# Steady Flow, Thermodynamics First Law

• Perodamh
In summary: I've too lazy to check your arithmetic. I've been pleased to help you with the conceptual...conceptual portion of the problem. :)

## 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]

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

Chestermiller said:
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?

Perodamh said:
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

Chestermiller said:
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

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?

Chestermiller said:
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".

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.

Chestermiller said:
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?

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

Perodamh said:
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.

Chestermiller said:
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.

## 1. What is steady flow?

Steady flow refers to a type of thermodynamic process in which the properties of a fluid, such as temperature, pressure, and velocity, remain constant at any given point in a system over a period of time.

## 2. What is the First Law of Thermodynamics?

The First Law of Thermodynamics, also known as the Law of Conservation of Energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another.

## 3. How does the First Law of Thermodynamics apply to steady flow?

In steady flow, the First Law of Thermodynamics applies by stating that the energy entering a system must equal the energy leaving the system, taking into account any changes in internal energy.

## 4. What are some real-life examples of steady flow?

Some common examples of steady flow include the flow of water through a pipe, the flow of air through a ventilation system, and the flow of blood through our circulatory system.