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Contents
Preface iv
1. Units and dimensions 1
2. Flow of fluids — energỵ and momentum relationships 16
3. Flow in pipes and channels 19
4. Flow of compressible fluids 60
5. Flow of multiphase mixtures 74
6. Flow and pressure measurement 77
7. Liquid mixing 103
8. Pumping of fluids 109
9. Heat transfer 125
10. Mass transfer 217
11. The boundarỵ laỵer 285
12. Momentum, heat and mass transfer 298
13. Humidification and water cooling 318
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SECTION 1
Units and Dimensions
PROBLEM 1.1
98% sulphuric acid of viscositỵ 0.025 N s/m2 and densitỵ 1840 kg/m3 is pumped at
685 cm3/s through a 25 mm line. Calculate the value of the Reỵnolds number.
Solution
Cross-sectional area of line D ⊲ /4⊳0.0252 D 0.00049 m2.
Mean velocitỵ of acid, u D ⊲685 ð 10 6⊳/0.00049 D 1.398 m/s.
∴ Reỵnolds number, Re D du / D ⊲0.025 ð 1.398 ð 1840⊳/0.025 D 2572
PROBLEM 1.2
Compare the costs of electricitỵ at 1 p per kWh and gas at 15 p per therm.
Solution
Each cost is calculated in p/MJ.
1 kWh D 1 kW ð 1 h D ⊲1000 J/s⊳⊲3600 s⊳ D 3,600,000 J or 3.6 MJ
1 therm D 105.5 MJ
∴ cost of electricitỵ D 1 p/3.6 MJ or ⊲1/3.6⊳ D 0.28 p/MJ
cost of gas D 15 p/105.5 MJ or ⊲15/105.5⊳ D 0.14 p/MJ
PROBLEM 1.3
A boiler plant raises 5.2 kg/s of steam at 1825 kN/m2 pressure, using coal of calorific
value 27.2 MJ/kg. If the boiler efficiencỵ is 75%, how much coal is consumed per daỵ?
If the steam is used to generate electricitỵ, what is the power generation in kilowatts assuming
a 20% conversion efficiencỵ of the turbines and generators?
1
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2 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Solution
From the steam tables, in Appendix A2, Volume 1, total enthalpỵ of steam at 1825 kN/m2 D
2798 kJ/kg.
∴ enthalpỵ of steam D ⊲5.2 ð 2798⊳ D 14,550 kW
Neglecting the enthalpỵ of the feed water, this must be derived from the coal. With an
efficiencỵ of 75%, the heat provided bỵ the coal D ⊲14,550 ð 100⊳/75 D 19,400 kW.
For a calorific value of 27,200 kJ/kg, rate of coal consumption D ⊲19,400/27,200⊳
D 0.713 kg/s
or:⊲0.713 ð 3600 ð 24⊳/1000 D 61.6 Mg/daỵ
20% of the enthalpỵ in the steam is converted to power or:
⊲14,550 ð 20⊳/100 D 2910 kW or 2.91 MW saỵ 3 MW
PROBLEM 1.4
The power required bỵ an agitator in a tank is a function of the following four variables:
(a) diameter of impeller,
(b) number of rotations of the impeller per unit time,
(c) viscositỵ of liquid,
(d) densitỵ of liquid.
From a dimensional analỵsis, obtain a relation between the power and the four variables.
The power consumption is found, experimentallỵ, to be proportional to the square of
the speed of rotation. Bỵ what factor would the power be expected to increase if the
impeller diameter were doubled?
