}_{tur}\underline{\textbf{1}}+{\underline{\varvec{T}}}_{tur}. \end{aligned}$$

(25)

From a formal point of view, we can assume the term $\mathcal {P}_{tur}$ as the homologous of the thermodynamic pressure p, and the term ${\underline{\varvec{T}}}_{tur}$ as the homologous of the viscous stress tensor ${\underline{\varvec{T}}}_{vis}$. By using Eq. (25), Eq. (19) can be written as:

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho \left[ \varvec{v}\right] \right) +\nabla \cdot \left( \rho \left[ \varvec{v}\right] \otimes \left[ \varvec{v}\right] \right) =-\rho g{\hat{\varvec{e}}}_z-\nabla \left[ p\right] -\nabla \cdot \left( \mathcal {P}_{tur}\underline{\textbf{1}}\right) \nonumber \\{} & {} -\nabla \cdot \underline{\varvec{\Gamma }}=-\rho g{\hat{\varvec{e}}}_z-\nabla \left[ p\right] -\nabla \mathcal {P}_{tur}-\nabla \cdot \underline{\varvec{\Gamma }}, \end{aligned}$$

(26)

where

$$\begin{aligned} \underline{\varvec{\Gamma }}=\left[ {\underline{\varvec{T}}}_{vis}\right] +{\underline{\varvec{T}}}_{tur}. \end{aligned}$$

(27)

The physical insights about $\mathcal {P}_{tur}$ and ${\underline{\varvec{T}}}_{tur}$ can be highlighted by using an alternative form of the RANS equations. For this purpose, we deduce the mean mechanical energy equation from Eq. (26) by scalar product by $\left[ \varvec{v}\right]$:

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho \left[ e_m\right] \right) +\nabla \cdot \left( \rho \left[ e_m\right] \left[ \varvec{v}\right] +\left[ p\right] \left[ \varvec{v}\right] +\mathcal {P}_{tur}\left[ \varvec{v}\right] +\underline{\varvec{\Gamma }}\cdot \left[ \varvec{v}\right] \right) \nonumber \\{} & {} \quad =\left[ p\right] \nabla \cdot \left[ \varvec{v}\right] +\mathcal {P}_{tur}\nabla \cdot \left[ \varvec{v}\right] +\underline{\varvec{\Gamma }}:\nabla \left[ \varvec{v}\right] . \end{aligned}$$

(28)

According to Eq. (28), the mean total energy equation can be given as:

$$\begin{aligned} \frac{\partial }{\partial t}\left( \rho \left[ e_t\right] \right) +\nabla \cdot \left( \rho \left[ e_t\right] \left[ \varvec{v}\right] +\left[ p\right] \left[ \varvec{v}\right] +\mathcal {P}_{tur}\left[ \varvec{v}\right] +\underline{\varvec{\Gamma }}\cdot \left[ \varvec{v}\right] \right) =0, \end{aligned}$$

(29)

whilst the mean internal energy equation can be expressed as:

$$\begin{aligned} \frac{\partial }{\partial t}\left( \rho \left[ u_i\right] \right) +\nabla \cdot \left( \rho \left[ u_i\right] \left[ \varvec{v}\right] \right) =-\left[ p\right] \nabla \cdot \left[ \varvec{v}\right] -\mathcal {P}_{tur}\nabla \cdot \left[ \varvec{v}\right] -\underline{\varvec{\Gamma }}:\nabla \left[ \varvec{v}\right] . \end{aligned}$$

(30)

By using Eq. (27), the comparison between Eqs. (22) and (30) leads to the relationship:

$$\begin{aligned} \nabla \cdot \left( \rho \left[ u_i^\prime \varvec{v}^\prime \right] \right) +\left[ p^\prime \nabla \cdot \varvec{v}^\prime \right] +\left[ {\underline{\varvec{T}}}_{vis}^\prime :\nabla \varvec{v}^\prime \right] =\mathcal {P}_{tur}\nabla \cdot \left[ \varvec{v}\right] +{\underline{\varvec{T}}}_{tur}:\nabla \left[ \varvec{v}\right] . \end{aligned}$$

(31)

According to Eq. (31), we conjecture that:

$$\begin{aligned}{} & {} \nabla \cdot \left( \rho \left[ u_i^\prime \varvec{v}^\prime \right] \right) +\left[ p^\prime \nabla \cdot \varvec{v}^\prime \right] =\mathcal {P}_{tur}\nabla \cdot \left[ \varvec{v}\right] , \end{aligned}$$

(32)

$$\begin{aligned}{} & {} \left[ {\underline{\varvec{T}}}_{vis}^\prime :\nabla \varvec{v}^\prime \right] ={\underline{\varvec{T}}}_{tur}:\nabla \left[ \varvec{v}\right] . \end{aligned}$$

(33)

By using Eqs. (27) and (33), the mean entropy equation (23) becomes:

$$\begin{aligned} \frac{\partial }{\partial t}\left( \rho T\left[ e_s\right] \right) +\nabla \cdot \left( \rho T\left[ e_s\right] \left[ \varvec{v}\right] +\rho T\left[ e_s^\prime \varvec{v}^\prime \right] \right) =-\underline{\varvec{\Gamma }}:\nabla \left[ \varvec{v}\right] , \end{aligned}$$

(34)

where the term $-\underline{\varvec{\Gamma }}:\nabla \left[ \varvec{v}\right]$ is the mean dissipation function $\left[ \phi \right]$. Thus, we can divide the mean dissipation function into two parts:

$$\begin{aligned} \left[ \phi \right] =-\underline{\varvec{\Gamma }}:\nabla \left[ \varvec{v}\right] =-\left[ {\underline{\varvec{T}}}_{vis}\right] :\nabla \left[ \varvec{v}\right] -{\underline{\varvec{T}}}_{tur}:\nabla \left[ \varvec{v}\right] , \end{aligned}$$

