Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David by Academic Minute

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Chapter 2 Solutions 2.1 Units of Measure

2.1.1 (a) (b) (c) (d) (e) (f) (g) (h)

2.1.2

2.2

N/mm or nm (nanometer) °C/M/s 100 kPa 273.15 K 1.50m, 45 kg 250°C J/s 250 N

a. The device measures equivalent masses when it is balanced (both pans aligned horizontally). b. The compressed spring measures the force under the influence of gravity. If this system were applied on the surface of the moon, the measurement would be much smaller. Unit Conversions

2.2.1

(a)

2.5 mi 1610 m mi

(b)

= 4.0 103 m or 4.0 km

52.66 Btu 1.055 kJ

= 55.56 kJ

Btu

(c)

5.00 hp 0.7457 kW hp

= 3.73 kW

(d)

50. gal 3.875 10-3 m3 min = 3.2 10−3 m3 s-1 60 s gal min

(e)

22.5 lbf 6.958 kPa

(f)

45.6 slug 32.174 lbm 0.4536 kg = 665 kg s-1 lbm slug s

(g)

17.0 hp h 745.7 J 3600 s = 4.56 106 J=4.56 103 kJ=4.56 MJ h hp s

(h)

35.9 gal 3.875 10-3 m3 ft 2 = 1.50 m3 m-2 s-1 0.30482 m2 gal s ft 2

in 2

lb f /in 2

= 157 kPa

(i) 357o F+459.67=816.67oR

816.67o R 5 K

= 454 K

9o R

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs 7.6 lbm 0.4536 kg

(j)

ft 3

0.02832 m3

lbm

= 120 kg m-3

2.2.2 (a)

6.000 ft 0.3048 m 106m = 1.829 106 m m ft

(b)

s 100 km 3.937 104 in h = 1.094 10−3 in s-1 6 3600 s 10 s km h 500K 9o R

(c)

= 900o R

900o R-459.67=440oF

(d)

1106 Btu 2.93 10-4 kW h h

= 293 kW

Btu m

-3

2.2046 lbm 16 oz = 9.3 oz bushel-1 kg m 28.3776 bushel lbm 15.367 lb ft 2 lb s 2 1.28510-3 Btu m f = 6.1374 10−4 Btu (f) 2 lbf ft s 32.174 lbm ft 7.5 kg

(e)

-3

(g)

0.2433 kg N s 2 ms2

(h)

10.1 A V

Pa m 2 1.41410-4 psi = 3.440 10−5 psi N Pa kg m

W kW = 0.0101 kW A V 1000 W

32.17 ft 36002 s 2 0.3048 m 1000 mm = 1.2711011 mm h-2 2 2 ft m h s 0.779 lb lb s 2 ft 3.76610-7 kW h 3600 s m f = 3.28 10−5 kW (j) 3 32.174 lb lb ft s f h f m (i)

2.2.3 Fractional reduction in miles driven = 1000/13500 = 0.07407 Gasoline saved = 0.07407(136.8) = 10.13 billion gallons 2.2.4 2000 Btu h

kW h $0.14 24 h 30 d = $18, 000 mo-1 11.2 Btu kW h d mo

2.2.5 4.24 ly 2.99792 108 m 3600 s 24 h 365 d mi AU = 2.68 106 AU 7 h s d ly 1610 m 9.2910 mi 2.2.6 8.6 ly 2.99792 108 m 3600 s 24 h 365 d pc = 2.6 pc 16 h s 3.086710 mi d ly

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs 2.2.7 400,000 bbl 42 gal 3.875 l 850 g lbm d = 5.075  106 lb m h-1 l gal 24 h bbl 454.3 g d 2.2.8 18 15  8 in 3 2.54 cm 3 3

297 s

gal

60 s

= 1.8 gal min-1

1000 cm 3 3.875 l min

2.2.9

12km 3.875 L mi = 28.9 mi/gal L gal 1.61 km 2.2.10

$1.29 CAD $0.79 USD 3.875 L L

$1 CAD

= $3.95 USD/gal

gal

2.2.11

8.314 kP a m3 kg mol K

atm ft3 35.31 ft3 kg mol 5 K = 0.7303 m3 2.2046 lbmol 9 R 101.3 kP a lbmol R atm

2.2.12

V = Ah =

3500 ft2 3 in ft 7.481 gal

2.2.13

12 in

ft3

= 6546 gal

Basis: 1 mi3

1mi3  5280 ft   12 in   2. 54 cm   1 m   1 mi   1 ft   1 in   100 cm  3

Basis: 1 ft3/s

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.2.14 a.

