Chapter 2 Solutions 2.1 Units of Measure 2.1.1
2.1.2
2.2
(a) (b) (c) (d) (e) (f) (g) (h)
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) (b) (c) (d) (e)
2.5 mi 1610 m 3 = 4.0 × 10 m or 4.0 km mi 52.66 Btu 1.055 kJ
= 55.56 kJ
Btu 5.00 hp 0.7457 kW
= 3.73 kW
hp -3
3
50. gal 3.875 × 10 m min 3 -1 −3 = 3.2 × 10 m s 60 s min gal
22.5 lb f 6.958 kPa in
2
lb f /in
2
= 157 kPa
(f)
45.6 slug 32.174 lb m 0.4536 kg -1 = 665 kg s lb s slug m
(g)
17.0 hp h 745.7 J 3600 s 6 3 = 4.56 × 10 J=4.56 × 10 kJ=4.56 MJ h hp s
(h)
-3
35.9 gal 3.875 × 10 m s ft
2
3
gal
(i) 357 o F+459.67=816.67 o R
ft 2 3 -2 -1 = 1.50 m m s 0.30482 m 2 o
816.67 R 5 K o
9 R
= 454 K
(j)
7.6 lb m 0.4536 kg 3
ft
lb m
ft 3
0.02832 m3
= 120 kg m
-3
2.2.2 (a) (b)
6.000 ft 0.3048 m 106 µ m 6 = 1.829 × 10 µ m m ft 4
100 km 3.937 × 10 in h
km o
500K 9 R
(c) (d) (e) (f) (g) (h)
o
5K 6
-4
h
Btu -3
m
-3
15.367 lb m ft s
2
lb f s
2
Pa m 2 1.414×10-4 psi −5 = 3.440 × 10 psi N Pa kg m
2
ms
10.1 A V
W
kW = 0.0101 kW 1000 W AV
2
h
0.779 lb m ft s
2
1.285×10-3 Btu −4 = 6.1374 × 10 Btu lb ft 32.174 lb m ft f
2
0.2433 kg N s
s
= 293 kW
2.2046 lbm 16 oz -1 = 9.3 oz bushel kg 28.3776 bushel lb m
7.5 kg m
o
900 R-459.67=440 F
1 × 10 Btu 2.93 × 10 kW h
2
(j)
o
= 900 R
32.17 ft 3600 s
(i)
s h -1 −3 = 1.094 × 10 in µ s 3600 s 106 µ s
3
2
2
0.3048 m 1000 mm 11 -2 = 1.271 × 10 mm h m ft 2
3.766×10-7 kW h 3600 s −5 = 3.28 × 10 kW lbf ft 32.174 lb m f h lb f s
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 -1 = $18, 000 mo 11.2 Btu kW h d mo
2.2.5 8
4.24 ly 2.99792 × 10 m 3600 s 24 h 365 d mi AU 6 7 = 2.68 × 10 AU h s d ly 1610 m 9.29×10 mi
2.2.6 8
8.6 ly 2.99792 × 10 m 3600 s 24 h 365 d s
h
d
ly
pc = 2.6 pc 3.0867×1016 mi
2.2.7 400,000 bbl 42 gal 3.875 l 850 g lb m d 6 -1 = 5.075 × 10 lb m h gal 24 h d bbl l 454.3 g
2.2.8 18 × 15 × 8 in
3
3
2.54 cm
297 s
in
3
3
gal
l
60 s
1000 cm 3.875 l min 3
2.2.9 12km / 3.875 L/
mi
3 8.314 kPa / m /
atm
35.31 ft 3
/ 101.3 kPa
3 m /
= 1.8 gal min
-1
= 28.9 mi/gal L/ gal 1.61 km / 2.2.10 $1.29 CAD $0.79 USD 3.875 L = $3.95 USD/gal L $1 CAD gal 2.2.11
kgmol K/ / 2.2.12 = V Ah =
2.2.13
/ 3500 ft/ 2 3 in
a.
b.
kgmol 5 K/ atm ft 3 / = 0.7303 2.2046 lbmol 9° R lbmol° R
/ ft
7.481 gal = 6546 gal / 12 in ft/ 3
Basis: 1 mi3
1mi 3 5280 ft 3 12 in 3 2. 54 cm 3 1 m 3 1 mi 1 ft 1 in 100 cm
Basis: 1 ft3/s
=
2.2.14 a.
b. c.
