Hi,
May I ask something about the Geothermal power cycle exercise? It's part two in week 2's exercises.
In a) it's asked to calculated the mass flow of the water steam, leaving the separator.
Since x_2 is 0.166 I'm pretty sure, that there is less saturated water than steam. Thought the solution suggests the opposite and says the mass flow of the liquid water is $\Dot{m}_l = 191 kg/s (=x_2*230kg/s)$ and the mass flow of the steam/gas is $\Dot{m}_g = 0.38kg/s (=(1-x_2)*230 kg/s$.
Has an error crept in the solutions or do I misinterpret the naming or an entire concept?
Secondly, I wasn't able to find the enthalpy for state 11 resp. state 11s.
I assumed the heat exchanger to be isobar and therefore set $p_{11} = p_{11s} = p8 = 0.4MPa$.
With $s_{11} = s_{10}$ I realized, that the state 11s is a under cooled liquid.
For water we can make the assumption, that the liquid is incompressable and use the saturation line for all liquids. Is there my error? Can't we make that assumption for isobutane?
If not is there an aditional table/setting to find the properties of under cooled liquids? Or how is $_{11s}$ found?
I thank you in advance for the help.
Kind regards
Christina
In reply to Christina Sarah Graf
Re: 2. Thermodynamics Exercise 2 Geothermal power cycle
par Shuo Liu,
Hi,
The x2 is the fraction of gas phase component, which describes the extent of the vaporization for the mixture. Please note the expression for x2 in the solution. Also, we notice that h2 is 990 kJ/kg, which is closer to the saturated liquid enthalpy (640 kJ/kg). This means the majority is in liquid rather than gas.
To know the properties of 11s, you need to know two parameters: you know the pressure is 3.25 MPa (same pressure as stream 8), in addition, you know the entropy is 1.24 kJ/(kg.K) (isentropic process from stream 10). Then by using the web tool, you can find the properties of 11s.
You can also assume liquid isobutane is incompressible and use the specific volume to calculate the work. The isentropic work done by pump is V*deltaP (It is 5.2 kJ/kg). You will get more or less the same answer as the solution.
Regards,
Shuo
The x2 is the fraction of gas phase component, which describes the extent of the vaporization for the mixture. Please note the expression for x2 in the solution. Also, we notice that h2 is 990 kJ/kg, which is closer to the saturated liquid enthalpy (640 kJ/kg). This means the majority is in liquid rather than gas.
To know the properties of 11s, you need to know two parameters: you know the pressure is 3.25 MPa (same pressure as stream 8), in addition, you know the entropy is 1.24 kJ/(kg.K) (isentropic process from stream 10). Then by using the web tool, you can find the properties of 11s.
You can also assume liquid isobutane is incompressible and use the specific volume to calculate the work. The isentropic work done by pump is V*deltaP (It is 5.2 kJ/kg). You will get more or less the same answer as the solution.
Regards,
Shuo
In reply to Shuo Liu
Re: 2. Thermodynamics Exercise 2 Geothermal power cycle
par Christina Sarah Graf,
Hi Shuo,
Thanks for your answer!
I just realized now that I confused the fractions of liquid and vapour in the graphical picture of the wet steam area with lever arms.
Of course closer to the liquid phase means more liquid.
And the state 11s which I never found in the plots, I found now in the table version of the online data bank.
Thanks again.
Regards,
Christina
Thanks for your answer!
I just realized now that I confused the fractions of liquid and vapour in the graphical picture of the wet steam area with lever arms.
Of course closer to the liquid phase means more liquid.
And the state 11s which I never found in the plots, I found now in the table version of the online data bank.
Thanks again.
Regards,
Christina
