Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING

SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
Episode 40 :

Calculation of Usalt
Calculation of ∆𝑷 𝑯𝟏
Calculation of ∆𝑷 𝑽
𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 ∆P Bends
Calculation of ΔPH2 and ΔP total
Calculation of Pg and ∆𝑷 𝒃𝒍𝒐𝒘𝒆𝒓
Conclusion
OUTLINE
Comparison of Results:

DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING
A plastics production plant wants to increase the capacity
through an existing conveying system. The existing system has
6 inch ID pipes and is configured as shown in the diagram
below.
The High Density Polyethylene (HDPE) particles have an
average size of 4 mm. The conveying gas is at 68oF. The
existing blower can produce 1375 SCFM.
The desired capacity increase is from 20,000 lbm/hr to 30,000
lbm/hr. Can the existing blower and pipe system meet this
increase in capacity?
Assume the pressure drop across the cyclone is 5 inches of
water. The pressure drop across the blower inlet pipe and
silencers is 0.3 psi. The pipe bends have R/D = 6. Pipe
roughness is k = 0.00015 ft. The particles have density pρ = 59
lbm/ft3. Terminal velocity of the particles is = 30.6 ft/s.

Conversion factors
LH1 325 ft 99 m
LH2 100 ft 30.48 m
LV 50 ft 15.24 m
ρf 0.0753 Ibm /ft 3 1.2 kg/m3
ρp 59 Ibm /ft 3 947.3 kg/m3
D 6 in 0.1524 m
Mp 30000 Ibm/hr 3.78 kg/s
x 4 mm 0.004 m
Gas
flowrate
1375 ft3/min 0.649 m3/s

Calculation of Usalt:
𝑈𝑠𝑎𝑙𝑡 =
4𝑀 𝑝10 𝛼 𝑔
𝛽
2 𝐷
𝛽
2 −2
𝜋𝜌 𝑓
1
𝛽+1
𝛼 = 1440 𝑥 + 1.96
𝛼 = 1440 0.004 + 1.96
𝛼 = 7.72
𝛽 = 1100 𝑥 + 2.5
𝛽 = 1100 0.004 + 2.5
𝛽 = 6.9
𝑈𝑠𝑎𝑙𝑡 =
4(3.78)107.72 9.81
6.9
2 0.1524
6.9
2 −2
𝜋(1.2)
1
6.9+1
𝑼 𝒔𝒂𝒍𝒕 = 𝟐𝟏. 𝟔𝟕 𝒎/𝒔 (71 𝑓𝑡/𝑠)
𝑈 = 1.5 𝑈𝑠𝑎𝑙𝑡
𝑈 = 1.5(21.67)
𝑼 = 𝟑𝟐. 𝟓 𝒎/𝒔 (𝟏𝟎𝟔 𝒇𝒕/𝒔)

𝑈 𝑃𝐻 = 𝑈(1 − 0.0638 𝑥0.3 𝜌 𝑝
0.5)
𝑈 𝑃𝐻 = 32.5(1 − 0.0638 (0.004)0.3(947.3)0.5)
𝑈 𝑃𝐻 = 20.322 𝑚/𝑠
𝐺 = 𝜌 𝑃(1 − 𝜀 𝐻) 𝑈 𝑃𝐻
𝜀 𝐻 = 1 −
𝑀 𝑃
𝐴𝜌 𝑝 𝑈 𝑃𝐻
𝜀 𝐻 = 1 −
3.78
(0.01823)(947.3)(20.322)
𝜀 𝐻 = 0.9892
𝑈𝑓𝐻 =
𝑈
𝜀 𝐻
𝑈𝑓𝐻 =
32.5
0.9892
𝑈𝑓𝐻 = 32.854 𝑚/𝑠

𝑅𝑒 𝑝 =
𝜌 𝑓 𝑈𝑓𝐻 − 𝑈 𝑃𝐻 𝑥
𝜇
𝑅𝑒 𝑝 =
1.2 32.854 − 20.322 (0.004)
18.4𝑥10−6
𝑅𝑒 𝑝 = 3269.2 (𝑁𝑒𝑤𝑡𝑜𝑛 𝑟𝑒𝑎𝑔𝑖𝑜𝑛)
𝐶 𝐷 = 0.44
𝑓𝑝 =
3
8
𝜌 𝑓
𝜌 𝑝
𝐷
𝑥
𝐶 𝐷
𝑈𝑓 − 𝑈 𝑃
𝑈 𝑝
2
𝑓𝑝 =
3
8
1.2
947.3
0.1524
0.004
(0.44)
32.854 − 20.322
20.322
2
𝑓𝑝 = 0.003
𝑓𝑔 = 0.005