Solution
If the power P D F⊲DN ⊳, then a tỵpical form of the function is P D kDaNb c d, where k
is a constant. The dimensions of each parameter in terms of M, L, and T are: power, P
D ML2/T3, densitỵ, D M/L3, diameter, D D L, viscositỵ, D M/LT, and speed of rotation,
NDT 1
Equating dimensions:
M: 1 DcCd
L: 2 D a 3c d
T: 3D b d
Solving in terms of d : a D ⊲5 2d⊳, b D ⊲3 d⊳, c D ⊲1 d⊳
∴ PDk D5 N3 d
D2d Nd d
or: P/D5N3 D k⊲D2N / ⊳ d
that is: NP D k Rem
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Contents
Preface iv
1. Units and dimensions 1
2. Flow of fluids — energỵ and momentum relationships 16
3. Flow in pipes and channels 19
4. Flow of compressible fluids 60
5. Flow of multiphase mixtures 74
6. Flow and pressure measurement 77
7. Liquid mixing 103
8. Pumping of fluids 109
9. Heat transfer 125
10. Mass transfer 217
11. The boundarỵ laỵer 285
12. Momentum, heat and mass transfer 298
13. Humidification and water cooling 318
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SECTION 1
Units and Dimensions
PROBLEM 1.1
98% sulphuric acid of viscositỵ 0.025 N s/m2 and densitỵ 1840 kg/m3 is pumped at
685 cm3/s through a 25 mm line. Calculate the value of the Reỵnolds number.
Solution
Cross-sectional area of line D ⊲ /4⊳0.0252 D 0.00049 m2.
Mean velocitỵ of acid, u D ⊲685 ð 10 6⊳/0.00049 D 1.398 m/s.
∴ Reỵnolds number, Re D du / D ⊲0.025 ð 1.398 ð 1840⊳/0.025 D 2572
PROBLEM 1.2
Compare the costs of electricitỵ at 1 p per kWh and gas at 15 p per therm.
Solution
Each cost is calculated in p/MJ.
1 kWh D 1 kW ð 1 h D ⊲1000 J/s⊳⊲3600 s⊳ D 3,600,000 J or 3.6 MJ
1 therm D 105.5 MJ
∴ cost of electricitỵ D 1 p/3.6 MJ or ⊲1/3.6⊳ D 0.28 p/MJ
cost of gas D 15 p/105.5 MJ or ⊲15/105.5⊳ D 0.14 p/MJ
PROBLEM 1.3
A boiler plant raises 5.2 kg/s of steam at 1825 kN/m2 pressure, using coal of calorific
value 27.2 MJ/kg. If the boiler efficiencỵ is 75%, how much coal is consumed per daỵ?
If the steam is used to generate electricitỵ, what is the power generation in kilowatts assuming
a 20% conversion efficiencỵ of the turbines and generators?
1
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2 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Solution
From the steam tables, in Appendix A2, Volume 1, total enthalpỵ of steam at 1825 kN/m2 D
2798 kJ/kg.
∴ enthalpỵ of steam D ⊲5.2 ð 2798⊳ D 14,550 kW
Neglecting the enthalpỵ of the feed water, this must be derived from the coal. With an
efficiencỵ of 75%, the heat provided bỵ the coal D ⊲14,550 ð 100⊳/75 D 19,400 kW.
For a calorific value of 27,200 kJ/kg, rate of coal consumption D ⊲19,400/27,200⊳
D 0.713 kg/s
or:⊲0.713 ð 3600 ð 24⊳/1000 D 61.6 Mg/daỵ
20% of the enthalpỵ in the steam is converted to power or:
⊲14,550 ð 20⊳/100 D 2910 kW or 2.91 MW saỵ 3 MW
PROBLEM 1.4
The power required bỵ an agitator in a tank is a function of the following four variables:
(a) diameter of impeller,
(b) number of rotations of the impeller per unit time,
(c) viscositỵ of liquid,
(d) densitỵ of liquid.
From a dimensional analỵsis, obtain a relation between the power and the four variables.
The power consumption is found, experimentallỵ, to be proportional to the square of
the speed of rotation. Bỵ what factor would the power be expected to increase if the
impeller diameter were doubled?
Solution
If the power P D F⊲DN ⊳, then a tỵpical form of the function is P D kDaNb c d, where k
is a constant. The dimensions of each parameter in terms of M, L, and T are: power, P
D ML2/T3, densitỵ, D M/L3, diameter, D D L, viscositỵ, D M/LT, and speed of rotation,
NDT 1
Equating dimensions:
M: 1 DcCd
L: 2 D a 3c d
T: 3D b d
Solving in terms of d : a D ⊲5 2d⊳, b D ⊲3 d⊳, c D ⊲1 d⊳
∴ PDk D5 N3 d
D2d Nd d
or: P/D5N3 D k⊲D2N / ⊳ d
that is: NP D k Rem
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