(35)

where $-\left[ {\underline{\varvec{T}}}_{vis}\right] :\nabla \left[ \varvec{v}\right]$ is the contribution due to the viscosity; $-{\underline{\varvec{T}}}_{tur}:\nabla \left[ \varvec{v}\right]$ is the contribution due to the turbulent fluctuations. This is the physical essence of the conjecture given by Eqs. (32) and (33): consistent with this line of reasoning, the scalar field $\mathcal {P}_{tur}$ is not responsible for entropy production; the term $\nabla \cdot \left( \mathcal {P}_{tur}\underline{\textbf{1}}\right) =\nabla \mathcal {P}_{tur}$ describes the effects of turbulent fluctuations on the momentum flux; the tensor field ${\underline{\varvec{T}}}_{tur}$ is such that the additional friction forces (per unit volume) due to turbulent fluctuations are given as $-\nabla \cdot {\underline{\varvec{T}}}_{tur}$. Hence, $\mathcal {P}_{tur}$ can be identified as a convective term, and ${\underline{\varvec{T}}}_{tur}$ as a dissipative term. According to Gibbs’s equation, setting:

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho \left[ u_i\right] \right) +\nabla \cdot \left( \rho \left[ u_i\right] \left[ \varvec{v}\right] \right) -\frac{\partial }{\partial t}\left( \rho T\left[ e_s\right] \right) -\nabla \cdot \left( T\rho \left[ e_s\right] \left[ \varvec{v}\right] \right) \nonumber \\{} & {} \quad =\frac{\partial }{\partial t}\left( \rho \left[ e_p\right] \right) +\nabla \cdot \left( \rho \left[ e_p\right] \left[ \varvec{v}\right] \right) , \end{aligned}$$

(36)

and using Eqs. (30) and (34), we deduce the following equation:

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho \left[ e_p\right] \right) +\nabla \cdot \left( \rho \left[ e_p\right] \left[ \varvec{v}\right] \right) \nonumber \\{} & {} \quad =-\left[ p\right] \nabla \cdot \left[ \varvec{v}\right] -\mathcal {P}_{tur}\nabla \cdot \left[ \varvec{v}\right] +\nabla \cdot \left( \rho T\left[ e_s^\prime \varvec{v}^\prime \right] \right) . \end{aligned}$$

(37)

The comparison between Eqs. (24) and (37) provides the relationship:

$$\begin{aligned} \nabla \cdot \left( \rho \left[ e_p^\prime \varvec{v}^\prime \right] \right) +\left[ p^\prime \nabla \cdot \varvec{v}^\prime \right] +\nabla \cdot \left( \rho T\left[ e_s^\prime \varvec{v}^\prime \right] \right) =\mathcal {P}_{tur}\nabla \cdot \left[ \varvec{v}\right] , \end{aligned}$$

(38)

whilst Eqs. (32) and (38) lead to the equation:

$$\begin{aligned} \nabla \cdot \left( \rho \left[ u_i^\prime \varvec{v}^\prime \right] \right) -\nabla \cdot \left( \rho T\left[ e_s^\prime \varvec{v}^\prime \right] \right) =\nabla \cdot \left( \rho \left[ e_p^\prime \varvec{v}^\prime \right] \right) . \end{aligned}$$

(39)

We stress that, by using the barotropic, the continuity equation becomes:

$$\begin{aligned} \frac{\partial \left[ p\right] }{\partial t}+\nabla \left[ p\right] \cdot \left[ \varvec{v}\right] +\varepsilon \nabla \cdot \left[ \varvec{v}\right] =0, \end{aligned}$$

(40)

whilst, by using the dissipation function, Eq. (28) can be written as:

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\left( \rho \left[ e_m\right] \right) +\nabla \cdot \left( \rho \left[ e_m\right] \left[ \varvec{v}\right] +\left[ p\right] \left[ \varvec{v}\right] +\mathcal {P}_{tur}\left[ \varvec{v}\right] +\underline{\varvec{\Gamma }}\cdot \left[ \varvec{v}\right] \right) \nonumber \\{} & {} \quad =\left[ p\right] \nabla \cdot \left[ \varvec{v}\right] +\mathcal {P}_{tur}\nabla \cdot \left[ \varvec{v}\right] -\left[ \phi \right] . \end{aligned}$$

(41)

We focus on the needed closure equations for the momentum and entropy equations. By splitting the (symmetric) tensor ${\underline{\varvec{T}}}_{tur}$ in the isotropic and deviatoric parts:

$$\begin{aligned} {\underline{\varvec{T}}}_{tur}=\left( {\underline{\varvec{T}}}_{tur}\right) _{iso}+\left( {\underline{\varvec{T}}}_{tur}\right) _{dev}, \end{aligned}$$

(42)

where the components of $\left( {\underline{\varvec{T}}}_{tur}\right) _{iso}$ are the turbulent normal stresses, whilst the components of $\left( {\underline{\varvec{T}}}_{tur}\right) _{dev}$ are the turbulent shear stresses, Eq. (25) becomes:

$$\begin{aligned} {\underline{\varvec{T}}}_{Re}=\mathcal {P}_{tur}\underline{\textbf{1}}+\left( {\underline{\varvec{T}}}_{tur}\right) _{iso}+\left( {\underline{\varvec{T}}}_{tur}\right) _{dev}. \end{aligned}$$

(43)

Within the framework of the eddy-viscosity model and by using the Boussinesq formulation [5], we generalize Eq. (11) for the viscous stresses, and we assume:

$$\begin{aligned}{} & {} \left( {\underline{\varvec{T}}}_{tur}\right) _{iso}=-\eta _{tur}\nabla \cdot \left[ \varvec{v}\right] \underline{\textbf{1}}, \end{aligned}$$

(44)

$$\begin{aligned}{} & {} \left( {\underline{\varvec{T}}}_{tur}\right) _{dev}=-2\mu _{1,tur}\left[ \underline{\varvec{D}}\right] _{dev}, \end{aligned}$$

(45)

where $\left[ \underline{\varvec{D}}\right] _{dev}$ is the deviatoric part of the main strain rate tensor $\left[ \underline{\varvec{D}}\right] =\frac{1}{2}\left( \nabla \left[ \varvec{v}\right] +{\nabla \left[ \varvec{v}\right] }^\textrm{T}\right)$; $\mu _{1,tur}$ the eddy viscosity; $\eta _{tur}$ the turbulent bulk viscosity [4]. Due to the constraint $\textrm{Tr}\left( {\underline{\varvec{T}}}_{Re}\right) >0$, it has to be stressed that $\mathcal {P}_{tur}-\eta _{tur}\nabla \cdot \left[ \varvec{v}\right] >0$. According to the adopted line of reasoning, we conjecture that:

$$\begin{aligned}{} & {} \mathcal {P}_{tur}=m_{tur}\rho \left( \left[ \varvec{v}\right] \cdot \left[ \varvec{v}\right] \right) =m_{tur}\rho \left[ v\right] ^2, \end{aligned}$$

(46)

$$\begin{aligned}{} & {} \rho T\left[ e_s^\prime \varvec{v}^\prime \right] =\alpha _{tur}\rho T\left[ e_s\right] \left[ \varvec{v}\right] , \end{aligned}$$

(47)

where $m_{tur}$ and $\alpha _{tur}$ are coefficients of turbulent diffusivity of momentum and of entropy, respectively.

