Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Thermodynamics, open system first law problem
Reply to thread
Message
[QUOTE="A330NEO, post: 5072094, member: 518890"] [h2]Homework Statement [/h2] >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] [h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2] 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. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Thermodynamics, open system first law problem
Back
Top