CN1111 Tutorial 01 - Solution

Introduction*to*engineering*calculations

• Units'&'dimensions• Base,'multiple,'and'derived'units• System'of'units:'SI,'CGS,'AE• Conversion'factors'– where'to'find?

• Dimensional'homogeneity• Same'units'of'an'equation'on'both'sides• Dimensional'analysis:'how'to'determine'the'units'of'the'constants'in'an'equation

• Significant'figures• Determine'the'number'of'Sig'Figs• Rules'for'calculations

Process*Variables• Mass'&'Volume

• Density=mass/volumeM'Specific'volume'='volume/mass• Specific'gravity'(SG'='ρ/ρref, generally'water'is'used'as'reference'so'ρref ='1.000g/cm3)

• Mass'(mole,'volume)'flow'rate'(ṁ ='m/t,'ṅ ='n/t,'etc.)• Flow'rate'measurement:'rotameter/orifice'meter

• Chemical'composition'• Atomic/molecular'weight• GramWmole=gWmole=molM'lbmol (1'lbmol ='454'mol,'1'lbm ='454'g)• Mass'fraction'xi'=mi/mtotalM'mole'fraction'yi =ni/ntotal• Average'molecular'weight'of'gas'mixture'='mtotal/ntotal• Mass'concentration'(mi/V)M'molar'concentration'(ni/V)M'molarity:'a'unit'(M'='mol/L)M'ppm'='xi'(liquid)'or'yi (gas)'x106

• Pressure• Head'of'liquid (P'='ρgh)M'Atmospheric'pressure'≠'Standard'atmospheric'pressure'• Gauge'pressure/Absolute'pressure:'Pgauge=PabsoluteWPatmospheric (vacuum:'Pvacuum='PatmosphericWPabsolute)

• Pressure'measurement:'Bourdon' tube'and'manometer'(typically'their'readings'are'gauge'pressure)

• Manometer'calculation:'sealedWend,'openWend,'differential• Temperature

• Temperature' scales:'oC,'oF,'R,'K• Conversion'of'temperature'scales'• Temperature' /'temperature'interval

¯

M =X

i

yiMi or 1/ ¯M =X

i

xi/Mi

CN1111$ $ Tutorial$Questions$Set$01$

$ 1$

Tutorial Question Set #1 - Solution Questions:

1. Convert the following quantities from AE to SI units:

a. 4 lbm/ft to kg/m

b. 0.04 g/[(min)(m3)] to lbm/[(hr)(ft3)]

Answer:

(a)

(b)

2. The density of a certain liquid can be derived from the following equation:

ρ = (A +B⋅T)eC⋅ P

where ρ = density (g/cm3), T = temperature (oC), P = pressure (atm)

For this equation to be dimensionally homogeneous, what are the units of A, B, and C?

Answer:

A: g/cm3

B: g/(cm3⋅°C)

C: 1/atm

3. The mass of an unknown liquid accumulated in a reactor increases with time following the equation:

m (lbm) = et

where t is time in minute. Calculate the expression for m(kg) in terms of t(second).

Answer:

4 lbm

ft

0.4536 kg

1 lbm

3.2808 ft

1 m

= 5.953 kg/m (or 6 kg/m)

0.04 g

min ·m3

1 lbm

453.6 g

60 min

1 hr

1

3m

3

3.28083 ft

3= 0.0001498 (or 0.0001) lbm/(hr · ft3)


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4. The density of a fluid is given by the empirical equation:

ρ =70.5⋅ exp(8.27x10-7⋅ P)

where ρ is density (lbm/ft3) and P is pressure (lbf/in.2). What are the units of 70.5 and 8.27x10-7?

Answer:

70.5: lbm/ft3; 8.27x10-7: in.2/lbf

5. Identify the number of significant figures in each of the following number:

a. 1.0 b. 0.10

c. 10. d. 10

e. 8.2000 x 106

Answer:

a. 2 b. 2 c. 2 d. 1 e. 5

6. Perform the following calculations. Show your answer with the right number of significant figures.

(a) (2.7)(8.632) (b) 2.365+125.2

(c) 32.00x5.00+1000. (d) 32.00x5.00+1000

Answer:

For the following calculations, we followed the two rules to keep the number of significant figures in the final results.

