An unsteady state balance differs from a steady state one by retaining
Answer: A
At steady state nothing changes with time and accumulation vanishes, which is what reduces the balance to input equals output. Retaining it turns an algebraic equation into a differential one. Time then becomes an explicit variable.
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Assertion (A): An unsteady state balance is a differential equation rather than an algebraic one.
Reason (R): The accumulation term is the rate of change of the inventory with time.
Answer: B
Both are true and the reason gives the reason for the change of character: a derivative appears, and solving the problem means integrating with an initial condition rather than merely rearranging. The answer is a function of time rather than a number.
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The accumulation term in a material balance is written as
Answer: C
It is a rate, so its units match those of the flow terms it is balanced against. Writing the inventory itself rather than its derivative is the commonest mistake in setting these problems up. A units check catches it at once.
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A tank is filled at 10 litres per minute and drained at 4 litres per minute. At what rate does its contents accumulate?
Answer: D
Correct answer: 6 litres per minute
\[ \frac{\mathrm{d}V}{\mathrm{d}t} = \dot{V}_{in} - \dot{V}_{out} = 10 - 4 = 6\ \mathrm{L/min}. \] Accumulation is what is left over when the outflow is subtracted from the inflow, and a positive value means the tank is filling. Adding the two rates gives the 14 offered, which would describe two streams entering rather than one entering and one leaving.
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State whether the following is true or false.
A tank whose level is constant is necessarily at steady state in every respect.
Answer: A
The total mass may be constant while the composition inside still changes with time, as in a washout. Steady state must hold for every species and for energy, not merely for the total inventory. A washout is exactly such a case.
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The well-mixed assumption means that
Answer: B
It is what allows the outlet stream to be written in terms of the single unknown inventory, which is what makes the differential equation solvable. Without it the vessel would need a spatial description as well as a temporal one.
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Assertion (A): The well-mixed assumption is what makes a stirred tank problem tractable.
Reason (R): It lets the outlet concentration be replaced by the concentration inside the tank.
Answer: C
Both are true and the reason gives the mechanism: one variable then describes the whole vessel, and the balance becomes an ordinary differential equation instead of a partial one in space and time. Good agitation is what justifies the assumption.
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A well-mixed tank holds 100 litres and is fed pure water at 5 litres per minute, with the same flow leaving. What is its time constant?
Answer: D
The time constant is the volume divided by the volumetric flow, that is a hundred over five. It is also called the residence time and sets how quickly the tank responds to any change at its inlet. Three of them cover most of any transient.
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Consider the following statements about the time constant of a stirred tank.
1. It equals the volume divided by the volumetric flow rate.
2. After one time constant about 63 per cent of a step change has occurred.
3. A larger tank at the same flow responds more quickly.
Answer: A
Correct answer: 1 and 2 only
Statements 1 and 2 are correct: \( \tau = V/\dot{V} \), and the exponential response reaches \( 1-e^{-1} = 63 \) per cent of a step change in one time constant. Statement 3 is inverted - a larger tank at the same flow has a longer time constant and responds more *slowly*, which is precisely why a surge vessel is fitted to damp fluctuations.
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A well-mixed 200 litre tank contains a salt solution and is flushed with pure water at 10 litres per minute. What fraction of the original salt remains after 20 minutes?
Answer: B
Twenty minutes is exactly one time constant, and the exponential decay leaves the reciprocal of e, which is about thirty-seven per cent. The decay never reaches zero, which is why a washout is specified to a tolerance.
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