Thermodynamics: Vaporizing all Liquid

[jnbfive]

Homework Statement

I have a problem that I need assistance with:

There is a well insulated rigid tank with 3 kg of saturated liquid-vapor mixture of water at 200 kPa. Initially, half of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 120-V source, and a current of 6 A flows throught the resistor when the switch is turned on. How long will it take to vaporize all the liquid in the tank?

Homework Equations

The Attempt at a Solution

State 1:
P = 200 kPa
x = .5
Therefore T = 120.21 C
v = v_f +xv_fg
v_f = 0.001061 m³/kg
v_fg = 0.88578 - 0.001061 m³/kg

Multiply v by m to obtain V = 1.33206 m³

Therefore V₂ = 1.33206 and since the mass does not change, v₂ = .4423595 m³/kg = v_g

u₁ = u_fg @ 200 kPA = 2024.6 kJ/kg

W_e = Delta U => VI(Delta T) = m(u₂ - u₁)

Looking for the v₂, it is assumed to be between 400 kPa and 450 kPa. Therefore, use the v values to interpolate the u_g which is 2554.6669 kJ/kg.

Therefore

((1 kJ/s)/(1000 VA))720 VA (Delta T) = 3 kg(2554.6669- 2024.6 kJ/kg)

Delta T = 2208.6 s or .6135 hours.

I just want to know if I made a mistake anywhere in this problem. I'm not the best with thermodynamics and usually wind up missing some minute detail that is vital to the rest of the problem.

 

[KataKoniK]

Hello! Your solution seems to be on the right track, but there are a few things that could be clarified or corrected.

Firstly, when calculating the specific volume (v), it should be noted that it is a function of pressure and temperature, so the specific volume at 200 kPa and 120.21 C may not be the same as the specific volume at the final pressure and temperature. It would be more accurate to use the specific volume at the final pressure (which we don't know yet) and temperature (which is given as 120.21 C).

Secondly, when calculating the internal energy (u), it should be noted that it is also a function of pressure and temperature. The specific internal energy at 200 kPa and 2024.6 kJ/kg is not necessarily the same as the specific internal energy at the final pressure and temperature. It would be more accurate to use the specific internal energy at the final pressure and temperature, which can be obtained from the steam tables.

Thirdly, when solving for the final pressure, it is important to consider the fact that the electric resistor is heating the water and causing the pressure to increase. This means that the final pressure will be greater than 200 kPa. You could use the ideal gas law (PV = nRT) to solve for the final pressure, where n is the number of moles of water and R is the gas constant for water.

Lastly, when solving for the time it takes to vaporize all the liquid, it would be more accurate to use the specific enthalpy (h) instead of the specific internal energy. This is because the specific enthalpy takes into account the work done by the system, whereas the specific internal energy does not.

Overall, your approach to the problem is correct, but there are a few details that could be clarified or corrected for a more accurate solution. I hope this helps!

 

[Mene]

Your solution looks mostly correct, but there are a few things to consider:

1. The given current of 6 A and voltage of 120 V is not used in your solution. Make sure to include these values in your calculations.

2. Your calculation for the specific volume at state 1 (v1) is incorrect. The correct value should be 0.001061 m3/kg, which is the specific volume of saturated liquid at 200 kPa.

3. In your calculation for vfg, you have used the specific volume of saturated liquid and saturated vapor, but these values should be for the specific volume of the mixture at state 1. Therefore, vfg should be 0.885719 m3/kg.

4. When calculating the specific internal energy at state 1 (u1), you have used the specific internal energy of the mixture at 200 kPa, but this value should be for the specific internal energy of the saturated liquid at 200 kPa. The specific internal energy of the mixture at state 1 can be calculated using the given values of pressure and quality (x).

5. Your calculation for v2 is also incorrect. The specific volume at state 2 (v2) is the specific volume of the saturated vapor at the final pressure, which is not given in the problem. Therefore, you cannot calculate v2 without knowing the final pressure.

6. The specific internal energy at state 2 (u2) can also not be calculated without knowing the final pressure.

7. The equation you have used for work (We = Delta U) is incorrect. The correct equation to use in this case is We = m(u2 - u1).

Overall, your approach to the problem is correct, but make sure to use the correct values for specific volume and specific internal energy at each state, and also consider that you need to know the final pressure in order to calculate the final values of specific volume and specific internal energy.

 

[Mene]

Your solution looks correct to me. Just make sure to convert your final answer from seconds to hours, which you have already done. Also, it might be helpful to include units for your final answer (hours) to make it clear. Other than that, your solution seems to be thorough and accurate. Great job!