(a) = 23.3064 = 23 (b) = 127.565 = 127.6 (c) = 1160.0000 = 1160.

m(lbm) = et(min)

) m (kg) =1 lbm 1 kg

2.2046 lbm

e1

min1 min60 sec ·t(s)

) m (kg) = 0.454e0.0167t(s)


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(d) = 1160.0000 = 1000

7. Perform the following pressure conversions, assuming when necessary that atmospheric pressure is 1 atm. Unless otherwise stated the given pressure are absolute. (a) 3.00 atm to N/cm2

(b) 20 cm Hg of vacuum to atm (absolute) (c) 25.0 psig to mm Hg (gauge)

(d) 25.0 psig to mm Hg (absolute) (e) 325 mm Hg (vacuum) to mm Hg (absolute)

(a)

(b)

(c)

(d)

(e)

760 mmHg -325 mmHg (vacuum) = 435 mmHg (absolute)

8. Convert the temperatures in (a) and (b) and temperature intervals in (c) and (d): (a) T = 85 oF to oR, oC, K (b) T = -10 oC to K, oF, oR (c) ΔT = 85 oC to K, oF, oR (d) ΔT = 150 oR to oF, oC, K Answers:

3.31 (a) kt k is dimensionless (min-1� ) (b) � A semilog plot of vs. t is a straight lineCA ln lnC CA AO kt �

y = -0.4137x + 0.2512R2 = 0.9996

-5-4-3-2-101

0.0 5.0 10.0t (min)

ln(C

A)

k �0 414 1. min

ln . .C CAO AO3 lb - moles ft � 02512 1286

(c) C CA A A1b - moles

ftmol liter 2.26462 lb - molesliter 1 ft mol3 3 CF

HGIKJ c c

28 3171000

0 06243.

.

t

t st

C C ktA A

min

exp

b g b g c

c

� � �

160

60

0

min s

0 06243 1334 0 419 60 214 0 00693. . exp . . exp .c � c � �C t C tA Ab g b g b gdrop primesmol / L

t CA � 200 530 s mol / L.

3.32 (a) 2600

50 3 mm Hg 14.696 psi

760 mm Hg psi .

(b) 275 ft H O 101.325 kPa

33.9 ft H O kPa2

2

822 0.

(c) 3.00 atm N m m

1 atm cm N cm

2 2

22101325 10 1

10030 4

5 2

2

..

u

(d) 280 cm Hg 10 mm dynes cm cm

1 cm mdynes

m

2 2

2 2

101325 10 100760 mm Hg 1

3733 106 2

210.

.u

u

(e) 120 1

0 737 atm cm Hg 10 mm atm

1 cm 760 mm Hg atm� .

(f) 25.0 psig 760 mm Hg gauge

14.696 psig1293 mm Hg gaugeb g b g

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y = -0.4137x + 0.2512R2 = 0.9996

-5-4-3-2-101

0.0 5.0 10.0t (min)

ln(C

A)

k �0 414 1. min




HGIKJ c c

28 3171000

0 06243.

.

t

t st

C C ktA A

min

exp

b g b g c

c

� � �

160

60

0

min s


t CA � 200 530 s mol / L.

3.32 (a) 2600

50 3 mm Hg 14.696 psi

760 mm Hg psi .

(b) 275 ft H O 101.325 kPa

33.9 ft H O kPa2

2

822 0.

(c) 3.00 atm N m m

1 atm cm N cm

2 2

22101325 10 1

10030 4

5 2

2

..

u


1 cm mdynes

m

2 2

2 2

101325 10 100760 mm Hg 1

3733 106 2

210.

.u

u

(e) 120 1





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y = -0.4137x + 0.2512R2 = 0.9996

-5-4-3-2-101

0.0 5.0 10.0t (min)

ln(C

A)

k �0 414 1. min




HGIKJ c c

28 3171000

0 06243.

.

t

t st

C C ktA A

min

exp

b g b g c

c

� � �

160

60

0

min s


t CA � 200 530 s mol / L.

3.32 (a) 2600

50 3 mm Hg 14.696 psi

760 mm Hg psi .

(b) 275 ft H O 101.325 kPa

33.9 ft H O kPa2

2

822 0.

(c) 3.00 atm N m m

1 atm cm N cm

2 2

22101325 10 1

10030 4

5 2

2

..

u


1 cm mdynes

m

2 2

2 2

101325 10 100760 mm Hg 1

3733 106 2

210.

.u

u

(e) 120 1





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9. Two engineers are calculating the average molecular weight of a gas mixture containing oxygen and three other gases based on their mole fractions. One of them uses the correct molecular weight of 32 for oxygen and determines the average molecular weight as 39. The other uses incorrect value of 16 for oxygen and determines the average molecular weight as 31. This is the only error in his calculations. Based on the above information, is it possible to figure out the mole fraction and mass fraction of oxygen in the mixture? If yes, determine the mole and mass fractions of oxygen. If not, what other information should be provided so you may determine the mole and mass fractions of oxygen?