2.2.15

a. Basis: 60.0 mile/hr

ft 60.0 mile 5280 ft 1 hr = 88 sec hr 1 mile 3600 sec b. Basis: 50.0 lbm/(in)2

50.0/lb 2

1(in)2

454 g 1 kg m

(in)

(100 cm)2 2

1 lb 1000 g (2.54 cm)

(1 m)

4 kg = 3.52  10 m2

c. Basis: 6.20 cm/(hr)2

6.20 cm 2

(hr)

109nm

1(hr)2 2

100 cm 1 m (3600 sec)

= 4.79

nm sec2

2.2.16

14.91 kW

, not enough power even at 100% efficiency; 68 kW = 91.2 hp.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.2.17

None: 20 gal more are needed. 2.2.18

Let tA be the time for A to paint one house; tB for B A does a house in 5 hours, or 1 house/5 hr. B does one house in 3 hours, or 1 house/3 hr.

1 house tA hr 1 house tB hr + = 1 house 5 hr 3 hr Also tA = tB so that

t = A

15 8

3 15

t + A

5 15

t = 1 or A

t =1 15 A

hr = t = 1.875 hr or 112.5 min B

2.2.19

2.2.20

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.2.21

Basis: 1 lb H2O a.

1 1 1 lbm  3 ft 3 (s 2 )(lbf ) KE= mv2 = = 0.14(ft)(lb f )   2 2 s 32.174(ft)(lb )   m

Let A = area of the pipe and v = water velocity. The flow rate is

2.2.22

Pump Work =

2 hp 2545 Btu

= 5090 Btu/hr hp-hr Rate of energy input for heating = PW - PE =5090 - 4818 = 272 Btu/hr

2.2.23

The object has a mass of 21.3 kg (within a precision of ± .1 kg). The weight is the force used to support the mass.

2.2.24

In American Engineering System Power =

800 lbf 300 ft 1 min

2.4 105

(lbf )(ft) or 7.27 hp min

In SI Power =

4000 N 1.5 m 1 (watt)(s) 1 s 1(N)(m) 6000 watts

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.2.25 KE

m v2

2 1 2300 kg 1 lbm  10.0 ft  2 2 0. 454 kg  s  = = 10.11 Btu

2.2.26

Basis: 10 tons at 6 ft/s 2

1 20,000 lb  6 ft  1(s2 )(lb ) m f KE = mv = = 11,200(ft)(lbf )   2 2  s  32.2(ft)(lb m) 1

2.2.27

Basis: 1 mRNA

Section 2.2.1 Temperature

2.2.28

TK = –10 + 273 = 263K T°F = –10 (1.8) + 32 = 14°F T°R = 14 + 460 = 474°R

2.2.29

Yes, if the temperature scale is a linear relative one (°C, °F), or a logarithmic scale (ln 1° is zero). No, if the scale is absolute, but read J. Wisniak, J. Chem. Educ., 77, 518-522 (2000) for a different conclusion.

2.2.30

Substitute

1.8 TK = T°R

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

8.41 + 2.4346  10–5 Cp

= =

8.41+1.35310−5 To

 TR 

(gmol )(K)2  1.8  R

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.2.31

10 C 1.8 F 1.0 C

+ 32 = 50 F

10 C 1.8 F 1.0C

+ 32 + 460°R = 510 R

150K 1.8R 1.0K

= 270 R

2.2.32 First multiply the RHS of the equation so that

Btu (lbmol)(°F)

1054.8J 1 Btu

1 lb mol 454 gmol

1.8°F 1.0°C

1°C 1K

= 4.182

(gmol)(K) and substitute T°F = 1.8T°C + 32

C =[8.488 + 0.5757 10−2 (1.8T + 32) − 0.2159 10−5 (1.8T + 32)2 + p C C J 9 3 − 0.3059 10 (1.8T + 32) ]4.182 C (gmol)(K) Simplifying, Cp = 36.05 + 0.0447T – 0.2874  10–4 T2 + 0.7424  10–8 T3

Section 2.3 Equations and Units 2.3.1 Solving for the constant term in the equation, 4.8 =

Q Wh

= 1.5

gpm

-2.5

in in1.5 = gpm in

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs 2.3.2 From the back inside cover, R =

10.73 psia ft 3 7.481 gal lbmol 9o R = 0.3180 psia gal gmol-1 K-1 o 3 lbmol R ft 454.3 gmol 5K

2.3.3 Solving for M the constant term in the equation recognizing that a radian is dimensionless, TL N m m m2 M = = = m4 N S 2.3.4 P = m˙ gh = -3 3 h TW 60 min 24 h 365 d 4.3106 gal 9.801 m 180 m 3.875 10 m 1000 kg J s 2 W s 2 2 3 kg m h min J 3600 s 1012 W d y s m gal

= 4.3 TW h y-1 2.3.5

152 ft 2 30 ft 7.481 gal car ft 3

= 2600 cars

15 gal

2.3.6 CD =

v 2 A

-1

2 -1 lbf ft 3 s2 Note that if the right hand side is multiplied by gc CD = lb s lb ft f m lbm ft 2 ft 2

becomes dimensionless. 2.3.7 G=F

r2 Mm

= N m

kg kg

= N m2 kg-2

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs 2.3.8 a1. a2. a3.

