Steady Flow, Thermodynamics First Law

[Perodamh]

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]

 

[Chestermiller]

You are aware that $V=1/\rho$, correct?

 

[Perodamh]

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?

 

[Chestermiller]

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

 

[Perodamh]

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

 

[Chestermiller]

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?

 

[Perodamh]

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".

 

[Chestermiller]

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.

 

[Perodamh]

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?

 

[Perodamh]

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

 

[Chestermiller]

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.

 

[Perodamh]

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.