# Why does this solution assume it's a steady-flow system?

• EastWindBreaks
In summary: I see now. Thank you so much for the help and explanation!In summary, the conversation discusses a problem in thermodynamics where the energy balance equation is used to find the amount of heat transferred in a non-steady-flow system. The solution approach involves finding the final mass, enthalpy, and internal energy using the ideal gas equation, and then using the equation Q=m(U2-U1) to solve for the amount of heat transferred. There is also a discussion about the difference between transient and equilibrium calculations in solving this type of problem.
EastWindBreaks

## Homework Equations

unreliable source of solution:

## The Attempt at a Solution

[/B]
I came across this problem while practicing for the final exam in thermodynamic,my attempt: i recall this type of problems are not steady-flow system. Since the change in kinetic and potential energy is neglected, and the system does not produce work, therefore, the energy balance equation should becomes:
dE/dt= Q_dot+ m_dot(h_inlet)
h=u+Pv
since the question asks for Q instead of Q_dot: ( m_dot can not be found from given information alone either)
ΔE=ΔU=m(U2-U1)
therefore, m(U2-U1) = Q+ m* (H_inlet)
at this point, I see I can find final mass from ideal gas equation PV=mRT,
i can also find H and U2 from the given temperature and pressure from ideal gas property table of air. (T1=T2, P1=P2)
and U1 of the evacuated bottle is just 0, Q can be finally solved.

what I do not understand is that the solution assuming it's a steady-flow system, even if it is, dE/dt should be 0, so i guess they made a typo there. also, steady or not, it does not seem to alter the solution because dE/dt is 0 under steady-flow, but ΔE is not 0, correct?

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EastWindBreaks said:
what I do not understand is that the solution assuming it's a steady-flow system, even if it is, dE/dt should be 0, so i guess they made a typo there. also, steady or not, it does not seem to alter the solution because dE/dt is 0 under steady-flow, but ΔE is not 0, correct?

I think there is a difference between the questions

1) What's the variation of transferred heat versus time/pressure/mass, ...?
2) What's the amount of heat transferred after reaching equilibrium?

In the first case you for sure have to treat the problem as transient, since you are interested in the dependency of the heat versus a variable. In the second case, you just compare the final and the initial states and calculate the transferred heat. You just compare the change of the state variables and you can obtain the integral of the changing process variable without knowing its evolution.

So I think you are correct to say it is a transient process, but since only the amount of transferred heat and not its evolution is of interest, you don't need to do the transient calculation (which is both, more complicated and more interesting). However, I agree that the statement in the text about the steady-flow system is misleading (at least in my understandig).

EastWindBreaks said:
i can also find H and U2 from the given temperature and pressure from ideal gas property table of air. (T1=T2, P1=P2)

Shouldn't it be possible to solve this problem without tables? All you need to know is the ideal gas constant of air and the ideal gas equation (and the energy balance of course) ...

EastWindBreaks
stockzahn said:
I think there is a difference between the questions

1) What's the variation of transferred heat versus time/pressure/mass, ...?
2) What's the amount of heat transferred after reaching equilibrium?

In the first case you for sure have to treat the problem as transient, since you are interested in the dependency of the heat versus a variable. In the second case, you just compare the final and the initial states and calculate the transferred heat. You just compare the change of the state variables and you can obtain the integral of the changing process variable without knowing its evolution.

So I think you are correct to say it is a transient process, but since only the amount of transferred heat and not its evolution is of interest, you don't need to do the transient calculation (which is both, more complicated and more interesting). However, I agree that the statement in the text about the steady-flow system is misleading (at least in my understandig).
Shouldn't it be possible to solve this problem without tables? All you need to know is the ideal gas constant of air and the ideal gas equation (and the energy balance of course) ...

Thank you very much for taking the time to read all that and answering my questions! I got it now.
I am not sure how to find H and U2 without the tables, are you saying using ideal gas equation to find volume and then H=U2+PV? and ΔU= mass*specific heat of air*change in temperature? but we do now know the initial temperature inside the bottle though...

EastWindBreaks said:
Thank you very much for taking the time to read all that and answering my questions! I got it now.
I am not sure how to find H and U2 without the tables, are you saying using ideal gas equation to find volume and then H=U2+PV? and U2= mass*specific heat of air*change in temperature?

