The gas tank contains a 0.6-m depth of gasoline and a 0.2-m depth of water as shown in Fig. 5–15. Determine the time needed to drain the water if the drain hole has a diameter of 25 mm. The tank is 1.8 m wide and 3.6 m long. The density of gasoline is ρg = 726 kg/m³, and for water, ρw = 1000 kg/m³

Question

The gas tank contains a 0.6-m depth of gasoline and a 0.2-m depth of water as shown in Fig. 5–15. Determine the time needed to drain the water if the drain hole has a diameter of 25 mm. The tank is 1.8 m wide and 3.6 m long. The density of gasoline is [latex] \rho_g=726 kg / m ^3 [/latex], and for water, [latex] \rho_w=1000 kg / m ^3 [/latex].

Accepted Answer

Fluid Description. The gasoline is on top of the water because its density is less than that of water. Because the tank is large relative to the drain hole, we will assume steady flow. Also, we will consider the two fluids to be ideal.

Bernoulli Equation. Here we will select the vertical streamline containing points B and C, Fig. 5–15. At any instant the level of water is h, as measured from the datum, and so the pressure at B is due to the weight of the gasoline above it, that is,

[latex] p_B=\gamma_g h_{A B}=\left(726 kg / m ^3 ight)\left(9.81 m / s ^2 ight)(0.6 m )=4.273\left(10^3 ight) N / m ^2 [/latex]

To simplify the analysis for using the Bernoulli equation, we will neglect the velocity at B since [latex] V_B \approx 0, ext { and so } V_B^2 [/latex] will be even smaller.
Since C is open to the atmosphere, [latex] p_C=0 [/latex]. Thus,

[latex] \begin{gathered} \frac{p_B}{\gamma_w}+\frac{V_B^2}{2 g}+z_B=\frac{p_C}{\gamma_w}+\frac{V_C^2}{2 g}+z_C \ \frac{4.273\left(10^3 ight) N / m ^2}{\left(1000 kg / m ^3 ight)\left(9.81 m / s ^2 ight)}+0+h=0+\frac{V_C^2}{2\left(9.81 m / s ^2 ight)}+0 \end{gathered} [/latex]

[latex] V_C=4.429 \sqrt{h+0.4356} [/latex] (1)

Continuity Equation. The continuity of flow at B and C will allow us to relate the actual nonzero [latex] V_B ext { to } V_C [/latex]. We will choose a control volume that contains all the water up to the depth h. Since [latex] V_B [/latex] is downward and h is positive upward, then at the top control surface, [latex] V_B=-d h / d t [/latex]. Thus,

[latex] \begin{gathered} \frac{\partial}{\partial t} \int_{ cv } ho d \sout{V}+\int_{ cs } ho V _{f / c s} \cdot d A =0 \ 0-V_B A_B+V_C A_C=0 \ 0-\left(-\frac{d h}{d t} ight)[(1.8 m )(3.6 m )]+V_C\left[\pi(0.0125 m )^2 ight]=0 \ \frac{d h}{d t}=-75.752\left(10^{-6} ight) V_C \end{gathered} [/latex]

Now, using Eq. 1,

[latex] \frac{d h}{d t}=-75.752\left(10^{-6} ight)(4.429 \sqrt{h+0.4356}) [/latex]

or

[latex] \frac{d h}{d t}=-0.3355\left(10^{-3} ight) \sqrt{h+0.4356} [/latex]

Notice that when [latex] h=0.2 m , V_B=d h / d t=0.268\left(10^{-3} ight) m / s [/latex], which is very slow compared to [latex] V_C=3.53 m / s [/latex], as determined from Eq. 1, and so indeed we were justified in neglecting [latex]V_B^2[/latex] in Eq. 1.
If [latex] t_d [/latex] is the time needed to drain the tank, then separating the variables in Eq. 2 and integrating, we get

[latex] \begin{aligned} & \int_{0.2 m }^0 \frac{d h}{\sqrt{h+0.4356}}=-0.3355\left(10^{-3} ight) \int_0^{t_d} d t \ & \left.(2 \sqrt{h+0.4356}) ight|_{0.2 m } ^0=-0.3355\left(10^{-3} ight) t_d \end{aligned} [/latex]

Evaluating the limits, we get

[latex] \begin{aligned} -0.2745 & =-0.3355\left(10^{-3} ight) t_d \ t_d & =818.06 s =13.6 min \end{aligned} [/latex]

Question 5.5: The gas tank contains a 0.6-m depth of gasoline and a 0.2-m ...

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