Question 10.2: The open tank shown contains water at 20 ̊ C and is being fi...
The open tank shown contains water at 20 ̊ C and is being filled through section 1 in Figure 10.10. Assume incompressible flow. First derive an analytic expression for the water-level change dh/dt in terms of arbitrary volume flows Q_1, Q_2, Q_3, and tank diameter d. Then, if the water level h is constant, determine the exit velocity V_2 for the given data V_1=3 m/s and Q_3 =0.01 m³/s.
Given: Tank inlet and outlet information.
Find: dh/dt, unknown exit velocity.
Assume: Inlet and outlet velocity profiles are uniform, one dimensional.
Fluid is incompressible; density is uniform.
We intend to consider the fluid in the tank shown in Figure 10.11 as the contents of a control volume.
We must have mass conservation,
0=\frac{∂}{∂t}\int\limits_{CV}{ρ d V}+\int\limits_{CS}{ρ \underline{V}\cdot d\underline{A}},
which for this CV can be written as
\frac{d}{dt}\left[ρ\frac{\pi d^2}{4}h\right]+ρ (Q_2-Q_1-Q_3)=0 ,
where we have changed the partial derivative to a total one, as time is the only dependence of the quantity in brackets, and where the signs on various flow rates depend on whether they are into or out of the CV: Q_2 is outflow, and thus positive, while Q_1 and Q_3 are both inflow and negative. We further simplify by canceling the common density and by removing the constants from the time derivative. We get
\frac{ \pi d^2}{4} \frac{dh}{dt}+(Q_2-Q_1-Q_3)= 0
\frac{dh}{dt} =\frac{4(Q_1+Q_3-Q_2)}{ \pi d^2} .
If h is constant, dh/dt = 0 and we must have Q_1 + Q_3 – Q_2 = 0 . Each Q_i=V_i A_i . We can thus solve for the requested value of V_2 , which which corresponds to dh/dt = 0:
Q_2 = V_2 A_2 = Q_1 + Q_3 = 0.01 \frac{\textrm{m}^3}{s}+\frac{\pi}{4}(0.05 \textrm{ m})^2(3 \frac{\textrm{m}}{\textrm{s}})=0.0159\frac{\textrm{m}^3}{s}
V_2=Q_2/A_2= \frac{0.0159\frac{\textrm{m}^3}{s}}{\frac{\pi}{4}(0.07 \textrm{ m})^2}=4.13 \frac{\textrm{m}}{s} .
