Recent content by canadiansmith

I am still a little confused. There are 2 equations for both first and second law? i only have delta U = Q-W and Q = integral of TodeltaS

Question is attached I have spent a great deal of time on this problem and I am hoping someone can help me out. My attempt at problem. first for the heat engine QL/Qw1 = TL/Twaste so efficiency = 1- Tl/Twaste efficiency = 0.93808 and I know that efficiency = Wout/ Qw1 This is as...

Nope. I guess it was to throw you off as well as the pressure at p2 after the valve was never used. I hate extra info that that. Thanks again

Ok I got it. So h3 will be 267.97 because it is sat vapor at 1Mpa and h2 will be 137.42 which is the sat-liquid value for 60 C at a pressure of 1681.3 Kpa. And because the hin = hout of a valve this can be used. Plug all this in the the formula we talked about and a got the right answer. Thanks...

hmm. I'm not sure. Like there shouldn't be. min to valve should = mout

Oh my bad you are very correct. I read the table wrong. I believe the pressure at inlet 2 is so high due to the valve.

Sorry is this better?

Actually I believe the h3 value is correct because r-134a at 1000kpa the saturation temp is only 39.39 C and the sat vapor value in the table hg= 267.97. The numbers are much less intuitive because it is not water. Still confused by this velocity given. It feels like It is a necessary piece of...

I have tied to use that method which would be m1h1 + m2h2 = (m1 + m2)h3 but then m2 = -5.716 and this is wrong I don't now how to relate the velocity into the equation. You said that outflow = m3*V right? what would outflow be? wouldn't the units be like kgm/s^2 or something like that?

Hi I was wondering if someone could help me with this question. Question 4 uploaded in picture. Ok, so far this is what I have found Pipe 1: m1=2kg/s P1=1000kpa T1=100 C h1=334.82kJ/kg Pipe 2: sat-liquid T2=60 C P2= 1000kpa h2=291.36KJ/kg Outlet: sat vapor P3=1000kpa V=20...

Thank you both very much for your help. I didn't think I would ever see the light at the end of the tunnel. Thanks again

In conclusion D2 is in fact off. Thank you very much. I got very confused with the negative potentials. But I realize now that they cannot make the potential at any point higher than the original potential.

va would equal 5-VR1 the voltage drop across R1 would be the current * R1. OH! VR1 would be the current induced by both the -10 draw and the +5 push. therefore the total current in the left side would be 5+10-.7 =14.3V/R1+R2 = 1.43mA VR1 = 1.43mA*R1 = 7.15 V and Va = 5-7.15V = -2.15V is this right?

ok what you are saying makes perfect sense but the voltage at Va will still be positive and Vb will be negative so that means D2 has the wrong polarity to turn it off

Is this wrong? If so please help me. I have been working on this all day.