Flash vaporization balance - ODEs, deviation variables, linearization

[halycos]

Homework Statement

Given the attached figure,

a) Develop an ordinary differential equation that describes the dynamic height h(t) in the flash tank in terms of $\dot{m}$$_{i}$, $\dot{m}$$_{l}$,$\dot{m}$$_{v}$, $\rho$$_{i}$, $\rho$$_{l}$, $\rho$$_{v}$, and A.

b) Given the fact that the process is isenthalpic, eliminate $\dot{m}$$_{v}$ from the equation in part (a).

c) Develop a nonlinear ordinary differential equation assuming $\dot{m}_{l}=C_{v}\sqrt{h(t)}$. Simplify as much as possible.

If the operating limit, $\Delta$, is 10 cm. What is the maximal change in $\dot{m}_i$, so steady state is reached before the operating limit is reached?

d) Determine the above using steady state conditions, deviation variables, and linearization.
Values are given for the following variables: $\dot{m}$$_i, T_{in}, P_{in}, T_{out}, P_{out}, 1-(H_i-H_l)/(H_v-H_l), A, \rho_l, C_v$

e) Solve the nonlinear DE, and compare with (d)

Homework Equations

$\dot{m}$$_{i}$-$\dot{m}$$_{l}$-$\dot{m}$$_{v}$=$\dot{m}$$_{acc}$

$m= \rho V = \rho Ah(t)$

For isenthalpic processes, mass fraction vaporized = $Y= (H_i-H_l)/(H_v-H_l)$

The Attempt at a Solution

Were told to include the effect of vapor mass to the height of the vessel liquid, so the equation for a should be

$\dot{m}$$_{i}$-$\dot{m}$$_{l}$-$\dot{m}$$_{v}$=$\rho$$_{l}Ah(t)$

$\frac{dh(t)}{dt}$=$\frac{1}{\rho_l A}($ $\dot{m}$$_{i}$-$\dot{m}$$_{l}$-$\dot{m}$$_{v}$)

For (b), given specific enthalpies, then $\dot{m}_v=Y\dot{m}_i = Y= \dot{m}_i (H_i-H_l)/(H_v-H_l)$.

My issues start at (c). I'm able to find the following equation

$\frac{dh(t)}{dt}+C_v \sqrt{h(t)}/\rho _l A = [m_i(H_i-H_l)/(H_v-H_l)]/ \rho _l A$

The only thing I see that I can simplify is saying that the mass fraction of liquid left unvaporized is $1-(H_i-H_l)/(H_v-H_l)$. Otherwise, I don't see any way of reducing it further. Also, there are no temperature or pressure dependencies, and I am given values for these at steady state to use for parts (d) and (e). I don't even know how to apply the deviation variables without these dependencies.
If anybody can offer some input, I'd be very grateful.

 

[KataKoniK]

Hello there! As a fellow scientist, I would like to offer my thoughts on your solution attempt.

Firstly, I would like to point out that your equation for (a) is incorrect. The equation should be:

\dot{m}_{i}-\dot{m}_{l}-\dot{m}_{v}=\frac{d}{dt}(\rho_l Ah(t))

This is because the dynamic height, h(t), is changing over time and needs to be included in the derivative.

For (b), you are correct in saying that for isenthalpic processes, the mass fraction vaporized can be calculated using Y= (H_i-H_l)/(H_v-H_l). However, to eliminate \dot{m}_v, you need to use the fact that the process is isenthalpic, which means that the total enthalpy is constant. This can be written as:

\dot{m}_i(H_i)+\dot{m}_l(H_l)+\dot{m}_v(H_v)=constant

Since we know that Y= (H_i-H_l)/(H_v-H_l), we can substitute this into the equation and solve for \dot{m}_v, giving us:

\dot{m}_v=\dot{m}_i(H_i-H_l)/(H_v-H_l)-\dot{m}_l

Substituting this into our equation from (a), we get:

\dot{m}_i-\dot{m}_l-(\dot{m}_i(H_i-H_l)/(H_v-H_l)-\dot{m}_l)=\frac{d}{dt}(\rho_l Ah(t))

Simplifying this, we get:

\frac{d}{dt}(\rho_l Ah(t))=\dot{m}_i(H_l-H_i)/(H_v-H_l)

For (c), you are on the right track. However, keep in mind that the mass fraction of liquid left unvaporized is not 1-(H_i-H_l)/(H_v-H_l), but rather 1-Y. So our equation becomes:

\frac{d}{dt}(\rho_l Ah(t))+C_v\sqrt{h(t)}=\dot{m}_i(1-Y)/(\rho_l A)

To simplify this further, you can substitute Y= (H_i-H_l)/(H_v-H_l) and use the fact that \rho_l= \dot