# Adiabatic filling of a tank - Minimum work required

1. Jul 23, 2015

### alystar

Hi guys, need an help about that problem
That's an adiabatic filling of a reservoir (tank) from a line.
Line is at 350 K and 8 MPa. The tank is, in the final state, at 365 K and 6 MPa.
Tank is insulated (no heat exchanged during the process), and only one stream enter in the tank.
How can i find the minimum work to be done to make the filling? And how can i find the "real work" to be done to make the filling?

Thanks a lot...

2. Jul 24, 2015

### Andrew Mason

Welcome to PF alystar!

How about telling us what you are filling the tank with?

AM

3. Jul 24, 2015

### alystar

Thanks for your welcome!
Sorry, thought the procedure (energy balance or kind of) was almost independent from the kind of substance involved. :) Substance is Ethane.
I've to do the same, also, with Acetylene, but on different pressures and temperatures conditions. (the one i've wrote above are about Ethane).

Thanks!

4. Jul 24, 2015

### Andrew Mason

We just need to know if you can apply the ideal gas law. Ethane can be modeled as an ideal gas in this temperature range. We also have to know its heat capacity (constant volume) CV

Start by using the first law of thermodynamics: $Q = \Delta U + W$. Since Q = 0 (adiabatic) what is the relationship between W and U? Using the ideal gas law and the CV for Ethane, can you determine the $\Delta U$? Can you then relate that to W?

AM

5. Jul 25, 2015

### alystar

Nope, ideal gas law cannot be applied. Data for Cp are available as polynomial as function of temperature, so i cannot give you an exact value right now.
The energy balance for the system is:
d(U)/d(t)= (Mass flow in*)Enthalpy In - (Mass flow out)*Enthalpy out + Heat exchanged + Work

where U is the internal energy and "t" is time.
The three last terms are = 0, because there's no mass flow out, no heat exchanged, and no work of deformation done (because the volume of the tank is constant, there's no work done through areas which are not crossed by streams). I'm sure about that energy balance because it had been already applied to another similar tank filling.
Therefore, that's why i'm really in trouble. No idea about a formula in order to get the minimum work.
Right now i calculated the Enthalpy difference H(f)-H(e), where H(f) is the final enthalpy of the tank at the end of the filling, and H(e) is the enthalpy of the line (costant during the filling). I calculated H(f)-H(e) because it was needed in order to get the final status of the tank at the end of the filling.
I'm pretty sure that difference it's necessary in order to calculate the minimum work. Still thinking about a simple relationship between the already found enthalpy and something else, like entrophy. (of course suitable for that case, an adiabatic and not-closed system)

6. Jul 25, 2015

### Andrew Mason

OK But the units are not quite right. Should it not be:

$d(U)/d(t)= (\Delta H_{in})dm_{in}/dt- (\Delta H_{out})dm_{out}/dt + dQ_{exch}/dt + dW/dt$ ?

This is simply a restatement of the first law:

$\Delta U = Q + W$ where W is the work done on the system

As far as the minimum work, under what conditions would the work be minimum (think of the process - what kind of process would require minimum work?)

Can you find $\Delta U$ per unit mass? and the $\Delta Q$ per unit mass (i.e. $\Delta H_{in}$) for the process?

The net change in internal energy has to equal the change in enthalpy + work done on the gas. So if you can find $\Delta U$? and $\Delta Q$ you can find W.

AM

Last edited: Jul 26, 2015