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Generate time-varying volumetric flow rate

**Library:**Simscape / Foundation Library / Moist Air / Sources

The Controlled Volumetric Flow Rate Source (MA) block
represents an ideal mechanical energy source in a moist air network. The volumetric flow
rate is controlled by the input physical signal at port **V**. The
source can maintain the specified volumetric flow rate regardless of the pressure
differential. There is no flow resistance and no heat exchange with the environment. A
positive volumetric flow rate causes moist air to flow from port **A**
to port **B**.

The volumetric flow rate and mass flow rate are related through the expression

$$\dot{m}=\{\begin{array}{ll}{\rho}_{B}\dot{V}\hfill & \text{for}\dot{V}\ge 0\hfill \\ {\rho}_{A}\dot{V}\hfill & \text{for}\dot{V}0\hfill \end{array}$$

where:

*$$\dot{m}$$*is the mass flow rate from port**A**to port**B**.*ρ*_{A}and*ρ*_{B}are densities at ports**A**and**B**, respectively.*$$\dot{V}$$*is the volumetric flow rate.

The equations describing the source use these symbols.

c_{p} | Specific heat at constant pressure |

h | Specific enthalpy |

h_{t} | Specific total enthalpy |

$$\dot{m}$$ | Mass flow rate (flow rate associated with a port is positive when it flows into the block) |

p | Pressure |

ρ | Density |

R | Specific gas constant |

s | Specific entropy |

T | Temperature |

Φ_{work} | Power delivered to the moist air flow through the source |

Subscripts A and B indicate the appropriate port.

Mass balance:

$$\begin{array}{l}{\dot{m}}_{A}+{\dot{m}}_{B}=0\\ {\dot{m}}_{wA}+{\dot{m}}_{wB}=0\\ {\dot{m}}_{gA}+{\dot{m}}_{gB}=0\end{array}$$

Energy balance:

$${\Phi}_{A}+{\Phi}_{B}+{\Phi}_{work}=0$$

If the source performs no work (**Power added** parameter is set to
`None`

), then $${\Phi}_{work}=0$$.

If the source is isentropic (**Power added** parameter is set to
`Isentropic power`

), then

$${\Phi}_{work}={\dot{m}}_{A}\left({h}_{tB}-{h}_{tA}\right)$$

where

$$\begin{array}{l}{h}_{tA}={h}_{A}+\frac{1}{2}{\left(\frac{{\dot{m}}_{A}}{{\rho}_{A}{S}_{A}}\right)}^{2}\\ {h}_{tB}={h}_{B}+\frac{1}{2}{\left(\frac{{\dot{m}}_{B}}{{\rho}_{B}{S}_{B}}\right)}^{2}\end{array}$$

The mixture-specific enthalpies, *h*_{A} =
*h*(*T*_{A}) and
*h*_{B} =
*h*(*T*_{B}), are constrained by
the isentropic relation, that is, there is no change in entropy:

$${\int}_{{T}_{A}}^{{T}_{B}}\frac{1}{T}}dh\left(T\right)=R\mathrm{ln}\left(\frac{{p}_{B}}{{p}_{A}}\right)$$

There are no irreversible losses.

There is no heat exchange with the environment.