Help please Problem 6 (Tii ocation: 1 hour) A. Consider the piston-cylinder arra
ID: 2074014 • Letter: H
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Help please Problem 6 (Tii ocation: 1 hour) A. Consider the piston-cylinder arrangement shown in the figure below. The cross-sectional area of the frictionless piston is 0.06 m2 and the piston is free to travel between the two sets of stops. When the piston is resting on the bottom stops, the enclosed volume is 0.03 m', and when the piston encounters the upper stops, the enclosed volume is 0.075 m . The mass of the piston is 1200 kg, the atmospheric pressure is 100 kPa and the gravitational acceleration, g.is 10 m/'s Initially, the cylinder is filled with 4 kg of saturated liquid-vapor mixture of water at 35 C. Heat is added to the water in a quasi-equilibrium manner. Heat addition is stopped when the pressure of the water reaches 7 MPa i) Find the initial state of the water (State 1). ii) Identify the processes that take place as heat is added to the water iii) Determine the P, v, T, x of the water after each process. iv) Plot the P-V diagram for the whole process showing the saturation dome (the diagram doesn't have to be to scale). v) Determine the work done during the overall process. vi) Determine the heat transfer to the water during the overall process. B. Following the heating process that increased the pressure to 7 MPa, the piston-cylinder arrangement is insulated (assume perfect insulation) and then a mass of 1200 kg is placed on the piston. What will be the final volume? WATER Department of Mechanical Engineering, Dalhousie University V.iUgursal MECH3805, 1t Law Problems Page 3 of 3Explanation / Answer
(1)
For State 1:
T=35 degrees Celcius
v(specific volume) = 0.03/m=0.03/4=7.5 * 10-3
The state 1 is in saturated State
Using the steam table ,
P= 0.0563 bar = 5.7KPa
Quality from steam table , x = 2.575 * 10-4
(2)
External pressure = Patm + pressure due to weight of piston
= 100 +1200*10/0.06/1000
=300kPa
Initial Internal Pressure =5.7KPa
Final Internal Pressure = 7Mpa
Since the Initial Internal Pressure is less than the external pressure, the process starts in an isochoric manner until the Internal Pressure becomes equal to the External pressure . Then the process becomes isobaric untill the piston hits the upper stop following which the process is isochoric untill the internal pressure rises to 7Mpa.
(5)Work done = Pext * (v2 - v1)
= 300*(0.075-0.03)
=13.5kJ
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