40 Separable Differential Equations
Learning Objectives
By the end of this section, you should be able to:
Provide initial conditions for well-mixed separable transient single-unit processes.
Solve separable transient balances to find a property of a system at any given time.
Separable Differential Equations
Separable differential equations are differential equations where the variables can be isolated to one side of the equation. Take the following differential equations:
1 – [latex]\frac{dx}{dy}=(x^{3}+x)*(y-y^{2})[/latex]
This equation is separable because you can completely isolate the x and y variables as follows:
[latex]\frac{dx}{x^{3}+x}=dy*(y-y^{2})[/latex]
2 – [latex]\frac{dx}{dy}=\frac{x}{x+y}[/latex]
This equation is non-separable because you cannot completely isolate the x and y variables:
[latex]dy*(x+y)=dx*x[/latex]
These non-separable equations we will discuss later.
Example: Chemical Reactor
Consider a “continuous stirred-tank reactor” (CSTR). CSTRs are reactors with continuous feed and exit streams and some kind of mixer.
Say we know the following information about this CSTR:
- The feed enters at a constant volumetric flowrate of [latex]\dot{V}_{0}[/latex] in L/s
- The volume of the tank is [latex]V[/latex] in L.
- Initially ([latex]t=0[/latex]), the tank is filled to [latex]V_i[/latex] in L
- The exit stream flows at a constant rate of [latex]\dot{V}[/latex] in L/s
- We can assume that the density of all streams in the system is constant at [latex]\rho[/latex] in g/L
We want to write a balance for the total (overall) mass in the system under transient conditions.
We start off by writing out the overall balance:
[latex]IN-OUT+GEN-CON=ACC[/latex]
Mass is not being consumed or generated, just changed from one substance to another. This means the [latex]GEN[/latex] and [latex]CON[/latex] terms are negligible. We get:
[latex]IN=\dot{V}_{0}*\rho[/latex]
[latex]OUT=\dot{V}*\rho[/latex]
[latex]ACC=\frac{dM}{dt}=\frac{d(V*\rho)}{dt}=\rho*\frac{dV}{dt}[/latex]
The units for the [latex]IN[/latex], [latex]OUT[/latex], and [latex]ACC[/latex] terms are kg/s. Simplifying the balance, we get:
[latex]\dot{V}_{0}*\rho-\dot{V}*\rho = \rho*\frac{dV}{dt}[/latex]
Since the densities are all constant, we can cancel them out:
[latex]\dot{V}_{0} - \dot{V} = \frac{dV}{dt}[/latex]
Using separation of variables (from calculus), we can integrate both sides. We want to find a given value about our system at a specific final time ([latex]t_{f}[/latex]), starting from an initial time (say [latex]t_{0} = 0[/latex]). We know the initial volume [latex]V_{i}[/latex] and want to find the final volume [latex]V_{f}[/latex].
[latex](\dot{V}_{0} - \dot{V})dt = dV[/latex]
[latex]\int^{t_{f}}_{t_{0}}(\dot{V}_{0} - \dot{V})dt =\int^{V_{f}}_{V_{i}} dV[/latex]
[latex](\dot{V}_{0} - \dot{V})*(t_{f}-t_{0}) =V_{f}-V_{i}[/latex]
[latex]V_{f} = V_{i}+(\dot{V}_{0} - \dot{V})*(t_{f}-t_{0})[/latex]
Let’s try substituting in some numbers to this equation. Say the rate of flow in is