Steady Flow, Thermodynamics First Law

  • #1
21
0

Homework Statement


12kg of a fluid per minute goes through a reversible steady flow process. The properties of fluid at the inlet are p1 = 1.4bar, ρ1 = 25kg/m3, C1= 120m/s and u1= 920kJ/kg and at the exit are p2= 5.6bar, ρ
2= 5 kg/m3, C2= 180m/s and u2

Homework Equations



u1 + P1V1 + (C1)2/2 + g.Z + Q = u2 + P2V2 + (C2)2/2 + g.Z + W

u1 + P1V1 and u2 + P2V2 are both enthalpies

The Attempt at a Solution


i)
h1-h2 = change in enthalpy
plugging in values gives (920kj/kg + 1.4*V) - (720kj/kg + 5.6*V)
920kj/kg + 1.4*V - 720kj/kg - 5.6*V
200kj/kg - 4.2*V
but how do I get what V is since it wasn't given and I believe it means specific volume. I probably got it wrong, educate me. And also for the second question how do I go about finding W if Q is not given. Thanks a lot [/B]
 

Answers and Replies

  • #3
21
0
You are aware that ##V=1/\rho##, correct?
OF COURSE, it is. I should take my time with formulas, thank you. Can I run the answer with you once I retry and i'd also try the second part again and see if I can figure that part out?
 
  • #4
20,878
4,549
OF COURSE, it is. I should take my time with formulas, thank you. Can I run the answer with you once I retry and i'd also try the second part again and see if I can figure that part out?
Of course
 
  • #5
21
0
Of course
Apologies for my late reply. I've been working on the question and ran into S.I unit issues and I got -93.6kJ/kg for the first one.
The second one on the other hand, using the relation W = - [(u2 + P2V2) - (u1 + P1V1) + ((C2)2/2) - ((C1)2/2) + (g.Z2 - g.Z1 ) - Q]
Correct me if I'm wrong but to get Q do I divide the fluid reject in the question being 60kJ/s by the mass rate and also how do I get Z2 or Z1, a fluid rise was given as 60m, does that mean the initial is 0 and the final is 60? I've really tried in finding these out, I'd still keep trying but any assistance would be appreciated. Thanks
 
  • #6
20,878
4,549
I really don’t understand what you are saying. Are there two separate and distinct problems? Is the process really supposed to be reversible, or is it adiabatic, not reversible?
 
  • #7
21
0
I really don’t understand what you are saying. Are there two separate and distinct problems? Is the process really supposed to be reversible, or is it adiabatic, not reversible?
This is the question "12kg of a fluid per minute goes through a reversible steady flow process. The properties of fluid of fluid at the inlet are p1 = 1.4bar, ρ1 = 25kg/m^3, C1 = 120m/s and u2 = 920 kJ/kg and the exit are p2 = 5.6bar, ρ2 = 5kg/m^3, C2 = 180m/s and u2 = 720kJ/kg. During the passage, the fluid rejects 60KJ/s and rises through 60m. Determine i) the change in enthalpy ii) work done during the process".
 
  • #8
20,878
4,549
The form of the equation you are using applies per unit mass passing through the system. You can solve your problem in this form, or you can multiply the entire equation by the mass flow rate to get the work and heat per unit time. I suggest keeping the current form, and working per unit mass. In that case, the 60 kJ/s is the same as 60/(12/60) kJ/kg. Your interpretation of the elevations z is correct.
 
  • #9
21
0
The form of the equation you are using applies per unit mass passing through the system. You can solve your problem in this form, or you can multiply the entire equation by the mass flow rate to get the work and heat per unit time. I suggest keeping the current form, and working per unit mass. In that case, the 60 kJ/s is the same as 60/(12/60) kJ/kg. Your interpretation of the elevations z is correct.
Thanks, i'd work towards an answer for the second question, but what do you think of the first, I got -93.6kJ/kg?
 
  • #10
21
0
I just solved for the second plugging in values I got w = -[ -93.6 * 10^3 + ((180^2)/2 - (120^2)/2) + ((60 * 9.81) - 0) - 300 * 10^3] which is equal to 384.0114 kJ/kg? Tell me where I've got it wrong if there are any errors. Thanks
 
  • #11
20,878
4,549
I just solved for the second plugging in values I got w = -[ -93.6 * 10^3 + ((180^2)/2 - (120^2)/2) + ((60 * 9.81) - 0) - 300 * 10^3] which is equal to 384.0114 kJ/kg? Tell me where I've got it wrong if there are any errors. Thanks
I've too lazy to check your arithmetic. I've been pleased to help you with the conceptual problem of how to apply the first law to an open system.
 
  • #12
21
0
I've too lazy to check your arithmetic. I've been pleased to help you with the conceptual problem of how to apply the first law to an open system.
Aah, that's awesome all the same, thanks a lot.
 

Related Threads on Steady Flow, Thermodynamics First Law

Replies
1
Views
223
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
0
Views
2K
Replies
4
Views
434
Replies
32
Views
5K
Replies
2
Views
7K
Replies
6
Views
684
Top