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Heat and Pipes
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Introduction Mechanisms of heat transfer Heat exchangers Pumps
Heat transfer and pipe �ow
Professor Eric S FragaRoom 2.05 Engineering Front BuildingDepartment of Chemical Engineering
UCL
c©2009
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Heat transfer and pipe �ow
Introduction Mechanisms of heat transfer Heat exchangers Pumps
Objectives
What is heat transfer?
How is heat transferred?
What equipment is used to transfer heat between two �uidsand how does it work?
What is the power required for pumping a �uid?
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Reading material
Coulson, J.M and Richardson J.F. �Chemical Engineering� Vol.1, Pergamon Press.
Pitts, D. and Sissom, L. �Heat Transfer�, Schaum's Outlines,McGraw-Hill.
Cengel, Y.A. �Heat transfer a practical approach�,McGraw-Hill.
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Heat transfer
Heat transfer is concerned with
temperature: represents the thermal energy, or heat, that isavailable, and
�ow of heat: represents the movement of thermal energy fromone place to another.
Heat transfer is thermal energy in transit due to a!temperature di�erence!.
Temperature di�erence is the driving force that causes heat tobe transferred.
Heat transfer plays a major role in the design of processequipment.
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How is heat transferred?
Heat may be transferred in three di�erent ways:
conduction transfer of heat as a result of molecular motionand the subsequent transfer of kinetic energy.Conduction is predominant in solid materials andin static �uids.
convection the �ow of heat as a result of macroscopicmovement of matter from a hot to a cool region
radiation transfer of energy in the form of rays or waves orparticles (α, β, γ)
We will consider only conduction and convection.
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Conduction through a wallConsider a wall of thickness x and surface area A that has auniform temperature T1 on one side and T2 on the other:
The heat transfer, q (W ), through the wall is
q =k
xA∆T (1)
where k (in WmK
) is the thermal conductivity of the materialwhich gives a measure of the ability of the material to conductheat and ∆T ≡ T1 − T2.
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Thermal conductivity
Materials with a large thermal conductivity, k , are goodthermal conductors. These will transfer large amounts of heatover time, e.g. copper.
Conversely, materials with low thermal conductivities are poorthermal conductors. These will transfer small amounts of heatover time, e.g. concrete.
The ratio kx is called the heat transfer coe�cient.
The inverse of the heat transfer coe�cient, xk , is the thermal
resistance.
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Composite wall example
Applying eq. 1 to each section of thecomposite wall:
q =ki
xiA(Ti − Ti+1) i = 1, . . . , 3
⇒ q =1∑xiki
A∆T
so overall thermal resistance is sumof the individual resistances(analogous to electrical circuits).
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Heat transfer through convection
Example: heating a pot of water.
At �rst, when the water is cold andstill, it behaves as a solid and heatis transferred by conductionthrough the bottom of the pot.
Bubbles, when created, will transferheat from the bottom to the topby convection due to buoyancy.
Cooler, more dense water at thetop will sink to the bottom.
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Types of convection
Natural convection occurs when the motion of �uid is due tobuoyancy e�ects. Example, the cooling of a heated pipe:
Forced convection is when the �uid motion is produced by anexternal agent.
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Convective heat transfer
Heat transfer by convection, q, is
q = hA∆T (2)
where
A (m2) is the characteristic area of contact.
∆T (K ) is the temperature di�erence between the solid andthe �uid.
h ( Wm2 K
) is the convective heat transfer coe�cient and is aproperty of the system, not a property of the �uid as is thethermal conductivity, k .
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Convective heat transfer coe�cient
The value of h depends on the surface geometry, theproperties of the �uid and the �uid motion regime:
Type of convection Material h(
Wm2 K
)Natural Gases 2-25
Liquids 50-1000Forced Gases 25-250
Liquids 100-20000
Liquids transfer greater amounts of heat than gases, which aregood thermal insulators and forced convection gives greaterheat transfer than natural convection for both gases andliquids.
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What a�ects convective heat
transfer?
h is a function of the properties of the system and depends on:
geometry of the system, i.e. a characteristic length L.
physical properties of the �uid: i.e. viscosity, µ, density, ρ,heat capacity, cp, and thermal conductivity, k .
