Materials-Handling Equipment—Design
and Costs
Md Shahinoor Islam
Professor
Department of Chemical Engineering, BUET
Ref: Chap-12: Peter, M.S., Timmerhaus, K. D., West R., E., Plant Design
and Economics for Chemical Engineers, 5th Ed.
Materials-Handling Equipment
 Liquid: Pump
 Gas: Compressors, fans and blowers.
 Solid: Conveyors, chutes and hoists.
 Other types of special equipment: Blenders, mixers,
crushers, and grinders.
Basic Concept of Fluid Transport
 Power/mechanical energy is required to transfer materials to overcome
frictional resistance, changes in elevation, changes in internal energy,
and other resistances encountered in the flow system.
 At steady state the total energy balance equation per unit mass:
Eqn-12.01:
V12
V22
Z1 g + p1 1 +
+ u1 + q + W = Z 2 g + p2 2 +
+ u2
2
2
Z is the vertical distance above datum plane
g the local gravitational acceleration
p the absolute pressure
v the specific volume of the fluid
V the average fluid velocity
α the correction factor (1 for turbulent flow and 0.5 for viscous flow) to account for
the use of the average velocity
u the internal energy of the fluid
q the heat energy transmitted across the fluid boundary from an outside source
Mechanical Energy balance
Eqn-12.02:
Z1 g − 
2
1
V12
V22
vdp +
+ W0 = Z 2 g +
+F
2
2
W the total shaft work provided from an outside source
W0 the mechanical work provided from an outside source, and
F the mechanical energy loss due to friction.
Evaluation of

2
1
vdp term is difficult for a compressible fluid
However, for a noncompressible fluid, the specific volume v remains
constant, and the integral term reduces simply to v(p2 — p1).
Newtonian Fluids
Dp
Fanning Friction Factor, f =
2  2VL
DV 
Reynolds number, Re =

Laminar flow (Hagen-Poiseuille eqn):
16
16 
f =
=
; when Re  2100
Re DV 
Turbulent flow (Blasius eqn):
0.079
5
f =
;
when
4000

Re

10
0.25
( Re )
The Colebrook relation is a good approximation
of the friction factor for rough pipe over the entire
turbulent flow range:
  D 1.256
1
= −4 log 
+

f
 3.7 Re f

 ; Re  4000

If the velocity, viscosity and density remains constant, and the pipe
diameter is uniform over the total pipe length, the mechanical
energy loss due to friction may be obtained from:
2 fLV 2
F=
D
Newtonian Fluids
Power required for noncompressible fluids:
V2 
W0 = g Z +  
 +  ( pv ) +  F
 2 
 V2 
P
F
As head W0 = Z +  
+  + 
g
 
