Unsteady-state material and energy balances

Why it matters

Start-ups, shutdowns, batch heating, tank filling and draining, flushing a vessel and responding to a feed upset are all transient. Unsteady-state balances tell you how long a batch takes to heat, how quickly a contaminant is washed out of a tank, or when a level reaches an alarm – and they are the starting point for process dynamics and control.

Key ideas

Accumulation is a rate. For any system, the general balance written per unit time becomes a differential equation: rate of accumulation = rate in − rate out + rate of generation − rate of consumption. At steady state the left-hand side is zero; in unsteady operation it is d(quantity in system)/dt.

Total mass balance. dM/dt = ṁ_in − ṁ_out. With constant flows, M changes linearly with time. A tank with equal inflow and outflow has constant mass even if its composition is changing.

Component balance in a well-mixed tank (CSTR assumption). The outlet has the same composition as the tank contents. For volume V and volumetric flow Q in and out: d(V·C)/dt = Q·C_in − Q·C (+ generation). With constant V and pure solvent entering (C_in = 0), the solution is an exponential washout: C = C₀·exp(−t/τ), where τ = V/Q is the residence (space) time. After one τ the concentration has fallen to 36.8 % of its initial value; after 3τ to about 5 %.

Energy balance. For a liquid-filled tank with no phase change and negligible kinetic/potential energy: d(m·c_p·T)/dt = Σ ṁ_in·c_p·(T_in − T_ref) − ṁ_out·c_p·(T − T_ref) + Q̇ − Ẇ_s. For a closed batch, m·c_p·dT/dt = Q̇.

• Constant heat input: T rises linearly.
• Heating by a coil or jacket at fixed temperature T_s: Q̇ = U·A·(T_s − T), giving an exponential approach to T_s with time constant m·c_p/(U·A).
• Heat loss to surroundings: Q̇ = −U·A·(T − T_amb).

Solving. Write the balance, separate variables, integrate between the initial condition (t = 0) and time t. Only initial conditions are needed for these lumped (well-mixed) models. If flows or volumes vary with time (filling while draining, gravity draining with outflow ∝ √h), carry that dependence into the integration.

Gravity draining. For a tank of cross-section A draining through an orifice, outflow ∝ √h (Torricelli), so A·dh/dt = −k·√h and the time to drain from h₀ to h is t = 2A(√h₀ − √h)/k.

Formulas

• dM/dt = ṁ_in − ṁ_out – kg/s.
• V·dC/dt = Q·(C_in − C) – well-mixed tank, constant V; C in kg/m³, Q in m³/s.
• C = C_in + (C₀ − C_in)·exp(−t/τ) with τ = V / Q – washout or approach to a new inlet concentration.
• t = τ·ln[(C₀ − C_in) / (C − C_in)]
• m·c_p·dT/dt = Q̇ – closed batch; constant Q̇ gives t = m·c_p·(T₂ − T₁) / Q̇.
• T_s − T = (T_s − T₀)·exp(−U·A·t / (m·c_p)) – batch heated by a coil at constant T_s.
• t = (m·c_p / (U·A))·ln[(T_s − T₀) / (T_s − T)]
• t = 2A·(√h₀ − √h) / k – gravity draining with outflow = k·√h.

Worked examples

Example 1 (standard, washout). A well-mixed 10 m³ tank holds brine at 20 kg/m³. Pure water enters at 0.5 m³/min and brine leaves at the same rate. Find the concentration after 30 min and the time to reach 2 kg/m³.

1. Volume is constant (in = out). τ = V/Q = 10 / 0.5 = 20 min.
2. Salt balance: V·dC/dt = −Q·C, so C = C₀·exp(−t/τ).
3. At 30 min: C = 20 × exp(−30/20) = 20 × 0.2231 = 4.46 kg/m³.
4. Time to 2 kg/m³: t = τ·ln(C₀/C) = 20 × ln(20/2) = 20 × 2.3026 = 46.1 min.

Answer: C(30 min) ≈ 4.46 kg/m³; 46.1 min to reach 2 kg/m³

Example 2 (GATE level, coil heating). 2000 kg of oil (c_p = 2.0 kJ/(kg·K)) in a well-stirred tank is heated from 25 °C to 120 °C by a coil with condensing steam at 150 °C; U·A = 2.0 kW/K. Heat losses are negligible. Find the heating time and the steam used (λ = 2113.7 kJ/kg at 150 °C).

1. Energy balance: m·c_p·dT/dt = U·A·(T_s − T).
2. Time constant: m·c_p / (U·A) = (2000 × 2.0 kJ/K) / (2.0 kW/K) = 2000 s.
3. Separate and integrate: t = 2000 × ln[(150 − 25) / (150 − 120)] = 2000 × ln(4.167) = 2000 × 1.4271 = 2854 s.
4. Heat transferred = m·c_p·ΔT = 4000 kJ/K × 95 K = 380 000 kJ.
5. Steam = 380 000 / 2113.7 = 179.8 kg.

Answer: t ≈ 2854 s (47.6 min); steam ≈ 180 kg

Common mistakes

• Setting accumulation to zero in a start-up or batch problem.
• Using the inlet concentration instead of the tank concentration for the outlet of a well-mixed tank.
• Treating V as constant when inflow and outflow differ.
• Using a constant temperature difference (T_s − T₀) for coil heating – the driving force falls as the batch warms.
• Losing the sign when separating variables, giving negative times.
• Mixing minutes and seconds between τ and t.

For GATE CH

Expect NAT questions on washout or concentration change in a stirred tank, time to heat or cool a batch with a coil or jacket, tank filling with unequal flows and gravity draining time. Practise writing the balance as an ODE, separating variables and checking that the answer tends to the right limit as t → ∞.

Quick check

1. In a tank with equal inflow and outflow, does the mass change?
2. A tank has V = 4 m³ and Q = 0.2 m³/min. What is τ?
3. After one time constant of washout, what fraction of the initial concentration remains?
4. What is the time constant of a batch heated by a coil?
5. With constant heat input to a batch, how does T vary with time?

Answers: 1. no (total mass is constant); 2. 20 min; 3. 36.8 %; 4. m·c_p/(U·A); 5. linearly.