Is There a Solution to My Pressure Cooker Problem?

[DiogoSousa9630]

Homework Statement

I am working on a thermodynamics problem involving a pressure cooker and have come across two different approaches to solve it. Surprisingly, both methods yield the same result. I am looking for some guidance and clarification on why this is the case, and if one of the methods is more appropriate than the other. Here's the problem:

A pressure cooker has a volume of 6 L and an operating pressure of 75 kPa (relative). Initially, the cooker contains 1 kg of liquid water. Heat is supplied at a rate of 500 W for 30 minutes after the operating pressure has been reached. Assuming the atmospheric pressure is 100 kPa, determine the amount of water contained in the pressure cooker at the end of the process.
----------

Method 1:

The energy change of the closed system (inside the cooker) is described as:

$$u_2 m_2 - u_1 m_1 = Q - m_\text{out} h_\text{out}$$

Where:

- $u_1, u_2$ are the specific internal energies at the initial and
final states, respectively
- $m_1, m_2$ are the masses of water at the initial and final states,
respectively
- $Q$ is the heat supplied to the system
- $m_\text{out}$ is the mass of the water that left the system
- $h_\text{out}$ is the specific enthalpy of the water that left the
system


----------


**Method 2:**

The energy change of the closed system can also be described as:

$$\Delta E = m_\text{out} \Delta h $$

Where:

- $m_\text{out}$ is the mass of the water
- $\Delta h$ is the change in specific enthalpy

Comparing both methods, I found that they yield the same result for the amount of water contained in the pressure cooker at the end of the process.

Relevant Equations

$$u_2 m_2 - u_1 m_1 = Q - m_\text{out} h_\text{out}$$
$$\Delta E = m_\text{out} \Delta h $$

**Question:**

Why do these two different approaches lead to the same result in this case? Is one of them more appropriate or valid than the other? I am looking for an explanation and clarification on this matter.Thank you in advance for your insights.

 

[Chestermiller]

Homework Statement: I am working on a thermodynamics problem involving a pressure cooker and have come across two different approaches to solve it. Surprisingly, both methods yield the same result. I am looking for some guidance and clarification on why this is the case, and if one of the methods is more appropriate than the other. Here's the problem:

A pressure cooker has a volume of 6 L and an operating pressure of 75 kPa (relative). Initially, the cooker contains 1 kg of liquid water. Heat is supplied at a rate of 500 W for 30 minutes after the operating pressure has been reached. Assuming the atmospheric pressure is 100 kPa, determine the amount of water contained in the pressure cooker at the end of the process.
----------

Method 1:

The energy change of the closed system (inside the cooker) is described as:

$$u_2 m_2 - u_1 m_1 = Q - m_\text{out} h_\text{out}$$

Where:

- $u_1, u_2$ are the specific internal energies at the initial and
final states, respectively
- $m_1, m_2$ are the masses of water at the initial and final states,
respectively
- $Q$ is the heat supplied to the system
- $m_\text{out}$ is the mass of the water that left the system
- $h_\text{out}$ is the specific enthalpy of the water that left the
system----------**Method 2:**

The energy change of the closed system can also be described as:

$$\Delta E = m_\text{out} \Delta h $$

Where:

- $m_\text{out}$ is the mass of the water
- $\Delta h$ is the change in specific enthalpy

Comparing both methods, I found that they yield the same result for the amount of water contained in the pressure cooker at the end of the process.
Relevant Equations: $$u_2 m_2 - u_1 m_1 = Q - m_\text{out} h_\text{out}$$
$$\Delta E = m_\text{out} \Delta h $$

**Question:**

Why do these two different approaches lead to the same result in this case? Is one of them more appropriate or valid than the other? I am looking for an explanation and clarification on this matter.Thank you in advance for your insights.

Welcome to Physics Forums!

