Thermo exam question help

Baartzy89

Hi everybody,

I've got a thermodynamics exam coming in a few weeks. I couldn't sit the exam when I took the subject 12 months ago because of work commitments so I'm sitting a deferred exam this semester. Trouble is I have forgotten most of it, and having difficulty getting help from lecturers and working it out myself. Below is a question with my answers, but I know I've made errors...just needing a bit of direction...thanks!

Question 1: A 1.8m3 rigid tank contains steam at 220°C. One third of the volume is in the liquid phase and the rest is in the vapor form. Determine:
a)the pressure of the steam, b)the quality of the saturated mixture, and c) the density of the mixture.

a)From steam table A-4, saturated vapor at 220°C has a pressure of 2319.6
kPa, or 2.312 MPa.

b)If volume of the tank is 1.8m3 and one third is liquid, two thirds are vapor:
1.2m3 is vapor 0.6m3 is liquid
from the steam tables the specific volume for the two states are: vf = 0.001190, vg = 0.086094
Therefore the mass of each phase must equal the volume of the phase " multiplied " by its specific volume:
mf = 0.6 x 0.001190 = 7.14 x 10-4, mg = 1.2 x 0.086094 = 0.1033128
Quality
= mg / mtotal, where mtotal = mg + mf
=7.14x10-4 +0.1033128
=0.1040268 = 0.1033128 / 0.1040268
= 0.993136384 or .99 or 99%

c) Using density formula: Density = Mass / Volume, and mtotal = 0.1040268 and V = 1.8= 0.1040268 / 1.8 = 0.057792666
= 0.058 kg/m3

Redbelly98

Welcome to Physics Forums! I have moved your post to Engineering, Comp Sci, & Technology from Introductory Physics. Physics majors can easily go through their entire education without ever seeing a steam table!

Looks good, though you probably mean 2.32 MPa?

Try including the proper units with the "0.6" and the "0.001190", and hopefully you'll spot the error in multiplying these quantities.

That's the right idea, you just need to recalculate the masses.

Yes, just need the correct mass here and you're all set.

Baartzy89

Ok, so im struggling a bit with the dimensional analysis of the question noqw. I have to admit this area is not one of my strong points.

So,
0.6m^3 x 0.001190 m^3/kg
= 7.14 x 10^-4 m^6/kg or
= sqrt(7.14 x 10^-4) m^3/sqrt(kg)

Is there a way of cancelling the m^6 or m^3?

Obviously this is not kg...In the text there is another equation using vavg for this type of question but no example in the text or course notes. Anyone experience with this property?

Redbelly98

Okay, that's good that at least you can spot there is a problem. How about if you take the 0.6 m^3 number and divide it by 0.001190 m^3/kg? The units should work out better this way.

Baartzy89

Ha, I feel like an idiot now. I was hell bent on multiplying them because I thought the result looked better. Typical case of looking for the answer, then to the map. Now when i look at the big picture it all makes sense.

Thanks for your help btw, this forum is incredible...

Redbelly98

You're welcome!

Baartzy89

Last question in the paper, stuck on this final section. I haven't been able to find a similar example in the Rankine Cycle section or the mass flow section of the text.

Consider a 210 MW steam power plant that operates on a simple ideal rankine cycle. Steam enters the turbine at 10 MPa and 500 degrees C, and is cooled in the condenser at a pressure of 10kPa. Determine:
a) the quality of the steam at the turbine exit
b) the thermal efficiency of the cycle
c) the mass flow rate of the steam

I have worked out the first two problems but I am at a loss in figuring out the last. Any pointers?

rock.freak667

If you are neglecting the pump work then the work output via the turbine is 210 MW. So if you apply the steady state steady flow equation to the turbine you will get

W_turbine=m_s(h₁-h₂)


You can find h₁ easily and you can get h₂ using the steam quality at the turbine exit.

Baartzy89

Ok, this is the equivalent of another equation that I found:
Wturbine= ms(hgf)

But i've also found another equation:
ms = 3600*P/he, where P is Load in kW, he specific enthalpy of evap.

The 3600 comes from adjustment to the units? And he is the same as hgf. For the first equation that you stated, what are the units for Wturbine and ms? (there seems to be quite a few different equations all essentially the same just using different units...)

rock.freak667

You are super heating it so h₁ is not the same as h_g and you are assuming the process is 100% isentropic so the h₂ is the same as h_f at the final pressure.



Usually in SI units (with the units they give you in the steam tables)

W_turbine= m_s(h₂-h₁) where m_s is in kg/s and h is in kJ/kg such that W_turbine is in kW.

Redbelly98

I don't know about those equations, but 3600 could be a conversion factor between seconds and hours. Does that make sense here?

Baartzy89

After a bit more research I found a worked example for the equation using 3600, and it was to convert back to kg/hr.

Here is what I've got together now, if there are mistakes please point them out:

Wturbine = ms*(h2-h1)
ms = Wturbine/(h2-h1)

If state 1 is superheated steam at 10 MPa, 500 deg c and state 2 is saturated steam at 10 kPa: h1 = 3375.1 kJ/ kg, h2 = 191.81 kJ/kg, (from steam tables).

ms = 210,000/(191.81 - 3375.1)
= 210,000/-3183.29
= -65.97 kg/s

The negative sign meaning that work has been done on the system.

I'm a little bit confused with your use of the term isentropic, does this mean simply that heat losses throughout the system were negligable? If their is work done on the system energy changes, therefore it is not isentropic?

rock.freak667

You need to have h1-h2

I believe h₂ should be h_g at 10 kPa, you really don't want liquid water flowing through your turbine.