�uid regime, i.e. a characteristic velocity v .
To determine h is therefore a complex task and we rely onexperiments.
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Prandtl and Nusselt numbers
The properties of the �uids and the di�erent forms of heattransfer can be described by some dimensionless numbers:
Prandtl the ratio between �uid ability to store heat and totransfer heat through conduction, independent ofthe system geometry:
Pr =µcpk
Nusselt ratio between heat transfer through convectionand conduction:
Nu =hL
k
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Reynolds and Grashof numbersOther numbers describe �uid �ow properties:
Reynolds The ratio between inertial and viscous forces inthe �uid which identi�es the �uid �ow regime:
Re =ρvL
µ(3)
High values correspond to turbulent �ow regime,and therefore to high convection h.t.c.
Grashof replaces Re when �uid motion is driven bythermal expansion of the �uid:
Gr =β∆Tρ2gL3
µ2
where β is the volumetric thermal expansioncoe�cient.
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Flow patterns and temperature
pro�les
Consider the �ow of air over an in�nite �at plate.
Ts is the temperature of the plate, while Tf is the air �owtemperature away from the plate, with Ts > Tf .
Regardless of the type of convection, we analyse the �owpattern and the temperature pro�les at the wall and awayfrom it.
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Flow layers
Adjacent to the wall, a laminar sub-layer forms with no �uidmixing. Heat transfer across the sub-layer is by conductiononly with large resistance to heat transfer and largetemperature change.
Away from the wall, turbulent �ow with large eddies and high�uid mixing. Heat transfer is by convection with lowresistance to heat transfer and small temperature change.
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Overall heat transfer coe�cientGiven a �at wall of uniform, homogeneous material havingconstant thermal conductivity, k , exposed to �uids h, attemperature Th, and c , at Tc , on either side:
with convective heat transfer coe�cients hh and hc , wallthermal conductivity, k , and wall thickness, x . What is theoverall heat transfer coe�cient, U ( W
m2 K), for the combined
conductive-convective heat transfer:
q = UA∆T (4)
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Assumptions
In steady state, the same amount of heat q must pass througheach section.
Heat transfer is by convection across the hot and cold �lmand by conduction through the solid wall.
The �uid temperatures su�ciently far from the wall, Th andTc , are una�ected by the heat transfer and are known.
The surface temperatures T1 and T2 are unknown.
The heat transfer area, A, is the same for all sections.
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Overall heat transfer
q = hhA(Th − T1) hot �lm
q =k
xA(T1 − T2) solid wall
q = hcA(T2 − Tc) cold �lm
⇓ (solve)
q =A(Th − Tc)(1
hh+ x
k+ 1
hc
) = UA∆T
Note: U can be calculated in a manner similar to resistancesin electrical networks:
1
U≡∑i
1
Ui
(5)
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Example: double-paned glass
window
Consider a 0.8m-high and 1.5m-wide double-pane glasswindow consisting of two 4 mm thick layers of glass(kg = 0.78 W
mK ) separated by a 10 mm wide stagnant air space
(ka = 0.026 WmK ).
Determine the steady rate of heat transfer, q, through thedouble-pane window and the temperature of its inner surface,T1, for a day during which the room is maintained at 20 ◦Cwhile the temperature outside is −10 ◦C.
Take the convection heat transfer coe�cients on the inner andouter surfaces of the window to be h1 = 10 W
m2Kand
h2 = 40 Wm2K
.
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Solution: Diagram
Assume that the heat transfer through the window is steadystate since the surface temperatures remain constant.