 2 g 
Eqn-12.12:
F
L V2
 g =  hL = f D 2 g ;α = 1for turbulent flow
The use of the mechanical energy balance is not recommended for
compressible fluids when large pressure drops in the flow are
involved. In such cases, the total energy balance should be used.
Rearrangement of the total energy balance is given by:
Eqn-12.13:
V2 
W0 = g Z +  
 + h − q
 2 
where h is the enthalpy and is equal to u + pv.
Table 12-1 Expressions for evaluating frictional losses in the flow of fluids through
pipeline systems
(For noncircular, cross-sectional area and turbulent flow, replace D by RH= 4 (crosssectional flow area/wetted perimeter)for an approximate frictional loss).
K CV2 2
Sudden contraction, FC =
;  = 0.5 for streamline flow
2
(V1 − V2 ) 2
Sudden enlargement, Fe =
;  = 1.0 for turbulent flow
2
Flow through long, straightpipe ofconstant cross-sectional area:
2 fV 2 L
F =
D
2 fV 2 Le
Fittings, valves, etc.: F =
D
(V0 − V2 ) 2
Rounded orifice: Fe =
2
Sharp-edged orifice, venturi: Fp=-p f
Pump Design and Costs Estimation
Example 12.1:
Water is pumped from a large reservoir into an open tank using a
standard steel pipe with an inside diameter of 0.0525 m. The
reservoir and the tank are open to the atmosphere. The difference in
vertical elevation between the water surface in the reservoir and the
discharge point at the top of the tank is 21.3 m. The length of the
pipeline is 305 m. Two gate valves and three standard 90° elbows
are part of the piping system. The efficiency of the pump is 40
percent and includes the losses at the entrance and exit of the
pump. If the flow rate of the water is to be maintained at 3.15 x I0-3
m3/s and the water temperature remains constant at 160C, estimate
the power rating of the motor required to drive the pump.
Use the mechanical energy balance between locations 1 (surface of
the water in the reservoir) and 2 (outside at the discharge point). By
rearranging Eq. (12-2), the mechanical work required is given by:
V22 V12
W0 = ( Z 2 − Z1 ) g + (
−
) + ( P2v2 − Pv
1 1) +  F
2 2
The selection of points 1 and 2 results:
V22/2α = 0, V12/2α = 0, and p2 = p1.
Water is noncompressible:
v2 = v1, P2v2 = P1v1. and Z2 — Z1=21.3 m.
g(Z2 – Z1) = (21.3)(9.806) = 208.7 N-m/kg
µ H2O (16 0C) = 1.12 x l0-3 Pa-s and ρ H2O (16 0C) = 997 kg/m3
Average velocity in the pipe, V =
Re =
DV 

3.15 10−3

(0.0525) 2
4
= 1.455 m / s
0.0525 1.455  997
=
= 68000
−3
1.12 10
 4.57 10−5
=
= 0.00087
D
0.0525
The friction factor from Fig. 12-1 is estimated as 0.0057.
The total Le for fittings and valves is
Le = 2(7)(0.0525) + 3(32)(0.0525) = 5.8 m
Friction due to flow through pipe and all fittings is :
2 f ( L + Le )V 2 2  0.0057  (305 + 5.8)  (1.455) 2
F=
=
= 143.4 N .m / kg
D
0.0525
Friction due to contraction and enlargement (from Table 12-1) is :
K CV2 2 (V1 − V2 ) 0.5  (1.455) 2 (1.455 − 0) 2
F=
+
=
+
= 1.6 N .m / kg
2
2
2 1
2 1
 F =(143.4 + 1.6) N .m / kg = 145.0 N .m / kg
The theoretical mechanical energy required from the pump is 208.7 + 145.0 = 353.7 N-m/kg.
353.7  3.15 10−3  997
W0 =
= 2780W  3kW
0.4
A software package designed to establish pumping requirements indicates that a 3.5-kW motor would
be needed. Assuming 85 percent efficiency for the motor, the software value for W0 , is also ~3 kW.
Problem 2:
Crude petroleum oil at a flow rate of 2×105 kg/h is pumped from bottom of a column to the
bottom of a storage tank through a 130 m long pipeline. The product having a viscosity of
0.6×10-3 N.s/m2 and a density of 800 kg/m3. The pipeline has 70 m of 200 mm ID at pump
suction and 60 m of 140 mm ID at pump discharge. The 200 mm pipe ID has 2(two) 900
standard radius elbows. The level of the liquid in the column is maintained at 4 m above
ground while the liquid level in the storage tank may vary from 5 m to 12 m, above the ground
level. If the pump operating efficiency is 70%, what will be the supply power in the pump?
Neglect the entrance and exit losses at the column and storage tank. Assume both distillation
column and storage tank were operated at atmospheric condition. Le/D for 900 standard radius
elbow is 32.