Calculation of ∆𝑷 𝑯𝟏:
∆𝑃 𝐻1 =
𝜌 𝑓 𝜀 𝐻 𝑈𝑓𝐻
2
2
+
𝜌 𝑝(1 − 𝜀 𝐻)𝑈 𝑃𝐻
2
2
+
2𝑓𝑔 𝜌 𝑓 𝑈2
𝐿 𝐻1
𝐷
+
2𝑓𝑝 𝜌 𝑝(1 − 𝜀 𝐻)𝑈 𝑃𝐻
2
𝐿 𝐻1
𝐷
∆𝑃 𝐻1
=
(1.2)(0.9892)(32.854)2
2
+
947.3(1 − 0.9892)(20.322)2
2
+
2(0.005)(1.2)(32.5)2(99)
0.1524
+
2(0.003)(947.3)(1 − 0.9892)(20.322)2(99)
0.1524
∆𝑷 𝑯𝟏 = 𝟐𝟕𝟒𝟓𝟓 𝑷𝒂 (𝟑. 𝟗𝟖𝟐 𝑷𝒔𝒊𝒈)

Calculation of ∆𝑷 𝑽:
𝑈 𝑇 = 1.74
𝑋 𝜌 𝑝−𝜌 𝑓
𝑔
𝜌 𝑓
0.5
= 1.74
0.004 947.3 − 1.2 9.81
1.2
0.5
𝑼 𝑻 = 𝟗. 𝟔𝟖 𝑴/𝑺
𝑈 𝑃 𝑉
=
𝑈
𝜀 𝑉
− 𝑈 𝑇
𝐺 = 𝜌 𝑃 1 − 𝜀 𝑉 𝑈 𝑃 𝑉
𝐺 =
𝑀𝑃
𝐴
=
3.78
0.01823
= 207.35
𝜀 𝑉
2
9.68 − 9.68 + 32.5 +
207.35
947.3
𝜀 𝑉 + 32.5 = 0
9.68𝜀2 − 42.4𝜀 𝑉 + 32.5 = 0
𝜺 𝑽 = 𝟎. 𝟗𝟗𝟎𝟓

𝑈 𝑃𝑉 =
𝑈
𝜀 𝑉
− 𝑈 𝑇
=
32.5
0.9905
− 9.68
𝑼 𝑷𝑽 = 𝟐𝟑. 𝟏𝟑
𝑴
𝑺
𝑃𝑉 =
2𝑓𝑔 𝜌 𝑓 𝑈2
𝐿 𝑉
𝐷
+
2𝑓𝑃 𝜌 𝑃 1 − 𝜀 𝑉 𝑈 𝑃𝑉
2
𝐿 𝑉
𝐷
+ 𝜌 𝑃 1 − 𝜀 𝑉 𝑔𝐿 𝑉 sin 𝜃 + 𝜌 𝑓 𝜀 𝑣 𝑔𝐿 𝑉 sin 𝜃
𝜃 = 0 → sin 𝜃 = 1
𝑃𝑉 =
2 0.005 1.2 32.5 2 15.24
0.1524
+
2 .003 947.3 1 − 0.9905 23.13 2 15.13
0.1524
+ 947.3 1 − 0.9905 9.81 15.24 + 1.2 0.9905 9.81 15.24
∆𝑷 𝑽= 𝟓𝟔𝟕𝟗. 𝟒 𝑷𝒂

𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 ∆P Bends:
Fro two bends in the system:
∆p Bend = 7.5 M Of vertical pipe
∆𝑃 𝑉
𝐿 𝑉
=
5679.4
15.25
= 372.66 Pa/m
∆PBend = 2 × 7.5 × 372.66
∆PBend= 5590 𝑃𝑎

Calculation of ΔPH2 and ΔP total :
∆𝑃 𝐻2 =
𝜌 𝑓 𝜀 𝐻 𝑈𝑓𝐻
2
2
+
𝜌 𝑝(1 − 𝜀 𝐻)𝑈 𝑃𝐻
2
2
+
2𝑓𝑔 𝜌 𝑓 𝑈2 𝐿 𝐻2
𝐷
+
2𝑓𝑝 𝜌 𝑝(1 − 𝜀 𝐻)𝑈 𝑃𝐻
2
𝐿 𝐻2
𝐷
∆𝑃 𝐻2
=
(1.2)(0.9892)(32.854)2
2
+
947.3(1 − 0.9892)(20.322)2
2
+
2(0.005)(1.2) 32.5 2(30.48)
0.1524
+
2(0.003)(947.3)(1 − 0.9892) 20.322 2
(30.48)
0.1524
∆𝑷 𝑯𝟐 = 𝟏𝟎𝟑𝟓𝟖. 𝟑𝟔 𝑷𝒂 (𝟏. 𝟓𝟎𝟐 𝑷𝒔𝒊𝒈)
∆𝑃 𝑇𝑜𝑡𝑎𝑙 = ∆𝑃 𝐻1 + ∆𝑃 𝐻2 + ∆𝑃𝑉 + ∆𝑃𝐵𝑒𝑛𝑑
∆𝑃 𝑇𝑜𝑡𝑎𝑙 = 27455 + 10358.36 + 5679.4 + 5590
∆𝑃 𝑇𝑜𝑡𝑎𝑙 = 49082.76 𝑃𝑎 (7.119 𝑃𝑠𝑖𝑔)

