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OlgaM077 [116]
3 years ago
11

A piston-cylinder device contains 0.58 kg of steam at 300°C and 0.5 MPa. Steam is cooled at constant pressure until one-half of

the mass condenses. Use data from the steam tables.
Determine (a) the final temperature; (b) the volume change; and (c) show the process on a T-V diagram with respect to saturation lines.

Engineering
1 answer:
Mumz [18]3 years ago
3 0

Answer:

a) Tբ = 151.8°C

b) ΔV = - 0.194 m³

c) The T-V diagram is sketched in the image attached.

Explanation:

Using steam tables,

At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.

Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C

b) The volume change

Using data from A-5 and A-6 of the steam tables,

The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume

(αբ) (which is calculated from the final quality and the consituents of the specific volumes).

ΔV = m(αբ - αᵢ)

αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)

q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg

αբ = 0.00109 + 0.5(0.3748 - 0.00109)

αբ = 0.187945 m³/kg

αᵢ = 0.5226 m³/kg

ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³

c) The T-V diagram is sketched in the image attached

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tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5

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\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2}  + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta  + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2}  + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56)  + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\

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