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Let there be a cylindrical vessel with diameter $D$ that only has liquid water and water vapor as the picture below shows (The water vapor is the grey portion and the liquid water is the blue portion). All contents of the cylinder are at temperature $T$. The water vapor is at pressure $P$. The contents of the container are heated using a burner. The power delivered from the burner to the cylinder is $\dot Q$. As a result, I believe that some of the water will evaporate.

If $m_v$ is the total mass of the water vapor in the container, then I believe that $m_v$ will be a function of the diameter $D$, $h$, $P$, $T$, and $\dot Q$. I don't know what this function is. I would be really glad if someone could give it to me or give me a reference.

enter image description here

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  • $\begingroup$ How long can you wait for the answer :)? I'm investigating a very similar problem in detail; I'll develop a simple theoretical model and test it experimentally in a few weeks. I'll get back to you then! $\endgroup$ – Physics Llama Nov 15 '14 at 15:57
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Are you studying basic thermodynamics? Because your question is exactly what people learn in the first few classes of this subject.

First Law of Thermodynamics tells us that every joule of energy from the burner will act to increase the system's internal energy.note 1 Also, the total volume of your system is constant (isochoric process). We need only to use the fact that the state of a pure, simple substance is uniquely determined by two independent parameters in order to predict the behavior of your system.

So, in a nutshell, our case in point is completely determined by: $$U=U_0+\dot Q\Delta t$$ $$V=\mathrm{constant}$$

What we want to find is the parameter named quality (symbol $x$), which is the mass of vapor divided by total mass. As with any other parameter, it can be expressed as a function of two others (so $x=x(U,V)$. Once we have it, we find the mass of vapor by: $$m_v=m\cdot x(U,V)$$

where $m$ is the total mass of water.

There's only one problem: finding an explicit formula for $x(U,V)$. Some exist, but finding them is not straightforward. However, there are many tables and graphs readily available on the internet, and you can find solutions for individuals points $(U,V)$ with them. Below I'll show you how to do it with a water table, but first check out these graphs. They give you an idea of how your process evolves with time.

isochors

¹ if your system was under constant pressure, we should use the enthalpy of vaporization, but since the cylinder is constant-volume, the correct one is indeed the energy of vaporization. All this comes from the First Law: $Q=\Delta U+W$


Example: finding how much water evaporated after $t$ seconds

Suppose it's a 1 m³ cylinder with 1 kg of water inside, receiving 2 kW from a burner. Initially, it's at 20 °C. We want to know how much water vaporized after 5 minutes.

We are going to use this table as our model of the behavior of water.note 2

The line at 20 ºC says that, at such temperature, liquid water would have specific volume 0.001 m³/kg, while whole vapor would be 57.757 m³/kg. We're in between, at 1.000 m³/kg. So, initially, our quality is given by: $$v=v_l+xv_g\Leftrightarrow x=\frac{v-v_l}{v_g}$$$$\Leftrightarrow x_1=\frac{1-0.001}{57.757}\approx 1.73\%$$

With this we can find our initial internal energy, which stands between those given at the table: $$u=u_l+xu_g$$$$\Leftrightarrow u_1=83.91+1.73\%\cdot2402.3\approx125.5\frac{\mathrm{kJ}}{\mathrm{kg}}$$

What do we do, now? We know that, after those minutes, we have a lot more internal energy: $$u_2=u_1+\frac{\dot Q\Delta t}{m}=125.5+\frac{2\cdot5\cdot60}{1}=725.5\frac{\mathrm{kJ}}{\mathrm{kg}}$$

We also know that the specific volume remained at 1 m³/kg. With this, we can apply trial and error to find our current state. Let's guess that the final temperature would be 70 °C. Then, we'd have: $$x_{2'}=\frac{1-0.00102}{5.0395}\approx19.8\%$$

This gives: $$u_{2'}=293.03+19.8\%\cdot2468.9=782.4\frac{\mathrm{kJ}}{\mathrm{kg}}$$