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Solution: overall thermal resistanceFrom eq. 5,
1
U≡ 1
h1+
L1
kg+
L2
k2+
L1
kg+
1
h2
where1
h1=
1
10 Wm2K
= 0.100m2K
W
L1
kg=
0.004m
0.78 WmK
= 0.00513m2K
W
L2
k2=
0.01m
0.026 WmK
= 0.385m2K
W
L3
kg=
0.004m
0.78 WmK
= 0.00513m2K
W
1
h2=
1
40 Wm2K
= 0.025m2K
W
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Solution: heat transferred
1
U= 0.52026
m2K
W
⇒ U = 1.92W
m2Kand A = 0.8m × 1.5m = 1.2m2
∴ q = UA(Ti − To)
= 1.92W
m2K× 1.2m2 × 30K
= 69.12W
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Solution: inner surface temperature
q = h1A(Ti − T1)
⇓
T1 = Ti −q
h1A
= (20 + 273)K − 69.12W
10 Wm2K× 1.2m2
= 287.24K
= 14.24◦C
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Non-uniform heat transfer area
In some cases, the area for transfer applicable to each mediacould di�er.
For example, the radial �ow of heat through a thick pipe wallor cylinder.
The area of transfer in these cases is a function of position.
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Example: curved wallConsider a curved solid wall with constant thermalconductivity k exposed to a convective hot outer �uid and aconvective cold inner �uid:
To and Ti are the temperatures at ro and ri respectively.
ho and hi are the convective heat transfer coe�cients in theouter and inner �lm, and impact of the thickness of these�lms is assumed to be negligible.
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Derivation
Start with eq. 1, q = kxA∆T , and let x ≡ r .
Consider heat transfer over a small part of the pipe and thecorresponding change in temperature:
dT =q
kAdr
Integrate over the pipe wall:∫ To
Ti
dT =
∫ ro
ri
q
kAdr =
∫ ro
ri
q
2πrLkdr
⇒ To − Ti =q
2πLk(log ro − log ri )
∴ q =k
log(ro/ri )2πL (To − Ti )
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Combined system
Assuming thin �lms on either side of the pipe, we can writethree equations:
q = ho2πroL(Th − To)
q =k
log(ro/ri)2πL(To − Ti)
q = hi2πriL(Ti − Tc)
which have three unknowns, To , Ti and q, so we can solve asusual.Note: the heat transfer coe�cient will often be given withrespect to a speci�c reference area.
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Engineers and heat transfer
In practice, engineers often have to design equipment to e�ectheat transfer, say to achieve a speci�c temperature change ina �uid stream of known mass �ow rate,
Such equipment will typically be in the form of a heatexchanger and the engineer will need to
determine the surface area to transfer heat at a given rate forgiven �uid temperatures and �ow rates.predict the outlet temperatures of hot and cold �uid streamsfor a speci�ed heat transfer.
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Heat exchangers
A heat exchanger is any device that e�ects transfer of thermalenergy between two �uids that are at di�erent temperature.
The two �uids do not come in direct contact but areseparated by a solid surface or tube wall.
Common heat exchangers include:
Shell-and-tube (single pass or multi-pass)Flat-plateFinned tubes
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Double pipe heat exchanger
The simplest form of an heat exchanger consists of twoconcentric cylindrical tubes, the double pipe heat exchanger:
Heat transfer involves convection in each �uid and conductionthrough the wall separating the two �uids.
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Parallel or co-current �ow
The �uids both �ow in the same direction:
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Counter-current �ow
The �uids �ow in opposite directions:
In both cases, the �uids are forced to �ow using pumps or fans.
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Shell and tube single pass
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Multi-pass shell and tube
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Multi-pass �ow arrangement
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Plate heat exchangers
Plate heat exchangers are built up from individual platesseparated by gaskets, assembled in a pack and clamped in aframe.
They are applied in the energy recovery section of manyprocesses because of low initial cost, high e�ciency and lowmaintenance costs.
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Plates
Thin sheet material, resulting in economicunits, particularly when expensive materialis involved.
Plates are especially corrugated topromote turbulence also at low Re,resulting in:
very high heat transfer coe�cientsreduces foulingfacilitates chemical cleaning
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Finned tube heat exchangers
Finned tubes exchangers areemployed in large air/liquid heatexchanger systems to give greaterheat transfer area because gases,which are good thermal insulators,transfer smaller amounts of heatthan liquids.
There are various types of �nnedtubes, depending on theapplication.
Applied in various systems:
large air conditioning systemsradiators for large truck.