2.3.9

g/cm3 A is in g/(cm3)(°C) B is in Since the exponent of e must be dimensionless C is in atm-1

3 3 1 lb lb m = 68.4 m3 A = 1.0963 g (30.48)3 cm 454g ft cm ft

lb 0.00086 g (30.48)3cm3 1 lbm 1∘C = 0.0298 3 mo 3 ∘ 3 ∘ (ft )( R) (cm )( C) ft 454 g 1.8 R

0.000953 1 atm 1 2 = 0.0000648 atm 14.7 lb f /in lbf /in2

Introduce the units. The net units are the same on both sides of the equation.

2.3.10 No.

   10-3m 50×103 (kg)(m) q = 0.6(2m2 )  (2) 9.8m s2 kg (s2 )(m2 )  

  1  2   2  1-   5  

1/ 2

The net units on the right-hand side of the equation are m2  4  s 



Consequently, the formula will not yield 80.8 m3/s; presumably in the formula the g should be gc for use in the AE system.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.3.11

(2Δp)/

Q = 0.61S

= 3,579 lbf /ft

=

14.7 l bH2O 62.4 ft3H 2O

0.703

Assume hole is open to atmosphere.

ft3

1 lb/ft3H O 2

=43.87

lbm

ft3

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.3.12

= 1+

B C D

Units cm3 / g mol (cm3/ g mol)2 (cm3/ g mol)3

=1+

B* C* D*

Units ft3 / lbm (ft3/ lbm)2 (ft3/ lbm)3

If B is the original coefficient, B* is obtained by multiplying B by conversion factors. Let MW be the molecular weight of the compound. B cm3 

 1 g mol 454 g 0.016 B B = = MW   MW g 1 lbm lbm g mol  30.48 cm  *

ft3

1 ft

 ft3   cm3   1 ft   1 g mol   454g  2.57 10-4 C  = C        =C   MW2  g mol   30.48 cm   MWg   1 lbm   lbm  6

 ft3  4.096×10-6 D  = D  MW3  lbm  *

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.3.13

To get u in ft/s, substitute for  and for  , and multiply both sides of the equation by 3.281 ft/1 m (k is dimensionless).

2.3.14

Place units for the symbols in the given equation, and equate the units on the left- and right-hand sides of the equation by assigning appropriate units to the coefficient 0.943. LHS

RHS 3   14   lb m  2 ft Btu (hr)(ft) Btu Btu ?   ( ) 2 ( ) ( )( )( )  3 ( )2  F  F ft hr lbm ft lbm hr ft  F  hr ft

( )

The units are the same on the right and left so that 0.943 has no units associated with it.

2.3.15



numerator denominator

Numerator

Y c  H c − x/s

b/e

 mol cell C   e− equiv.  energy   mole substrate C  mol cell C mol e− equiv.    

= 

energy mol substrate C

(1)

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Denominator

= Hc

cat

energy mol substrate C

(1) (2)

(2)

energy energy

There is no missing conversion factor. What the author claimed about the units is correct.

2.3.16

For dimensional consistency: B – absolute temperature in either ºR or K C – absolute temperature in either ºR or K A – dimensionless In most cases, the argument of a logarithm function should be dimensionless, but in this case it will not be. Therefore, the numerical values of A, B, and C will depend upon the units used for temperature and the units used for p*.

2.3.17

The equation is

The units on the right-hand side (with f dimensionless) in SI are

Hence, the equation is not dimensionally consistent because p has the units of N/m2. If g is replaced with g, the units would be correct.

2.3.18 First multiply the RHS of the equation so that

Btu

1054.8J

1 lb mol

1.8°F

1°C

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs (lbmol)(°F)

1 Btu

454 gmol

1.0°C

= 4.182

(gmol)(K) and substitute T°F = 1.8T°C + 32

2.3

Measurement Errors and Significant Figures

2.4.1 1

(a) 1.24×10 : 3 significant figures -3 (b) 2.3×10 : 2 significant figures 2 (c) 1. ×10 : 1 significant figure 2 Or 1.0×10 : 2 significant figures 2 Or 1.00×10 : 3 significant figures 4 (d) 2.234×10 : 4 significant figures 4 Or 2.2340×10 : 5 significant figures -1 (e) 2.0001×10 : 5 significant figures

