Problem 4.18Richard Felder Chemical Process Analysis 3rd ed.

Links in the description!

Standard Deviation of Mass Balance Data Word

Problem

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

a) Taking a basis of 100 kg feed, calculate(i) 𝑤𝑊, the mass fraction of water in the wet sugar leaving the evaporator, and (ii) the ratio (kg H O vaporized/kg wet sugar leaving the evaporator).

b) If 1000 tons/day of wet sugar is fed to the evaporator, how much additional water must be removed from the outlet sugar to dry it completely, and what annual revenue can be expected if dry sugar sells for $. 15/𝑙𝑏𝑚?

c) The evaporator is built to achieve the production rate of part (b), installed, and started up, and the water content of the partially dried sugar is measured on successive days of operation. The results follow. In subsequent runs, the evaporator is to be shut down for maintenance if falls more than three standard deviations from the mean of this series of runs. Calculate the endpoints of this range. Considering the results of parts (a) and (c) together, what can you conclude about the recently installed evaporator?

Day ww

1 0.05132 0.04863 0.054 0.05075 0.05416 0.04987 0.05128 0.04749 0.0511

10 0.0494

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

a) Taking a basis of 100 kg feed, calculate(i) 𝑤𝑊3, the mass fraction of water in the wet sugar leaving the

evaporator, and (ii) the ratio (kg H O vaporized/kg wet sugar leaving the evaporator).

wet sugar

Some water is vaporized

This stream represents gas phase water

We’re assuming this stream is pure water

Water vapor

This stream represents dry sugar.

We’re assuming this stream is mostly dry but there is still some water

Dry sugar

Break it down.

Fill it what you know now…

𝑚1 = 100 𝑘𝑔

𝑤𝑊1= 0.2 𝑘𝑔 𝑊/𝑘𝑔

𝑤𝑠1= 0.8 𝑘𝑔 𝑠/𝑘𝑔

𝑚2 =? ? ? 𝑘𝑔𝑤𝑊2

= 100%𝑚𝑊2

= .85(𝑚𝑊1)

Mass Fraction

𝑤𝑤 =𝑚𝑖

𝑚

𝑚3 =? ? ? 𝑘𝑔𝑤𝑊3

=? ? ?𝑤𝑠3

=? ? ? 𝑘𝑔 𝑠/𝑘𝑔

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

a) Taking a basis of 100 kg feed, calculate(i) 𝑤𝑊3, the mass fraction of water in the wet sugar leaving the

evaporator, and (ii) the ratio (kg H O vaporized/kg wet sugar leaving the evaporator).

wet sugar

Water vapor

Dry sugar

𝑚1 = 100 𝑘𝑔

𝑤𝑊1= 0.2 𝑘𝑔 𝑊/𝑘𝑔

𝑤𝑠1= 0.8 𝑘𝑔 𝑠/𝑘𝑔

𝑚2 =? ? ? 𝑘𝑔𝑤𝑊2

= 100%𝑚𝑊2

= .85(𝑚𝑊1)

𝑚3 =? ? ? 𝑘𝑔𝑤𝑊3

=? ? ?𝑤𝑠3

=? ? ?

Pause and try by yourself…

First thing we can do is solve for 𝑚𝑊2using variables

𝑚𝑊2= .85(𝑚𝑊1

)𝑚𝑊1

= 𝑤𝑊1𝑚1

𝑚𝑊1= 0.2 100

Or just to be clear it can be written as one equation…𝑚𝑊2

= .85( 0.2 100)𝑚𝑊2

= 17 𝑘𝑔 𝑊

Overall Mass BalanceΣmin = Σmout

m1 = m2 + m3100 = 𝑚2 + 𝑚3

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

a) Taking a basis of 100 kg feed, calculate(i) 𝑤𝑊3, the mass fraction of water in the wet sugar leaving the

evaporator, and (ii) the ratio (kg H O vaporized/kg wet sugar leaving the evaporator).

wet sugar

Water vapor

Dry sugar

𝑚1 = 100 𝑘𝑔

𝑤𝑊1= 0.2 𝑘𝑔 𝑊/𝑘𝑔

𝑤𝑠1= 0.8 𝑘𝑔 𝑠/𝑘𝑔

𝑚3 =? ? ? 𝑘𝑔𝑤𝑊3

=? ? ?𝑤𝑠3

=? ? ?

𝑚2 = 17 𝑘𝑔

100 = 𝑚2 + 𝑚3100 = 17 + 𝑚3

𝑚3 = 83 𝑘𝑔

Sugar Mass BalanceΣw𝑆i

min = ΣwSimout

wS1m1 = wS3

m3

80 = wS3m3

Note that this is the amount of sugar in both the feed and product: in = out.