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Heat transfer in a heat exchanger
Heat exchangers operate for long periods of time with nochange in the operating conditions, thus they can be modelledas steady-�ow devices:
1 The overall heat transfer coe�cient, U, is constantthroughout the exchanger.
2 The mass �ow rate of each �uid remains constant.
3 The speci�c heats of the �uids are constant.
4 The temperature of the two �uids are constant over a speci�ccross-section.
5 The outer surface is perfectly insulated, so that any heattransfer occurs between the two �uids only.
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Energy balance
Under these assumptions, it follows that the rate of heattransfer from the hot �uid to be equal to the rate of heattransfer to the cold one.
The basic design equations for heat exchangers are thereforethe energy balance for each �uid:
q = −mhcph(Tho − Thi ) (Energy given up by hot �uid)
q = mccpc(Tco − Tci ) (Energy gained by cold �uid)
where m (kg/s) is the mass �ow rate and cp ( kJkg K ) is the
speci�c heat.
But what is the driving force, ∆T?
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Temperature pro�le: co-current �ow
The temperature di�erence ∆T islarge at the inlet but decreasesexponentially towards the outlet.
Temperature of the hot �uiddecreases and the temperature ofthe cold �uid increases along theheat exchanger.
The outlet temperature of the cold�uid can never exceed that of thehot �uid, no matter how long theheat exchanger.
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Temperature pro�le: counter-current
�ow
The hot and cold �uids enter theheat exchanger from opposite ends.
The outlet temperature of the cold�uid may exceed the outlettemperature of the hot �uid,temperature cross.
The outlet temperature of the cold�uid can never exceed the inlettemperature of the hot �uid.
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Design
Recall eq. 4, q = UA∆T .
Need to calculate ∆T for eitherco-current or counter-currentexchange.
However, the temperaturedi�erence varies across the range ofoperation.
We introduce the log meantemperature di�erence (LMTD),∆TLM, as an expression of ∆T .
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Log mean temperature di�erence
derivation � I
At any point along theexchanger:
δq = −mhcphδTh
δq = mccpcδTc
δq = U(Th − Tc)δA (6)
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LMTD � IIRearrange for δT in each equation and �nd the di�erence:
δTh = − δq
mhcph
δTc =δq
mccpc
⇒ δTh − δTc = −δq(
1
mhcph+
1
mccpc
)+(6) : δ(Th − Tc) = −UδA(Th − Tc)
(1
mhcph+
1
mccpc
)⇒ δ(Th − Tc)
(Th − Tc)= −UδA
(1
mhcph+
1
mccpc
)
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LMTD � IIILet the δ terms be di�erentials and integrate along the lengthof the exchanger:
log (Tho − Tco)− log (Thi − Tci) = −UA(
1
mhcph+
1
mccpc
)which, when combined with overall energy balance on each�uid (solve for mcp term and substitute):
log (Tho − Tco)− log (Thi − Tci) = −UA(Thi − Tho) + (Tco − Tci)
q
⇓
log∆T2 − log∆T1 = UA∆T2 −∆T1
q
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LMTD � IV
∴ q = UA∆T2 −∆T1
log∆T2 − log∆T1
or q = UA∆T2 −∆T1
log ∆T2
∆T1
The term in the box is ∆TLM, the log mean temperaturedi�erence (LMTD).
For counter-current exchangers, the same result is obtainedalthough for di�erent ∆T1 and ∆T2.
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Correction factor for complex heat
exchangers
Determination of the average temperature di�erence, ∆TLM,is di�cult for complex heat exchangers.
It is practice to introduce a correction factor, Ft. The heattransfer rate is then given by:
q = UAFt ∆TLM
where ∆TLM is that for the counter �ow double-pipe heatexchangers with the same �uid inlet and outlet temperaturesas in the more complex design.
Ft values for several con�gurations are given in referencebooks (cf. Perry or Kern).
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Example: area comparison
Suppose we wish to exchange heat between a hot stream, withThi = 200 ◦C and Tho = 150 ◦C, and a cold stream, withTci = 80 ◦C and Tco = 120 ◦C.
Assuming the same heat transfer coe�cient, U, in both casesand the same amount of heat, q, to be exchanged, what is theratio of area required for co-current exchange to area requiredfor counter-current exchange?
