Question 2.3: A fuel tank is shown in Figure 2.3. If the tank is given a u...
A fuel tank is shown in Figure 2.3. If the tank is given a uniform acceleration to the right, what will be the pressure at point B?
From equation (2-3) \boldsymbol{\nabla} P=\rho(\mathbf{g}-\mathbf{a})
the pressure gradient is in the g – a direction, therefore the surface of the fluid will be perpendicular to this direction. Choosing the y axis parallel to g – a, we find that equation (2-3) may be integrated between point B and the surface. The pressure gradient becomes d P / d y \mathbf {e}_{y} with the selection of the y axis parallel to g – a as shown in Figure 2.4. Thus,
\frac{d P}{d y} \mathbf{e}_{y}=-\rho|\mathbf{g}-\mathbf{a}| \mathbf{e}_{y}=-\rho \sqrt{g^{2}+a^{2}} \mathbf{e}_{y}Integrating between y = 0 and y = d yields
P_{\mathrm{atm}}-P_{B}=\rho \sqrt{g^{2}+a^{2}}(-d)or
P_{B}-P_{\mathrm{atm}}=\rho \sqrt{g^{2}+a^{2}}(d)The depth of the fluid, d, at point B is determined from the tank geometry and the angle θ.
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