(f) 2.000×10 : 4 significant figures -3 (g) 9.×10 : 1 significant figure -3 (h) 2.30×10 : 3 significant figures

(a) 3 significant figures:  0.1% (b) 2 significant figures:  1% (c) 1 significant figure:  10% Or 2 significant figures:  1% Or 3 significant figures:  0.1% (d) 4 significant figures:  0.01% Or 5 significant figures:  0.001% (e) 5 significant figures:  0.001%

(f) 4 significant figures:  0.01% (g) 1 significant figure:  10% (h) 3 significant figures:  0.1%

(i) 7.78×10 : 3 significant figures 2

(j) 7.780×10 : 4 significant figures

2.4.2

(i) 3 significant figures:  0.1% (j) 4 significant figures:  0.01%

2.4.3 -3

(a) 1.1 ×10 : 2 significant figures

(b) 5. ×10 : 1 significant figure

(f) 4.354×10 : 4 significant figures 5 Or 4.3540×10 : 5 significant figures 5 Or 4.35400×10 : 6 significant figures 2 (g) 1.0010×10 : 5 significant figure

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs 3

Or 5.0×10 : 2 significant figures

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs 3

Or 5.00×10 : 3 significant figures 3 Or 5.000×10 : 4 significant figures 0 (c) 2.3001×10 : 5 significant figures 5 (d) 8.877×10 : 4 significant figures 5 Or 8.8770×10 :5 significant figures 5 (e) 8.87700×10 : 6 significant figures

(h) 5.61×10 : 3 significant figures 0 (i) 1.000×10 : 4 significant figures 1

(j) 1.×10 : 1 significant figures 1 Or 1.0×10 : 2 significant figures

2.4.4 (a) 2 significant figures:  1%

(b) 1 significant figure:  10% Or 2 significant figures:  1% Or 3 significant figures:  0.1% Or 4 significant figures  0.01% (c) 5 significant figures:  0.001% (d) 4 significant figures:  0.01% Or 5 significant figures:  0.001% (e) 6 significant figures:  0.0001%

(f) 4 significant figures :  0.01% Or 5 significant figures:  0.001% Or 6 significant figures:  0.0001% (g) 5 significant figures:  0.001%

(h) 3 significant figures:  0.1% (i) 4 significant figures:  0.01% (j) 1 significant figures:  10% Or 2 significant figures:  1%

2.4.5

Two because any numbers added to the right-hand side of the decimal point in 10 are irrelevant.

2.4.6

The sum is 1287.1430. Because 1234 has only 4 significant figures to the left of an implied decimal point, the answer should be 1287 (no decimal point).

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.4.7

The number of significant figures to the right of the decimal point is 1 (from 210.0m); hence, the sum of 215.110 m should be rounded to 215.1 m.

2.4.8

A calculator gives 569.8269 000, but you should truncate to 4 significant figures, or 569.8 cm2.

2.4.9

Two significant figures (based on 6.3). Use 4.8  103.

2.4.10

Step 1: The product 1.3824 is rounded off to 1.4. Step 2: Calculate errors. For absolute error, the product 1.4 means 1.4 + 0.05. 0.05

Thus

100% = 3.6% error

1.4 Similarly, 3.84 has

0.005

100% = 0.13% error

3.84 0.005

and 0.36 has

100% = 1.4% error

0.36 Total 2.7% error

2.5 Validating Solutions 2.5.1

a. 6  4/ 3 = 8 Exact solution = 7.704 b. 1.45 The second term is much smaller than the first term. Exact solution = 1.423 c. 3 4  28 / 3 = 4  8 = 32 Exact solution = 31.54 d. 8  3 (16  50 − 25) / 3 = 8(800 − 25) = 6400 − 200 = 6200 Exact solution = 6767 e. 88.8  (1 + 1) / 2 = 88.8 Exact solution = 87.47 2.6 Mass, Moles, and Density 2.6.1 Choosing a Basis

2.6.1

(a) A gas requires a convenient basis of 1 or 100 g moles or kg moles (if using SI units). (b) A gas requires a convenient basis of 1 or 100 lb moles (if using AE units).

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

(c) Use 1 or 100 kg of coal, or 1 or 100 lb of coal because the coal is a solid and mass is a convenient basis. (d) Use 1 or 100 moles (SI or AE) as a convenient basis as you have a gas. (e) Use 1 or 100 moles (SI or AE) as a convenient basis as you have a gas.

2.6.2

Since the mixture is a gas, use 1 or 100 moles (SI or AE) as the basis.

2.6.3

Pick one day as a basis that is equivalent to what is given—two numbers: (a) 134.2 lb C1

(b) 10.7  106 gal water.

