Question

Thermodynamics

Question regarding conservation of energy equation.

Answer 1

Looking at an open system we have transient, steady state, and uniform flow uniform state. I was told/taught that transient is when the mass in the control space/volume is changing with some time. Steady state is when the mass flow in is at the same rate as the mass flow out. Finally, UFUS is when the mass flow entering isn't the same as the mass flow exiting or visa-versa. I'd like to discuss a little bit about conservation of equations for each one. Discuss conservation of mass (COM) and conservation of energy (COE) analysis with anyone.

Answer 2

most definitely up for talking/ learning about about this one

post an equation or some source.

Answer 3

Wow, raffle, that's some question ! I'm going guess, again... Open system implies that the thing is linked to the "outside world" with all its glorious "perturbations" and various inputs. So, presumably, it's going to be an APPROXIMATE steady state, since a perturbation will, possibly, make it ... unsteady - steady state bridges collapsing in possibly bridge resonant wind turbulence.
In your open system, you can't be sure about the in out mass flow rate - there may be leaks.
For some reason, my what's left of my brain keeps thinking about the Pitot tube for measuring gas/air speeds here (maybe because it's the only bit of "simple" fluid mechanics I learned - a couple of weeks ago ... :( )
Any paper balls coming my way, let me know and I'll try to duck

Answer 4

Now looking at system to closed we have transient. Where it says dm/dt = 0 makes sense to me because the system is closed no mass is flowing from the outside into the control volume so then m2 - m1 = 0. BTW COM equation is dm/dt = summation(mi_dot) - summation(me_dot).

dE/dt = Q_dot - W_dot where Q dot and W dot are the heat flow and work. This equation can expand to the following...

E2 - E1 = Q_dot - W_dot
U2 - U1 + delta_KE + delta_PE = Q_dot -W_dot_shaft - W_dot_electrical - W_dot_boundary.

In my class I will be taking we won't be dealing with electrical or shaft quite yet so those terms go away. You know what I think I've might have just answered one of my questions after looking at the formula sheet carefully. Here is one of my main questions. If you look at the open system and looking at transient, notice there is a mi_dot(hi + vi^2/2 + gzi) - me_dot(he + ve^2/s + gze) ?

I know that hi and he represent enthalpy, middle term is kinetic energy, and last term is potential energy. So if you expand the COE equation under COM equation which sits under open we get this...

dE/dt = Q_dot - W_dot + mi_dot(hi + vi^2/2 + gzi) - me_dot(he + ve^2/2 + gze)

E2 - E1 = Q_dot - W_dot_boundary + W_dot_shaft + W_dot_ele + mi_dot(hi + vi^2/2 + gzi) - me_dot(he + ve^2/2 + gze)

W_dot_ele and W_dot_shaft go away.
U2 - U1 + delta_PE + delta_KE = Q_dot - W_dot_boundary + mi_dot(hi + vi^2/2 + gzi) - me_dot(he + ve^2/2 + gze)

So looking here I was told that the PE and KE on the left side of the equation is the control volume itself, while the PE and KE on the right of the equal sign is the volume itself, moving across the volume control.

Answer 5

You agree with the following:

isobaric: constant pressure (for example unconstrained piston cylinder where we have mas less piston)

isothermal: constant temperature

isometric: constant volume (rigid tank)

adiabatic: no heat flow, Q = 0.

polytropic: where pressure and volume change. P = const./V^n, where n = some constant. This one would be a good discussion.

isentropic: where we have a reversible + adiabatic process. This comes from second law.

Answer 6

BTW this is more engineering than physics.

