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Link to image for problem. Show all steps to recieve all points

http://imgur.com/gallery/rtF1cFQ/new

For the system shown, water is heated at a constant pressure from its initial state to state a (where the piston hits the stop). At this point, the water continues to heat (but at a constant volume) until its final pressure is 3 bar.

1) Find the pressure at state 1 using a force balance on the piston (include the force due to gravity and atmospheric pressure).

2)Using the pressure from 1) and the quality find the specific volume at its initial state.

3)Using geometry, relate the specific volume of the water at its initial state to its specific volume when it hits its stop. Compare the specific volume at the stop to vg at 3 bar. If it is > vg find the mass.

4)Find the work due to expansion to the stop.

5)Find internal energy change from the initial state to final state.

6)What is the heat transferred to the water?

Explanation / Answer

1) balancing the forces :

Patm x area + M.g = P1 x area

so 1x 10^5 x (pi/4 x 0.1^2) + 40 x 9.81 = P1 x (pi/4 x 0.1^2)

so pressure ar 1 = P1 = 1.49 x 10^5 Pa = 1.49 bar

2) at 1.49 bar from steam tabel :

specific volume = v1 = vf + x.vfg

              so v1 = 0.00105 + 0.25 x 1.158 = 0.29 m^3/Kg

3) let m = mass of water

then Volume at initail state = specific volume x mass = 0.29 m

now volume = area x length

so area x 1 cm = 0.29 m

at the stop :

length = 4.5 cm

so volume = area x 4.5 (here area reaisn constant)

so new volume = area x 4.5 = V2

and V1 = area x1

so V2 = 4.5 V1

so v2 = 4.5v1 (mass remains constant)

so specific volume at stop =4.5 times specific volume of inital = 4.5 x0.29 =              1.305 m^3/Kg

4) work done = P1.(V2 - V1)

Now as said above : are x 1 = volume

so (pi/4 x 0.1^2) x 0.1 m^3 = 0.29 m

so amss of water = m = 2.7 x 10^-3 Kg

so work done = 1.49 x 10^5 x 2.7 x 10^-3 x (1.305 - 0.29)

                     = 408.33 J

5) at P1 = 1.49 bar and 0.29 m^3/Kg specifc volume : interanl enegyr from steam table = u1 = 467 + 0.25 x 2052.5 = 980.125 KJ/Kg

at P1 = 1.49 bar and specifrc volume = 1.305 m^3/Kg from steam table :

internal enrgy = u2 = 2100 KJ/Kg

so change in intetrnal enrgy = m.(u2 - u1)

               = 2.7 x 10^-3 x (2100 - 980.125)

                 = 3.023 KJ = 3023 J

6) heat added = W + dU

             408.33 + 3023 = 3431.99 J

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