80 = wS3(83 𝑘𝑔)𝑤𝑆3

= .963

Now we can do a whole mass balance on water again or just recall

Σ𝑥𝑖 = 1𝑤𝑊3

= .036

𝑅 =𝑘𝑔 𝑊 𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑒𝑑

𝑘𝑔 𝑤𝑒𝑡 𝑠𝑢𝑔𝑎𝑟 𝑙𝑒𝑎𝑣𝑖𝑛𝑔=

𝑚2

𝑚3=

17

100 − 17

𝑅 = .204𝑛𝑜𝑡𝑒 𝑜𝑛 𝑤𝑜𝑟𝑑𝑖𝑛𝑔

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

a) Taking a basis of 100 kg feed, calculate(i) 𝑤𝑊3, the mass fraction of water in the wet sugar leaving the

evaporator, and (ii) the ratio (kg H O vaporized/kg wet sugar leaving the evaporator).

wet sugar

Water vapor

Dry sugar

𝑚1 = 100 𝑘𝑔

𝑤𝑊1= 0.2 𝑘𝑔 𝑊/𝑘𝑔

𝑤𝑠1= 0.8 𝑘𝑔 𝑠/𝑘𝑔

𝑚2 = 17 𝑘𝑔

𝑚3 = 83 𝑘𝑔𝑤𝑊3

= .036𝑤𝑠3

= .963

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

𝑚1 = 100 𝑘𝑔

𝑤𝑊1= 0.2 𝑘𝑔 𝑊/𝑘𝑔

𝑤𝑠1= 0.8 𝑘𝑔 𝑠/𝑘𝑔

𝑚2 = 17 𝑘𝑔

𝑚3 = 83 𝑘𝑔𝑤𝑊3

= .036𝑤𝑠3

= .963

a) Ta

b) If 1000 tons/day of wet sugar is fed to the evaporator, how much additional water must be removed from the outlet sugar to dry it completely, and what annual revenue can be expected if dry sugar sells for $0.15/lb ?

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

𝑤𝑊1= 0.2 𝑡𝑜𝑛𝑠 𝑊/𝑡𝑜𝑛

𝑤𝑠1= 0.8 𝑡𝑜𝑛𝑠 𝑆/𝑡𝑜𝑛

ሶ𝑚2 =? ? ? 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

ሶ𝑚3 =? ? ? 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦𝑤𝑊3

= .036 𝑡𝑜𝑛𝑠 𝑊/𝑡𝑜𝑛𝑠𝑤𝑠3

= .963 𝑡𝑜𝑛𝑠 𝑆/𝑡𝑜𝑛𝑠

a) Ta

b) If 1000 tons/day of wet sugar is fed to the evaporator, how much additional water must be removed from the outlet sugar to dry it completely, and what annual revenue can be expected if dry sugar sells for $. 15/𝑙𝑏𝑚?

𝑚1 = 1000 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

Try yourself…

OMB1000 = ሶ𝑚2 + ሶ𝑚3

Sugar Mass BalancewS1

ሶm1 = wS3ሶ𝑚3

800 = .963 ሶ𝑚3ሶ𝑚3 = 830.74 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

How much more water do we need to evaporate? ሶ𝑚𝑊3

= 𝑤𝑊3ሶ𝑚3

ሶ𝑚𝑊3= 0.036(830.74)

ሶ𝑚𝑊3= 29.9 𝑡𝑜𝑛𝑠 𝑊/𝑑𝑎𝑦

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

𝑤𝑊1= 0.2 𝑡𝑜𝑛𝑠 𝑊/𝑡𝑜𝑛

𝑤𝑠1= 0.8 𝑡𝑜𝑛𝑠 𝑆/𝑡𝑜𝑛

ሶ𝑚2 = 1000 − 830 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

ሶ𝑚3 = 830.74 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦𝑤𝑊3

= .036 𝑡𝑜𝑛𝑠 𝑊/𝑡𝑜𝑛𝑠𝑤𝑠3

= .963 𝑡𝑜𝑛𝑠 𝑆/𝑡𝑜𝑛𝑠

a) Ta

b) If 1000 tons/day of wet sugar is fed to the evaporator, how much additional water must be removed from the outlet sugar to dry it completely, and what annual revenue can be expected if dry sugar sells for $. 15/𝑙𝑏𝑚?

𝑚1 = 1000 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

Alternatively,𝑚1 = 100 𝑘𝑔 →

𝑚𝑊3= .036 83𝑘𝑔 = 2.99 𝑘𝑔 𝑊

ሶ𝑚𝑊3=? ? ? →

ሶ𝑚1 = 1000 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

𝑚𝑊3

𝑚1=

ሶ𝑚𝑊3

ሶ𝑚1

2.99

100=

ሶ𝑚𝑊3

1000 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

ሶ𝑚𝑊3= 29.9 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

𝑤𝑊1= 0.2 𝑡𝑜𝑛𝑠 𝑊/𝑡𝑜𝑛

𝑤𝑠1= 0.8 𝑡𝑜𝑛𝑠 𝑆/𝑡𝑜𝑛

ሶ𝑚2 = 1000 − 830 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

ሶ𝑚3 = 830.74 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦𝑤𝑊3

= .036 𝑡𝑜𝑛𝑠 𝑊/𝑡𝑜𝑛𝑠𝑤𝑠3

= .963 𝑡𝑜𝑛𝑠 𝑆/𝑡𝑜𝑛𝑠

a) Ta

b) If 1000 tons/day of wet sugar is fed to the evaporator, how much additional water must be removed from the outlet sugar to dry it completely, and what annual revenue can be expected if dry sugar sells for $. 15/𝑙𝑏𝑚?