Answer 7

I'm not following how your integrations are going. For example you state that:
$$\frac{dE}{dt} = \dot Q - \dot W$$
(\frac{dE}{dt} = \dot Q - \dot W)

But this is nothing except:
$$\frac{dE}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} $$
(\frac{dE}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} )

which you managed to "expand" (which suspiciously looks like integrating):
$$E_2 - E_1 = \dot Q - \dot W$$
(E_2 - E_1 = \dot Q - \dot W)

which isn't correct (not really strictly correct), because if you are going to just go and integrate the equation you have to do it to both sides, something like:
$$\frac{dE}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} \large/ \cdot dt \\
\int_{t_1}^{t_2}dE = \int_{t_1}^{t_2}dQ- \int_{t_1}^{t_2}dW \\
E_2-E_1 = Q_2 - Q_1 - (W_2 - W_1)$$
( \frac{dE}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} \large/ \cdot dt \\
\int_{t_1}^{t_2}dE = \int_{t_1}^{t_2}dQ- \int_{t_1}^{t_2}dW \\
E_2-E_1 = Q_2 - Q_1 - (W_2 - W_1) )

In the end this kind of thing doesn't quite really get you all that far. At best you can say that the total change of energy is the difference of the total change of Q and W. But if you don't know the Q and W total changes (if they're not measured at least or expressable as their primitive functions you won't be able to use this). in fact just integrating it like this might not work all the time because Q and W could be rather complex functions at times dependent even on each other. What happens when:

$$\int_{t_1}^{t_2}dE(Q, W, t) = \int_{t_1}^{t_2}dQ(T, m(t), t)- \int_{t_1}^{t_2}dW(\Delta T(t, m))$$
( \int_{t_1}^{t_2}dE(Q, W, t) = \int_{t_1}^{t_2}dQ(T, m(t), t)- \int_{t_1}^{t_2}dW(\Delta T(t, m)) )

How do you integrate that? (T is temp, delta T is some temp difference between two containers, m is mass which can be dependant on time t).

You've kind of also skipped the integration of mass for the open transient systems too (on your picture) even though there's a dependence on velocity v in there (which I presume is implicitly dependent on time).

It'd be much easier if I knew at what to focus exactly when having this discussion here. Like this the best I can point out is some inconsistencies in math.

Answer 8

Oh, the think in "( )" under the formulas are the formulas above, it's their latex notation. It'd be much much easier to read your equations if you could write them like that and instead of "(" and ")" just put "$$" double dollar signs around them.

If you want to use some of my equations just right click them -> "show math as" -> "TeX commands" and copy paste the formula in the popup window.

Answer 9

late to the party as usual, but how much of this can we condense into single continuity equations for energy, mass, etc?

So there's only 1 equation and a whole lot of science.

Answer 10

I intentionally skipped the integration. I was taught to skip the integration part. My instructor showed up the integration part in class, but we were not expected to show it. She showed us the whole entire derivation. My cheat sheet isn't mine. My instructor made it for her class. There are 3 more equation sheets like this one. But thank you for pointing the integration part out. Can you please answer my question about the KE and PE on either side of the equation? What about the mass flow entering and leaving on the right side of the equation?

Answer 11

Maybe we could start with some of the components discussed in my class.

Answer 12

These are old notes. Here we have a turbine and a pump. I have 6 more components I'd like to discuss.

Answer 13

In steady state dE/dt = 0 and Q = 0, why? Q = 0 makes sense to me, but why does energy go to zero? Because there is no energy transfer?

Answer 14

Can you shed some light on what is h? I can see it written as u+Pv but I can't make out what that is either
From the equation it's one of the things I still can't make out
$$\dot E = \dot Q - \dot W + \left(\dot m h + E_{kin}+E_{pot}\right)_{in} - \left(\dot m h + E_{kin}+E_{pot}\right)_{out}$$
except that the m in the kinetic and potential energies is time dependent and with the steady state constraint the total change of kinetic and potential energies going in should be connected with total kinetic and potential energies going out such that:
$$ dE_{\text{tot, in}} - dE_{\text{tot, out}} = 0$$
which means that the dm/dt and the speed changes have to be the same in and out. Which is what the seems to happen in your task.

Except there's a dm/dt * h member I can't figure out the meaning of that seems to survive to create some work W.

Answer 15

h is specific enthalpy

Answer 16

this is a nightmare...
@raffle_snaffle nice handwriting :)

Answer 17

Yes, h is specific enthalpy.

Answer 18

So Q_dot, delta_KE, and delta_PE are negligible because those components are super small so we can ignore them.

Answer 19

hello

Answer 20

so enthalpy is a conserved property throughout?

Answer 21

Yes, I believe so.