A 2.0 kg of water at 90 C and 1 bar is placed in a vertical cylinder and sealed

Question

A 2.0 kg of water at 90 C and 1 bar is placed in a vertical cylinder and sealed with a massless piston. The piston is perfectly insulated and is able to slide freely without friction. Both the piston and the cylinder have a diameter of 3.0 m. After placing the piston on the water at 90C and 1 bar, what is the height difference between the bottom of the cylinder and the piston? B. If you place an additional 4 bar of external pressure on the piston as detailed in question 1, what is the enthalpy of the water in the cylinder-piston system? Assume the temperature remains constant at 90C. C. Heat is pumped into the piston-cylinder as described in Question 1, until the the piston is 0.49 m above the cylinder bottom. The external pressure acting on the piston remains constant at 1 bar. Heat is then released from the system until the piston drops to a height of 0.07 m. How much heat is removed in this process? Answer in kJ and do not use a negative sign in your answer (it is inferred that that the amount of heat transferred must be negative).

Explanation / Answer

1) at 90 C and 1 bar, from steam table : specifc volume = v1 = 0.001036 m^3/Kg

so volume of enclosed space = V = m x v = 2 x 0.001036 = 0.002072 m^3

so distance = V/Area = 0.00272/(pi/4 x 3^2) = 2.93 x 10^-4 m = 0.293 mm

2) after 4 bar pressure uis added : pressure inside becomes + 1 + 4 = 5 bar

now we look into the Compressed water table :

so, at 90 C and 5 bar, we get enthalpy =376.92 KJ/Kg

so total enthalpy = 2 x 376.92 = 753.84 KJ

3)

at 0.49 m , Volume = area x 0.49 = pi/4 x 3^2 x 0.49 = 3.46 m^3

so, specific volume = v1 = V/m = 3.46/2 = 1.73 m^3 Kg

so, at 1 bar and 1.73 m^3/Kg, from steam tabel :

enthalpy = h1 = 2680 kJ/Kg

and at 0.07 m : volume = pi/4 x 3^2 x 0.07 =

0.49355 m^3

so specific volume = V/m = v2 = 0.49355/2 = 0.247 m^3/Kg

so at P = 1 bar na v2 = 0.247 m^3/Kg from steam table : h2 = 750.855 Kj/Kg

so heat transfer = Q = m. (h2 - h1)

= 2 x (750.855 - 2680) = 3858.3 KJ

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