2.2.15
a. Basis: 60.0 mile/hr
60.0 mile 5280 ft 1 hr ft = 88 hr 1 mile 3600 sec sec b. Basis: 50.0 lbm/(in)2
50.0/lb m 454 g 1 kg 1(in) 2 (100 cm) 2 kg = 3.52 × 104 2 2 2 2 (in) 1 lb 1000 g (2.54 cm) (1 m) m c. Basis: 6.20 cm/(hr)2
6.20 cm 1 m 109 nm 1(hr) 2 nm = 4.79 2 2 2 (hr) 100 cm 1 m (3600 sec) sec 2.2.16
14.91 kW , not enough power even at 100% efficiency; 68 kW = 91.2 hp.
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
tA =
2.2.19
2.2.20
3 5 8 tA + tA = 1 or tA =1 15 15 15
15 hr = tB = 1.875 hr or 112.5 min 8
2.2.21
Basis: 1 lb H2O 3
a.
1 1 1 lb m 3 ft (s 2 )(lb f ) KE= mv 2 = = 0.14(ft)(lb f ) 2 2 s 32.174(ft)(lb m )
b.
Let A = area of the pipe and v = water velocity. The flow rate is
2.2.22
2 hp 2545 Btu = 5090 Btu/hr hp-hr Rate of energy input for heating = PW - PE =5090 - 4818 = 272 Btu/hr = Pump Work
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 =
=
FV
800 lb f 300 ft 1 min
=
2.4 ×105
In SI Power = =
4000 N 1.5 m 1 (watt)(s) 1 s 1(N)(m) 6000 watts
(lb f )(ft) or 7.27 hp min
2.2.25 KE
=
=
1 m v2 2 1 2300 kg 1 lb m 10.0 ft 2 1 1 Btu 2 0. 454 kg s 32. 2 lb m ft 778.2( ft )(lb f ) 2 lb f sec
= 10.11 Btu 2.2.26
Basis: 10 tons at 6 ft/s 2
1 1 20,000 lb m 6 ft 1(s 2 )(lb f ) KE = m v2 = = 11,200(ft)(lb f ) 2 2 s 32.2(ft)(lb m ) 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
Cp
= =
8.41 + 2.4346 × 10–5
J T°R 2 (gmol )(K) 1.8
8.41 + 1.353 ×10−5 To R
2.2.31
a)
10°C 1.8°F 1.0°C
b)
10°C 1.8°F 1.0°C + 32 + 460°R = 510°R
+ 32 = 50°F
c)
d)
150K 1.8°R 1.0K = 270°R
2.2.32 First multiply the RHS of the equation so that
Btu (lbmol)(°F) = 4.182
1054.8J 1 Btu
1 lb mol 454 gmol
1.8°F 1.0°C
J (gmol)(K)
and substitute T°F = 1.8T°C + 32
C= [8.488 + 0.5757 ×10−2 (1.8T°C + 32) − 0.2159 ×10−5 (1.8T°C + 32)2 + p J 0.3059 ×10−9 (1.8T°C + 32)3 ]4.182 (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 gpm -2.5 = = gpm in 1.5 in in1.5 Wh
1°C 1K
2.3.2 From the back inside cover, R =
10.73 psia ft 3 7.481 gal lbmol 9o R -1 -1 = 0.3180 psia gal gmol K 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 4 M = = = m N θS 2.3.4 = = P mgh -3
3
4.3×106 gal 9.801 m 180 m 3.875 × 10 m 1000 kg J s 2 W s TW 60 min 24 h 365 d h 2 3 min kg m 2 J 3600 s 1012 W d y h s gal m = 4.3 TW h y
-1
2.3.5 2
π 15 ft
2
30 ft 7.481 gal car = 2600 cars ft 3 15 gal
4 2.3.6
CD =
2F -1 lbf ft 3 s 2 2 -1 lb f s lb m ft Note that if the right hand side is multiplied by gc CD = = 2 2 2 lb ft ft m ρv A
becomes dimensionless. 2.3.7 G=F
r
2
Mm
=
N m2
2
kg kg
= N m kg
-2
2.3.8
2.3.9
a1. a2. a3.