𝑓=

𝑓=
16
𝑁𝑅𝑒
for Re < 2100, and
0.04
𝑁𝑅𝑒 0.16
for Re > 2100
Pumps types and selection
➢
➢
Gas Pumps
Liquid Pumps
Selection of Pumps
 the total dynamic head required
 the suction and discharge heads
 temperature, viscosity, vapor pressure, and density of the fluid
 solids content in the liquid
 liquid corrosion characteristics
Table 12-4 General guidelines in the preliminary selection of liquid pumps
Type of
pump
Max
system
P, kPa
Max
Δp/stage,
kPa
Approx
cap limit,
m3/s
Pump
efficiency
range, %
Advantages or limitations
Centrifu
gal
(radial)
48000
2000
10
40-80
Simple, inexpensive, low maintenance cost,
viscosities <0.1 Pa-s, require priming, possible
cavitation, limited peak efficiency.
Axial
35000
200
10
50-80
Moderate cost, low maintenance cost, viscosities
<0.1 Pa-s, possible cavitation, high speed,
and low head.
Regener 5000
ative
(turbine)
3500
<1
20-40
Moderate cost and maintenance, handles
volatile liquids, viscosities <0.1 Pa-s, low
capacity, high head.
Gear
20000
0.1
40-85
Moderate cost, low maintenance cost, wide
range of viscosities to 400 Pa-s, low capacity,
high head, low noise, overpressure protection
recommended.
35000
Centrifugal Pumps
 Most widely used in the chemical industry for transferring
liquids.
 Capacity range 0.5 to 2 x 104 m3/h and discharge heads from a
few meters to approximately 4.9 x 103 m.
QN
Hp  N2
W  N3
Q = volumetric flow rate, N = speed of rotation,
Hp = pump head
W = power required (prime mover)
The preceding equations apply for the ideal case in which there
are no friction, leakage, or recirculation losses.
Net Positive Suction Head (NPSH)
VS2
− hS =
+
+  hL

 2g
Patm
PS
datum
hs
VS2 Patm
+
=
− hS −  hL
 2g

PVapor
PS VS2 PVapor Patm
+
−
=
− hS −  hL −
 2g



PVapor
Patm
( NPSH ) A =
− hS −  hL −
PS


For no cavitation:
(NPSH)A > (NPSH)R
Specific Speed
 Pump types may be more explicitly defined by the parameter
called specific speed (Ns) expressed by:
Ns =
N
H
Q
3
Where:
4
Q = discharge (m3/s, or l/s).
H = pump total head (m).
N = rotational speed (rpm).
Table 12.5: specific speeds of different pumps
Pump type
Ns range (Q – m3/s, H-m)
Process and feed
up to 2000
Turbine
Mixed flow
Axial flow
2000-5000
4000 to 10000
9000 to 15000
Pump Efficiency
Poweroutput Wo  QH t
p=
= =
Powerinput Wi
Wi
or
 QH t
Wi =
p
Which is the power input delivered from the motor to the impeller of
the pump.
Change in pump speed (constant size)
 If a pump delivers a discharge Q1 at a head H1 when running at
speed N1, the corresponding values when the same pump is running
at speed N2 are given by the similarity (affinity) laws:
Q2 N 2
=
Q1
N1
H2  N 2 
=

H1  N1 
2
where Q = discharge (m3/s, or l/s).
H = pump head (m).
N = pump rotational speed (rpm).
Wi = power input (hP, or kw).
Wi 2  N 2 

