40 Separable Differential Equations

Solution

Step 1: Determine what terms in the general balance are zero or negligible.

$IN - OUT + GEN - CON = ACC$

Since there are no feed streams and no reactions (no mass generated or consumed), the $IN$, $GEN$, and $CON$ terms can be omitted from the balance.

$-OUT = ACC$

Step 2: Write out the mass balance for species A in the system.

$(-\dot{m}_{out,A}) = \frac{dM_{A}}{dt}$

Step 3: Expand the mass terms.

Recall that $Mass=Volume*density$, so we can use that to express both terms in the mass balance:

$(-\dot{V}_{out,A}*\rho) = \rho*\frac{dV_{A}}{dt}$

Since the density is constant, the density terms on both sides cancel each other out

$(-\dot{V}_{out,A}) = \frac{dV_{A}}{dt}$

The volume terms can be further expanded. Recall that the volume of a cylinder $Area*Height$ or $A*h$. In differential terms, $\frac{dV}{dt}=A\frac{dh}{dt} + h\frac{dA}{dt}$. Since the area is constant in this case, this simplifies to: $\frac{dV}{dt} = A\frac{dh}{dt}$.

$(-\dot{V}_{out,A}) = A_{tank}*\frac{dh}{dt}$

$A_{tank} = \frac{\pi}{4}*{D_{tank}}^{2}=\frac{\pi}{4}*{2 m}^{2}=3.14 m^{2}$

Step 4: Solve the integral from t = 0 to t = 10 mins

$(-\dot{V}_{out,A})dt = A_{tank}*dh$

$\int^{t=10 min}_{0 min}(-\dot{V}_{out,A})dt =\int^{h_{final}}_{h_{intial}} A_{tank}*dh$

$\int^{t=10 min}_{0 min}(-0.10\frac{m^3}{min})dt =\int^{h_{final}}_{h_{intial}}3.14m^{2}*dh$

$(-0.1\frac{m^{3}}{min})*t\;\bigg|^{t=10 min}_{0 min} = (3.14m^{2})*h\;\bigg|^{h_{final}}_{5m}$

$(-0.1\frac{m^{3}}{min})*(10min-0min) = (3.14m^{2})*(h_{final}-5m)$

$-1 m^{3}*\frac{1}{3.14m^{2}} = h_{final}-5m$

$-0.32 m = h_{final}-5m$

$(-.32+5) m = h_{final}$

$h_{final}=4.68 m$

Solution

Step 1: Determine what terms in the general balance are zero or negligible.

$IN - OUT + GEN - CON = ACC$

Since there is no inlet stream and no chemical reactions in the tank, there is no $IN$, $GEN$ or $CON$ terms:

$-OUT = ACC$

Step 2: Write out the mass balance for species A in the system.

$(-\dot{m}_{out}) = \frac{dM}{dt}$

Step 3: Express each term using the given quantities:

Replace $\text{Mass}$ by $\text{Volume}×\text{density}$ for both terms:

$(-\dot{V}_{out}×\rho) = \frac{dV×\rho}{dt}$

Because $\rho$ are multiplied by both sides and we assume it is constant throughout the system, we can cancel $\rho$ in this step:

$(-\dot{V}_{out}) = \frac{dV}{dt}$

The volume of fluid flowing out of a pipe at any instant can be calculated using $\dot{V}=u×A$, where u is the instantaneous velocity of the fluid and A is the cross-sectional area of the pipe. We can use this to express the outlet flow. For the accumulation term, it is not beneficial to do this because we don't know the expression for $u$.

For the accumulation term, we can espress dV as $A_{1}×dh$ because we know that $A_{1}$ is constant, and this leaves only one variable (h) for us to solve for in the mass balance:

$-u×A_{2} = \frac{A_{1}dh}{dt}$

We can replace $u$ using $u=\sqrt{19.6×h}$ given in the question. T

$-\sqrt{19.6×h}×A_{2} = \frac{A_{1}dh}{dt}$

Step 4: Solve the integral to find $h_{f}$:

$\int_{t_{0}}^{t_{f}}-\sqrt{19.6×h}×A_{2}\;dt= \int_{h_{0}}^{h_{f}}A_{1}dh$

Separate the equation by moving all terms involving h to one side:

$\int_{t_{0}}^{t_{f}}-\frac{A_{2}}{A_{1}}dt = \int_{h_{0}}^{h_{f}}\frac{1}{\sqrt{19.6×h}}dh$

Replace the variables by the given information: $A_{2}=10cm^2=10×10^{-4}m^2$, $A_{1}=1m^2$, $t_{0}=0s$, $t_{f}=5min=300s$.

$$\begin{align*} \int_{0}^{t=300}-\frac{10×10^{-4}}{1} dt&= \int_{h_{0}=1}^{h_{f}}\frac{1}{\sqrt{19.6}}×h^{-\frac{1}{2}}dh \\ -10×10^{-4}× t\;\bigg|_{0}^{t=300}& = \frac{2}{\sqrt{19.6}}×h^{\frac{1}{2}}\;\bigg|_{h_{0}=1}^{h_{f}}\\ -10×10^{-4}×(300-0)& = \frac{2}{\sqrt{19.6}}×h_{f}^{\frac{1}{2}}-\frac{2}{\sqrt{19.6}}×(1)^{\frac{1}{2}}\\ \bigg(\frac{-10×10^{-4}×(300-0)+\frac{2}{\sqrt{19.6}}×(1)^{\frac{1}{2}}}{\frac{2}{\sqrt{19.6}}}\bigg)^2&=h_{f} \\ h_{f}&=0.11m \end{align*}$$