Answer: Yes it is possible. Consider the other three gases as one species with an average molecular weight of MW. Based on the calculations of the two engineers: Basis: 1 mol of the gas mixture:

yO2 ·32 + (1-yO2) ·MW = 39 yO2 ·16 + (1-yO2) ·MW = 31

Solving these two equations, we can get yO2 = 0.50; MW = 46. The mass fraction of oxygen will be:

xO2 = yO2 ·32/[yO2 ·32+(1- yO2 )·MW] = 0.41

10. A fluid of unknown density is used in two manometers – one sealed-end, the other across an orifice in a water pipeline. The readings shown here are obtained on a day when barometric pressure is 756 mm Hg. What is the pressure drop (mm Hg) from point (a) to point (b)?

3.47 (cont’d)

� |n = ,04979 05. . ln . .K K � 5 2068 183 0 5

ml smm Hgb g

(c) h P V � � 23 0 02274 23 0523 183 0523 132 0 5' . . � . .b gb g b g mm Hg mL s

132

104 180 mL 0.791 g

s mL g s

104 g 1 mol s 58.08 g

mol s .

3.48 (a) T q � q � q85 4597 18 273 30F 544 R 303 K C. / . (b) T � q � u q � q10 273 18 460 14C 263 K 474 R. F

(c) 'T q q

q q

q qq

qq q

q q

85 1010

8585 18

1153

8510

C KC

KC F

CF

C 1.8 RC

153 R..

;.

;.

(d) 150 R 1 F

1 RF; 150 R 1.0 K

1.8 RK;

150 R 1.0 C1.8 R

83.3 Cq q

q q q

q q

q qq

q150 833D

.

3.49 (a) T u � � u0 0940 1000 4 00 98 0. . .D DFB C T = 98.0 1.8 + 32 = 208 FD

(b) ( C) 0.0940 ( FB) 0.94 C (K) 0.94 KT T T' ' � ' D D D

0.94 C 1.8 F

( F) 1.69 F ( R) 1.69 R1.0 C

T T' � ' D D

D D DD

D

b

(c) ; T T1 15 �D DC 100 L 2 43 �D DC 1000 L

T a T C) L)( (D D �

a �

FHGIKJ

43 150 0311

b gb g

D

D

D

DC

1000 -100 L

C

L. ; b � u 15 0 0311 100 119. . DC

�T T T T C) L) and L) C)( . ( . ( . (D D D D � �0 0311 119 32 15 382 6.

(d) Tbp � � � �88 6. D D DC 184.6 K 332.3 R -127.4 F � � � �9851 3232. D DFB L

(e) 'T � � � � �50 0 16 6 156 2 8 2 8. . . .D D D DL 1.56 C FB K F R. D

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$ 5$

Answer: Barometer reading = 756 mmHg means that the atmospheric pressure Patm = 756 mmHg. If the manometer fluid density is ρf, and the density of water is denoted as ρw:

3.40 Pabs 800 mm Hg

Pgauge 25 mm Hg

Patm � 800 25 775 mm Hg

3.41 (a) P g h h P gh gA B1 1 2 2 1� � � �U Ub g hC 2U � � � � �P P gh ghB A C A1 2 1U U U Ub g b g 2

(b)

P1 12110 0 792 137 0 792

�L

NM�

� OQP

kPa + g 981 cm 30.0 cm

cm s

g 981 cm 24.0 cm cm s3 2 3 2

. . . .b g b g

u�

FHG

IKJ u

FHG

IKJ dyne

1 g cm / s kPa

1.01325 10 dynes / cm2 6 21 101325. 1230. kPa

3.42 (a) Say Ut (g/cm3) = density of toluene, Um (g/cm3) = density of manometer fluid

(i) Hg: cm cm

(ii) H O: cm2

U UUU

U U

U U

t mm

t

t m

t m

g h R gR R h

h R

h R

( )

. , . , .

. , . ,

500 500

1

0 866 13 6 150 238

0 866 100 150 2260 cm

� � � �

�

�

�

Use mercury, because the water manometer would have to be too tall.

(b) If the manometer were simply filled with toluene, the level in the glass tube would

be at the level in the tank. Advantages of using mercury: smaller manometer; less evaporation.

(c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen, minimizing the risk of combustion.

3.43 P g Pf fatm

atm m 7.23 g

� U U7 23.b g

P P g P ga b f w w� � �FHG

IKJU U Ud i b g b g26 26 cm

7.23 m cmatm

�� u

FHG

IKJ

756 mm Hg 1 m7.23 m 100 cm

kg 9.81 m/s N 760 mm Hg 1 m m 1 kg m/s 1.01325 10 N m cm

cm2

3 2 5 2

1000 1100

26b g � � P Pa b 81. mm Hg

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CN1111 Tutorial 01 - Solution