A is in g/cm3 B is in g/(cm3)(°C) Since the exponent of e must be dimensionless C is in atm-1
b1
A=
lb 1.096 g (30.48)3cm3 1 lb m = 68.4 m3 3 3 cm ft 454g ft
b2
B=
lb 0.00086 g (30.48)3cm3 1 lbm 1 C = 0.0298 3 mo 3 3 (cm )( C) ft 454 g 1.8 R (ft )( R)
b3
C=
0.000953 1 atm 1 = 0.0000648 2 atm 14.7 lb f /in lb f /in 2
Introduce the units. The net units are the same on both sides of the equation.
1/ 2
2.3.10 No.
-3 3 1 2 (2) 9.8m 10 m 50×10 (kg)(m) q = 0.6(2m ) 2 s2 kg (s 2 )(m 2 ) 2 1- 5
The net units on the right-hand side of the equation are 1/2
m3 m 4 s 2
≠
m3 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.
2.3.11
(2Δp)/ρ
Q = 0.61S
Assume hole is open to atmosphere.
= 3,579 lb f /ft 2
ρ=
lb ft 3
0.703 3
1 lb/ft H 2 O
l bH 2 O 14.7 62.4 ft 3 H 2 O
=43.87
lb m N ft 3
2.3.12
a.
b.
Z
= 1+
B C D
Units cm3 / g mol (cm3/ g mol)2 (cm3/ g mol)3
Z
=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. 3
ft 3 B cm3 1 ft 1 g mol 454 g 0.016 B = = B lb m g mol 30.48 cm MW g 1 lb m MW *
2
2
6
2
2
ft 3 cm3 1 ft 1 g mol 454g 2.57 × 10-4 C* = C = C MW 2 g mol 30.48 cm MWg 1 lb m lb m 3
ft 3 4.096×10-6 D* = D MW 3 lb m
2.3.13
To get u in ft/s, substitute for τ and for ρ , 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 lb m 2 ft Btu ( hr)(ft ) Btu Btu ? (hr ) ft 2 (∆ ° F) (hr )(ft )(∆ ° F) ft 3 (hr)2 lb m ft lb m ∆ ° F
1
( )
The units are the same on the right and left so that 0.943 has no units associated with it. 2.3.15
η
=
numerator denominator
Numerator
=
Yxc/ s γ b ∆H bc/ e− − energy mol cell C e equiv. − mole substrate C mol cell C mol e equiv.
=
=
energy mol substrate C
(1)
4
Denominator
= ∆H ccat =
energy mol substrate C
=
(1) energy = (2) energy
(2)
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 (lbmol)(°F)
1054.8J 1 Btu
1 lb mol 454 gmol
1.8°F 1.0°C
1°C 1K
= 4.182
J (gmol)(K)
and substitute T°F = 1.8T°C + 32
C= [8.488 + 0.5757 ×10−2 (1.8T°C + 32) − 0.2159 ×10−5 (1.8T°C + 32)2 + p J 0.3059 ×10−9 (1.8T°C + 32)3 ]4.182 (gmol)(K) Simplifying, Cp = 36.05 + 0.0447T – 0.2874 × 10–4 T2 + 0.7424 × 10–8 T3
2.3
Measurement Errors and Significant Figures
2.4.1 (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%
(a) 1.1 ×10 : 2 significant figures
(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
1
2.4.2
3
(i) 7.78×10 : 3 significant figures 2
(j) 7.780×10 : 4 significant figures 2
(i) 3 significant figures: ± 0.1% (j) 4 significant figures: ± 0.01%
2.4.3 -3
(b) 5. ×10 : 1 significant figure 3 Or 5.0×10 : 2 significant figures 3
5
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 3
(h) 5.61×10 : 3 significant figures 0 (i) 1.000×10 : 4 significant figures -
(j) 1.×10 : 1 significant figures 1 Or 1.0×10 : 2 significant figures 1
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).
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. Thus
0.05 × 100% = 3.6% error 1.4
Similarly, 3.84 has
0.005 × 100% = 0.13% error 3.84
and 0.36 has
0.005 × 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).
(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
a)
4 g mol mg Cl 2 (95.23)g M gCl2 = 380.9 g gmol M gCl2
b)
2 lb mol C3H8 (44.09 )lb C 3H8 454g C3 H8 4 = 4 × 10 g C3 H8 lb mol C3 H8 1 lb C3 H8
c)
(d)
16 g N2
gmol N 2 1 lb mol N 2 = 1.26 ×10−3 lb mol N 2 (28.02)g N2 454 gmol N 2
3 lb C2 H 6 O 1 lb mol C2 H 6 O 454 g mol = 29.56 g mol C2 H 6 O (46.07) lb C2 H 6 O 1 lb mol
2.6.5
(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)
(d)
2.6.6
11.9 g mol NaNO3
85 g NaNO3 2.20 × 10-3lb = 2.23 lb NaNO3 1 g mol NaNO3 1g
164 g mol SiO 2 60.1 g SiO 2 2.20 × 10-3lb = 21.7 lb SiO 2 1 g mol SiO 2 1g
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)
2.6.7
Vitamin A, C20O H30, Mol Wt.: 286
a.