=
Wi1  N1 
3
Characteristic curves and flow controls
Pump groups
Multistage pumps
Costs for Pumps and Motors
Figures 12-19 through 12-24 provide approximate costs for different
types of pumps and motors that can be used for preliminary design
estimates; firm estimates should be based on vendors' quotations.
Example 1
For the following pump, determine the required pipes diameter to pump
60 L/s and also calculate the needed power.
Minor losses 10 V2/2g
Pipe length 10 km
Roughness = 0.15 mm
Head Hs = 20 m
Q
L/s
70
60
50
40
30
20
10
0
Ht
31
35
38
40.6
42.5
43.7
44.7
45
P
40
53
60
60
57
50
35
-
To get 60 L/s from the pump Hs + hL must be < 35 m
Assume the diameter = 300 mm
Then:
A = 0.070m 2 ,V = 0.85m / s
Re = 2.25 105 ,  / D = 0.0005, f = 0.019
0.019 10000  ( 0.85 )
hf =
= 23.32m
0.3 19.62
2
10  V 2 10  (0.85)
hm =
=
= 0.37m
2g
2g
2
hs + h f + hm = 43.69m  35m
Assume the diameter = 350mm
Then:
A = 0.0962m 2 ,V = 0.624m / s
Re = 1.93 105 ,  / D = 0.00043, f = 0.0185
h f = 10.48m,
10  V 2 10  (0.624)
hm =
=
= 0.2m
2g
2g
2
 hs + h f + hm = 30.68m 35m
The pump would deliver approximately 70 l/s through the 350 mm pipe and to
regulate the flow to 60 l/s an additional head loss of 4.32 m by valve closure
would be required.
60
 35
 QH t 1000  9.81 1000
Pi =
=
= 38869.8W = 38.87kW
p
0.53
Problem
 A centrifugal pump running at 1000 rpm gave the following relation between
head and discharge:
Discharge (m3/min)
Head (m)
•
0
4.5
9.0
13.5 18.0 22.5
22.5 22.2 21.6 19.5 14.1
0
The pump is connected to a 300 mm suction and delivery pipe the
total length of which is 69 m and the discharge to atmosphere is 15 m
above sump level. The entrance loss is equivalent to an additional 6
m of pipe and f is assumed as 0.024.
1. Calculate the discharge in m3 per minute.
2. If it is required to adjust the flow by regulating the pump speed,
estimate the speed to reduce the flow to one-half.
1) System curve:
 The head required from pump =
static + friction + velocity head
H t =H stat +h f d +  hmd
Vd2
+h f s +  hms +
2g
 Hstat = 15 m
 Friction losses (including equivalent entrance losses) =
8 f LQ 2
 h fs + hms +  h fd +hmd = 2 g D 5
8  0.024  (69 + 6) 2
=
Q
2
5
 g (0.3)
= 61.21Q 2
where Q in m3/s
2
 Velocity head in delivery pipe =
where Q in m3/s
Vd2
1 Q
2
=
  = 10.2Q
2g 2g  A 
Thus:
 H t = 15 + 71.41Q 2
where Q in m3/s
or
−3
2
H
=
15
+
19
.
83

10
Q

where Q in m3/min
t
 From this equation and the data given in the problem the following table
is compiled:
Discharge (m3/min)
0
4.5
9.0
13.5 18.0 22.5
Head available (m)
22.5 22.2 21.6 19.5 14.1
0
Head required (m)
15.0 15.4 16.6 18.6 21.4 25.0
Head,Ht (m)
Pump and Sytem Curves
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
Pump Curve
System Curve
0
2
4
6
8
10
12
14
Discharge, Q (m3 /min)
16
18
20
22
24
From the previous Figure, The operating point is:
 QA = 14 m3/min
 HA = 19 m
14
W0 = QH t = 1000  9.81 1000
19 = 2609.5W = 2.61kW
 At reduced speed: For half flow (Q = 7 m3/min) there
will be a new operating point B at which:


QB = 7 m3/min
HB = 16 m
Head,Ht (m)
Pump and Sytem Curves
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
A
B
Pump Curve
System Curve
A
B
0
2
4
6
8
10
12
14
Discharge, Q (m3 /min)
16
18
20
22
24
Q2 N 2
=
Q1
N1
H Q 

=
H B  QB 
H2  N 2 
=

H1  N1 
2
2
Based on the conditions of affinity laws, frictional losses
are not considered here. Static head and frictional losses
will be automatically accounted in the total head required.
16 2
H = 2 Q = 0.327Q 2
7
This curve intersects the original curve for N1 = 1000 rpm at C where
Qc= 8.2 m3/ h and Hc= 21.9 m, then
QB N 2
=
QC N1
7 N2
=
8.2 1000
N2 = 855rpm
Head,Ht (m)
Pump and Sytem Curves
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
C
A
B
Pump Curve
System Curve
A
B
C
0
2
4
6
8
10
12
14
Discharge, Q (m3 /min)
16
18
20
22
24