Certainly! Here’s the **complete derivation** of the transfer function for the 3-tank system, presented as a unified, step-by-step process: --- ### **Step 1: System Overview and Flow Sequence** * The **input flow** enters the **central tank (Tank 1)**. * Liquid moves sequentially from **Tank 1 → Tank 2 (left) → Tank 3 (right)**. * The liquid then exits from Tank 3 as the **output flow** $q_o(t)$. **Flow Direction:** $$ \text{Tank 1} \longrightarrow \text{Tank 2} \longrightarrow \text{Tank 3} \longrightarrow q_o(t) $$ **Definitions:** * $h_1(t), h_2(t), h_3(t)$: liquid levels in tanks 1, 2, and 3 * $C_1, C_2, C_3$: capacitance (cross-sectional area) for each tank * $R_1, R_2, R_3$: flow resistances between tanks and at the outlet --- ### **Step 2: Formulate Mass Balance Equations** For each tank, the general balance is: $$ C \cdot \frac{dh(t)}{dt} = \text{Inflow} - \text{Outflow} $$ and the flow through a resistance is: $$ q = \frac{h_{\text{up}} - h_{\text{down}}}{R} $$ #### Tank 1 (Middle): * **Inflow:** $q(t)$ * **Outflow to Tank 2:** $q_1(t) = \frac{h_1(t) - h_2(t)}{R_1}$ $$ C_1 \frac{dh_1(t)}{dt} = q(t) - \frac{h_1(t) - h_2(t)}{R_1} $$ #### Tank 2 (Left): * **Inflow:** $q_1(t)$ * **Outflow to Tank 3:** $q_2(t) = \frac{h_2(t) - h_3(t)}{R_2}$ $$ C_2 \frac{dh_2(t)}{dt} = \frac{h_1(t) - h_2(t)}{R_1} - \frac{h_2(t) - h_3(t)}{R_2} $$ #### Tank 3 (Right): * **Inflow:** $q_2(t)$ * **Outflow:** $q_o(t) = \frac{h_3(t)}{R_3}$ $$ C_3 \frac{dh_3(t)}{dt} = \frac{h_2(t) - h_3(t)}{R_2} - \frac{h_3(t)}{R_3} $$ --- ### **Step 3: Apply the Laplace Transform** Let: $$ \frac{dh(t)}{dt} \rightarrow sH(s), \quad q(t) \rightarrow Q(s) $$ After Laplace transformation: #### Tank 1: $$ C_1 s H_1(s) = Q(s) - \frac{H_1(s) - H_2(s)}{R_1} $$ #### Tank 2: $$ C_2 s H_2(s) = \frac{H_1(s) - H_2(s)}{R_1} - \frac{H_2(s) - H_3(s)}{R_2} $$ #### Tank 3: $$ C_3 s H_3(s) = \frac{H_2(s) - H_3(s)}{R_2} - \frac{H_3(s)}{R_3} $$ #### Output Flow: $$ Q_o(s) = \frac{H_3(s)}{R_3} \implies H_3(s) = R_3 Q_o(s) $$ --- ### **Step 4: Assume Identical Parameters for Simplicity** Assume all tanks and resistances are identical: * $R_1 = R_2 = R_3 = R$ * $C_1 = C_2 = C_3 = C$ Then, the **time constant** is: $$ \tau = RC $$ --- ### **Step 5: Recognize Series of First-Order Systems** Each tank behaves as a first-order system: $$ \frac{\text{Output}}{\text{Input}} = \frac{1}{\tau s + 1} $$ With the tanks connected in **series**, their transfer functions multiply: $$ \frac{Q_o(s)}{Q(s)} = \left[\frac{1}{\tau s + 1}\right]^3 = \frac{1}{(RCs + 1)^3} $$ --- ### **Final Transfer Function:** $$ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{1}{(RCs + 1)^3} } $$ This is the **overall transfer function** for the 3-tank system assuming all tanks and resistances are identical.