Fluid Flow Loss Analysis Using Real Gas Properties

ScottAllenRauch

I am trying to model fluid flow and losses through a component (e.g., pipe) using REAL GAS properties since 1) I have access to RefProp from NIST, and 2) I am dealing with refrigerants, which are far from ideal.

I have seen nowhere an analysis of fluid flow losses (e.g., drag, friction, incidence, pressure drop) using real gas properties. In every discussion and book I have examined, somewhere in the analysis the gas is assumed to be ideal or perfect or steam, or no velocity change is allowed (which happens due to density change, if nothing else).

However, I have access to real gas properties (as I am sure many of us do), and I would greatly appreciate someone with a firmer grasp on reality than I have double-checking my 1D analysis.

The procedure in general is:
Given inlet fluid conditions, and pressure out, change mass flow rate until h0_out = h0_in (i.e., specific enthalpy stagnation at component exit = specific enthalpy stagnation at component inlet).

The way I figure it, all the energy to accelerate the fluid (due to the change in density) and overcome the losses (shear forces due to viscosity) must come from the internal energy, resulting in a reduction in fluid temperature.

I know this is long, but all the formulas and results (including units) are given, so I hope it is at least clear. Thanks in advance.
Code:
Inputs
Fluid                 = R134A

T_in_C                Temperature at component inlet
= 5.0000  C

Cd                    Drag coefficient
= 0.50000

Area                  Characteristic area
= 0.10000  m²

PressureRatio         p_out / p_in
= 0.90000

Calculations       ƒ()  =>  fluid function RefProp from NIST
Calculations Not Iterated
Fluid Properties at Component Inlet

T_in                  Temperature at component inlet
= T_in_C converted to K
= 278.15  K

p_in                  Pressure at component inlet
= Saturation pressure @ (T_in - 0.01 K) to ensure pure vapor = ƒ(T)
= 349.54  kPa

rho_in                Density at component inlet
= ƒ(P_in,T_in)
= 17.124  kg/m³

s_in                  Specific entropy at component inlet
= ƒ(P_in,T_in)
= 1.7245  m²/(s²·K)

h_in                  Specific enthalpy at component inlet
= ƒ(P_in,T_in)
= 401.50  m²/s²

u_in                  Specific internal energy at component inlet
= ƒ(P_in,T_in)
= 381.08  m²/s²

u_in_reality_check    u_in calculated differently
= h_in - p_in / rho_in
= 381.08  m²/s²

Fluid Properties at Component Exit

p_out                 Pressure at component exit
= p_in · PressureRatio
= 314.58  kPa

u_out_s               Specific internal energy at component exit isentropic
= ƒ(P_out,s_in)
= 379.05  m²/s²

Iterated Calculations       Solve for mdot

mdot                  Mass flow rate. Note circular iterative self-reference
= mdot · h0_target / h0_out_calculated
= 3.9505  kg/s

rho_out               Density at component exit. Uses u_out calculated below.
= ƒ(P_out,u_out)
= 15.366  kg/m³

Vdot_in               Volumetric flow rate at component inlet
= mdot / rho_in
= 0.23069  m³/s

c_in                  Relative fluid velocity at component inlet
= Vdot_in / Area
= 2.3069  m/s

ke_in                 Specific kinetic energy at component inlet
= c_in² / 2
= 2.6610  m²/s²

h0_in                 Specific enthalpy stagnation at component inlet
= h_in + ke_in
= 404.16  m²/s²

h0_out_target         Specific enthalpy stagnation at component exit
= h0_in
= 404.16  m²/s²

Vdot_out              Volumetric flow rate at component exit
= mdot / ρout
= 0.25709  m³/s

c_out                 Relative fluid velocity at component exit
= Vdot_out / Area
= 2.5709  m/s

ke_out                Specific kinetic energy at component exit
= c_out² / 2
= 3.3047  m²/s²

F_drag                Drag force
= Cd · rho_in · c_in² · area / 2
= 2.2784  kg·m/s²

Pwr_drag              Drag power
= F_drag · c_in
= 5.2561  kg·m²/s³

Delta_u               Δ specific internal energy due to drag
= Pwr_drag / mdot
= 1.3305  m²/s²

u_out                 Specific internal energy at component exit
= u_out_s + Delta_u
= 380.38  m²/s²

h0_out_calculated
= u_out + p_out / rho_out + ke_out
= 404.16  m²/s²/code]

Last edited:
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Chestermiller

Mentor
In order to answer the OP's questions, we need to know more about the specific system being analyzed, including geometry and imposed operating conditions. Offhand, it looks like the OP is trying to use the thermal energy balance when he should be using the mechanical energy balance equation.

"Fluid Flow Loss Analysis Using Real Gas Properties"

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