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 =
1.00 lb mol 176 lb 454 g = 79,900 g 1 lb mol 1 lb
For both
12 lb 454 g = 5450 g 1 lb
2.6.8
2.6.9
Mass fraction to mole fraction:
Mole fraction to mass fraction
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 =
2.6.12
MW 44 28 32 18
g 849.2 2018.8 208.0 37.8 3113.8
3113.8 = 31.138 100.0 Basis: 100 lb mol
Comp. CO2 CO N2
% = mol 16 10 30
MW 44 28 28
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
lb 2640 280 840 3760
Section 2.6.3 Density and Specific Gravity
2.6.14 100 lb m
2.6.15
ft 3 = 0.0305 ft 3 62.4 × 5.25 lb m
Basis: 1000 lb oil
lb oil lb H2 O 62.4 3 ft 3 = 57.78 lb oil ft lb H2 O ft 3 oil 1.00 3 ft 0.926
1000 lb oil
2.6.16
1 ft 3 oil 7.48 gal 57.78 lb oil 1 ft 3 = 129.5 gal
Basis: 10,010 lb
10,010 lb 1 gal 0.134 ft3 = gal 8.80 lb 152 ft 3 Basis: 1 g mol each compound
2.6.17
g mol
mw
g
ρ (g/cm3)
V (cm3)
Pb
1
207.21
207.21
11.33
18.3
Zn
1
65.38
65.38
7.14
9.16
C
1
12.01
12.01
2.26
5.31
V̂ =
mass (g) density (g/cm3 )
Set 2 is the correct one.
Section 2.7.2 Pressure and Hydrostatic Head 2.7.1
Basis: 15cm3 water a.
b.
m = ρV=
1000 kg 10 m 10 m 0.15 m = 15,000 kg m3
F m g 15,000 kg 9.80 m 1N 1 Pa = = = 1470 Pa 2 10 m 10 m 1 (kg)(m) N A A s s2 m2 = 1.47 kPa
1.470 kPa
or
1 atm 14.7 psi = 0.21 psi 101.3 kPa atm
15 cm H 2 O
1 in. 1 ft 14.696 psi = 0.21 2.54 cm 12 in. 33.91 ft H 2 O
2.7.2
Volume of concrete: Sides
Vs =
5 m 30.2 m 200 mm
1m 2 sides = 60.4 m 3 1000 mm
5 m 27.2 m 200 mm
1m 2 ends = 54.4 m3 1000 mm
Ends
Ve = Floor
Vf =
27.4 m 30.4 m 200 mm
1m 1000 mm
= 166.592 m
3
Total volume = 281.392 m3 Mass of concrete =
281.392 m3 2080 kg = 585,295 m3
Volume of displaced water required to float:
586,543.36 kg 1 m 3 H2 O = 586.543 m 3 1000 kg H2 O V = LWh → h = h= 2.7.3
V LW
586.54 m 3 27.4m 30.4 m
= 0.703 m
The pressure is a gauge pressure. Basis: 50.0 psig a.
50.0 psig 33.91 ft H2 O = 115 ft 14.7 psia (difference)
b.
No. Insufficient height.
Alternate solutions can be applying ∆p = ρ∆hg . 2.7.4
lb f in.2 = lb f → lb m in the AE system in.2 so the procedure is ok, although the unit conversion is ignored.
60°F
148 ft
2.7.5
Basis: Dept = 1000 m
p = p o + ρgh
p = 1 atm +
1000 m 1.024 g 9.8 m 1 kg 100 cm s2 103 g 1 m cm 3
=
3
1 kPa 1 atm 1 N (kg)(m ) 1N 1.013 × 10 5 N / m 2 (kg)(m ) s2 s2
100.1 atm 101.3 kPa 4 = 1.014 × 10 kPa 1 atm
Alternative solution: ATop = 730.62 m 2 kPa 2.7.6
(a) p ABS = p Gauge + p atm
22.4 lb f 144 in.2 28.6 in. Hg 14.7 psia 144 in.2 2 = 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 2 O 28.6 in. Hg 33.91 ft H 2 O + = 84.1 ft H 2 O 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.