3 One-dimensional unsteady turbulent flow

3.1 System of equations

We specialize the RANS equation to unsteady turbulent flow in pipes. We work within the framework of hydraulic approximation, according to which all quantities are constant in the pipe cross-section, i.e., the model is one-dimensional and time-dependent. We identify $\left[ \varvec{v}\right]$ in U; $\left[ p\right]$ in $\varPi$; $\left[ e_m\right] =\left[ e_k\right] =k=\frac{U^2}{2}$; $\left[ e_s\right]$ in $\sigma$; the nabla operator $\nabla$ in $\frac{\partial }{\partial x}$, where x is the longitudinal coordinate along the pipe axis; the term $-\nabla \cdot \underline{\varvec{\Gamma }}$ in $-\frac{\partial \tau }{\partial x}=-\rho \varpi \frac{U\left| U\right| }{2D}+\eta \frac{\partial ^2U}{\partial x^2}+\frac{\partial }{\partial x}\left( \eta _{tur}\left( \frac{\partial U}{\partial x}\right) \right)$ [4], with $\varpi$ indicating the Darcy friction factor [12], and D the (constant) pipe diameter. In first approximation, we can take $m_{tur}=constant$, $\eta _{tur}=constant$, and $\alpha _{tur}=constant$. Thus, for a horizontal cylindrical pipe, the momentum equation reads as:

$$\begin{aligned} \frac{\partial U}{\partial t}+U\frac{\partial U}{\partial x}= & {} -\frac{1}{\rho }\frac{\partial \varPi }{\partial x}-\frac{m_{tur}}{\rho }\frac{\partial \left( \rho U^2\right) }{\partial x}-\varpi \frac{U\left| U\right| }{2D}\nonumber \\{} & {} \quad +\,\frac{\eta +\eta _{tur}}{\rho }\frac{\partial ^2U}{\partial x^2}, \end{aligned}$$

(48)

whilst the kinetic energy is given by:

$$\begin{aligned} \frac{\partial k}{\partial t}+U\frac{\partial k}{\partial x}= & {} -\frac{U}{\rho }\frac{\partial \varPi }{\partial x}-2\frac{m_{tur}}{\rho }U\frac{\partial \left( \rho k\right) }{\partial x}-\varpi \frac{k\left| U\right| }{D} \nonumber \\{} & {} \quad +\,\frac{\eta +\eta _{tur}}{\rho }\left( \frac{\partial ^2k}{\partial x^2}-\left( \frac{\partial U}{\partial x}\right) ^2\right) . \end{aligned}$$

(49)

Equation (49) is deduced from Eq. (48) by scalar product by U, and by using the differential identity $U\frac{\partial ^2U}{\partial x^2}=\frac{\partial }{\partial x}\left( U\frac{\partial U}{\partial x}\right) -\left( \frac{\partial U}{\partial x}\right) ^2=\frac{\partial ^2k}{\partial x^2}-\left( \frac{\partial U}{\partial x}\right) ^2$. In 1D flow, the barotropic equation (17) reads as:

$$\begin{aligned} \rho =\rho _0\textrm{e}^\frac{\varPi -p_0}{\hat{\varepsilon }}, \end{aligned}$$

(50)

where $\hat{\varepsilon }$ is the bulk modulus of the pipe-fluid system [13,14,15], whilst the entropy equation reduces to:

$$\begin{aligned} T\frac{\partial \sigma }{\partial t}+TU\frac{\partial \sigma }{\partial x}=\varpi \frac{k\left| U\right| }{D}+\frac{\eta +\eta _{tur}}{\rho }\left( \frac{\partial U}{\partial x}\right) ^2-\alpha _{tur}\frac{T}{\rho }\frac{\partial \left( \rho U\sigma \right) }{\partial x}. \end{aligned}$$

(51)

Hence, the dissipation function is expressed as $\phi _{1D}=-\tau \frac{\partial U}{\partial x}=\rho \varpi \frac{k\left| U\right| }{D}+\left( \eta +\eta _{tur}\right) \left( \frac{\partial U}{\partial x}\right) ^2$. We stress that, in line with Eq. (50), $\hat{\varepsilon }$ is assumed as a constant and, for liquid-filled pipes, $\hat{\varepsilon }\gg 1$. In agreement with the barotropic equation (50), the wave celerity c is given as [4]:

$$\begin{aligned} c=\sqrt{\frac{\textrm{d}\varPi }{\textrm{d}\rho }}=\sqrt{\frac{\hat{\varepsilon }}{\rho }}, \end{aligned}$$

(52)

whilst the continuity equation (40) reads as:

$$\begin{aligned} \frac{\partial \varPi }{\partial t}+U\frac{\partial \varPi }{\partial x}+\hat{\varepsilon }\frac{\partial U}{\partial x}=0. \end{aligned}$$

(53)

The greater the value of $\hat{\varepsilon }$, the greater the value of c. We stress that in the usual hydraulic engineering applications, $U=O\left( 1\ \frac{m}{s}\right)$ [16]; thus the flow is under a low Mach number limit, $\frac{U}{c}\ll 1$, and the governing equations can be rewritten in a simpler form. For this purpose, we put these equations in dimensionless form. Following [17], we introduce, in addition to the velocity scale $U_0$ and to the length scale $L_0=D$, a pressure scale $\varPi _0$ defined as:

$$\begin{aligned} \varPi _0=\rho _0U_0c_0, \end{aligned}$$

(54)

where $c_0=\sqrt{\frac{\hat{\varepsilon }}{\rho _0}}$ is the reference celerity value. We define the time scale $t_0$ as:

$$\begin{aligned} t_0=\frac{D}{c_0}. \end{aligned}$$

(55)