2.6.2 The Mole and Molecular Weight 2.6.4

4 g mol mg Cl2 (95.23)g M gCl2

= 380.9 g

gmol M gCl2 454g C3H8 b) 2 lb mol C3H8 (44.09)lb C3H8 = 4  104g C3H8 lb mol C3H8 1 lb C3H8 c)

(d)

16 g N2

gmol N 2 1 lb mol N 2 = 1.26 10−3 lb mol N (28.02)g N2 454 gmol N 2 2

3 lb C2 H6O 1 lb mol C2H6O 454 g mol = 29.56 g mol C2H6O (46.07) lb C2H6O 1 lb mol

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.6.5

2.6.6

(a)

16.1 lb mol HCl 36.5 lb HCl = 588 lb HCl 1 lb mol HCl

(b)

19.4 lb mol KCl 74.55 lb KCl = 1466 lb KCl 1 lb mol KCl

(c)

11.9 g mol NaNO3 ___ 85 g NaNO3 2.20 10-3lb = 2.23 lb NaNO3 1 g mol NaNO3 1g

(d)

164 g mol SiO2 60.1 g SiO2 2.20 10-3lb = 21.7 lb SiO 2 1g 1 g mol SiO2

Basis: 100 g of the compound comp. C O H

g 42.11 51.46 6.43

MW 12 16 1.008

g mol 3.51 3.22 6.38 13.11

Ratio of Atoms 1.09 1 2

Multiply by 11 to convert the ratios into integers The formula becomes C12O11H22. Checking MW: 12(12) + 11(16) + 22(1.008) = 342 (close enough)

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.6.7

Vitamin A, C20O H30, Mol Wt.: 286

Vitamin A =

Vitamin C, C6 H8O6, mol. wt: 176

2.00 g mol 286 g

1 lb

= 1.26 lb

1 g mol 454 g 16 g 1 lb

= 0.0352 lb

454 g Vitamin C =

2.00 g mol 176 g 1 lb

= 0.775 lb

g mol 454 g 16 g 1 lb

= 0.0352 lb

454 g b.

Vitamin A =

1.00 lb mol 286 lb 454 g

= 130,000 g

1 lb mol 1 lb Vitamin C =

For both

1.00 lb mol 176 lb 454 g = 79,900 g 1 lb mol 1 lb 12 lb 454 g

= 5450 g

1 lb

2.6.8

2.6.9

Mass fraction to mole fraction:

Mole fraction to mass fraction

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.610

Basis: 100 kg mol gas Comp. CH4 H2 N2 Total

Mol % = mol 30 10 60 100

MW 16 2 28

kg 480 20 1680 2180

21.8 kg kg mol

2.6.11

Basis: 100 g mol gas Comp. CO2 N2 O2 H2O

mol = % 19.3 72.1 6.5 2.1 100.0

Avg. mol. wt =

3113.8

MW 44 28 32 18

g 849.2 2018.8 208.0 37.8 3113.8

= 31.138

100.0 2.6.12

Basis: 100 lb mol Comp. CO2 CO N2

% = mol 16 10 30

MW 44 28 28

lb 2640 280 840 3760

Avg. MW = 37.6 lb/lb mol 2.6.13 (a) 10×1000×6.022×1023 = 6.022×1027 molecules of Bz (b) 10×2.2046=22.046 lb mol Bz (c) 78×22.046=1719.6 lb Bz (d) 10,000 gmol Bz (e) 60,000 gmol H (f) 60,000 gmol C

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Section 2.6.3 Density and Specific Gravity

2.6.14 100 lbm

2.6.15

ft 3 = 0.0305 ft3 62.4  5.25 lbm

Basis: 1000 lb oil

0.926

lb oil

62.4

ft3 lb H2O 1.00 ft3 1000 lb oil

2.6.16

lb H2O ft3

= 57.78

lb oil ft3 oil

1 ft3 oil 7.48 gal 57.78 lb oil 1 ft3 = 129.5 gal

Basis: 10,010 lb

10,010 lb 1 gal 0.134 ft3 = 3 8.80 lb gal 152 ft 2.6.17

Basis: 1 g mol each compound g mol

 (g/cm3)

V (cm3)

207.21

11.33

18.3

65.38

7.14

9.16

12.01

2.26

5.31

V̂ =

mass (g)

Set 2 is the correct one.

density (g/cm3 )

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs Section 2.7.2 Pressure and Hydrostatic Head

2.7.1

Basis: 15cm3 water a.

m = ρV=

= 15,000 kg

F m g = 15,000 kg 9.80 m 1 N 1 Pa = 1470 Pa = 2 s 10 m 10 m 1 (kg)(m) N A A s2 m2 = 1.47 kPa

