
#1
Jun711, 08:14 AM

P: 14

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 A4, 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 104, mg = 1.2 x 0.086094 = 0.1033128 Quality = mg / mtotal, where mtotal = mg + mf =7.14x104 +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 



#2
Jun711, 08:15 PM

Mentor
P: 11,988

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!




#3
Jun811, 08:15 PM

P: 14

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? 



#4
Jun811, 10:48 PM

Mentor
P: 11,988

Thermo exam question help 



#5
Jun911, 12:28 AM

P: 14

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... 



#7
Jun1211, 01:58 AM

P: 14

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? 



#8
Jun1211, 03:17 AM

HW Helper
P: 6,213

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}(h_{1}h_{2}) You can find h_{1} easily and you can get h_{2} using the steam quality at the turbine exit. 



#9
Jun1211, 06:11 PM

P: 14

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...) 



#10
Jun1211, 07:18 PM

HW Helper
P: 6,213

W_{turbine}= m_{s}(h_{2}h_{1}) where m_{s} is in kg/s and h is in kJ/kg such that W_{turbine} is in kW. 



#11
Jun1211, 08:08 PM

Mentor
P: 11,988

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




#12
Jun1211, 11:22 PM

P: 14

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*(h2h1) ms = Wturbine/(h2h1) 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? 



#13
Jun1311, 12:35 AM

HW Helper
P: 6,213

I believe h_{2} should be h_{g} at 10 kPa, you really don't want liquid water flowing through your turbine. 


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