𝑚1 = 1000 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

ሶ𝑚𝑊3= 29.9 𝑡𝑜𝑛𝑠/𝑑𝑎𝑦

To sell at $0.15/lb, annually, of dry sugar we need…

830.74 − 29.9 𝑡𝑜𝑛𝑠 = 800.84

800.84 𝑡𝑜𝑛𝑠

𝑑𝑎𝑦×

𝑙𝑏𝑚

5 × 10−4𝑡𝑜𝑛𝑠×

365 𝑑𝑎𝑦𝑠

𝑦𝑒𝑎𝑟×

$0.15

𝑙𝑏𝑚

$87.7 𝑀𝐼𝐿𝐿𝐼𝑂𝑁/𝑦𝑒𝑎𝑟

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized.

b) The evaporator is built to achieve the production rate of part (b), installed, and started up, and the water content of the partially dried sugar is measured on successive days of operation. The results follow. In subsequent runs, the evaporator is to be shut down for maintenance if falls more than three standard deviations from the mean of this series of runs. Calculate the endpoints of this range. Considering the results of parts (a) and (c) together, what can you conclude about the recently installed evaporator?

I know I haven’t talked much about statistics yet, so let’s start by plotting the data on excel and I’ll go over the concepts. If you don’t care about stats then you’re done with this problem but I recommend you check it out because we’re gonna need these concepts in future classes and the rest of our lives prolly!

Day ww

1 0.05132 0.04863 0.054 0.05075 0.05416 0.04987 0.05128 0.04749 0.0511

10 0.0494

Relevant Statistics Crash Course

Our task is to solve for the standard deviation of the data and see if our mean lies within 3 s.ds of the set.

𝑆𝐷 =1

𝑁 − 1

𝑖=1

𝑁

𝑋𝑖 − ത𝑋 2

where 𝑁 is the number of data points, 𝑋𝑖 is the sample measurement, and ത𝑋 is the sample mean given by

ത𝑋 =1

𝑁

𝑖=1

𝑁

𝑋𝑖 0.046

0.047

0.048

0.049

0.05

0.051

0.052

0.053

0.054

0.055

0 1 2 3 4 5 6 7 8 9 10

ww

Days

Given Data Set

The evaporator is built to achieve the production rate of part (b), installed, and started up, and the water content of the partially dried sugar is measured on successive days of operation. The results follow. In subsequent runs, the evaporator is to be shut down for maintenance if falls more than three standard deviations from the mean of this series of runs. Calculate the endpoints of this range. Considering the results of parts (a) and (c) together, what can you conclude about the recently installed evaporator?

We calculated the SD to be 1.81 × 10−3 and the mean to be 0.05036.

The endpoints become0.05036 ± 3(1.81 × 10−3)

Upper limit: 0.0558

Lower limit: 0.0449

But the problem is…𝑤𝑤3 = 0.0306

This is below the lower limit which suggests that the evaporator is not working according to the design specifications

0.046

0.047

0.048

0.049

0.05

0.051

0.052

0.053

0.054

0.055

0 1 2 3 4 5 6 7 8 9 10

ww

Days

Given Data Set

PHEW… RECAP

First break down the problem by making sure you understand what the system diagram looks like

Use the conservation of mass to write and solve your mass balance. Careful with careless mistakes Σmin = Σmout

Use statistics to analyze your results

𝑆𝐷 =1

𝑁 − 1

𝑖=1

𝑁

𝑋𝑖 − ത𝑋 2

ത𝑋 =1

𝑁

𝑖=1

𝑁

𝑋𝑖

Tips● Try the problem by yourself as much as you can and try to not look at

the solutions!● Practice makes perfect● If you’re stuck, take a break and come back to it later. Sleep on it!● Understand that there are many ways to doing a problem. You just have

to find your own. Sometimes learning other solutions are more confusing to you than your own. That’s why you really need to try working it out yourself

● Careful with how you prime your brain to avoid mistakes due to notation, confusing variables, or misreading the question

● Focus on understanding rather than memorizing ● Try different studying habits (e.g pen and paper vs whiteboard. Alone vs

studying w/ friends)● Simulate the exam as much as possible (calculator allowed? Textbook

allowed?)

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