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Suppose that I'm boiling water in an open pot. As the pot is heated by the stove, it loses mass in the form of vapour.

The pot is my control volume. That system could be modeled by the following equations:

Mass balance:

$$ \frac{dm}{dt} = -\dot m_{out} $$

Energy balance: $$ \frac{dU}{dt} = -\dot H_{out} + \dot Q $$

Considering $U = mc_v^{liq}T$, if the boiling process occurs at constant temperature $T_b$,

$$ c_v^{liq} T_b\frac{dm}{dt} = -\dot m_{out} c_p^{vap} T_b + \dot Q $$

Let's throw in some numbers:

  • $c_v^{liq} = 4.0$ kJ/kg°C, the specific heat capacity of liquid water at constant volume

  • $c_p^{vap} = 2.0$kJ/kg°C, the specific heat capacity of vapour at constant pressure

  • $T_b = 100$ °C

  • $\dot Q = 400$ kW (the stove supplies a positive heat inflow to the system, whereas it also loses heat to the surroundings by convection at a constant rate $q = hA(T_{surr}-T_b)$, but I figure that the overall $\dot Q$ must be a positive number, in order to supply the latent heat necessary to vaporize the water.)

$$ -400\cdot\dot m_{out} = -200\cdot\dot m_{out} + 400 $$ $$ (-200\text{ kJ/kg})\cdot\dot m_{out} = 400\text{ kW} $$ $$ \dot m_{out} = -2.0\text{ kg/s} $$

So... I found a negative mass outflow... as if it was actually an inflow, with vapour condensing into liquid water. Moreover, 2.0 kilograms/second is a lot! Even though that's just a toy model, I believe the numbers I provided are at least in the ballpark of a realistic situation...

And is it even realistic that the rate of vapour outflow would be constant?

Where did I screw up my model? Can anyone show me a better, more realistic, one?

EDIT 1: Thank you Transistor for reminding me in the comments to account for heat of vaporization, what a blunder! Heating rate should be in watts, rather than kilowatts as well.

Redoing the calculations

$$ c_v^{liq} T_b\frac{dm}{dt} = -\dot m_{out} (c_p^{vap} T_b + \Delta h_{vap}) + \dot Q $$

$$ -400\cdot\dot m_{out} = -(200 + 2250)\cdot\dot m_{out} + 0.40 $$ $$ (1650\text{ kJ/kg})\cdot\dot m_{out} = 0.40\text{ kW} $$ $$ \dot m_{out} = 2.4\times10^{-4}\text{ kg/s} $$

That's more like it!

EDIT 2: I thought a little bit further, and I believe there were more things wrong with my previous reasoning.

The internal energy of the system during the phase change is $U = m_Lu_L + m_Vu_V$ (liquid + vapour). If $m_L + m_V = m_0$,

$$ U = m_L u_L + (m_0 - m_L)\cdot (u_L + \Delta u_{vap}) $$ $$ \frac{dU}{dt} = -\Delta u_{vap} \frac{dm_L}{dt} $$

Moreover, I was messing up with the reference for calculating the enthalpies. If my reference is liquid water at 0°C, then the enthalpy of the vapour leaving the system is actually $-\dot m_{out} (c_p^{liq} T_b + \Delta h_{vap})$, with $c_p^{liq}$ insetad of $c_p^{vap}$.

Hence,

$$ -\Delta u_{vap} \frac{dm_L}{dt} = -\dot m_{out} (c_p^{liq} T_b + \Delta h_{vap}) + \dot Q $$

The mass balance would be:

$$ \frac{d}{dt} (m_L + m_V) = 0 $$

$$ \frac{dm_L}{dt} = -\frac{dm_V}{dt} = -\dot m_{out} $$

$\Delta u_{vap} \approx 2090$ kJ/kg is a little bit lower than $\Delta h_{vap} \approx 2250$ kJ/kg at 1 atm.

$$ \dot m_{out} = \frac{\dot Q}{c_p^{liq}T_b+\Delta h_{vap}+\Delta u_{vap}} $$

Using the values above, $\dot m_{out} \approx 8\times10^{-5}$ kg/s. Now it's quite a bit smaller than I would expect. Maybe I got my signs wrong and it's actually $\Delta h_{vap} - \Delta u_{vap}$? That would make more sense, and $\dot m \approx 7\times10^{-4}$kg/s, which seems more reasonable (around 23 minutes to boil 1 kg of water).

EDIT 3: I've got the answer I was looking for in the Physics StackExchange. Thank you for all the help!

https://physics.stackexchange.com/questions/767996/mass-flow-of-vapour-produced-while-boiling-water-in-an-open-system/768019#768019

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  • $\begingroup$ 400 kW stove? My hob has 2 kW rings on it. Are you boiling elephants? $\endgroup$
    – Transistor
    Jun 11, 2023 at 13:19
  • $\begingroup$ Can you edit in an explanation of what $c_v^{liq}$ and $c_p^{vap}$ are. If the first one is specific heat capacity of water then 4.2 (or 4.184) will be a bit more accurate. The second seems to be the SHC of steam but that's not relevant as the vapour won't be heated further as it leaves the pot. You should be using the latent heat of evaporation, 2250 kJ/kg. $\endgroup$
    – Transistor
    Jun 11, 2023 at 13:29
  • $\begingroup$ Hmm... that makes sense, I may have mixed kilowatts and watts. That would explain the excessive 2.0 kg/s. But not the minus sign. Any positive heat rate results in a negative mass outflow, and that is what is bugging me the most. $\endgroup$
    – GFonseca
    Jun 11, 2023 at 13:32
  • $\begingroup$ You just solved my problem -- see edit. Give me an answer and I'll give you proper credit. $\endgroup$
    – GFonseca
    Jun 11, 2023 at 13:41
  • $\begingroup$ It turns out I don't have enough reputation to cast a vote yet, sorry. But to anyone reading this, please upvote Transistor's answer below. $\endgroup$
    – GFonseca
    Jun 11, 2023 at 13:48

2 Answers 2

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It's not clear what $c_v^{liq}$ and $c_p^{vap}$ are.

If the first one is specific heat capacity of water then 4.2 (or 4.184) kJ/kg/K will be a bit more accurate.

The second seems to be the SHC of steam but that's not relevant as the vapour won't be heated further as it leaves the pot. You should be using the latent heat of evaporation, 2250 kJ/kg.

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The governing equations are mass and energy balances. We can use an enthalpy balance rather than an internal energy balance because the process is at constant temperature and pressure.

$$\dot{m}_{vap} + \frac{dm_{liq}}{dt} = 0$$

$$\dot{m}_{vap}\tilde{H}_{vap}(T_b) + \frac{d(m_{liq}\tilde{H}_{liq}(T_b))}{dt} = \dot{q}_{stove}$$

The enthalpy balance is derived for a system with no heat loss. All stove heat goes directly to vaporize the liquid. A simple adjustment in a non-ideal case is to decrease the stove heat value by the expected (constant) heat loss, for example to the surroundings.

The vapor and liquid are at their normal boiling point $T_b$. The change is the vaporization enthalpy. Substitutions give a form as below.

$$\dot{m}_{vap} \Delta_{vap}\tilde{H} = \dot{q}_{stove}$$

Suppose that you need to heat the liquid from a lower temperature and boil it. The simplest adjustment is to neglect vaporization until you reach the normal boiling point. Calculate an initial time needed to heat $\Delta t_o$.

$$m_{liq,o}\tilde{C}_{p,liq} \left(T_b - T_o\right) = \dot{q}_{stove} \Delta t_o$$

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