Chapter 3 Unit Operations Problems
1.
Manometer with liquid hydrocarbon
Specific gravity 0.74
Pressure below atmospheric by 83cm
(a) Vacuum in the evaporator
Density of liquid hydrocarbon = 0.74 x 1000 kgm-3
Z = 83cm = 0.83 m
Z
=
P/g
0.83 =
P/(0.74 x 1000 x 9.81)
P
=
(0.83 x 0.74 x 1000 x 9.81)
[m x kg m-3 x ms-2]
-1 -2
=
6025 kg m s
=
6025 Pa
=
6.025 kPa
And this is the pressure below atmospheric so is equal to the vacuum.
The absolute pressure
=101.3 – 6.025
= 95.3kPa
= 95kPa
(b) Boiling Temperature of water at 95kPa
From Appendix 8 steam tables:
Pressure
Temperature
80kPa
93.5oC
100kPa
99.6oC
Interpolating linearly:
Therefore
95kPa
98.1oC
2.
Temperature of milk = 20 oC
From Appendix 4, for milk,
Density
=
1030 kg m-3
Viscosity
=
2.12 x 10-3 Nsm-2 = 2.12 x 10-3 kg s-1 m-1
Diameter
=
4cm =
0.04m
Length
=
130m
Area
=
 D2/4 =
3.14 (0.04)2 /2
= 1.26 x 10-3m2
-1
Velocity
=
2.7 ms
Mass rate of flow
=
Avρ
=
1.26 x 10-3x 2.7 x 1030
[m2 x ms-1 x kg m-3]
=
3.49 kgs-1
Ec
=
v2/2 =
(2.7)2/2
= 3.65 J kg-1
Power to provide kinetic energy
=
Ec x Mass rate of flow
=
3.65 x 3.49
=
12.74 J s-1
For friction energy:
(Re) =
(Dv/)
=
(0.04 x 2.7 x 1030)/ (2.12 x 10-3)
=
52,470
Therefore the flow is turbulent.
From Figure 3.8, the roughness factor, f = 0.006
Power to provide friction energy
=
(4fv2/2)(L/D)
=
(4 x 0.006 x (2.7)2/2)(130/0.04)
=
284.31 J s-1
Total power needed
=
12.74 + 284.31
=
297.05 J s-1
1 horse power
=
7.46 x 102 J s-1
Required power
=
0.40 hp
3.
22% sodium chloride at 10oC
Density
=
1160 kg m-3
Z
=
40m
D
=
0.015m
Area
=
1.77 x 10-4 m2
Rate
=
8.1 m3h-1
Mass rate
=
2.25 x 10-3 x 1160 kg s-1
= 2.25 x 10-3 m3s-1
= 2.61 kg s-1
(a)
Velocity
=
Velocity head =
=
(2.25 x 10-3m3s-1)/ (1.77 x 10-4 m2) =
v2 /2g = (12.7)2 /(2 x 9.81)
8.2m
12.7ms-1
(b)
Potential head =
Total head
Z
40m
48.2m
Ec
Zg
48.2 x 9.81
[m x ms-2]
-1
472.84Jkg
(Can also be calculated from Ec =
Ec x Mass rate of flow
472.84 x 2.61
1234 J s-1
7.46 x 102 J s-1
1.65 hp
68%
2.42hp
=
=
=
Total power
=
=
=
1 horsepower =
Required power =
Efficiency
=
Pump power =
4.
=
=
Milk Cooler
Diameter
Mass
Density
Viscosity
=
=
=
=
4 cm
=
-1
10,000kgh
=
1030kgm-3
2.12x10-3 Nsm-2
(Re)
(4000)
Velocity v
=
=
=
(Dv /µ )
(0.04 x v x 1030)/(2.12 x 10-3)
0.21 ms-1
Overall volume flow rate
=
=
=
[1Nm =1J]
Zg + v2 /2)
0.04m
2.8 kgs-1
=
mass /(density x time)
2.8/1030 m3 s-1
2.7 x 10-3 m3 s-1
2.12x10-3 kg s-1 m-1
[ m x ms-1x kgm-3/kgs-1m-1]
Pipe x-section
=
=
=
Volume flow rate in one tube =
=
=
Number of tubes in parallel =
=
=
5.
D2/4
0.042/4
1.256 x 10–3 m2
Axv
1.256 x10–3 m2 x 0.21 ms-1
0.26 x10–3 m3 s-1
2.7 x10-3 / 0.26 x10–3
10.38
11
Soya Bean Oil
From Appendix 4.
Viscosity
=
40x10-3 Nsm-2 = 40 kg s-1 m-1
Density
=
910 kgm-3
Length
=
148 m
Diameter
=
0.05m
6 RH bends, 2 gate valves, 1 globe valve
Equivalent Z =
3m
Mass
=
20 tonnes per hour
= 20x1000/60x60 = 5.56kgs-1
V
=
m/
=
(20 x 1000)/910
=
21.98 m3 h-1
=
6.10 x 10-3 m3 s-1
v
=
V/A
=
(6.10 x 10-3) /( π 0.052/4)
=
3.11ms-1
(Re) =
(Dv/µ )
=
(0.05 x 3.11x 910)/(40 x 10-3)
=
3.53 x 103
=
3530
Therefore flow is transitional and can take f = 0.15
Ef
=
(2f v2)(L/D) + 6 k1 v2 /2 + 2k2 v2 /2+ 1k3 v2 /2
=
v2{ 2f((L/D) + 6 k1 /2 + 2k2 /2+ 1k3 /2}
=
3.112 (2x 0.15 x 148/0.05 + 6 x 0.74/2 + 2 x 0.13/2 + 1 x 6 /2)
=
9.67 (888 + 2.22 + 0.13 + 3)
=
8638.7 Jkg-1
But fall is Z =
3m
P
=
Zg
=
3 x 9.81
Power in potential energy
=
Zg
=
3 x 9.81 x 910
=
26.78 x103Js-1
Power in total frictional energy
=
mass x Ef
=
5.56 x 8639
=
48.03 x 103 Js-1
Theoretical power needed
=
48.03x103 – 26.78 x 103
=
21.25 x 103Js-1
Efficiency
=
70%
Therefore actual power needed
=
21.25 x 103/ 0.7
=
=
=
6.
30.35 x 103Js-1
30.35 x 103 /7.46 x 102
41 hp
Air dryer at 80oC
Volume
=
100 m3min-1 =
100/60
= 1.67 m3s-1
Rectangular cross section area
=
0.6m x 1.4m = 0.84 m2
For 4 right angle bends of long radius:
Pressure drop = 0.25 x 4
=
1.0 velocity heads
Pressure drop in bed
=
4 velocity heads
Pressure drop in heater coils =
1.2 velocity heads
Total pressure drop
=
6.2 velocity heads
Velocity
=
1.67/0.84
=
1.99 ms-1
Velocity head
=
v2/2g
=
1.992/(2 x 9.81)
=
0.20m
Total pressure drop =
6.2 x 0.20
=
1.24m
Zair air
Zwater
=
Zwater water
=
Zair air/water
=
1.24 x 1 / 1,000
=
0.00124m
=
0.124cm
Therefor the fan will work in this circuit.