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I have a contained and thermally insulated cylinder of radius r. Inside this cylinder, there is water that reaches up to a certain height inside the cylinder (hw). There is a heater underwater that will supply thermal energy to the water at a certain wattage. The rest of the space is filled with air at atmospheric pressure (ha is the height the air occupies so that ha + hw = htotal). I start supplying thermal energy at a given wattage to the water. My goal is to keep the water bubbling at a fixed rate for as long as I can control the pressure inside the vessel. I want to know: How can I determine at what rate does the pressure increase? How do I determine the increase boiling point of water? How do I relate the input heat in wattage to bubbling rate? (evaporation rate?) I want to keep the water bubbling at a constant rate, how do I determine how the heater wattage should increase to match the increasing boiling point as the pressure builds up? Is there a reference for what a bubbling rate (evaporation rate?) looks like?

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  • $\begingroup$ How will you control the pressure ? A safety valve sounds like it will be a good idea... Combined with a pressure switch that cuts the supply to the heater. $\endgroup$ – Solar Mike Jul 6 '17 at 20:44
  • $\begingroup$ Is this something that you're actually doing, or a thought experiment? If it's something real then you need to step back a bit and think about the design and fabrication of the vessel, including the design pressure, the materials of construction, and the safety features to be implemented - beginning with a relief valve and interlock as @SolarMike suggests. Otherwise what you're building is a deadly bomb. $\endgroup$ – Mark Jul 7 '17 at 12:52
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Nukiyama had a similar setup in his experiment, which failed to account for the leidenfrost effect. The net result is he fried his heater. This is why boiling with electrical wiring is such a tricky matter.

However, there are several equations that relate the heat transfer to the water. The main question you ask is how an enclosed vessel will react. This is pretty easy to see when you look at a Pv diagram for water:

enter image description here

The air above the container is never just air, but a mix of air and water vapor, or steam. Thus we will only be moving up in pressure as we heat to newer and newer constant temperatures. The temperature throughout the chamber can be roughly approximated as constant in both phases throughout this, if you're heating so fast this assumption isn't valid, then Lidenfrost effects will indicate that most heat is being transferred via radiation anyways and you're probably melting your heater with most materials.

The specific volumes of either phase change (as the liquid drops or more mass moves to the vapor phase), but the total specific volume of the entire water must remain the same. In other words, given any temperature as we climb, there is a specific saturation pressure that the vessel remains at, and:

$$v_gm_g +v_lm_l +v_am_a = V_{vessel}$$

This equation has two unknowns, the mass in the liquid phase and the mass in the gas phase, but the specific volumes of each phase are known at a given temperature, using steam tables. The mass of the air is known from the start, and the specific volume of the air at a given temperature is given via the ideal gas laws $v_a = \frac{RT}{MP}$ However, since $m_g + m_l = m_w$, which is known at the start, the masses of each at that temperature can be known.

Finding what time you will reach a specific temperature, and therefore pressure, is more complicated, but not terrible. This involves finding the enthalpy of each phase, subtracting the starting enthalpy, and dividing by the wattage.

Note that above the critical point, the liquid acts like a gas and the gas acts like a liquid and you can't find a line anymore. Then the line goes straight up. You may not reach the critical point - if your specific volume isn't identical to the critical point's specific volume than the path simply goes straight up from there until you reach maximum pressure.

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  • $\begingroup$ Thank you, very helpful! I'm going through all the details you provided $\endgroup$ – gummibear Jul 14 '17 at 19:22

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