Thermo Heat Transfer: Water Boiling at 1 Atm Pressure

[bird91]

Hello all,

I am a little confused on this problem and want to be clear on it.

Given: Water is boiled at 1 atm pressure in a 20-cm i.d. pan. Water level in pan drops by 10cm in 30 min, determine the rate of heat transfer to the pan.

Now correct me if I'm wrong but water boils at 100 (degrees C) at 1 atm pressure, and 1 atm pressure is equal to 101.3 kPa.

Heres where I get confused, do I need to find the internal energy of saturated water at that pressure? If so than do i just take the difference between the sat. vapor and the saturated liquid? Where does the water level dropping come into effect? I know the larger the temp difference, the higher rate of heat transfer. Can you point me in the right direction, I would ask the teacher but I can't make the office hours.Thanks for your time.

 

[ehild]

Hello all,

I am a little confused on this problem and want to be clear on it.

Given: Water is boiled at 1 atm pressure in a 20-cm i.d. pan. Water level in pan drops by 10cm in 30 min, determine the rate of heat transfer to the pan.
...
Where does the water level dropping come into effect?

You need simply the latent heat of vaporization of water at its boiling point. It is 2260 kJ/kg. This latent heat is needed to convert 1 kg water into 1 kg water vapour. You can determine the amount of water evaporated in 30 min, don't you? I hope you can proceed from here.


ehild

 

[wais]

Hi there,

I understand your confusion with this problem. Let me try to break it down for you.

Firstly, you are correct in stating that water boils at 100 degrees Celsius at 1 atm pressure. This is the boiling point of water at standard atmospheric pressure. However, for this problem, we are not concerned with the temperature of the water, but rather the rate of heat transfer to the pan.

To determine the rate of heat transfer, we need to use the equation Q = m * h, where Q is the heat transfer rate, m is the mass flow rate, and h is the specific enthalpy of the fluid. In this case, the fluid is water.

To find the mass flow rate, we need to use the equation m = ρ * V, where m is the mass flow rate, ρ is the density of water, and V is the volume of water that has evaporated in 30 minutes. In this case, V is equal to the cross-sectional area of the pan (20 cm in diameter) multiplied by the change in water level (10 cm), which gives us a volume of 314.16 cm^3.

Now, to find the specific enthalpy of water at 1 atm pressure, we can use the steam tables. The enthalpy of saturated water at 1 atm pressure is 419 kJ/kg, and the enthalpy of saturated vapor at the same pressure is 2676 kJ/kg. To find the specific enthalpy of the water that has evaporated, we can use the equation h = hf + x * (hv - hf), where hf is the enthalpy of saturated liquid, hv is the enthalpy of saturated vapor, and x is the quality of the fluid (ratio of vapor mass to total mass).

Since we know the mass flow rate (m) and the specific enthalpy (h) of the water that has evaporated, we can plug these values into the equation Q = m * h to find the rate of heat transfer to the pan.

The water level dropping is a factor in this problem because as water evaporates, it takes energy (in the form of heat) from the pan. This energy transfer is what we are trying to calculate.

I hope this helps clarify the problem for you. If you have any further questions, please don't hesitate to ask. Good luck!