By adopting the notation:

$$\begin{aligned} b=b_0\left\{ b\right\} , \end{aligned}$$

(56)

where $\left\{ b\right\}$ indicates the relevant nondimensional variable, the dimensionless equations read as:

$$\begin{aligned} \begin{aligned}&\frac{1}{M}\left\{ \frac{\partial U}{\partial t}\right\} +\left\{ U\frac{\partial U}{\partial x}\right\} \\&\quad =-\frac{1}{M}\left\{ \frac{1}{\rho }\frac{\partial \varPi }{\partial x}\right\} -m_{tur}\left\{ \frac{1}{\rho }\frac{\partial \left( \rho U^2\right) }{\partial x}\right\} -\frac{\varpi }{2}\left\{ U\left| U\right| \right\} \\&\qquad +\left( \frac{1}{RS}+\frac{1}{TRS}\right) \left\{ \frac{1}{\rho }\frac{\partial ^2U}{\partial x^2}\right\} , \end{aligned} \end{aligned}$$

(57)

$$\begin{aligned} \frac{1}{M}\left\{ \frac{\partial \varPi }{\partial t}\right\} +\left\{ U\frac{\partial \varPi }{\partial x}\right\} +\frac{1}{M}\left\{ \frac{\partial U}{\partial x}\right\} =0, \end{aligned}$$

(58)

$$\begin{aligned} \left\{ \rho \right\} =\textrm{e}^{M\left\{ \varPi -p_0\right\} }, \end{aligned}$$

(59)

$$\begin{aligned} \begin{aligned}&\frac{1}{M}\left\{ \frac{\partial k}{\partial t}\right\} +\left\{ U\frac{\partial k}{\partial x}\right\} =-\frac{1}{M}\left\{ \frac{U}{\rho }\frac{\partial \varPi }{\partial x}\right\} \\&-2m_{tur}\left\{ \frac{U}{\rho }\frac{\partial \left( \rho k\right) }{\partial x}\right\} -\varpi \left\{ k\left| U\right| \right\} \\&+\left( \frac{1}{RS}+\frac{1}{TRS}\right) \left\{ \frac{1}{\rho }\left( \frac{\partial ^2k}{\partial x^2}-\left( \frac{\partial U}{\partial x}\right) ^2\right) \right\} , \end{aligned} \end{aligned}$$

(60)

$$\begin{aligned} \begin{aligned}&\frac{1}{M}\left\{ T\frac{\partial \sigma }{\partial t}\right\} +\left\{ T\frac{\partial \sigma }{\partial x}\right\} \\&\quad =\varpi \left\{ k\left| U\right| \right\} \\&\qquad +\left( \frac{1}{RS}+\frac{1}{TRS}\right) \left\{ \frac{1}{\rho }\left( \frac{\partial U}{\partial x}\right) ^2\right\} -\alpha _{tur}\left\{ \frac{T}{\rho }\frac{\partial \left( \rho U\sigma \right) }{\partial x}\right\} , \end{aligned} \end{aligned}$$

(61)

where $M=\frac{U_0}{c_0}$ is the Mach number, $RS=\frac{\varPi _0D}{\eta c_0}$ the Russo Spena number [17], $TRS=\frac{\varPi _0D}{\eta _{tur}c_0}$ the turbulent Russo Spena number [17]. In obtaining Eq. (61), we set $T\sigma =U_0^2\left\{ T\sigma \right\}$. Due to very important energy dissipation in unsteady pipe flows with wave damping [18], we expect that $\eta _{tur}\gg \ \eta$, thus $\frac{1}{RS}\ll \frac{1}{TRS}$. Accordingly, in low Mach approximation, Eqs. (57)–(61) reduce to:

$$\begin{aligned}{} & {} \left\{ \frac{\partial U}{\partial t}\right\} +\left\{ \frac{\partial \varPi }{\partial x}\right\} =\frac{1}{\mathrm {\Lambda }}\left\{ \frac{\partial ^2U}{\partial x^2}\right\} , \end{aligned}$$

(62)

$$\begin{aligned}{} & {} \left\{ \frac{\partial \varPi }{\partial t}\right\} +\left\{ \frac{\partial U}{\partial x}\right\} =0, \end{aligned}$$

(63)

$$\begin{aligned}{} & {} \left\{ \rho \right\} =1, \end{aligned}$$

(64)

$$\begin{aligned}{} & {} \left\{ \frac{\partial k}{\partial t}\right\} =-\left\{ \frac{U}{\rho }\frac{\partial \varPi }{\partial x}\right\} +\frac{1}{\mathrm {\Lambda }}\left\{ \frac{1}{\rho }\left( \frac{\partial ^2k}{\partial x^2}-\left( \frac{\partial U}{\partial x}\right) ^2\right) \right\} , \end{aligned}$$

(65)

$$\begin{aligned}{} & {} \left\{ T\frac{\partial \sigma }{\partial t}\right\} =\frac{1}{\mathrm {\Lambda }}\left\{ \frac{\partial U}{\partial x}\right\} ^2, \end{aligned}$$

(66)

where the dimensionless number $\mathrm {\Lambda }$ is given as $\mathrm {\Lambda }=\frac{TRS}{M}=\frac{\rho _0c_0D}{\eta _{tur}}$. According to Eqs. (62)–(66), once the initial and boundary conditions are given, the flow behavior is governed by $\mathrm {\Lambda }$. In dimension form, Eqs. (62)–(66) take the form:

$$\begin{aligned}{} & {} \frac{\partial U}{\partial t}+\frac{1}{\rho _0}\frac{\partial \varPi }{\partial x}=\frac{\eta _{tur}}{\rho _0}\frac{\partial ^2U}{\partial x^2}, \end{aligned}$$

(67)

$$\begin{aligned}{} & {} \frac{1}{\rho _0}\frac{\partial \varPi }{\partial t}+c_0^2\frac{\partial U}{\partial x}=0, \end{aligned}$$

(68)

$$\begin{aligned}{} & {} \rho =\rho _0, \end{aligned}$$

(69)

$$\begin{aligned}{} & {} \frac{\partial k}{\partial t}=-\frac{U}{\rho _0}\frac{\partial \varPi }{\partial x}+\frac{\eta _{tur}}{\rho _0}\left( \frac{\partial ^2k}{\partial x^2}-\left( \frac{\partial U}{\partial x}\right) ^2\right) , \end{aligned}$$

(70)

$$\begin{aligned}{} & {} T\frac{\partial \sigma }{\partial t}=\frac{\eta _{tur}}{\rho _0}\left( \frac{\partial U}{\partial x}\right) ^2, \end{aligned}$$