Your starting equation

$$u_2 m_2 - u_1 m_1 = Q - m_\text{out} h_\text{out}$$ is not the end result of the analysis. It is the starting equation. It represents the application of the open system (control volume) version of the 1st law of thermodynamics to the pressure cooker system. The m's in the equation are the total amounts of water present in the cooker (both liquid and vapor), not just liquid. The u's are the internal energies per unit mass of the cooker contents, and can be expressed as $$u=u_L(1-x)+u_Vx=u_L+(u_V-u_L)x$$where $u_L$ is the internal energy per unit mass the saturated liquid at 175 kPa, $u_V$ is the corresponding internal energy per unit mass of the saturated vapor at 175 kPa, and x is the mass fraction vapor. Another equation which must be satisfied is that the volumes of liquid and vapor in the cooker must add up to the total volume of the cooker:$$m[(1-x)v_L+xv_V]=m[v_L+(v_V-v_L)x]=V$$ where the v's are the saturated volumes. In addition, $m_{out}=m_1-m_2$

OK so far?

 

[DiogoSousa9630]

Yes, thank you! So far so good!

 

[Chestermiller]

OK. Please note also that $h_{out}=h_V=u_V+Pv_V$, where P is the saturation pressure. The next step is to solve for x as a function of m, and then express $x_1$ as a function of $m_1$ and $x_2$ as a function of $m_2$. What do you get? Then substitute these values into the equation for u to get $u_1$ and $u_2$. What do you get?

 

[DiogoSousa9630]

Okay. So this is what I got:
$$x_{1} = \frac{m_{v1}}{m_{v1}+m_{l1}} = \frac{m_{v1}}{m_{1}} $$
$$x_{2} = \frac{m_{v2}}{m_{v2}+m_{l2}} = \frac{m_{v2}}{m_{2}} $$

$$u_{1} = u_{l}+\frac{(u_{v}-u_{l})*m_{v1}}{m_{1}}$$
$$u_{2} = u_{l}+\frac{(u_{v}-u_{l})*m_{v2}}{m_{2}}$$

So now I guess I should replace the u's in the first formula?

 

[Chestermiller]

Okay. So this is what I got:
$$x_{1} = \frac{m_{v1}}{m_{v1}+m_{l1}} = \frac{m_{v1}}{m_{1}} $$
$$x_{2} = \frac{m_{v2}}{m_{v2}+m_{l2}} = \frac{m_{v2}}{m_{2}} $$

$$u_{1} = u_{l}+\frac{(u_{v}-u_{l})*m_{v1}}{m_{1}}$$
$$u_{2} = u_{l}+\frac{(u_{v}-u_{l})*m_{v2}}{m_{2}}$$

So now I guess I should replace the u's in the first formula?

Sorry,, my mistake. What I was looking for was $$x=\frac{\frac{V}{m}-v_L}{v_V-v_L}$$$$x_1=\frac{\frac{V}{m_1}-v_L}{v_V-v_L}$$$$x_2=\frac{\frac{V}{m_2}-v_L}{v_V-v_L}$$

 

[DiogoSousa9630]

Okay so the general formula for the u will be given by:
$$u = u_L + \frac{(u_V - u_L) (V/m - v_L)}{v_V - v_L}$$
What exactly is the next approach? Should I consider the given formula for entalphy and get something similar to this?
$$h = u_L + P v_L + \frac{v - v_L}{v_V - v_L} (u_V - u_L) + \frac{v - v_L}{v_V - v_L} (P v_V - P v_L)$$
Starting from this equation:
$$h = (u_L + P v_L) + x ((u_V + P v_V) - (u_L + P v_L))$$

 

[Chestermiller]

Okay so the general formula for the u will be given by:
$$u = u_L + \frac{(u_V - u_L) (V/m - v_L)}{v_V - v_L}$$
What exactly is the next approach? Should I consider the given formula for entalphy and get something similar to this?
$$h = u_L + P v_L + \frac{v - v_L}{v_V - v_L} (u_V - u_L) + \frac{v - v_L}{v_V - v_L} (P v_V - P v_L)$$
Starting from this equation:
$$h = (u_L + P v_L) + x ((u_V + P v_V) - (u_L + P v_L))$$

No. We are not done with this approach yet. The next step is to write out the equations for $u_1$ and $u_2$, then $m_1u_1$ and $m_2u_2$, and then $m_2u_2-m_1u_1$. What to you get?

