Can one use the following formula for calculating the flow energy rate of a fluid?

$Q = Cp_{fluid} T \dfrac{dm}{dt}$

where dm/dt is the mass flow, T the temperature and Cp the specific heat.

And how would it be related to the flow energy rate $\dfrac{dm}{dt}Pv$?


  • $\begingroup$ What is the term "energy rate of a flowing fluid" supposed to mean? $\endgroup$ – Chet Miller Oct 14 '16 at 20:21
  • $\begingroup$ The flow energy of a fluid is Pv (pressure x specific volume) in J/kg. The energy rate is Pv dm/dt, in W. $\endgroup$ – Physther Oct 14 '16 at 20:34
  • $\begingroup$ Let me guess: You are studying the open system (control volume) version of the first law of thermodynamics and you are trying to get an idea of how to interpret physically the Pv terms for the flow streams entering and leaving the control volume. $\endgroup$ – Chet Miller Oct 14 '16 at 20:51
  • $\begingroup$ Yes. That is correct. I use the definition of enthalpy (h = u + Pv) and what confuses me is the Pv = h - u. For a liquid h = CpT and u = CvT but for liquids Cp is close to Cv. In this case the flow energy would be almost 0. That's part of the confusion that I have. And then there is that first equation that I saw in open systems $\endgroup$ – Physther Oct 14 '16 at 21:03
  • $\begingroup$ For an incompressible liquid, $h=C_v(T-T_R)+Pv= C_p(T-T_R)+Pv$ where $T_R$ is the reference temperature datum for zero internal energy. $\endgroup$ – Chet Miller Oct 14 '16 at 22:11

only if the flow is isothermal. You derive this from the state equation (ideal gas law) but when you take the derivative wrt time each side you have to use the chain rule. And if temperature can change, then you need to include its derivative.

  • $\begingroup$ Hmm. Okay. But that is for gases only. I found a similar expression if I derive Q = d(m Cp T)/dt. But I get an extra term dCp/dt m T. $\endgroup$ – Physther Oct 14 '16 at 21:15

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