1.470 kPa

1 atm 14.7 psi = 0.21 psi 101.3 kPa atm

15 cm H2O

1 in. 1 ft 14.696 psi = 0.21 2.54 cm 12 in. 33.91 ft H2O

2.7.2

Volume of concrete: Sides Vs =

5 m 30.2 m 200 mm

1m 2 sides 3 = 60.4 m 1000 mm

Ve =

5 m 27.2 m 200 mm

1m 2 ends = 54.4 m3 1000 mm

Vf =

27.4 m 30.4 m 200 mm

Ends

Floor 1m 1000 mm

= 166.592 m3

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Total volume = 281.392 m3 Mass of concrete =

281.392 m3 2080 kg m3

= 585,295

Volume of displaced water required to float: 586,543.36 kg 1 m3 H2O = 586.543 m 3 1000 kg H2O V = LWh → h =

V LW

2.7.3

586.54 m3 27.4m 30.4 m

= 0.703 m

The pressure is a gauge pressure. 60°F Basis: 50.0 psig a. 50.0 psig 33.91 ft H2O = 115 ft 14.7 psia

148 ft

(difference) b.

No. Insufficient height.

Alternate solutions can be applying ∆p = hg.

2.7.4

lbf in.2 = lbf → lbm in the AE system in.2 so the procedure is ok, although the unit conversion is ignored.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.7.5

Basis: Dept = 1000 m

p = p o + gh

p = 1 atm + 1000 m 1.024 g 9.8 m 1 kg  100 cm 3 1 N 1 kPa 1 atm 3 2 3   1N (kg)(m) 1.013  105 N / m2 cm s 10 g 1 m (kg)(m) s2 s2 100.1 atm 101.3 kPa 4 = = 1.014  10 kPa 1 atm Alternative solution: ATop = 730.62 m2 kPa

2.7.6

(a) pABS = pGauge + patm

 22.4 lb 144 in.2   28.6 in. Hg 14.7 psia 144 in.2  2 f  = 5250 lb f /ft   + 2 2 2 1 ft   1 29.92 in. Hg 1 ft   1 in.

 22.4 psig 33.91 ft H 2O   28.6 in. Hg 33.91 ft H 2O   +  = 84.1 ft H 2O 1 14.7 psia   1 29.92 in. Hg  

(d) 

2.7.7

Neither John is necessarily right. The pressure at the top of Pikes Peak is continually changing.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.7.8

p = gh

3 1 g 9.8 m 13.1 m 1 kg (100 cm) Pa 1 kPa 2 1 kg 1000 Pa 1000 g m3 cm3 s m s2

= 128. 4 kPa b.

F mg Vg (S.A.)(t)g = = = A ABottom ABottom ABottom

A Bottom =

 4

D =

 (30.5 m)2 4

= 730.62 m

ATOP = 730.62 m2 Aside = Dh = (30.5 m)(13.1m) = 1255.2 m2 S.A.= ATop + A Bottom + ASide

= (730.62 + 730.62 + 1255.22) m2 = 2716.46 m2 (S.A.)(t)g = A

(

)

7.86 g 100 cm3 2,716.46 m2 9.35  10–3 m 9.8 m 3 3 2 cm m s2 730.6 m 1 kg 1 Pa 1 kPa = 2 1000 g 1 kg / ms 1000 Pa

2.68 kPa

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.7.9

pA + ρ1gh1 + ρ2gΔh = PD pB + ρ1gh1 + ρ2gΔh = PD

pA -pB = ρ1gh1 +ρ2gΔh − ρ1gh1 − ρ1gΔh = ρ2gΔh − ρ1gΔh = gΔh(ρ2 − ρ1 ) =

9.8 m 0.78in 1m 13.546 − 0.91g 1kg s2 3.28 ft 1000 g cm3

1 N 1 ft  100 cm  3 m2 1 Pa 760 mm H2 12 in  1 m  1 kg 1 N 101.3  103Pa (m)(s)2 m2 =18.4 mm Hg

The pressure at A is higher than the pressure at B. Alternate solution:

Section 2.7.4 Flow rate 2.7.10 2.7.11 2.7.12 Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs 2.7.13

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Section 2.7.5 Concentration

2.7.14

Component Na C1 O

g mol 1 1 3 5

mol fraction 0.20 0.20 0.60

MW 23 35.45 16

g 23 35.45 48 106.45

mass fraction 0.22 0.33 0.45 1.00

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.7.15

Basis: 1 gal of solution. Mass of solution:

1.0824 lb soln lb H O 62.4 3 2 3 ft H 2O ft H2O 1 ft3 1 gal = 9.03 lb soln . lb H2O 7. 481 gal 1.00 ft3H 2O Mass fraction KOH =

