Liquid-level dynamics: for a single well-mixed tank with linear outlet resistance, what is the transfer function from inlet flow rate to liquid level?

Introduction / Context:
Level control loops often involve a single tank with inflow and a gravity-driven outflow through a valve/orifice exhibiting approximately linear resistance around an operating point. Deriving the transfer function is a fundamental modeling exercise in process control.

Given Data / Assumptions:

• Tank cross-sectional area = A (constant).
• Outflow qout ≈ h/R (linearized around an operating point, R is resistance).
• Inflow = qin; state variable = level h.
• Time constant T = AR.

Concept / Approach:
Mass balance on the liquid: A dh/dt = qin − qout. With qout = h/R, the linear ODE becomes A dh/dt + (1/R) h = qin. Taking Laplace transforms and rearranging yields a first-order transfer function from Qin(s) to H(s).

Step-by-Step Solution:
Write dynamic balance: A dh/dt + (1/R) h = qin.Laplace: A s H(s) + (1/R) H(s) = Qin(s).Factor H(s): H(s) [ A s + 1/R ] = Qin(s).Transfer function: H(s)/Qin(s) = 1 / (A s + 1/R) = R / (A R s + 1) = R / (T s + 1).

Verification / Alternative check:
Step inflow of magnitude Δq causes level to approach a new steady state Δh = RΔq with time constant T = AR. This matches the DC gain (R) and dynamics of the derived transfer function.

Why Other Options Are Wrong:

• 1/(T s) or R/(T s): missing the +1 term; would imply an integrator, not a first-order lag.
• 1/(T s + 1): incorrect DC gain (should be R).

Common Pitfalls:
For large deviations, outflow is nonlinear (orifice law q ∝ sqrt(h)); linearization around operating point is required to retain first-order form and constant R.

Final Answer:
R / (T*s + 1)