2.7.8
∆p = ρgh a.
=
1 g 9.8 m 13.1 m 1 kg (100 cm )3 Pa 1 kPa 1 kg 1000 Pa m3 cm 3 s2 1000 g m s2
= 128. 4 kPa b.
mg ρVg ρ(S.A.)(t )g F = = = ABottom A ABottom A Bottom
π 2 π (30.5 m )2 2 A Bottom = D = = 730.62 m 4 4 ATOP = 730.62 m 2 Aside = πDh = π (30.5 m )(13.1m ) = 1255.2 m 2 S.A.= ATop + A Bottom + ASide 2
= (730.62 + 730.62 + 1255.22 ) m = 2716.46 m
2
ρ(S.A.)(t )g = A
(
7.86 g 100 cm m3 cm 3
3
)2,716.46 m 2 9.35 × 10 –3 m
1 kg 1 Pa 1 kPa = 2 1000 g 1 kg / ms 1000 Pa
2.68 kPa
9.8 m 730.6 m s2 2
2.7.9
p A + ρ1gh1 + ρ 2 gΔh = PD p B + ρ1gh1 + ρ 2 gΔh = PD
p A -p B = ρ1gh1 +ρ 2 gΔh − ρ1gh1 − ρ1gΔh = ρ 2 gΔh − ρ1gΔh = gΔh(ρ 2 − ρ1 ) =
9.8 m 0.78in 1m 13.546 − 0.91g 1kg 3.28ft 1000 g s2 cm3
1 N 1 ft 100 cm m 2 1 Pa 760 mm H2 12 in 1 m 1 kg 1 N 101.3 × 10 3 Pa (m )(s)2 m 2 3
=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 2.7.13
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
2.7.15
Basis: 1 gal of solution. Mass of solution:
lb H O 1.0824 lb soln 62.4 3 2 3 ft H 2O ft H2 O 1 ft 3 1 gal = 9.03 lb soln . lb H2 O 7. 481 gal 1.00 3 ft H 2O
Mass fraction KOH =
0.813 lb 9.03 lb
Mass fraction H2O = 1 – 0.09
2.7.16
0.91 1.00 Total
Basis: 30 lb gas Comp. CO2 N2
2.7.17
=
= 0.09
lb 20 10 30
a)
1000
b)
No
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.
2.7.18
On a paper-free basis, the total ppm are: Brand A: 6060 ppm
Brand B: 405 ppm
The respective mass fractions are: The other entries are similar
Fe
Cu
Brand A:
1310 = 0.216 6060
2000 = 0.330 6060
Brand B:
350 = 0.864 405
50 = 0.123 405
2750 = 0.454 6060
5 = 0.0123 405
Vs =
5 m 30.2 m 200 mm
1m 2 sides = 60.4 m 3 1000 mm
5 m 27.2 m 200 mm
1m 2 ends = 54.4 m3 1000 mm
Ends
Ve =
Pb
Floor
Vf =
27.4 m 30.4 m 200 mm
1m 1000 mm
= 166.592 m 3
Total volume = 281.392 m3 Mass of concrete =
281.392 m3 2080 kg = 585,295 kg m3
Volume of displaced water required to float:
586,543.36 kg 1 m 3 H2 O 3 = 586.543 m 1000 kg H2 O V = LWh → h = h=
2.7.19
V LW
586.54 m 3 27.4m 30.4 m
Basis: 190,000 ppm
190,000 g PCB × 100 = 19% 106
= 0.703 m
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) + 3CO 2 (g) a.
b.
2.7.22
2×107 disintegrations 1 min min
60
1 curie
106 μ curie
3×1010 disintegrations/s
1 curie
2 ×107 disintegrations 0.80 = min
= 11μ curie
1.6 ×107 cpm
The relation to use is t1/ 2 = n(2) /(k)(OH − ) T with (OH −= ) 1.5 ×106 k
t1/2 = (seconds)
Methanol Ethanol MTBE
30.8 ×105 4.6 ×105 7.7 ×105
0.15 ×10−12 1×10−12 0.60 ×10−12
The order is in increasing persistence ethanol, MTBE, and methanol.
