Fluid Flow Loss Analysis Using Real Gas Properties

In summary, the conversation discusses the use of real gas properties in modeling fluid flow and losses through a component, specifically in the case of refrigerants. The OP has access to RefProp from NIST and has observed that most discussions and books assume an ideal gas or steam, which does not account for changes in density and viscosity. They are looking for someone to double-check their 1D analysis, which involves changing the mass flow rate until the specific enthalpy stagnation at the component exit is equal to the specific enthalpy stagnation at the component inlet. The conversation also includes calculations and formulas for fluid properties at the component inlet and exit, as well as drag force and power. The conversation concludes with a suggestion that the OP may need
  • #1
ScottAllenRauch
1
0
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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  • #2
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.
 

FAQ: Fluid Flow Loss Analysis Using Real Gas Properties

What is "Fluid Flow Loss Analysis Using Real Gas Properties"?

"Fluid Flow Loss Analysis Using Real Gas Properties" is a scientific method used to analyze and understand the behavior of fluids (such as gases) as they flow through different systems. It takes into account the real gas properties, such as compressibility and viscosity, which can greatly affect the fluid flow and result in losses or inefficiencies.

Why is it important to use real gas properties in fluid flow analysis?

Using real gas properties in fluid flow analysis is important because it allows for a more accurate and realistic understanding of how the fluid will behave. Real gases deviate from ideal gas behavior at high pressures and low temperatures, which can greatly impact the fluid flow and result in significant losses. By considering these properties, engineers and scientists can design more efficient and effective systems.

What types of systems can benefit from fluid flow loss analysis using real gas properties?

Any system that involves the flow of fluids, especially gases, can benefit from this type of analysis. Some common examples include pipelines, compressors, heat exchangers, and turbines. It can also be applied to various industries such as oil and gas, chemical processing, and power generation.

What are some common techniques used in fluid flow loss analysis using real gas properties?

Some common techniques used in this type of analysis include equations of state (such as the Peng-Robinson or Redlich-Kwong equations), numerical simulations (such as computational fluid dynamics), and experimental measurements. These techniques can provide insights into the fluid flow behavior and help identify areas for improvement.

What are the potential benefits of using fluid flow loss analysis using real gas properties?

The potential benefits of using this type of analysis include increased efficiency, reduced losses, and improved system design. By understanding how the real gas properties affect fluid flow, engineers and scientists can make more informed decisions and optimize their systems to achieve better performance. This can result in cost savings, improved reliability, and more sustainable operations.

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