The final equation in the picture you posted is ## Q = m_{2} \left(u_{2} - h_{i}\right) ##, which is correct. It says that the final energy in the bottle is the sum of the entered enthalpy flux plus the heat transferred. Since the ambiance is very large and the filling of the bottle doesn't change the ambiance's properties, I'd say ##\varphi_2 = \varphi_a = \varphi_i##, where ##\varphi## stands for each specific state variable. So one could express the equation as ## Q = m_{i} \left(u_{i} - h_{i}\right) = m_{i} \left[u_{i} - \left(u_{i} + p_i v_i\right)\right] ##. You know ##p_i##, ##v_i## and ##m_i## by using the ideal gas equation (as you've stated yourself).

EastWindBreaks
stockzahn said:
The final equation in the picture you posted is ## Q = m_{2} \left(u_{2} - h_{i}\right) ##, which is correct. It says that the final energy in the bottle is the sum of the entered enthalpy flux plus the heat transferred. Since the ambiance is very large and the filling of the bottle doesn't change the ambiance's properties, I'd say ##\varphi_2 = \varphi_a = \varphi_i##, where ##\varphi## stands for each specific state variable. So one could express the equation as ## Q = m_{i} \left(u_{i} - h_{i}\right) = m_{i} \left[u_{i} - \left(u_{i} + p_i v_i\right)\right] ##. You know ##p_i##, ##v_i## and ##m_i## by using the ideal gas equation (as you've stated yourself).

ohhh I see! so in another words, the internal energy in the inlet is the same as the internal energy inside the bottle? its 4 am here in New York, I might be brain dead, would this problem be different if the inlet is a cylindrical pipe? sorry if I am asking stupid questions

Assume an adiabatic bottle. After it is filled (mechanical equilibrium) you close it, therefore ##Q=0##. The final internal energy then would correspond to the incoming enthalpy flux: ##m_i u_2 = m_i h_i##. Since the enthalpy is larger than the internal energy, the final internal energy of the air in the bottle would be higher than the one of the ambiance.

Now you open the bottle again and wait until the contained air and the ambiance is in thermal and mechanical equilbrium, which means that their properties are identical, so the internal energies are indentical after finding the equilibrium.

So the internal energy can change during the process, but in the final state it corresponds to ambient internal energy.

EastWindBreaks said:
ohhh I see! so in another words, the internal energy in the inlet is the same as the internal energy inside the bottle?

At the end, after equilibrium, but not necessarily during the entire process.

EastWindBreaks said:
its 4 am here in New York, I might be brain dead, would this problem be different if the inlet is a cylindrical pipe? sorry if I am asking stupid questions

Very motivated and it's not a stupid question. Of course the local and temporary internal energy states change with the geometry of the system. But if you investigate just the final state, that doesn't need to matter.

EastWindBreaks
I think it's even simpler than all this. You have simply $$Q=-m_ip_iv_i=-p_iV$$So there is no need to even use the ideal gas law. It's just the outside pressure times the volume of the tank.

EastWindBreaks and stockzahn

## 1. Why is it important to assume a steady-flow system in scientific solutions?

Assuming a steady-flow system in scientific solutions allows for simplification of complex processes and enables the use of fundamental equations and principles. This assumption is necessary for many engineering and scientific calculations to be valid.

## 2. How do we determine if a system is steady-flow or not?

A steady-flow system is one in which the mass and energy flow rates remain constant over time. This can be determined by analyzing the system's inputs and outputs, and observing if they remain consistent over a period of time.

## 3. What are the limitations of assuming a steady-flow system?

Assuming a steady-flow system may not accurately represent real-world scenarios, as many processes in nature are dynamic and constantly changing. Additionally, it may not account for transient effects and can lead to errors in calculations if the system is not truly steady.

## 4. How does the assumption of a steady-flow system affect the accuracy of scientific solutions?

The assumption of a steady-flow system can affect the accuracy of scientific solutions in both positive and negative ways. While it simplifies calculations, it may also introduce errors if the system is not truly steady. However, in many cases, the assumption is necessary and results in accurate solutions.

## 5. Are there any exceptions where assuming a steady-flow system is not necessary?

In some cases, assuming a steady-flow system may not be necessary or appropriate. For example, in situations where the process being studied is highly dynamic or transient, or when the effects of unsteady flow cannot be neglected, the assumption may not be valid. In these cases, a more complex and accurate approach may be required.

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