(71)

where the dissipation function is defined as $\phi _{1D}=\eta _{tur}\left( \frac{\partial U}{\partial x}\right) ^2$. By combining Eqs. (67)–(68), we obtain the linear damped wave equation:

$$\begin{aligned} \frac{\eta _{tur}}{\rho _0}\frac{\partial ^3U}{\partial x^2\partial t}+c_0^2\frac{\partial ^2U}{\partial x^2}-\frac{\partial ^2U}{\partial t^2}=0. \end{aligned}$$

(72)

We stress that, in agreement with Eqs. (70)–(71), the conversion of kinetic energy into entropy is governed by the equation:

$$\begin{aligned} \frac{\partial k}{\partial t}+T\frac{\partial \sigma }{\partial t}=-\frac{U}{\rho _0}\frac{\partial \varPi }{\partial x}+\frac{\eta _{tur}}{\rho _0}\frac{\partial ^2k}{\partial x^2}. \end{aligned}$$

(73)

3.2 Model application

To test the model reliability, we consider the classical water hammer phenomenon (without cavitation) which arises in a liquid-filled pipe as a consequence of flow disturbances; its distintive feature is the propagation of finite amplitude pressure waves [6]. We consider the standard reservoir-pipe-valve system, where the pipe connects the upstream pressurized tank to a downstream maneuver valve. The flow disturbances are identified in the velocity variations at the end of the pipe due to maneuvers of a closure valve. Starting from a steady-state configuration, the initial conditions are given by the normal flow formulas:

$$\begin{aligned}{} & {} \left. U\right| _{x,0}=U_0, \end{aligned}$$

(74)

$$\begin{aligned}{} & {} \left. \varPi \right| _{x,0}=\left. \varPi \right| _{\mathcal {L},0}+\ \rho _0\varpi \frac{U_0^2}{2D}\left( \mathcal {L}-x\right) , \end{aligned}$$

(75)

where $\mathcal {L}$ is the pipe length. The boundary conditions are assigned as:

$$\begin{aligned} \left. U\right| _{\mathcal {L},t}=U_0\left. \mathcal {H}\right| _{-t}, \end{aligned}$$

(76)

$$\begin{aligned} \left. \varPi \right| _{0,t}=constant, \end{aligned}$$

(77)

where $\mathcal {H}$ is the Heaviside step function. Equation (76) simulates an instantaneous closure of the valve. According to Eqs. (74)-(77) the linear damped equation:

$$\begin{aligned} \frac{\eta _{tur}}{\rho _0}\frac{\partial ^3U}{\partial x^2\partial t}+c_0^2\frac{\partial ^2U}{\partial x^2}-\frac{\partial ^2U}{\partial t^2}=0 \end{aligned}$$

(78)

is subjected to the initial conditions:

$$\begin{aligned}{} & {} \left. U\right| _{x,0}=U_0, \end{aligned}$$

(79)

$$\begin{aligned}{} & {} \left. \frac{\partial U}{\partial t}\right| _{x,0}=0, \end{aligned}$$

(80)

and to the boundary conditions:

$$\begin{aligned}{} & {} \left. \frac{\partial U}{\partial x}\right| _{0,t}=0, \end{aligned}$$

(81)

$$\begin{aligned}{} & {} \left. U\right| _{\mathcal {L},t}=U_0\left. \mathcal {H}\right| _{-t}. \end{aligned}$$

(82)

As shown in “Appendix”, the analytical solution of the initial boundary value problem is given by:

$$\begin{aligned} U= & {} U_0\left. \mathcal {H}\right| _{-t}+\sum _{n=0}^{\infty }\textrm{cos}\left( \lambda _nx\right) \frac{4U_0\left( -1\right) ^n\textrm{e}^{-\frac{\varphi _nt}{2}}}{\pi \left( 1+2n\right) \chi _n}\nonumber \\{} & {} \left( \chi _n\textrm{cosh}\left( \frac{t}{2}\chi _n\right) -\varphi _n\textrm{sinh}\left( \frac{t}{2}\chi _n\right) \right) , \end{aligned}$$

(83)

where

$$\begin{aligned} \lambda _n=\frac{\pi }{\mathcal {L}}\left( n+\frac{1}{2}\right) , \end{aligned}$$

(84)

$$\begin{aligned} \omega _n=c_0\lambda _n, \end{aligned}$$

(85)

$$\begin{aligned} \varphi _n=\frac{\eta _{tur}}{\rho _0}\lambda _n^2, \end{aligned}$$

(86)

$$\begin{aligned} \chi _n=\sqrt{\varphi _n^2-4\omega _n^2}. \end{aligned}$$

(87)

In principle, Eq. (83), completed by Eqs. (84)–(87) and by the (semi-empirical) relationships for $\eta _{tur}$, and $c_0$, can be used to find the field functions $U\left( x,t\right)$. On the other hand, we can calibrate the parameters $\eta _{tur}$ and $c_0$ matching numerical results to experimental data. As experimental data, we consider the thermodynamic pressure time-series collected at the downstream section of smooth steel pipe. The geometrical and physical parameters of the experimental setup are $\mathcal {L}=72\ m$; $D=42\ mm$; $\left. \varPi \right| _{\mathcal {L},0}=5.04\times {10}^5\ Pa$; $U_0=0.405\ \frac{m}{s}$; $Re=\frac{\rho _0U_0D}{\mu _1}=1.7\cdot 10^4$ (for water, we can assume $\mu _1={10}^{-3}\ \frac{kg}{ms}$; $\rho _0={10}^3\ \frac{kg}{m^3})$. The readers are referred to [7] for all details about experimental procedures. The Darcy friction factor $\varpi$ is computed by the Blasius formula:

$$\begin{aligned} \varpi =\frac{0.3164}{{Re}^{0.25}}. \end{aligned}$$

(88)

The Blasius formula can be used for smooth pipes in the range $4\times 10^3<Re<10^5$ [12]. At $x=\mathcal {L}$ the pressure field is given by (see “Appendix”):

$$\begin{aligned} \left. \varPi \right| _{\mathcal {L},t}=\left. \varPi \right| _{\mathcal {L},0}+\frac{4U_0\rho _0c_0^2}{\mathcal {L}}\sum _{n=0}^{\infty }\frac{\textrm{e}^{-\frac{\varphi _nt}{2}}\textrm{sinh}\left( \frac{t}{2}\chi _n\right) }{\chi _n}. \end{aligned}$$

(89)

