A rigid 5.00-liter vessel contains 4.00 L of liquid water in equilibrium with 1.

Question

A rigid 5.00-liter vessel contains 4.00 L of liquid water in equilibrium with
1.00 L of water vapor at 25oC. Heat is transferred to the water by means of an immersed
electrical coil. The volume of the coil is negligible. Use the steam tables to calculate the
final temperature and pressure (bar) of the system and the mass of water vaporized (g) if
2915 kJ is added to the water and no heat is transferred from the water to its
surroundings.
Hint 1: The rigid container does not change in volume during the process
Hint 2: A trial-and-error calculation is requiredeh

Explanation / Answer

At 25 deg C, u_f = 105 kJ/kg and u_g = 2410 kJ/kg, v_f = 0.001 m3/kg and v_g = 43.3 m3/kg

Mass of water = (4*10^-3)/0.001 = 4 kg

Mass of steam = (1*10^-3)/43.3 = 0.000023 kg

Quality x1 = 0.000023/(4+0.000023) = 0

Q - W = (U2 - U1)

Since, W = p dV and dV = change in volume = 0. W = 0.

So, Q = (U2 - U1)

U2 = 2915/4 + 105 = 833.8 kJ/kg

From tables, we now look against v2 = v1 = 0.001 m3/kg and u2 = 833.8 kJ/kg, we get P = 3.67 bar and T = 232 deg C.

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