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Mathematics 16 Online
OpenStudy (anonymous):

can someone help me to solve this problem .

OpenStudy (anonymous):

OpenStudy (anonymous):

Thermodynamics

OpenStudy (anonymous):

For this I have to study Thermodynamics for you.. :)

OpenStudy (anonymous):

thans :D

OpenStudy (anonymous):

What thans?? And I am not going to study that... :P

OpenStudy (anonymous):

@jim_thompson5910 good at thermodynamics?

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

Can you help me to him?? Whom to whom??

OpenStudy (anonymous):

to jim :)

OpenStudy (anonymous):

I did not study this subject when there was time.. :) Now there is time but again no mood of studying it.. :P

OpenStudy (agreene):

if I remember correctly you should be using \[q + (h_{in} +0.5c^2_{in}) – (h_{out}+0.5c^2_{out}) =0\] where: \(q\)=heat transferred per unit mass \(h_{in}\)= specific enthalpy of inlet fluid \(h_{out}\)= specific enthalpy of outlet fluid \(c_{in}\)= velocity of inlet fluid \(c_{out}\)= velocity of outlet fluid

OpenStudy (anonymous):

and then sir?

OpenStudy (agreene):

\[\rho_{in}*c_{in}*A_{in}=\rho_{out}*c_{out}*A_{out}\\p=\rho R T\] And you can reorganize the 1st i posted to this: \[q + (h_{in} - h_{out}) + 0.5(c^2_{in} – c^2_{out}) =0\]

OpenStudy (agreene):

\(A\)= section area \(ρ\)=density \(p\)=pressure \(R\)=specific gas constant \(T\)=temperature

OpenStudy (agreene):

yes, use those 3 to solve for what you need to figure out.

OpenStudy (anonymous):

@SolomonZelman

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