Water is contained in a piston-cylinder device as shown in the figure below. The
ID: 1717993 • Letter: W
Question
Water is contained in a piston-cylinder device as shown in the figure below. The atmospheric pressure outside of the device is 101.3 kPa. You may neglect friction between the piston and the wall and assume that the piston does not leak. Also, neglect the thermal interaction between the walls of the device and the water it contains. The mass of the water in the cylinder is m = 0.05 kg and the area of the piston face is A_c = 0.3 m^2. Initially, water is at T_1 = 110 degree C with a quality of x_1 = 0.85. Water is heated and the piston begins to rise; as this occurs, the spring is compressed and so the pressure in the cylinder begins to rise. The force F_spring exerted by the spring on the piston is related to the elevation, z, of the piston measured from the bottom of the cylinder by where K = 5000 N/cm is the spring constant and z_0 is the piston position at which the spring exerts no force. The heating stops when the water temperature reaches T_2 = 225 degree C and the elevation is Z_2= 0.2 m Is this a steady-state process? Justify What is the initial pressure in the cylinder? What is the initial elevation of the piston? Use a force balance on the piston to find a relation between the pressure in the cylinder and the elevation of the piston z. What is the pressure in the cylinder at the end of the heating process ? What is the phase of the water in the cylinder at the end of the heating process ? Justify How much work is done by the water in this process? What is the change in internal energy of the water? Determine the amount of heat that was transferred to the water in this process. You may choose to solve this problem with EES. If you do so, you should first write down by hand the equations that will lead to the solution before you implement them in EES. Turn in both your hand-written workings AND your EES printouts.Explanation / Answer
P1 = Patm = 101.3kpa
P2 = Psp + Patm .......... ( equation 1 )
Psp = Kx/A = (5000×.85)/(.3) = 14.166 kpa
From equation 1, P2 = 14.166+ 101.3 = 115.466 kpa
Work done = 1/2 (p1+p2) = 1/2 (101.3+115.466) = 108.383 KJ
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