Single well-mixed tank (first-order) – transfer functions to inlet flow rate In a standard constant-area, linear-resistance single-tank system, the transfer function of both (i) outlet flow rate and (ii) liquid level to the inlet flow rate has the form 1/(T S + 1). Which option is correct?

Introduction / Context:The single-tank liquid system is the archetypal first-order process in process control. With linear flow resistance at the outlet and small perturbations, both the level and the outlet flow respond to inlet flow changes through identical first-order dynamics with the same time constant T.

Given Data / Assumptions:

• Tank cross-sectional area is constant.
• Outlet obeys linear resistance: q_out ∝ head (level).
• Small-signal linearisation is valid around an operating point.
• Transfer functions are with respect to inlet flow as input.

Concept / Approach:Mass balance gives A dh/dt = q_in − q_out. Linear resistance gives q_out = k h in deviation form. Taking Laplace transforms yields H(s)/Q_in(s) = 1/(T s + 1), where T = A/k. Because q_out = k h, the outlet flow also follows first-order dynamics with the same T relative to q_in, i.e., Q_out(s)/Q_in(s) = k * H(s)/Q_in(s) = (k)/(T s + 1). After normalisation, its dynamic form is also 1/(T s + 1) in shape.

Step-by-Step Solution:

Verification / Alternative check:The step responses of level and outlet flow to inlet flow are exponential with time constant T; only the steady-state gains differ by k scaling.

Why Other Options Are Wrong:

• Outlet only / Level only: Each alone is incomplete; both share first-order form.
• Neither: Contradicts the standard derivation.

Common Pitfalls:Confusing dynamic shape with static gain; q_out has a different gain but identical pole location.

Final Answer:Both (a) & (b)