A standard test-and-try procedure is used to find the best values $c_0=1234\ \frac{m}{s}$; $\frac{\eta _{tur}}{\rho _0}=2.84\times {10}^3\frac{m^2}{s}$; as a result, $\hat{\varepsilon }=1.52\times {10}^9\, \hbox {Pa}$; $\varPi _0=\rho _0U_0c_0=5\times {10}^5\, \hbox {Pa}$; $M=3.28\times {10}^{-4}$; $TRS=5.99\times {10}^{-6}$; $\mathrm {\Lambda }=1.82\times {10}^{-2}$. Figure 1 shows the comparison between the experimental and analytical results. The proposed model provides satisfactory reproduction of smoothing and damping pressure waves; the wave celerity is also well reproduced. The rate of dissipation of kinetic energy per unit mass $\mathrm {\Phi }_{1D}\left( t\right)$ reads as (see “Appendix”):

$$\begin{aligned} \begin{aligned} \mathrm {\Phi }_{1D}&=\frac{1}{\mathcal {L}}\int _{0}^{\mathcal {L}}{\frac{\eta _{tur}}{\rho _0}\left( \frac{\partial U}{\partial x}\right) ^2\textrm{d}x}\\&=2\frac{\eta _{tur}U_0^2}{\rho _0\mathcal {L}^2}\sum _{n=0}^{\infty }\left( \frac{\textrm{e}^{-\frac{\varphi _nt}{2}}}{\chi _n}\left( \chi _n\textrm{cosh}\left( \frac{t}{2}\chi _n\right) -\varphi _n\textrm{sinh}\left( \frac{t}{2}\chi _n\right) \right) \right) ^2. \end{aligned} \end{aligned}$$

(90)

The evolution of $\mathrm {\Phi }_{1D}$ is provided in Fig. 2. We underline that $\mathrm {\Phi }_{1D}$ tends to infinity for $t\rightarrow 0$ because of the instantaneous closure of the valve. According to Eq. (90), the monotonic increase in entropy production $\mathrm {\Omega }_{1D}\left( t\right)$ is defined as:

$$\begin{aligned} \begin{aligned} \mathrm {\Omega }_{1D}&=\int _{0}^{t}{\mathrm {\Phi }_{1D}\textrm{d}t}=\int _{0}^{t}\left( \frac{1}{\mathcal {L}}\int _{0}^{\mathcal {L}}{\frac{\eta _{tur}}{\rho _0}\left( \frac{\partial U}{\partial x}\right) ^2\textrm{d}x}\right) \textrm{d}t\\&=\frac{\eta _{tur}U_0^2}{\rho _0\mathcal {L}^2}\sum _{n=0}^{\infty }\frac{\textrm{e}^{-\varphi _nt}}{\varphi _n\chi _n^2}\left( 4\omega _n^2+\textrm{e}^{\varphi _nt}\chi _n^2-\varphi _n^2\textrm{cosh}\left( t\chi _n\right) \right. \\&\quad \left. +\varphi _n\chi _n\textrm{sinh}\left( t\chi _n\right) \right) . \end{aligned} \end{aligned}$$

(91)

Figure 3 shows the behavior of the entropy production versus time (we stress that $\mathrm {\Omega }_{1D}\left( 0\right) =0$). As expected, for large values of time, the entropy production equals the initial kinetic energy, i.e., for $t\rightarrow \infty$ the following relationship holds (the readers may refer to [19] for the convergence of the series):

$$\begin{aligned} \begin{aligned} \mathrm {\Omega }_{1D}&=\frac{\eta _{tur}U_0^2}{\rho _0\mathcal {L}^2}\sum _{n=0}^{\infty }\frac{1}{\varphi _n}=\frac{\eta _{tur}U_0^2}{\rho _0\mathcal {L}^2}\sum _{n=0}^{\infty }{\frac{4\rho _0\mathcal {L}^2}{\eta _{tur}\pi ^2}\frac{1}{\left( 2n+1\right) ^2}}\\&=\frac{4U_0^2}{\pi ^2}\sum _{n=0}^{\infty }\frac{1}{\left( 2n+1\right) ^2}=\frac{U_0^2}{2}. \end{aligned} \end{aligned}$$

(92)

Fig. 1

Pressure time-series at the valve $x=\mathcal {L}$. The continuous line depicts analytical results, whereas the dotted line represents experimental data [7]. Parameters are $c_0=1234\ \frac{\hbox {m}}{\hbox {s}}$; $\frac{\eta _{tur}}{\rho _0}=2.84\times {10}^3\frac{\hbox {m}^2}{\hbox {s}}$

Fig. 2

Evolution of the averaged dissipation function

Fig. 3

Evolution of the entropy production. At very large time $t\rightarrow \infty$, $\mathrm {\Omega }_{1D}\rightarrow \frac{U_0^2}{2}=8.2\times {10}^{-2}\ \frac{\hbox {m}^2}{\hbox {s}^2}$

4 Conclusions

In this paper we derived a linear damped equation specializing the RANS equations for fluids with low compressibility to 1D unsteady pipe flows. We clarified the physical meaning of the terms arising in the proposed RANS model, and discussed the closure relationships for the momentum and entropy equations. We adopted the low Mach number approximation to derive a basic version of 1D model equations. By means of a dimensionless procedure, we have deduced the dimensionless number that plays a predominant role in flow behavior. After reducing the basic 1D model equations to the linear damped wave equation, we compared the analytical solution of the achieved initial boundary value problem against experimental results. The comparison allows us to test the reliability of the proposed solution. As experimental data, we employed the thermodynamic pressure time series acquired during a water hammer phenomenon in steel pipe. A satisfactory reliability of the proposed model has been observed. The kinetic energy dissipation and the entropy production accompanying the wave propagation have also been investigated. Due to its simplicity, the proposed 1D turbulent model can provide a useful reference compared to more complex turbulent models.

