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Let's say we had an empty insulated tank with no non PV work involved and a liquid/gas is injected into it via a pipeline whose temperature and pressure remain constant.

So via energy conservation equation we can say that

U(tank) = H(pipeline gas/liquid)

my question is how can we use the tabulated values of internal energy and enthalpy because they are about a reference whereas this equation involves H and U in absolute way. So if we take a reference we have to do the following

(U(tank) - U(ref) ) + U(ref) = (H(line) - H(ref)) + H(ref)

now since U(ref) + Pv(ref) = H(ref)

on substituting we get

(U(tank) - U(ref) ) = (H(line) - H(ref)) + Pv(ref)

now the (U(tank) - U(ref) ) = U tabulated (H(line) - H(ref)) = H tabulated U tabulated = H tabulated + Pv(ref)

but this doesn't seem to give the right answer

why?

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I don't know what you did, but I get the following:

$$U_{tank}=H_{line}=U_{line}+P_{line}V_{line}$$ so

$$U_{tank}-U_{line}=P_{line}V_{line}$$

You should note that $$H_{line}=(U_{line}-U_{ref})+P_{line}V_{line}$$

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  • $\begingroup$ Sir all I wanted to ask was how can we use reference values of Enthalpy and internal energy when via equation we arrived at absolute values which are unknown as we cannot determine absolute values of Enthalpy and entropy. $\endgroup$ – Yash Awasthi Sep 15 at 17:17
  • $\begingroup$ The tabulated value of U is relative to the reference state, so it should really be $U-U_{ref}$. The value of H should be consistent with this as $H=U-U_{ref}+PV$ $\endgroup$ – Chet Miller Sep 15 at 17:27
  • $\begingroup$ Can you give me an example via readings on steam table of what you mean. $\endgroup$ – Yash Awasthi Sep 15 at 17:29
  • $\begingroup$ In the steam tables, just add PV to the corresponding U values in the table, and you will get H. $\endgroup$ – Chet Miller Sep 15 at 17:30
  • $\begingroup$ here PV is the reference state Pressure x molar volume of the substance? $\endgroup$ – Yash Awasthi Sep 15 at 17:37

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