0.813 lb

= 0.09

9.03 lb Mass fraction H2O = 1 – 0.09

2.7.16

1.00 Total

Basis: 30 lb gas Comp. CO2 N2

2.7.17

0.91 =

lb 20 10 30

1000

MW 44 28

lb mol 0.455 0.357 0.812

mol fr. 0.56 0.44 1.00

c) Yes, because for solids and liquids the ratio in ppb is mass, whereas for gases the ratio is in moles.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

5 m 30.2 m 200 mm

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Floor Vf =

27.4 m 30.4 m 200 mm

1m 1000 mm

= 166.592 m3

Total volume = 281.392 m3 3 Mass of concrete = 281.392 m 20803 kg = 585,295 kg m

Volume of displaced water required to float: 586,543.36 kg

1 m3 H2O = 586.543 m3 1000 kg H2O

V = LWh → h =

2.7.19

V LW

586.54 m3 27.4m 30.4 m

Basis: 190,000 ppm

19%

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.7.20

Basis: 100 g of the sample The biomass sample is g = % dry weight of cells C O

50.2 20.1

H P

8.2 3.0 95.5 4.5 100.0

other Total

10.5 g cells 50.2 g C 1 g mol C = 0.439 g mol C/g mol ATP g mol ATP 100 g cells 12 g C

2.7.21

MMM(s) → NN(s) + 3CO2 (g) 1 curie

10 μ curie

3×1010disintegrations/s

1 curie

2×10 disintegrations 1 min min

2.7.22

= 11μ curie

2 107 disintegrations 0.80 = 1.6 10 7 cpm min

The relation to use is − t1/ 2 = ℓ n(2) /(k)(OH ) T with (OH− ) = 1.5106 k Methanol Ethanol MTBE

0.1510−12 110−12 0.60 10−12

t1/ 2 = (seconds) 30.8105 4.6 105 7.7 105

The order is in increasing persistence ethanol, MTBE, and methanol.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

2.7.23

Yes.

Bases are first entries.

4800 g mol CC1 103 g mol air kg mol air 154 g CC1 103 mg CC1 4

10 g mol air

kg mol air

22.8 m

g mol CC14

g CC14

32.4 mg CC14 /m3which exceeds the NIOSH standards 2.7.24

25,600 ton P 2.6 yr 1 1 gal 2000 lb 454 g 106  g yr 1g 1.2  1014 gal 3.785 L 1 ton 1 lb

= 133.1 μg/L b.

19,090 lb(P)municipal = 63. 4% 30,100 lb(P)total

100 lb(P )total 19,090 lb(P)municipal 0.70 lb(P)det. = 44.4% 30,100 lb(P)total 1 lb(P)municipal 30,100 lb(P)total

For d. and e. assume that (P)outflow remains unchanged. d.

P retained/yr = 2,240 + 6,740 + 0.7  19,090 – 4,500 = 17,843 ton/yr.

P conc. in ppb =

2000 lb 1 g / cm3 109 g 17,843 ton 2.6 yr 1 = 92.6 ppb yr 1.2  1014 gal ton 8.345 lb / gal 1billion g This is greater than 10 ppb. Eutrophication will not be reduced. e.

P retained/yr = 2,240 + 6,740 + 0.3  19,090 – 4,500 = 10,207

ton yr

P conc. in ppb =

10,207 ton 2.6 yr 2000 lb 1 g / cc 109 g = 53.2 ppb 1.2  1014 gal ton 8.345 lb / gal 1 billion g yr This will not help.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Basis: 106 g mol gas

2.7.25

350 g mol H2S 1 g mol gas 34g H2S 1 g mol CO2 6 10 g mol gas 1 g mol CO 2 1 g mol H 2O 44 g CO 2 =

270 g H2S 270 g H2S = 106 g CO 2 106 g total liquid

mass fraction H2S = 2.70 10−4

2.7.26

1 mol O2

(a)

10 4mol O2



100 mol gas (b)

or 104 ppm

10 mol gas

Basis: 100 mol gas answer

Comp. SO3 SO2 O2 Total 2.7.27

% = mol 55 3 1 59

mol fr. 0.932 0.051 0.017 1.000

or mol % 93.2 5.1 1.7 100.0

CaCO3: 100.06 Ca 40.05 Mg 24.3 C 12.01 O 16.00

100.06 g CaCO3 1 g mol CaCO3 1 g mol Ca = 2.50 g CaCO3 g Ca g mol Ca CO3 1 g mol Ca 40.05 g Ca Similarly

g CaCO3 g Mg

= 4.118

g CaCO3 g Mg

Total alkalinity = 2.50 (56.4) + 4.118 (8.8) = 177 mg CaCO3 L

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

1 molecule in 1023 or more is not 13-20 ppb

2.7.28

No.

