Thermodynamics: An Engineering Approach 6th Edition, Yunus A. Cengel, Michael A.

Question

Thermodynamics: An Engineering Approach 6th Edition, Yunus A. Cengel, Michael A. Boles, problem 3-43;
The spring-loaded piston-cylinder device is filled with 0.5 kg of water vapor that is initially at 4 MPa and 400 degrees Celsius. Initially, the spring exerts no forces against the piston. The spring constant in the spring forces relation F = kx is k = 0.9 kN/cm and the piston diameter is D = 20 cm. The water now undergoes a process until its volume is one-half of the original volume. Calculate the final temperature and the specific enthalpy of the water.
Book

Explanation / Answer

Forget book solution. Its confusing.

see at initial state:

at p1 = 4 Mpa and T1 = 400 C,

v1= 0.07343 m3/kg

at this stage spring exerts no force. hence pressure is balanced by the mass of the piston

D = 20cm = 0.2 m

hence on balancing the forces acting on the piston

W piston = P1.A = 4x106 x (.D2/4) = P1.A = 4x106 x ( x 0.22/4)

   = 125.663 KN

Now at state 2 :

v2 = v1/2 =  0.07343 / 2 = 0.03672 m^3/Kg

at this stage spring is extended and exerts a force on piston in upward direction = K.x

K =  0.9 kN/cm = 90 KN/m

Now, A.X = dV = change in volume = m. (v1 - v2), m = mass of water

(.D2/4) . X = 0.5 x ( 0.07343 - 0.03672   )

putting , D = 0.2 m, we get

X = 0.58425 m

Now at state 2., balancing the force:

W piston = P2.A + K.x

125.663 = P2 x  ( x 0.22/4)  + 90 x  0.58425

P2 = 2326.224 Kpa

now, v2 = 0.03672 m^3/Kg = vf + x. vfg

0.03672 = 0.001190 + x . 0.084904

x = quality = 0.4185

h2 = hf + x .hfg

   =943.55 + (0.4185)(1857.4) = 1720.8719 KJ/Kg

also T2 = T saturation at P2 =  2326.224 Kpa

hence T2 = 220 C

a word of advice : I know that a guy has mailed you solutions of this book. Let me tell you this is against cramster policy and the solution manual is copyright and cannot be owned by a student. The book solution will only confuse you. Pls try do it yourselg.

Best !!!

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