Calculation of Pg:
Pressure drop in cyclone = 5 inches water
= 0.127 m water
= 1254.87 Pa
𝑃𝑏 = 𝑃𝑎 + ∆𝑃𝑐𝑦𝑐𝑙𝑜𝑛𝑒
𝑃𝑏 = 101325 + 1254.87
𝑃𝑏 = 102570.87 𝑃𝑎
𝑃𝑔 = 𝑃𝑏 + ∆𝑃 𝑇𝑜𝑡𝑎𝑙
𝑃𝑔 = 49082.76 + 102570.78
𝑷 𝒈 = 𝟏𝟓𝟏𝟔𝟓𝟑. 𝟔𝟑 𝑷𝒂

Calculation of ∆𝑷 𝒃𝒍𝒐𝒘𝒆𝒓:
Inlet pressure to blower Pin
𝑃𝑖𝑛 = 𝑃𝑎𝑡𝑚. − ∆𝑃𝑠𝑖𝑙𝑒𝑛.
𝑃𝑖𝑛 = 101325 − 2068.43
𝑃𝑖𝑛 = 99256.57 𝑃𝑎
∆𝑃𝑏𝑙𝑜𝑤𝑒𝑟= 𝑃𝑔 − 𝑃𝑖𝑛
∆𝑃𝑏𝑙𝑜𝑤𝑒𝑟= 151653.63 − 99256.57
∆𝑷 𝒃𝒍𝒐𝒘𝒆𝒓= 𝟓𝟐𝟑𝟗𝟕. 𝟎𝟔 𝑷𝒂 𝟕. 𝟏𝟏𝟗 𝑷𝒔𝒊𝒈

Conclusions:
Calculation of gas velocity supplied by blower:
𝑈𝑔 =
𝑄 𝑔
𝐴
𝑈𝑔 =
0.649
0.01823
= 35.6 𝑚/𝑠
The velocity in the pipeline supplied by blower is (𝟑𝟓. 𝟔 𝒎/𝒔 ),
hence the velocity everywhere in the pipeline exceeds the
saltation velocity (21.67 m/s).
Assuming the blower is capable of the 52.397 kPa pressure
increase, the velocity provided by the flow rate of 1375
SCFM exceeds the saltation velocity everywhere in the pipe
line; hence the blower and pipe system is capable of
conveying 3.78 kg/s of solids.

Item Results by Martin
Rhods equations
Solved example results
BTU SI unit
U salt 𝟐𝟏. 𝟔𝟕 𝒎/𝒔 71.2 ft/s 21.7 m/s
ΔPH1 𝟐𝟕𝟒𝟓𝟓 𝑷𝒂 1.729 psig 11921 Pa
ΔPV 𝟓𝟔𝟕𝟗. 𝟒 𝑷𝒂 0.465 psig 3206 Pa
ΔPH2 𝟏𝟎𝟑𝟓𝟖. 𝟑𝟔 𝑷𝒂 0.473 psig 3261.2 Pa
ΔP Bend 𝟓𝟓𝟗𝟎 𝑷𝒂 0.559 psig 3845.2 Pa
ΔP Total 𝟒𝟗𝟎𝟖𝟐. 𝟕𝟔 𝑷𝒂 3.226 psig 22242.5 Pa
ΔP blower 𝟓𝟐𝟑𝟗𝟕. 𝟎𝟔 𝑷𝒂 4.44 pig 30612.7 Pa
Comparison of Results:

It is obvious from the above table that the
pressure loss calculated by Martin Rhods
equations more than the results in the
solved example. That difference may
because the Martin Rhods equations depend
on the saltation velocity and the equations
in the solved example depend on gas
velocity in each point.

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Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING

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Episode 40 : DESIGN EXAMPLE – DILUTE PHASE PNEUMATIC CONVEYING