Physics Help Forum Heat powered elevator

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Nov 13th 2015, 11:51 AM #11 Junior Member   Join Date: Nov 2015 Posts: 10 Assuming that ; Air = ideal gas The system is well insulated so there is no heat loss to the surroundings deltaKE = deltaPE = 0 I have described my states as ; Sate 1 = the initial conditions (T1, P1, V1) State 2 = when the casting is added on (T1, P2, V1) State 3 = when the air heats up and is lifted (T2, P2, V2) State 4 = casting is off (T2, P1, V2) State 5 returning to 1 = cools down (T1, P1, V1) Is it possible to have two constants? I've never had an example with two. Ie can something be Isometric and Isobaric? Or Isothermal and Isobaric? From what I understand and have calculated ; (i)Temperature & Pressure P1 =100 T1=25 P2 =128 T2=27.16 V1=30 V2=32.6 CV=0.919 CP-1.206 (ii)Would the work done be ; w=Pressure X deltaV w=128000 x 2.6 w=332.8j? (iii)Would the heat transfer be ; Q= m x cp x deltaT if so is m1=m2 or are they different? Using the below method they are different so which would i substitute in? m1, m2 or mT? My friend and I were having a discussion about this as we found an example of both. using m=P1 x V1/Rair x T1 m=35kg m=P2 x V2/Rair x T2 m=48.5kg mT=m1+m2 mT=83.5kg?
Nov 13th 2015, 11:59 AM   #12
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 State 2 = when the casting is added on (T1, P2, V1)
Molyticus, a couple of questions.

1) How does the pressure magically jump upwards when the casting is added?

2) At what point in the calculations does it make a difference whether the system is insulated or not?

Nov 13th 2015, 12:04 PM   #13
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 Originally Posted by studiot Molyticus, a couple of questions. 1) How does the pressure magically jump upwards when the casting is added? 2) At what point in the calculations does it make a difference whether the system is insulated or not?
Would the casting not add to the pressure? So the initial pressure would be 100kpa and when the casting is added on it is 128kpa, and when the casting is taken off it would return to 100kpa?

Nov 13th 2015, 12:18 PM   #14
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 Would the casting not add to the pressure? So the initial pressure would be 100kpa and when the casting is added on it is 128kpa, and when the casting is taken off it would return to 100kpa?
This is what I meant by understanding the mechanics of the elevator.

Initially the tank is capped by the piston.

Is the upward force of the air pressure on the underside of the piston greater than, less than or equal to the combined weight of the piston plus the external air pressure?

Nov 13th 2015, 12:35 PM   #15
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 Originally Posted by studiot This is what I meant by understanding the mechanics of the elevator. Initially the tank is capped by the piston. There is no casting added. Is the upward force of the air pressure on the underside of the piston greater than, less than or equal to the combined weight of the piston plus the external air pressure?
I understand that the tank is capped by the piston but it says that a casting is slid on before heat exchanges, would that not add to the pressure? How can a casting not be added if it is slid on?

Myself and a group of friends are working on this and we always assumed that the casting is slid on, the air is heated up gradually which causes the piston+casting to rise and as soon as it reaches the top the casting is slid off before it is cooled down to be repeated again, but we have looked everywhere for examples that are similar and there are none. You said we need to understand the mechanics of the elevator but that's what we understood from what was being asked.

 Nov 13th 2015, 12:38 PM #16 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Why don't you try to answer my question? You never know it may lead to a Eureka moment. Let me rephrase it. What would happen if the upward force of the tank air on the piston was greater than the combined weight of the piston and the force of the external air pressure? Force balances is mechanics 101.
Nov 13th 2015, 01:15 PM   #17
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 Originally Posted by studiot Is the upward force of the air pressure on the underside of the piston greater than, less than or equal to the combined weight of the piston plus the external air pressure?
Equal to each other?
So the force undearneath = piston + Patm?

Tbh, I'm just getting more confused lol.

 Nov 13th 2015, 02:03 PM #18 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 OK I make the area of the piston = 0.4418 square metres. So the upward force on the piston is initially 0.4418 x 10^5 N. = 44180N The weight of the piston is 230 x 9.81 = 2256N Standard air pressure of 101325 N/sqm times piston area gives a downward force of 44765N. Combine this with the piston weight gives 54575 N. So even to raise the piston without the casting the tank pressure would have to increase. (Which is as it should be) The difference between the upward force and the downward force is provided by the reaction from the nibs at the side which are clearly shown in the diagram. The piston rests on the nibs closing the tank, in the quiescient condition. Add the weight of the casting and you can calculate the pressure at which the piston plus casting just starts to move. Adding heat up to this point merely raises the pressure, the volume does not change. The support for the piston and casting is gradually transferred from the nibns to the air and the nib reaction falls to sero as the tank pressure rises. So the line from point 1 on your PV diagram to point 2 is vertical straight line from the intial tank pressure to the moment the piston starts to rise. Once the piston is rising (slowly and reversibly) the pressure will not increase but remain at the pressure until the piston hits the stops at the top, where a downward reaction will occur. So the expansion is under constant pressure infinitessimally above that required to just lift the piston plus casting. So the line from point 2 to point 3 on your PV diagram is a horizontal line. Can you continue with the cooling phase of the cycle to return to point 1 on the PV diagram?
 Nov 14th 2015, 04:50 AM #19 Junior Member   Join Date: Nov 2015 Posts: 1 [QUOTE=studiot;30779]OK 'I make the area of the piston = 0.4418 square metres. So the upward force on the piston is initially 0.4418 x 10^5 N. = 44180N The weight of the piston is 230 x 9.81 = 2256N Standard air pressure of 101325 N/sqm times piston area gives a downward force of 44765N. Combine this with the piston weight gives 54575 N' I get how it works now but can i ask where did get those numbers from? How did you get the value for the area of the piston? How did you get the value for the downward force?
 Nov 14th 2015, 08:09 AM #20 Junior Member   Join Date: Nov 2015 Posts: 10 Area of the piston = pi/4(d)^2 x 10^5 =44180N 1N = 1pa 100kpa = 100000pa Downward force = Patm x piston area = 44765N Weight of piston = m x g = 2245N No idea where "54575 N" is coming from though lol.

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