Let's solve the problem step by step: --- ## **Problem Restatement** A cylindrical tank (diameter \( D_o \), height \( H \)) is filled with water. An orifice (diameter \( D \)) is at the bottom. Find the time required for the tank: - (a) to empty halfway - (b) to empty completely Assume negligible losses (smooth entrance). --- ## **Step 1: Cross-sectional Areas** - **Tank Area:** \( A_{\text{tank}} = \frac{\pi}{4} D_o^2 \) - **Orifice Area:** \( A_{\text{orifice}} = \frac{\pi}{4} D^2 \) --- ## **Step 2: Outflow Velocity (Torricelli’s Law)** At any instant, the water depth above the orifice is \( h \). By Torricelli’s theorem (neglecting velocity of approach & losses): \[ v = \sqrt{2gh} \] where \( g \) is the acceleration due to gravity. --- ## **Step 3: Outflow Rate** \[ Q = A_{\text{orifice}} \cdot v = \frac{\pi}{4} D^2 \sqrt{2gh} \] --- ## **Step 4: Rate of Change of Water Level** The volume of water in the tank at height \( h \): \[ V = A_{\text{tank}} \cdot h \] The rate of decrease of water volume: \[ \frac{dV}{dt} = A_{\text{tank}} \frac{dh}{dt} \] But this equals the negative of the outflow rate: \[ A_{\text{tank}} \frac{dh}{dt} = - Q \] \[ A_{\text{tank}} \frac{dh}{dt} = - \frac{\pi}{4} D^2 \sqrt{2gh} \] \[ \frac{dh}{dt} = - \frac{ \frac{\pi}{4} D^2 \sqrt{2g h} }{ A_{\text{tank}} } \] \[ \frac{dh}{dt} = - \frac{ D^2 }{ D_o^2 } \sqrt{2g h } \] --- ## **Step 5: Separate Variables and Integrate** \[ \frac{dh}{\sqrt{h}} = - \frac{ D^2 }{ D_o^2 } \sqrt{2g} \, dt \] Integrate both sides: - From \( h = H \) to \( h = h_f \) (final height) - From \( t = \) to \( t = t_f \) \[ \int_{H}^{h_f} \frac{dh}{\sqrt{h}} = - \frac{ D^2 }{ D_o^2 } \sqrt{2g} \int_^{t_f} dt \] \[ \left[ 2\sqrt{h} \right]_{H}^{h_f} = - \frac{ D^2 }{ D_o^2 } \sqrt{2g} \cdot t_f \] \[ 2\left(\sqrt{h_f} - \sqrt{H}\right) = - \frac{ D^2 }{ D_o^2 } \sqrt{2g} \cdot t_f \] \[ t_f = \frac{2 D_o^2 }{ D^2 \sqrt{2g} } \left( \sqrt{H} - \sqrt{h_f} \right) \] --- ## **Step 6: Particular Cases** ### (a) **Time to Empty Halfway (\( h_f = H/2 \))** \[ t_{\text{half}} = \frac{2 D_o^2 }{ D^2 \sqrt{2g} } \left( \sqrt{H} - \sqrt{H/2} \right) \] \[ t_{\text{half}} = \frac{2 D_o^2 }{ D^2 \sqrt{2g} } \left( \sqrt{H} - \frac{\sqrt{H}}{\sqrt{2}} \right) \] \[ t_{\text{half}} = \frac{2 D_o^2 \sqrt{H} }{ D^2 \sqrt{2g} } \left( 1 - \frac{1}{\sqrt{2}} \right) \] --- ### (b) **Time to Empty Completely (\( h_f = \))** \[ ttext{empty}} = \frac{2 D_o^2 }{ D^2 \sqrt{2g} } (\sqrt{H} - ) \] \[ t_{\text{empty}} = \frac{2 D_o^2 \sqrt{H} }{ D^2 \sqrt{2g} } \] --- ## **Final Answers** ### **(a) Time to empty halfway:** \[ \boxed{ t_{\text{half}} = \frac{2 D_o^2 \sqrt{H}}{ D^2 \sqrt{2g} } \left( 1 - \frac{1}{\sqrt{2}} \right) } \] ### **(b) Time to empty completely:** \[ \boxed{ t_{\text{empty}} = \frac{2 D_o^2 \sqrt{H}}{ D^2 \sqrt{2g} } } \] --- If you need a numeric example or further explanation, let me know!