This is more than our real $u_2$, but no too much. We overshot a bit, but we're close! Let's check if it isn't 65 °C instead (calculations suppressed): $$x_{2^{\ \prime\ \prime}}=16.1\%\rightarrow u_{2^{\ \prime\ \prime}}=669.3\frac{\mathrm{kJ}}{\mathrm{kg}}$$

$u_{2^{\ \prime\ \prime}}<u_2<u_{2'}$ tells us that $65\mathrm{°C}<T<70\mathrm{°C}$, and also $x_{2^{\ \prime\ \prime}}<x_2<x_{2'}$. There's no better way to settle this deal than with a linear interpolation, so we do: $$x_2=x_{2'}+\frac{x_{2^{\ \prime\ \prime}}-x_{2'}}{u_{2^{\ \prime\ \prime}}-u_{2'}}\left(u_2-u_{2'}\right)\approx17.9\%$$

Thus, the amount of water that evaporated is given by: $$\Delta m_v=x_2m-x_1m\approx161.7\ \mathrm g$$

² in the table, properties ending in "f" are for completely liquid water, while those ending in "g" are for completely vapor water (normally it's "ℓ" and "v" though)

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  • $\begingroup$ Why do you assume that all the power provided by the burner is used to vaporize the water ? Why can't it be the case that only a fraction of the burner power is used to evaporate some of the water and the remaining fraction of the burner's power is used to raise the temperature of the liquid water ? $\endgroup$ – Amr Nov 15 '14 at 14:14
  • $\begingroup$ You're right. I started assuming constant temperature, but as I concluded afterwards, it's not! I'll fix it. $\endgroup$ – André Chalella Nov 15 '14 at 15:01
  • $\begingroup$ @Amr done. Maybe I overdid it a bit, but it'll definitely get you going. $\endgroup$ – André Chalella Nov 15 '14 at 17:07
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I measured evaporation in an open container to be about 0.6 grams per hour. Room temperature average about 72 degrees. Surface diameter of water is about 6 inches.

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As the liquid and vapor temperature increases, the liquid density goes down, and the liquid expands. Also, as the temperature goes up, the vapor density goes up, due to the higher vapor pressure of water at the higher temperature. Both liquid and vapor density can be obtained from steam tables, or from a quadratic or cubic curve fit on steam table data.

Regarding the volume of liquid and vapor, as the liquid expands in a closed container, it takes up more volume than it originally did, so the "liquid" space expands while the "vapor" space contracts. I don't know if an analytic solution exists for this process, but a numeric simulation, and a resulting curve fit, yielded the following for liquid and vapor volumes:

Given conditions - for the temperature range from 10-200 deg C, starting with 50% liquid water at 10 deg C;

Fraction of liquid = 1.656E-06(T^2) + 4.198E-05(T) + 0.5

Fraction of vapor = -1.656E-06(T^2) - 4.198E-05(T) + 0.5

With the volume data and the density data, you can easily calculate the mass of liquid and vapor as a function of temperature. In addition, steam table enthalpy data will allow you to calculate temperature as a function of heating rate, if you need that particular information.

Note: the particular equations above apply ONLY to the 50% liquid starting fraction. For other liquid starting fractions, the equations will be different.

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The equilibrium vapor pressure $P$ of liquid water at a temperature $T$ can be accurately estimated using the Tetens equation:

$$ \  P\ =\ 0.61078\ \text{exp}\left( {\frac{17.27\ T}{T+237.3}}\right)$$

where
$P$ = equilibrium vapor pressure in kPa
$T$ = temperature in °C

If $m_v$ is the total mass of the water vapor in the container, then I believe that $m_v$ will be a function of the diameter $D$, $h$, $P$, $T$, and $\dot Q$.

The amount of water vapor is given by the ideal gas law $PV = nRT$ once it has been rearranged to $n = PV/RT$. Use the heat capacity of the water along with the amount of heat input $\dot Q$ to the system by the burner to figure out the final temperature $T$ of the water after heating. Once you have the final temperature of the water $T$, the volume of the container $V$, and the vapor pressure $P$ (from the equation above) the problem becomes a plug-and-chug. If using $R= 0.082 \frac{L\ atm}{mol\ K}$ then to convert moles water vapor to mass water vapor all you need to do is multiply by the molar mass.

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