2.7.29

On a mol basis, the carbon dioxide concentration in air is about 350 parts per million (ppm), while that of oxygen is about 209,500 ppm. If the atmospheric concentration of carbon dioxide is increasing at about 1% per year (i.e., from 350 ppm this year to 353.5 ppm next year), and not to 1%, the 3.5-ppm change in dioxide concentration causes the oxygen concentration to fall from 209,500 to about 209,497 ppm, which is less than a 0.002% decrease. So, there is no need to worry about an oxygen deficit at present.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Chapter 3 Solutions 3.2

Introduction to Material Balances

3.2.1

Boundary:

Around both pumps and include the soil at the end of the pipes. It’s an open system.

It’s at steady state except at startup (note system boundary limits fluid), or if fluid enters, unsteady state.

3.2.2

Either open (flow) or closed (batch) is acceptable if explanation is given. (1)

flow—Material comes in and out continuously over a suitably long period of time.

(2)

batch—Material is injected into the system, and then in a short period of time, a reaction occurs with the system valves closed.

(a)

open if you have to replace water, and water evaporates; otherwise closed

(b)

open

3.2.3

3.2.4

If the overall flows in and out over a time period for several batches are considered, and the local batches ignored, the process can be treated as continuous.

3.2.5

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

3.2.6

The system is the radiator. Open System

Closed System

Steady StateUnsteady State

X (Before filling)

X (After filling)

X (Before filling)

3.2.7

system contains

Ice

ice + H2O

boundary

boundary (c)

Unsteady state for any assumptions

b open assume water (melted ice) leaves the system   a flow b closed assume water stays in system because the ice and water   a batch remain on melting

3.2.8

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

3.2.9

(a)

(b)

(c)

(d) x

(e)

(f) x

1. 2.

3. 4.

x x Depends on the time period considered

5. 6. 7.

(g)

x x

3.2.10

If the overall flows in and out over a time period for several batches are considered, and the local batches are ignored, the process can be treated as continuous.

3.2.11

Basis: 1 minute Accumulation (kg)

In (kg)

mfinal − 0

= 300 + 100

Out (kg) −

380

nfinal = 20 kg 20 kg 60 min = 1200 kg min 1 hr

3.2.12

Basis: 600 kg solution Accumulation (kg)

In (kg)

nfinal − 0

= 100 + 500

Out (kg) −

300

nfinal = 300 kg if no water evaporates in which case less than 300 kg would remain

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

3.2.13

(c)

3.2.14

Based on the process measurements, there are 5,000 lb/h more flow for the process leaving the heat exchanger than the feed rate to the heat exchanger; therefore, the material balance for the process fluid does not close. The reason for this discrepancy could be faulty flow sensor readings or possibly a leak of the condensate or steam into the process stream.

3.2.15

Basis: Data shown on flowsheet units are MTA. In 2.5  106

Out 404 103 228 103 152 103 101 103 67 103 36 103 100 103 1190 103 2,278  103

Mass in does not equal to mass out Reasons:

(1) Some of the material was burned as fuel. (2) Some of the gaseous material formed was exhausted to atmosphere (such as H2O, CO2). (3) Errors in measurement. (4) Some process streams are not shown.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

3.2.16

In = 250,000 ton/yr. In (ton/yr) 250,000

Out = 244,500 ton/yr. Out (ton/yr) Heavy Combustibles 3,800 Light Combustibles 39,500 Polyethylene 30,000 Polystyrene 5,000 PVC 40,000 Acrylonitorile 20,000 DB 8,000 Phenol 10,000 Acetone 5,750 Rubber 10,000 LPG 24,000 Aromatics 48,000 Total 244,050

Balance is not exact but is very good for an operating plant. 3.2.17

Basis: 1 week in (tons) = 920 + 0.6 = 920.6 out (tons) = 3.8 + 620 + 0.01 + 0.01 + 0.08 + 1.1 + 275 + 20 = 920 Yes

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

3.2.18

The density of the crystalline silicon in the cylinder is 2.4 g/cm3. Basis: 62 kg silicon The system is the melt, and there is no generation or consumption. Let mt be the accumulation. Accumulation mt

Input 0

Output 0.5(62 kg)

mt = − 31 kg

Let t be the time in minutes to remove one-half of the silicon. 2.4 g π(17.5cm)2 0.3 cm t min = 1 (62,000 g) cm3 4 min 2 t = 179 min

3.2.19

Basis: 1 hr Overall material balance (kg): 13,500 + 26,300 ? 39,800 Yes. The balance is satisfied. NaC1 balance: 0.25 (13,500) + 0.05 (26,300) ? 0.118 (39,800) 4690  4696 The balance is closely satisfied but not exactly. The closure is good for industrial practice.

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs