Botswana International University of Science and

Technology

College of Engineering and Technology

CHEE 210 – PROCESS ENGINEERING I

COURSE CO-ORDINATOR: Dr. TA Mamvura

1st SEMESTER ASSIGNMENT MEMO

Due date: 30 NOVEMBER 2020

Total Marks: 165

INSTRUCTIONS
1. Answer all questions, and clearly number your solutions using the same

numbering format as in the question.

2. Answer in groups of 4 students per group

3. Submit in my office deadline is 12:00 pm on Monday 30 November 2020. Late

submissions will have marks deducted.

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QUESTION 1

Assume that you add 1.2 kg of ethylene glycol (HOCH2CH2OH) as an anti-freeze to

4.0 kg of water in the radiator of your car. Anti-freeze is a chemical additive that lowers

the freezing point of a water-based liquid. What is the molar composition of the

solution?

Solution:

Molecular mass for ethylene:

𝑀𝑟 = 1 + 16 + 12 + 2 × 1 + 12 + 2 × 1 + 16 + 1 = 62 g/mol (1/2)

Moles of ethylene:

𝑚 1200
𝑛= = = 19.35 mol (1/2)
𝑀𝑟 62

Molecular mass for water: 18 g/mol

Moles of water:

𝑚 4000
𝑛 = 𝑀𝑟 = = 222.22 mol (1)
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Total moles:

𝑛𝑇 = 19.35 + 222.22 = 241.57 moles (1)

Mol fraction of ethylene:

19.35
= 241.57 𝑥100% = 8.01% (1)

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Mole fraction of water:

222.22
= 241.57 𝑥100% = 91.99% (1)

QUESTION 2

Most processes for producing high-energy-content gas from coal include some type

of gasification step to make hydrogen. Pressure gasification is preferred because of

its greater yield of methane and higher rate of gasification. Given that a 50 kg test runs

of gas has an average content of 10% H2, 40% CH4, 30% CO, and 20% CO2:

a) Estimate the average molecular weight of the gas.

b) Determine the moles of the gas in the test run.

Solution:

a) Average molecular mass:

Molecular mass for H2 = 2 g/mol (1)

Molecular mass for CH4 = 16 g/mol (1)

Molecular mass for CO = 28 g/mol (1)

Molecular mass for CO2 = 44 g/mol (1)

Average molecular mass:

𝑀𝑟𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 0.1(2) + 0.4(16) + 0.3(28) + 0.2(44) (2)

𝑀𝑟𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 23.8 g/mol (1)

b) Moles in test run:

𝑚 50
𝑛 = 𝑀𝑟 = 23.8 = 2.10 kmols (3)
𝑎𝑣𝑒𝑟𝑎𝑔𝑒

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QUESTION 3

A Batswana royal treasure was found 40.0 m below the surface of the ocean. A group

of process engineering 1 student divers wanted to secure the treasure. However,

before diving deep into the sea, they need to know the pressure they will be dealing

with so they can prepare the proper equipment.

What is the pressure 40.0 m below the surface of the lake (in N/m) and in

atmospheres?

They know that the atmospheric pressure is 10.4 m H2O and the density of water is

63.0 lbm/ft3.

Use acceleration due to gravity as 9.81 m/s2.

Solution:

Convert density to SI units:

63 lbm 453.6 g 35.31 ft3 1 kg

ft3 1 lbm 1 m3 1000 g

(3)

Therefore:

63(453.6)(35.31)
Density = = 1009.05 kg/m3 (1)
1000

Pressure below the surface:

𝑃 = 𝑃𝑎𝑡𝑚 + 𝜌𝑔ℎ (1)

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But:

𝑃𝑎𝑡𝑚 = 𝜌𝑔ℎ (1)

Therefore:

𝑃 = 1009.05(9.81)(10.4) + 1009.05(9.81)(40) (2)

𝑃 = 498898.54 Pa (1)

𝑃 = 4.92 atm (1)

QUESTION 4

During the semester break, Thato, a 2nd year BIUST chemical engineering student

managed to get a temporary job at a plant that manufacture isotonic drinks. Isotonic

drink has the same concentration as the solute concentration in a human cell. As a

result, when we sweat a lot (e.g. after sports activity or heavy physical activity),

drinking isotonic drink can help us to quickly replace the fluid lost.

While working in the quality control unit, Thato learned that the isotonic drink should

contain 0.154 mol/litre of sodium chloride and 6 to 8% of glucose by mass of isotonic

drink. The quality of the product is checked using a densitometer; to pass quality

control, the product specific gravity should be between 1.07 and 1.097. The drink is

made in a continuous process, where glucose and sodium chloride (both in powder

form), and water are added continuously in a vessel to produce the isotonic drink.

One day, a new production supervisor came in and changed the settings of the flow

rates of glucose, sodium chloride and water to those shown in Figure 1. Given the new

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settings, will the isotonic drink produced meet the product quality requirement? Justify

your answer. (5)

Figure 1: New setting of isotonic drink production flow rates

Solution:

Basis: 1 minute

Overall mass balance:

Accumulation = In – Out + Generation – Consumed

But there is no reaction and accumulation, so: (1)

In = Out (2)

Use density of water as 1 g/cm3 = 1000 kg/m3 (1)

For water:

100 L 1 m3 1000 kg

1000 L m3

(1)

100(1000)
Mass of water = = 100 kg (1)
1000

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Molecular mass of NaCl: 𝑀𝑟 = 23 + 35.5 = 58.5 g/mol (1)

Mass of NaCl: 𝑚 = 𝑛𝑀𝑟 = 15.4(58.5) = 900.9 g = 0.9009 kg (1)

Overall balance:

8 + 0.9009 + 100 = 𝑂𝑢𝑡 (1)

Therefore: 𝑂𝑢𝑡 = 108.9009 g (1)

Density of product:

108.9009 kg 1 1000 L

100 L m3

(1)

108.9009(1000)
𝜌𝑝𝑟𝑜𝑑𝑢𝑐𝑡 = = 1089.009 kg/m3 (1)
100

Specific gravity of product:

𝜌𝑝𝑟𝑜𝑑𝑢𝑐𝑡 10089.009
𝑆𝐺 = = = 1.089 (1)
𝜌𝑤𝑎𝑡𝑒𝑟 1000

SG is within the required range of 1.07 to 1.097 (1)

Mass fraction for glucose:

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 8
= 108.9009 = 0.073 = 7.3% w/w (1)

Glucose is within the required range of 6 to 8% w/w (1)

Concentration of glucose:

𝑛 15.4
𝐶=𝑉= = 0.154 mol/L (1)
100

NaCl concentration is the required concentration (1)

Yes, the isotonic drink produced will meet the product quality required (2)

QUESTION 5

Convert the temperatures in parts (a) and (b) and temperature intervals in parts (c)
and (d) showing all your calculation steps:

a) T = 85oF to oR, oC, K

b) T = -10oC to K, oF, oR
c) T = 85oC to K, oF, oR
d) T = 150oR to oF, oC, K

Solution:

Temperature conversion:

a) T in oR

𝑡(𝑜 𝐹) = 𝑇(𝑅) − 459.67

85 = 𝑇(𝑅) − 459.67 and 𝑇 = 544.67oR (2)

T in oC

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 𝑡(𝑜 𝐹) = 1.8𝑡(𝑜 𝐶) + 32

85 = 1.8𝑇 + 32 and 𝑇 = 29.4oC (2)

T in K:

𝑡(𝑜 𝐶) = 𝑇(𝐾) − 273.15

29.4 = 𝑇 − 273.15 and 𝑇 = 302.55 K (2)

b) T in K

𝑡(𝑜 𝐶) = 𝑇(𝐾) − 273.15

−10 = 𝑇 − 273.15 and 𝑇 = 263.15 K (2)

T in oF

𝑡(𝑜 𝐹) = 1.8𝑡(𝑜 𝐶) + 32

𝑇 = 1.8(−10) + 32 and 𝑇 = 14oF (2)

T in oR

𝑡(𝑜 𝐹) = 𝑇(𝑅) − 459.67

𝑇 = 14 + 459.67 and 𝑇 = 473.67oR (2)

c) T in K

𝑡(𝑜 𝐶) = 𝑇(𝐾)

∆𝑇 = 85 K (1)

T in oF

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 𝑡(𝑜 𝐹) = 1.8𝑡(𝑜 𝐶)

∆𝑇 = 1.8(85) = 153oF (2)

T in oR

𝑡(𝑜 𝐹) = 𝑇(𝑅)

∆𝑇 = 183oR (1)

d) T in oF

𝑡(𝑜 𝐹) = 𝑇(𝑅)

∆𝑇 = 150oF (1)

T in oC

𝑡(𝑜 𝐹) = 1.8𝑡(𝑜 𝐶)

150 = 1.8𝑇 and ∆𝑇 = 83.3oC (2)

T in K

𝑡(𝑜 𝐶) = 𝑇(𝐾)

∆𝑇 = 83.3 K (1)

QUESTION 6

It is required to prepare 1250 kg of a solution composed of 12% w/w ethanol and 88%

w/w water. Two solutions are available, the first contains 5% w/w ethanol, and the

second contains 25% w/w ethanol. How much of each solution are mixed to prepare

the desired solution?

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Solution:

Overall mass balance:

Accumulation = In – Out + Generation – Consumed

But there is no reaction and accumulation, so: (1)

In = Out (1)

The diagram for both cases is as follows:

(2)

Overall balance for 1st case:

𝐸𝑡ℎ𝑎𝑛𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛(𝐴) + 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝐵) = 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛(𝑀) (2)

Balance on ethanol:

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0.05𝐴 + 0.25𝐵 = 0.12𝑀 (2)

And:

0.05𝐴 + 0.25𝐵 = 0.12(1250) and 𝐴 + 5𝐵 = 3000 kg eqn 1 (2)

Balance on water:

0.95𝐴 + 0.75𝐵 = 0.88𝑀 (2)

0.95𝐴 + 0.75𝐵 = 0.88(1250) and 19𝐴 + 15𝐵 = 22000 kg eqn 2 (2)

Making A subject of the formula in (1) and substituting in (2), we have:

19(3000 − 5𝐵) + 15𝐵 = 22000 (2)

Solving for B:

22000−57000
𝐵= = 437.5 kg (2)
−95+15

Substituting for B in (1):

𝐴 = 3000 − 5(437.5) = 812.5 kg (2)

QUESTION 7

A gaseous mixture (F) consisting of 16 mol% CS2 and 84 mol% air is fed to the

absorption column at a rate of 1000 Ibmole/hr. Most of the CS2 input is absorbed by

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liquid benzene (L) which is fed to the top of the column. 1% of benzene input is

evaporated and goes out with the exit gas stream which consists of 96 mol% air, 2

mol% CS2 and 2 mol% benzene. The product liquid stream (P) consists of benzene

and CS2.

a) Draw a diagram to represent the situation.

b) Calculate the mole flow rates of (G), (L) and (P) and their compositions.

Solution:

a) Diagram to represent the situation:

(5)

b) Mole flow rates of (G), (L) and (P) and their compositions:

Basis: 1 hour (1)

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Balance on Air:

0.84𝐹 = 0.96𝐺 (2)

Substituting:

0.84(1000) = 0.96𝐺 and 𝐺 = 875 lbmole (2)

Balance on Benzene:

Benzene in G: = 0.02(875) = 17.5 lbmole (2)

1% Benzene is lost in G, so:

0.01𝐿 = 17.5 and 𝐿 = 1750 lbmole (2)

99% Benzene goes to P, so:

Benzene in P = 0.99𝐿 = 0.99(1750) = 1732.5 lbmole (2)

Let x be mole fraction of benzene in P, so:

𝑥𝑃 = 1732.5 eqn 1 (2)

Balance on CS2:

0.16𝐹 = 0.02𝐺 + (1 − 𝑥)𝑃

And:

0.16(1000) = 0.02(875) + 𝑃 − 𝑥𝑃 eqn 2 (2)

Solving simultaneously to get x and P by substituting for xP from (1) into (2):

160 = 17.5 + 𝑃 − 1732.5 and 𝑃 = 1875 lbmole (2)

And from (1):

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 1732.5
𝑥𝑃 = 1732.5 and 𝑥= = 92.4 mol% (2)
1875

Mole fraction of CS2 in P:

= 1 − 0.924 = 7.6 mol% (1)

Checking the final answers:

In: 𝐹 + 𝐿 = 1000 + 1750 = 2750 lbmole/hr

Out: 𝐺 + 𝑃 = 875 + 1875 = 2750 lbmole/hr

QUESTION 8

A labelled flowchart of a continuous steady-state two-unit process is shown below.

Each stream contains two components A and B in different proportions. Three streams

whose flowrates and/or compositions are not known are labelled 1, 2 and 3.

Calculate the unknown flowrates and compositions of streams 1, 2 and 3.

Solution:

Basis: 1 hour (1)

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The boundaries used are shown below:

Acc = Mass in – Mass out + Maa generated – Mass consumed (1)

However, there is no accumulation and no chemical reactions, so:

Mass in = Mass out (1)

Consider outer boundary: (1)

Overall balance:

10 + 30 = 40 + 30 + 𝑚̇3 (2)

Therefore:

𝑚̇3 = 60 kg (1)

Balance on A:

0.5(100) + 0.3(30) = 0.9(40) + 0.6(30) + 𝑥3 (60) (2)

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Therefore:

𝑥3 = 0.0833 (1)

Consider the boundary for the first unit: (1)

Overall balance:

100 = 40 + 𝑚̇1 (2)

Therefore:

𝑚̇1 = 60 kg (1)

Balance on A:

0.5(100) = 0.9(40) + 𝑥1 (60) (2)

Therefore:

𝑥1 = 0.233 (1)

Consider boundary on mixing point: (1)

Overall balance:

60 + 30 = 𝑚̇2 (2)

Therefore:

𝑚̇2 = 90 kg (1)

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Balance on A:

0.233(60) + 0.3(30) = 𝑥2 (90) (2)

Therefore:

𝑥2 = 0.255 (1)

QUESTION 9

Methane is burned with air in a continuous steady-state combustion reactor to yield a

mixture of carbon monoxide, carbon dioxide and water. The reactions taking place are:

3
𝐶𝐻4 + 2 𝑂2 → 𝐶𝑂 + 2𝐻2 𝑂 (1)

𝐶𝐻4 + 2𝑂2 → 𝐶𝑂2 + 2𝐻2 𝑂 (2)

The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2 and 72.8 mol% N2.

The percentage conversion of methane is 90% and the gas leaving the reactor

contains 8 mol CO2/mol CO.

(a) Draw a diagram to represent the situation

(b) Calculate the molar composition of the product stream.

Solution:

(a) Diagram of the situation:

(4)

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(b) Molar compositions and extent of reactions:

Basis: 100 mol Feed (1)

Acc = Mass in – Mass out + Mass generated – Mass consumed

But there is no accumulation, so:

Mass in + Mass generated = Mass out + Mass consumed (1)

Moles in + Moles generated = Moles out + Mols consumed

Do a CH4 balance:

This is used up but not generated, so:

Moles in = Moles out + Moles consumed (1)

From reaction 1: 1 mol CH4 makes 1 mol CO (1)

From reaction 2: 1 mol CH4 makes 1 mol CO2 (1)

7.80%(100) = 0.78 + 𝑛𝐶𝐻4,𝑟𝑥𝑛 1 + 𝑛𝐶𝐻4, 𝑟𝑥𝑛2

7.80%(100) = 0.78 + 𝑛𝐶𝑂 + 𝑛𝐶𝑂2 eqn 1 (1)

But:

𝑛𝐶𝑂2 = 8𝑛𝐶𝑂 eqn 2

So:

7.80 − 0.78 = 𝑛𝐶𝑂 + 8𝑛𝐶𝑂 (1)

And:

𝑛𝐶𝑂 = 0.78 mols (1)

19
From eqn 2:

𝑛𝐶𝑂2 = 8(0.78) = 6.24 mols (1)

Do a H2O balance:

This is generated, so:

Moles generated = Moles out (1)

From reaction 1: 1 mol CH4 makes 2 mols H2O (1)

From reaction 2: 1 mol CH4 makes 2 mols H2O (1)

𝑛𝑔𝑒𝑛 = 2(0.78) + 2(6.24) = 14 mols (2)

Do an O2 balance:

This is consumed, so:

Moles in = Moles out + Moles consumed (1)

From reaction 1: 1 mol CH4 requires 1.5 mols O2 (1)

From reaction 2: 1 mol CH4 requires 2 mols O2 (1)

19.4%(100) = 𝑛𝑂2, 𝑜𝑢𝑡 + 1.5(0.78) + 2(6.24) (2)

𝑛𝑂2, 𝑜𝑢𝑡 = 5.75 mols (1)

In summary, the stack gas contains:

0.78 mols CH4; 0.78 mols CO; 6.24 mols CO2; 14 mols H2O; 5.75 mols O2 and 72.8

mols N2

0.78% CH4; 0.78% CO; 6.2% CO2; 14% H2O; 5.75% O2 and 72.5% N2 (6)

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TOTAL MARKS: 165
© BIUST 2020

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