Let's break down the problem step by step: --- ## (a) Differential Equation for Water Height Decrease Let: - \( H \): Height of water at time \( t \) - \( Q_i \): Inlet volumetric flow rate - \( Q_o \): Outlet volumetric flow rate - \( A_o \): Outlet area - \( R \): Radius of sphere - \( V(H) \): Volume of water in the tank at height \( H \) ### **Step 1: Volume of Water at Height \( H \)** For a sphere of radius \( R \), the volume of liquid up to height \( H \) from the bottom is: \[ V(H) = \pi H^2 \left( R - \frac{H}{3} \right) \] This comes from the geometry of a spherical cap. ### **Step 2: Mass Conservation** \[ \frac{dV}{dt} = Q_i - Q_o \] ### **Step 3: Outlet Flow Rate (Torricelli's Law)** \[ Q_o = A_o \sqrt{2gH} \] where \( g \) is the acceleration due to gravity. ### **Step 4: Chain Rule for \( \frac{dV}{dt} \)** \[ \frac{dV}{dt} = \frac{dV}{dH} \cdot \frac{dH}{dt} \] Calculate \( \frac{dV}{dH} \): \[ \frac{dV}{dH} = \frac{d}{dH} \left[ \pi H^2 \left( R - \frac{H}{3} \right) \right] = \pi \left[ 2H \left( R - \frac{H}{3} \right) + H^2 \left( -\frac{1}{3} \right) \right] = \pi \left( 2RH - \frac{2}{3} H^2 - \frac{1}{3} H^2 \right) = \pi \left( 2RH - H^2 \right) \] ### **Step 5: Final Differential Equation** \[ \frac{dV}{dt} = \frac{dV}{dH} \cdot \frac{dH}{dt} = Q_i - A_o\sqrt{2gH} \] So, \[ \pi (2RH - H^2) \frac{dH}{dt} = Q_i - A_o\sqrt{2gH} \] --- ## (b) Relation Between Time and Height of Water (\( Q_i = \)) Set \( Q_i = \): \[ \pi (2RH - H^2) \frac{dH}{dt} = -A_o\sqrt{2gH} \] \[ (2RH - H^2) \frac{dH}{dt} = -\frac{A_o}{\pi}\sqrt{2gH} \] \[ \frac{dH}{dt} = -\frac{A_o}{\pi (2RH - H^2)}\sqrt{2gH} \] Rearrange for separation of variables: \[ (2RH - H^2) dH = -A \sqrt{H} dt, \quad \text{where} \quad A = \frac{A_o \sqrt{2g}}{\pi} \] Integrate both sides: \[ \int_{H_}^{H} (2RH - H^2) H^{-1/2} dH = -A \int_^t dt \] Expand the integrand: \[ (2RH - H^2) H^{-1/2} = 2R H^{1/2} - H^{3/2} \] Integrate: \[ \int (2R H^{1/2} - H^{3/2}) dH = 2R \int H^{1/2} dH - \int H^{3/2} dH = 2R \cdot \frac{2}{3} H^{3/2} - \frac{2}{5} H^{5/2} = \frac{4R}{3} H^{3/2} - \frac{2}{5} H^{5/2} \] So, \[ \left[ \frac{4R}{3} H^{3/2} - \frac{2}{5} H^{5/2} \right]_{H_}^{H} = -A t \] \[ \frac{4R}{3} (H^{3/2} - H_^{3/2}) - \frac{2}{5} (H^{5/2} - H_^{5/2}) = -A t \] --- ## (c) Time to Empty the Tank (\( Q_i = \)) To find time to empty (\( H = \)), set \( H = \): \[ \frac{4R}{3} ( - H_^{3/2}) - \frac{2}{5} ( - H_^{5/2}) = -A t_{empty} \] \[ -\frac{4R}{3} H_^{3/2} + \frac{2}{5} H_^{5/2} = -A t_{empty} \] \[ \frac{4R}{3} H_^{3/2} - \frac{2}{5} H_^{5/2} = A t_{empty} \] Recall \( A = \frac{A_o \sqrt{2g}}{\pi} \): \[ t_{empty} = \frac{1}{A} \left( \frac{4R}{3} H_^{3/2} - \frac{2}{5} H_^{5/2} \right) \] \[ t_{empty} = \frac{\pi}{A_o \sqrt{2g}} \left( \frac{4R}{3} H_^{3/2} - \frac{2}{5} H_^{5/2} \right) \] --- ## **Summary of Final Answers** ### (a) **Differential Equation:** \[ \boxed{ \pi (2RH - H^2) \frac{dH}{dt} = Q_i - A_o\sqrt{2gH} } \] --- ### (b) **Time-Height Relation (\( Q_i = \)):** \[ \boxed{ \frac{4R}{3} (H^{3/2} - H_^{3/2}) - \frac{2}{5} (H^{5/2} - H_^{5/2}) = -\frac{A_o \sqrt{2g}}{\pi} t } \] --- ### (c) **Time to Empty (\( Q_i = \)):** \[ \boxed{ t_{empty} = \frac{\pi}{A_o \sqrt{2g}} \left( \frac{4R}{3} H_^{3/2} - \frac{2}{5} H_^{5/2} \right) } \]