Once we are done with this approach, you will recognize immediately how the other approach works.

 

[DiogoSousa9630]

Okay. So after making the substitutions and simplifying I got the following:
$$m_2u_2-m_1u_1 = -m_{out} \frac{(u_Vv_L - u_Lv_V)}{v_V - v_L}$$

With:
$$m_1u_1 = m_1(u_L + \frac{(u_V - u_L) (V/m_1 - v_L)}{v_V - v_L})$$
$$m_2u_2 = m_2(u_L + \frac{(u_V - u_L) (V/m_2 - v_L)}{v_V - v_L})$$

 

[Chestermiller]

Okay. So after making the substitutions and simplifying I got the following:
$$m_2u_2-m_1u_1 = -m_{out} \frac{(u_Vv_L - u_Lv_V)}{v_V - v_L}$$

With:
$$m_1u_1 = m_1(u_L + \frac{(u_V - u_L) (V/m_1 - v_L)}{v_V - v_L})$$
$$m_2u_2 = m_2(u_L + \frac{(u_V - u_L) (V/m_2 - v_L)}{v_V - v_L})$$

Very nice. Now bring the enthalpy term to the left side of the equation and solve for Q, reducing everything on the left to a lcd.

 

[DiogoSousa9630]

Thank you! After doing that I got:
$$
Q = \frac{m_\text{out} \left[u_L v_V + u_V v_V - 2 u_V v_L + P v_V^2 - P v_L v_V\right]}{v_V - v_L}
$$
I'm not sure about this one. I think I might have missed something.

 

[Chestermiller]

Thank you! After doing that I got:
$$
Q = \frac{m_\text{out} \left[u_L v_V + u_V v_V - 2 u_V v_L + P v_V^2 - P v_L v_V\right]}{v_V - v_L}
$$
I'm not sure about this one. I think I might have missed something.

Try the algebra again. I get $$Q=m_{out}[(u_V+Pv_V)-(u_L+Pv_L)]\frac{v_V}{(v_V-v_l)}$$$$=m_{out}\Delta h_{vap}\frac{v_V}{(v_V-v_L)}$$

There is one more step after this I'll lead you through it once you confirm that you have corrected the algebra and we now match.

 

[DiogoSousa9630]

I just realized that I forgot to include the negative sign before the parentheses - my mistake! Now that I've corrected it, I have the same expression as yours.
I suppose that $\Delta h_{vap}\frac{v_V}{(v_V-v_L)}$ will be equivalent to $\Delta h$?

 

[Chestermiller]

I just realized that I forgot to include the negative sign before the parentheses - my mistake! Now that I've corrected it, I have the same expression as yours.
I suppose that $\Delta h_{vap}\frac{v_V}{(v_V-v_L)}$ will be equivalent to $\Delta h$?

No. The thing to do is associate $\frac{v_V}{(v_V-v_L)}$ with $m_{out}$, not with $\Delta h_{vap}$. We will show that $m_{out}\frac{v_V}{(v_V-v_L)}$ is the mass of liquid water in the cooker that evaporates when Q is added.

We know that $m_{out}=m_1-m_2$. Let $m_{L1}$ be the amount of liquid water in the cooker initially and $m_{L2}$ be the amount of liquid water in the cooker finally. We know that $$m_{L1}=m_1(1-x_1)$$ and $$m_{L2}=m_2(1-x_2)$$. Use these equations together with our equations for $x_1$ and $x_2$ to show that $$m_{L1}-m_{L2}=(m_1-m_2)\frac{v_V}{(v_V-v_L)}$$From this, it follows that $$Q=[m_{L1}-m_{L2}]\Delta h_{vap}$$In other words, the amount of liquid water which vaporizes times the heat of vaporization is equal to the heat you have to add.

 

[DiogoSousa9630]

Ah, I finally understand it now. All makes sense. Thank you so much for your guidance.