2.7.23
Yes.
Bases are first entries.
4800 g mol CC14 103 g mol air kg mol air 154 g CC14 103 mg CC14 = 109 g mol air kg mol air 22.8 m3 g mol CC14 g CC14 32.4 mg CC14 /m3 which exceeds the NIOSH standards 2.7.24
a.
25,600 ton P 2.6 yr 1 1 gal 2000 lb 454 g 106 µ g yr 1.2 × 1014 gal 3.785 L 1 ton 1 lb 1 g
= 133.1 μg/L
b.
19,090 lb(P )municipal = 63. 4% 30,100 lb(P)total
c.
19,090 lb(P )municipal 0.70 lb(P)det. 100 lb(P )total = 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 =
17,843 ton 2.6 yr 1 2000 lb 1 g / cm 3 109 g = 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
P conc. in ppb =
ton yr
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.
Basis: 106 g mol gas
2.7.25
350 g mol H 2S 1 g mol gas 34g H 2S 1 g mol CO 2 6 10 g mol gas 1 g mol CO 2 1 g mol H 2 O 44 g CO 2 =
270 g H 2S 270 g H 2S = 6 6 10 g CO 2 10 g total liquid
mass fraction H2S = 2.70 × 10−4 2.7.26
(a)
1 mol O 2 104 mol O 2 or 104 ppm ⇒ 6 100 mol gas 10 mol gas
(b)
Basis: 100 mol gas
Comp. SO3 SO2 O2 Total 2.7.27
MW
% = mol 55 3 1 59
answer 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 g CaCO3 = 2.50 g mol Ca CO3 1 g mol Ca 40.05 g Ca g Ca
Similarly
g CaCO3 g CaCO3 = 4.118 g Mg g Mg
Total alkalinity = 2.50 (56.4) + 4.118 (8.8) = 177
mg CaCO3 L
2.7.28
No.
1 molecule in 1023 or more is not 13-20 ppb
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.
3.2
Chapter 3 Solutions
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
3.2.3
3.2.4
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
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
3.2.6
The system is the radiator. Open System
Closed System
Steady State Unsteady State
a)
X
X
b)
X
X
c)
X (Before filling)
d)
X (After filling)
X (After filling)
X
X
X (Before filling)
3.2.7
system contains ice + H2O
Ice
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
3.2.9
(a)
(b)
(c)
(d) x
(e)
(f) x
(g)
1. 2. x x ____________________________________________________ 3. x x 4. Depends on the time period considered ____________________________________________________ 5. x x 6. x x x ____________________________________________________ 7. 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)
m final − 0
= 300 + 100
Out (kg)
−
380
n final = 20 kg 20 kg 60 min = 1200 kg min 1 hr 3.2.12
Basis: 600 kg solution Accumulation (kg)
n final − 0
In (kg) = 100 + 500
Out (kg)
−
300
n final = 300 kg if no water evaporates in which case less than 300 kg would remain
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
Mass in does not equal to mass out Reasons:
Out 404 ×103 228 ×103 152 ×103 101 ×103 67 ×103 36 ×103 100 ×103 1190 ×103 2,278 × 103
(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.
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
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)
∆m t = − 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) 3 cm 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.
3.2.20
Basis: 1 hour The overall material balance is In (lb) 106,000
?
Out (lb) 74,000 + 34,000 = 108,000
The error in the overall material balance is 2000 lb/h or 1.9%; therefore, the overall material balance is within the expected error for industrial flow sensors. Propylene balance:
0.7 × 106,000 ? (0.997) (74,000) + (0.1) (34,000) 74,200 ? 73,778 + 3,400 = 77,178 (3.8% error)
Propane balance:
0.4 × 106,000 ? (0.003) (74,000) + (0.9) (34,000) 31,800 ? 222 + 30,600 = 30,622 (3.8% error)
Note that the relative error for the component balances is twice as large as the relative error for the overall material balance, indicating that there is additional error in the composition measurements used for the component balances.
3.2.21
Basis: 100 kg wet sludge
The system is the thickener (an open system). No accumulation, generation, or consumption occurs. The total mass balance is In Out = 100 kg 70 kg + kg of water
Consequently, the water amounts to 30 kg.
3.3
A General Strategy for Solving Material Balance Problems
3.3.1
(a) Water and air. (b) Insulation, air, and what is in the atmosphere. (c) Yes (cold water in, hot water out) (d) Yes