Machine-translated by Google. Questions are authored in English.
Top

Material Balance Without Reaction Questions and Answers

Answer: C

The salt appears in only two streams, so its balance contains the fewest unknowns and often solves directly. Choosing the species with fewest appearances is the general tactic. The water balance then follows from the total.

Please log in or register to report this question.

12.

An evaporator is fed 1000 kilograms per hour of a 5 per cent salt solution and produces a 25 per cent product. How much water is evaporated?

Answer: D

Correct answer: 800 kilograms per hour

The salt is the tie component, since all of it leaves in the product: \[ 1000\times0.05 = 50\ \mathrm{kg/h\ salt}, \qquad \frac{50}{0.25} = 200\ \mathrm{kg/h\ product}, \] so the water evaporated is \( 1000-200 = 800 \) kg/h. Working from the salt rather than the water is what makes this a one-line calculation.

Please log in or register to report this question.

13.

Arrange the steps of solving a material balance problem in the recommended order.
1. count degrees of freedom
2. draw and label a flow diagram
3. write and solve the balances
4. choose a basis of calculation

Answer: A

Correct answer: 2 - 4 - 1 - 3

Draw and label the flowsheet first, because the diagram defines the unknowns. Choose a basis next, since without one the problem has no scale and cannot be solved numerically. Count degrees of freedom then, to confirm it is determinate and to find where to start. Only then write and solve the balances. Working in this order turns most problems into arithmetic.

Please log in or register to report this question.

Answer: B

Its amount is the same on both sides, so the ratio of the two streams follows at once. Ash in a coal problem and nitrogen in a combustion problem are the standard examples. Inert species are the natural place to look for one.

Please log in or register to report this question.

15.

All of the following can serve as a tie component in a drying operation EXCEPT

Answer: C

Water is the species being transferred, so its amount changes between streams and it cannot tie them together. The other three all pass through in unchanged quantity. A tie component must not take part in the transfer.

Please log in or register to report this question.

16.

A wet solid containing 20 per cent water is dried to 5 per cent water. How much water is removed per 100 kilograms of wet feed?

Answer: D

Eighty kilograms of dry solid at ninety-five per cent gives a product of about eighty-four point two kilograms, so about fifteen point eight kilograms of water leaves. Subtracting the percentages directly gives fifteen and is wrong.

Please log in or register to report this question.

17.

Assertion (A): Drying calculations are done on a bone dry basis rather than a wet basis.
Reason (R): The mass of dry solid is unchanged through the drier while the total mass is not.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the advantage. The bone dry solid passes through unchanged, so it serves as a tie component and every stream can be referred to it. Working on a wet basis means both the numerator and the denominator change as water leaves, which makes the algebra needlessly awkward - and it is why moisture content in drying is quoted as kilograms of water per kilogram of dry solid.

Please log in or register to report this question.

18.

A recycle-free absorber removes a solute from a gas into a liquid. The balance that is usually written first is

Answer: B

The carrier gas is inert and appears in only two streams, which makes it the natural tie component. The solute is the species of interest but its balance carries the most unknowns. Solving the easy balance first exposes the harder one.

Please log in or register to report this question.

19.

Three of the following units change stream composition. Which is the odd one out?

Answer: C

A splitter divides a stream without altering what it contains, so its balances are far simpler. Recognising this class of unit early saves writing species balances that carry no information. Its outlets differ only in how much flows through each.

Please log in or register to report this question.

20.

Consider the following statements about a two-product distillation column.
1. Overall and component balances give two independent equations for a binary.
2. Specifying the feed and both product compositions fixes the two flows.
3. The energy balance is needed to find the product flow rates.

Answer: D

The first two are correct and are the basis of the familiar lever rule. The third is wrong for the flows themselves, since the material balances alone determine them; the energy balance is needed for the reboiler and condenser duties.

Please log in or register to report this question.

Loading…
Tags: Material Balance Without Reaction Questions and Answers || Material Balance Without Reaction MCQ Questions and Answers || Material Balance Without Reaction GK Questions and Answers || Material Balance Without Reaction GK MCQ Questions || Material Balance Without Reaction Multiple Choice Questions and Answers || Material Balance Without Reaction GK || GK on Material Balance Without Reaction || Chemical Process Calculations Questions and Answers || Chemical Process Calculations MCQ Questions and Answers || Chemical Process Calculations GK Questions and Answers || GK on Chemical Process Calculations