Appendix

We consider the following initial boundary value problem:

$$\begin{aligned}{} & {} \alpha \frac{\partial ^3U}{\partial x^2\partial t}+\beta \frac{\partial ^2U}{\partial x^2}-\frac{\partial ^2U}{\partial t^2}=0, \end{aligned}$$

(A1)

$$\begin{aligned}{} & {} \left. U\right| _{x,0}=f, \end{aligned}$$

(A2)

$$\begin{aligned}{} & {} \left. \frac{\partial U}{\partial t}\right| _{x,0}=g, \end{aligned}$$

(A3)

$$\begin{aligned}{} & {} \left. \frac{\partial U}{\partial x}\right| _{0,t}=A, \end{aligned}$$

(A4)

$$\begin{aligned}{} & {} \left. U\right| _{\mathcal {L},t}=B, \end{aligned}$$

(A5)

where $U=U\left( x,t\right)$ is the unknown scalar field, with $x\in \left[ 0,\mathcal {L}\right]$, and $t>0$; $\alpha =\frac{\eta _{tur}}{\rho _0}>0$; $\beta =c_0^2>0$. The functions $f=f\left( x\right)$ and $g=g\left( x\right)$ define the initial conditions (A2)–(A3); the functions $A=A\left( t\right)$ and $B=B\left( t\right)$ specify the boundary conditions (A4)–(A5). Initial conditions (A2)–(A3), and boundary conditions (A4)–(A5) refer to water hammer phenomena in pipes. Solution of the boundary value problem (A1)–(A5) can be found by applying a similar procedure as in [20] based on a separation of variables method. We split the unknown function $U=U\left( x,t\right)$ in two functions:

$$\begin{aligned} U=W+K, \end{aligned}$$

(A6)

where $K=K\left( x,t\right)$ satisfies the boundary conditions (A4)–(A5):

$$\begin{aligned} K=A\left( x-\mathcal {L}\right) +B, \end{aligned}$$

(A7)

whereas $W=W\left( x,t\right)$ satisfies the initial boundary value problem:

$$\begin{aligned}{} & {} \alpha \frac{\partial ^3W}{\partial x^2\partial t}+\beta \frac{\partial ^2W}{\partial x^2}-\frac{\partial ^2W}{\partial t^2}=\frac{\textrm{d}^2A}{\textrm{d}t^2}\left( x-\mathcal {L}\right) +\frac{\textrm{d}^2B}{\textrm{d}t^2}, \end{aligned}$$

(A8)

$$\begin{aligned}{} & {} \left. \frac{\partial W}{\partial x}\right| _{0,t}=0, \end{aligned}$$

(A9)

$$\begin{aligned}{} & {} \left. W\right| _{\mathcal {L},t}=0, \end{aligned}$$

(A10)

$$\begin{aligned}{} & {} \left. W\right| _{x,0}=f-\left. A\right| _{0}\left( x-\mathcal {L}\right) -\left. B\right| _{0}, \end{aligned}$$

(A11)

$$\begin{aligned}{} & {} \left. \frac{\partial W}{\partial t}\right| _{x,0}=g-\left. \frac{\textrm{d}A}{\textrm{d}t}\right| _{0}\left( x-\mathcal {L}\right) -\left. \frac{\textrm{d}B}{\textrm{d}t}\right| _{0}. \end{aligned}$$

(A12)

We assume the following eigenfunction expansion:

$$\begin{aligned} W=\sum _{n=0}^{\infty }{T_n\textrm{cos}\left( \frac{\pi x}{\mathcal {L}}\left( n+\frac{1}{2}\right) \right) }=\sum _{n=0}^{\infty }{T_n\textrm{cos}\left( \lambda _nx\right) }, \end{aligned}$$

(A13)

with $T_n=T_n\left( t\right)$ denoting the unknown expansion coefficients, $\lambda _n=\frac{\pi }{\mathcal {L}}\left( n+\frac{1}{2}\right)$. Equation (A13) satisfies Eqs. (A9)–(A10) and the x-dependence of the homogeneous counterpart of Eq. (A8). By using the orthogonality property of the cosine eigenfunctions, we obtain the following non-homogeneous differential equation:

$$\begin{aligned}{} & {} \frac{\textrm{d}^2T_n}{\textrm{d}t^2}+\alpha \lambda _n^2\frac{\textrm{d}T_n}{\textrm{d}t}+\beta \lambda _n^2T_n\nonumber \\{} & {} \quad =-\frac{2}{\mathcal {L}}\int _{0}^{\mathcal {L}}\left( \frac{\textrm{d}^2A}{\textrm{d}t^2}\left( x-\mathcal {L}\right) +\frac{\textrm{d}^2B}{\textrm{d}t^2}\right) \textrm{cos}\left( \lambda _nx\right) \textrm{d}x=Q_n, \end{aligned}$$

(A14)

and the corresponding initial conditions:

$$\begin{aligned} \left. T_n\right| _{0}= & {} \frac{2}{\mathcal {L}}\int _{0}^{\mathcal {L}}\left( f-\left. A\right| _{0}\left( x-\mathcal {L}\right) -\left. B\right| _{0}\right) \nonumber \\{} & {} \quad \textrm{cos}\left( \lambda _nx\right) \textrm{d}x=F_n, \end{aligned}$$

(A15)

$$\begin{aligned} \left. \frac{\textrm{d}T_n}{\textrm{d}t}\right| _{0}= & {} \frac{2}{\mathcal {L}}\int _{0}^{\mathcal {L}}\left( g-\left. \frac{\textrm{d}A}{\textrm{d}t}\right| _{0}\left( x-\mathcal {L}\right) -\left. \frac{\textrm{d}B}{\textrm{d}t}\right| _{0}\right) \nonumber \\{} & {} \quad \textrm{cos}\left( \lambda _nx\right) \textrm{d}x=G_n. \end{aligned}$$

(A16)

By using the variation of the parameters method [21], the solution of the Cauchy problem (A14)–(A16) is readily given by:

$$\begin{aligned} \begin{aligned} T_n&=\textrm{e}^{-\frac{\alpha \lambda _n^2t}{2}}\left( F_n\textrm{cosh}\left( \frac{t\lambda _n}{2}\sqrt{\alpha ^2\lambda _n^2-4\beta }\right) \right. \\&\quad +\left. \frac{2G_n+F_n\alpha \lambda _n^2}{\lambda _n\sqrt{\alpha ^2\lambda _n^2-4\beta }}\textrm{sinh}\left( \frac{t\lambda _n}{2}\sqrt{\alpha ^2\lambda _n^2-4\beta }\right) \right) \\&\quad +\frac{2}{\lambda _n\sqrt{\alpha ^2\lambda _n^2-4\beta }}\int _{0}^{t}\textrm{e}^{-\frac{\alpha \lambda _n^2\left( t-\tau \right) }{2}}\\&\quad \textrm{sinh}\left( \frac{\left( t-\tau \right) \lambda _n}{2}\sqrt{\alpha ^2\lambda _n^2-4\beta }\right) Q_n\textrm{d}\tau . \end{aligned} \end{aligned}$$