# Step-by-Step Solution ## Problem Overview We have a spherical vessel with water that has an outlet at the bottom. The flow rates of water entering and exiting the vessel influence the height of the water. We need to derive a differential equation for the height of water, find the relation between time and height when the inlet flow is zero, and determine the time required to empty the tank. ## (a) Differential Equation for Water Height Decrease ### Definitions - \( H \): Height of water at time \( t \) - \( Q_i \): Inlet volumetric flow rate - \( Q_o \): Outlet volumetric flow rate - \( A_o \): Outlet area - \( g \): Acceleration due to gravity ### Step 1: Volume of Water at Height \( H \) The volume of water in a spherical cap is given by: \[ V(H) = \pi H^2 \left( R - \frac{H}{3} \right) \] ### Step 2: Mass Conservation Using conservation of mass, we write: \[ \frac{dV}{dt} = Q_i - Q_o \] ### Step 3: Outlet Flow Rate (Torricelli's Law) The outlet flow rate is given by: \[ Q_o = A_o \sqrt{2gH} \] ### Step 4: Chain Rule Applying the chain rule, we can express the change in volume as: \[ \frac{dV}{dt} = \frac{dV}{dH} \cdot \frac{dH}{dt} \] ### Step 5: Derivative of Volume Calculating \( \frac{dV}{dH} \): \[ \frac{dV}{dH} = \pi (2RH - H^2) \] ### Step 6: Final Differential Equation Substituting into the mass conservation equation gives: \[ \pi (2RH - H^2) \frac{dH}{dt} = Q_i - A_o\sqrt{2gH} \] Thus, the differential equation governing the decrease in water height is: \[ \boxed{\pi (2RH - H^2) \frac{dH}{dt} = Q_i - A_o\sqrt{2gH}} \] --- ## (b) Relation Between Time and Height of Water (\( Q_i = 0 \)) Assuming \( Q_i = 0 \): \[ \pi (2RH - H^2) \frac{dH}{dt} = -A_o\sqrt{2gH} \] Rearranging gives: \[ \frac{dH}{dt} = -\frac{A_o}{\pi (2RH - H^2)}\sqrt{2gH} \] Separating variables: \[ (2RH - H^2) dH = -A \sqrt{H} dt, \quad \text{where} \quad A = \frac{A_o \sqrt{2g}}{\pi} \] Integrating both sides leads to: \[ \int (2RH^{1/2} - H^{3/2}) dH = -A \int dt \] The left-hand side integration results in: \[ \left[ \frac{4R}{3} H^{3/2} - \frac{2}{5} H^{5/2} \right]_{H_0}^{H} = -At \] Thus, the relation between time and height is: \[ \boxed{\frac{4R}{3} (H^{3/2} - H_0^{3/2}) - \frac{2}{5} (H^{5/2} - H_0^{5/2}) = -At} \] --- ## (c) Time Required to Empty the Tank (\( Q_i = 0 \)) To find the time to empty the tank when \( H \) reaches \( 0 \): \[ \frac{4R}{3} H_0^{3/2} - \frac{2}{5} H_0^{5/2} = A t_{empty} \] Rearranging gives: \[ t_{empty} = \frac{1}{A} \left( \frac{4R}{3} H_0^{3/2} - \frac{2}{5} H_0^{5/2} \right) \] Substituting \( A \): \[ t_{empty} = \frac{\pi}{A_o \sqrt{2g}} \left( \frac{4R}{3} H_0^{3/2} - \frac{2}{5} H_0^{5/2} \right) \] Thus, the time required to empty the tank is: \[ \boxed{t_{empty} = \frac{\pi}{A_o \sqrt{2g}} \left( \frac{4R}{3} H_0^{3/2} - \frac{2}{5} H_0^{5/2} \right)} \]