Thermodynamics, open system first law problem

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Homework Statement



>A gas flows through a small orifice in a pipe as shown above. On the higher pressure side, the gas is at 1Mpa and temperature of 300K. The pressure reduces to 1kPa after it flows through the orifice. The equation of state for the gas is
[tex]v=\frac{RT}{P}+10^{-6}T^2[/tex]
If the specific heat at constant pressure is constant at [itex]C_{p} = 1kJ/kg-k[/itex], determine the temperature on the lower pressure side of the orifice assuming no heat loss from the pipe and steady-state operation. Use the following relationship for generalized equation for enthalpy.
[tex]dh=C_{p}dT+\left [ v-T\left \{ \frac{\partial v}{\partial T} \right \}_{p}\right ]dP[/tex]

Homework Equations




The Attempt at a Solution


What I tried is to solve [itex]dh=C_{p}dT+\left [ v-T\left \{ \frac{\partial v}{\partial T} \right \}_{p}\right ]dP[/itex]. Since [itex]v=\frac{RT}{P}+10^{-6}T^2[/itex], [itex]\frac{\partial v}{\partial T}[/itex] should be [itex]\frac{R}{P}+2*10^{-6}T^2[/itex]. Thus, [itex]\frac{dh}{dT} = C_{p}-\left [ 10^{-6} \right ]\frac{dP}{dT}[/itex]. And since [itex]\frac{dh}{dT[/itex]=[itex]C_{p}[/itex], [itex]0 = \left [ 10^{-6} \right ]\frac{dP}{dT}[/itex]. No problem until this, but I can't go further because dP/dT suddenly became 0. Any help would be appreciated.
 

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Answers and Replies

  • #2
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Let me guess. You as studying the version of the first law of thermodynamics applicable to open systems. What does this version of the first law tell you regarding the change in enthalpy of a gas passing through an adiabatic nozzle, porous plug, or oriface?

Chet
 

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