(A17)

The solution for U finally reads as:

$$\begin{aligned} \begin{aligned} U&=A\left( x-\mathcal {L}\right) +B\\&\quad +\sum _{n=0}^{\infty }\textrm{cos}\left( \lambda _nx\right)\left( \frac{2}{\lambda _n\sqrt{\alpha ^2\lambda _n^2-4\beta }}\int _{0}^{t}\textrm{e}^{-\frac{\alpha \lambda _n^2\left( t-\tau \right) }{2}}\right.\\&\quad \textrm{sinh}\left( \frac{\left( t-\tau \right) \lambda _n}{2}\sqrt{\alpha ^2\lambda _n^2-4\beta }\right) Q_n\textrm{d}\tau \\&\quad +\textrm{e}^{-\frac{\alpha \lambda _n^2t}{2}}\left( F_n\textrm{cosh}\left( \frac{t\lambda _n}{2}\sqrt{\alpha ^2\lambda _n^2-4\beta }\right) \right. \\&\quad +\left.\left. \frac{2G_n+F_n\alpha \lambda _n^2}{\lambda _n\sqrt{\alpha ^2\lambda _n^2-4\beta }}\textrm{sinh}\left( \frac{t\lambda _n}{2}\sqrt{\alpha ^2\lambda _n^2-4\beta }\right) \right)\right) . \end{aligned} \end{aligned}$$

(A18)

Equation (A18) can be also written in the following more convenient form:

$$\begin{aligned} \begin{aligned} U=&\,A\left( x-\mathcal {L}\right) +B\\&+\,\sum _{n=0}^{\infty }\textrm{cos}\left( \lambda _nx\right) \left( \frac{2}{\chi _n}\int _{0}^{t}{\textrm{e}^{-\frac{\varphi _n\left( t-\tau \right) }{2}}\textrm{sinh}\left( \frac{t-\tau }{2}\chi _n\right) Q_n\textrm{d}\tau }\right. \\&\left. +\,\textrm{e}^{-\frac{\varphi _nt}{2}}\left( F_n\textrm{cosh}\left( \frac{t}{2}\chi _n\right) +\frac{2G_n+F_n\varphi _n}{\chi _n}\textrm{sinh}\left( \frac{t}{2}\chi _n\right) \right) \right) , \end{aligned} \end{aligned}$$

(A19)

where

$$\begin{aligned} \omega _n=\sqrt{\beta }\lambda _n=\sqrt{\beta }\frac{\pi }{\mathcal {L}}\left( n+\frac{1}{2}\right) , \end{aligned}$$

(A20)

$$\begin{aligned} \varphi _n=\alpha \lambda _n^2 \end{aligned}$$

(A21)

represent the eigenfrequency and the damping rate of the nth mode, respectively. We stress that the behavior of the solution (A19) depends on $\chi _n^2=\varphi _n^2-4\omega _n^2$. If this term is positive, we obtain decaying solutions, if it is negative, we obtain damped oscillations. This means that the supercritical damping case corresponds with the condition $\frac{\pi }{2\mathcal {L}}>\frac{2\sqrt{\beta }}{\alpha }$, i.e., $\alpha >\frac{4\mathcal {L}\sqrt{\beta }}{\pi }$, and solution decays exponentially without oscillations. The case of instantaneous closure can be analysed by assuming $A=0$ and the valve law closure ${B=U}_0\left. \mathcal {H}\right| _{-t}$. By noting that $\frac{\textrm{d}^2B}{\textrm{d}t^2}=-U_0\frac{\textrm{d}\delta }{\textrm{d}t}$, where $\delta =\delta \left( t\right)$ is the Dirac Delta function, the analytical solution (A19) reduces to:

$$\begin{aligned} U= & {} U_0\left. \mathcal {H}\right| _{-t}+\sum _{n=0}^{\infty }\textrm{cos}\left( \lambda _nx\right) \frac{4U_0\left( -1\right) ^n\textrm{e}^{-\frac{\varphi _nt}{2}}}{\pi \left( 1+2n\right) \chi _n}\nonumber \\{} & {} \quad \left( \chi _n\textrm{cosh}\left( \frac{t}{2}\chi _n\right) -\varphi _n\textrm{sinh}\left( \frac{t}{2}\chi _n\right) \right) . \end{aligned}$$

(A22)

Accordingly, we obtain the following explicit expressions:

$$\begin{aligned} \left. \frac{\partial U}{\partial x}\right| _{\mathcal {L},t}= & {} -\sum _{n=0}^{\infty }\frac{2U_0\textrm{e}^{-\frac{\varphi _nt}{2}}}{\mathcal {L}\chi _n}\left( \chi _n\textrm{cosh}\left( \frac{t}{2}\chi _n\right) \right. \nonumber \\{} & {} -\left. \varphi _n\textrm{sinh}\left( \frac{t}{2}\chi _n\right) \right) , \end{aligned}$$

(A23)

$$\begin{aligned} \left. \varPi \right| _{\mathcal {L},t}= & {} \left. \varPi \right| _{\mathcal {L},0}\nonumber \\{} & {} \quad +\frac{4U_0\rho _0c_0^2}{\mathcal {L}}\sum _{n=0}^{\infty }\frac{\textrm{e}^{-\frac{\varphi _nt}{2}}\textrm{sinh}\left( \frac{t}{2}\chi _n\right) }{\chi _n}. \end{aligned}$$

(A24)

By using the orthogonality property of cosine eigenfunctions, $\mathrm {\Phi }_{1D}\left( t\right)$ reads as:

$$\begin{aligned} \begin{aligned}&\mathrm {\Phi }_{1D}=\frac{1}{\mathcal {L}}\int _{0}^{\mathcal {L}}{\frac{\eta _{tur}}{\rho _0}\left( \frac{\partial U}{\partial x}\right) ^2\textrm{d}x}\\&=2\frac{\eta _{tur}U_0^2}{\rho _0\mathcal {L}^2}\sum _{n=0}^{\infty }\left( \frac{\textrm{e}^{-\frac{\varphi _nt}{2}}}{\chi _n}\left( \chi _n\textrm{cosh}\left( \frac{t}{2}\chi _n\right) -\varphi _n\textrm{sinh}\left( \frac{t}{2}\chi _n\right) \right) \right) ^2. \end{aligned} \end{aligned}$$

(A25)