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Solution
co-current ∆T1 = 200 ◦C− 80 ◦C = 120 ◦C
∆T2 = 150 ◦C− 120 ◦C = 30 ◦C
⇒ ∆Tco-current =30 ◦C− 120 ◦C
log 30 ◦C120 ◦C
≈ 65 ◦C
counter-current ∆T1 = 200 ◦C− 120 ◦C = 80 ◦C
∆T2 = 150 ◦C− 80 ◦C = 70 ◦C
⇒ ∆Tcounter-current =70 ◦C− 80 ◦C
log 70 ◦C80 ◦C
≈ 75 ◦C
∴Aco-current
Acounter-current=
qU∆Tco-current
qU∆Tcounter-current
=∆Tcounter-current
∆Tco-current
≈ 1.15
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Solution
co-current ∆T1 = 200 ◦C− 80 ◦C = 120 ◦C
∆T2 = 150 ◦C− 120 ◦C = 30 ◦C
⇒ ∆Tco-current =30 ◦C− 120 ◦C
log 30 ◦C120 ◦C
≈ 65 ◦C
counter-current ∆T1 = 200 ◦C− 120 ◦C = 80 ◦C
∆T2 = 150 ◦C− 80 ◦C = 70 ◦C
⇒ ∆Tcounter-current =70 ◦C− 80 ◦C
log 70 ◦C80 ◦C
≈ 75 ◦C
∴Aco-current
Acounter-current=
qU∆Tco-current
qU∆Tcounter-current
=∆Tcounter-current
∆Tco-current
≈ 1.15
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Solution
co-current ∆T1 = 200 ◦C− 80 ◦C = 120 ◦C
∆T2 = 150 ◦C− 120 ◦C = 30 ◦C
⇒ ∆Tco-current =30 ◦C− 120 ◦C
log 30 ◦C120 ◦C
≈ 65 ◦C
counter-current ∆T1 = 200 ◦C− 120 ◦C = 80 ◦C
∆T2 = 150 ◦C− 80 ◦C = 70 ◦C
⇒ ∆Tcounter-current =70 ◦C− 80 ◦C
log 70 ◦C80 ◦C
≈ 75 ◦C
∴Aco-current
Acounter-current=
qU∆Tco-current
qU∆Tcounter-current
=∆Tcounter-current
∆Tco-current
≈ 1.15
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Example: Cooling a hot process �uid
in a counter-�ow heat exchangerA counter-�ow double-pipe heat exchanger is used to cool ahot process �uid using water. The process �uid �ows at 18kg
sand is cooled from 105 ◦C to 45 ◦C. The water �owscounter-currently to the process �uid, entering at 25 ◦C andleaving at 50 ◦C.
1 Assuming no heat losses, calculate the required �ow-rate forthe cooling water. The speci�c heat for water is 4.2 kJ
kg K and
that of the process �uid is 3.4 kJkg K .
2 Neglecting the tube wall curvature, calculate the required areafor heat exchange. Under these conditions, the process �uidside �lm heat transfer coe�cient is 2500 W
m2K, the cooling
water side heat transfer coe�cient is 1200 Wm2K
. The tube wall
thickness is 3 mm and the thermal conductivity is 220 WmK .
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Solution: Diagram
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Solution: 1. required water �ow rate
q = −Gf cpf (Tho − Thi) = 3672kW (heat from hot stream)
= Gwcpw (Tco − Tci) (energy balance)
⇓
Gw = 34.97kg
s(water �ow rate required)
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Solution: 2. heat exchanger area
q = UA∆TLM
∆TLM =∆T2 −∆T1
log ∆T2
∆T1
= 34.6K
1
U=
1
hf+
x
k+
1
hw(thermal resistance)
⇒ U = 802W
m2K(overall htc)
∴ A =q
U∆TLM= 132.3m2
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Fouling
The performance of a heat exchanger depends upon surfacesbeing clean but deposits form over time.
The layer of deposits presents additional resistance to heattransfer and must be accounted for by a fouling factor, Rf (cf.Perry, Kern).
1
U= Rconv,hot �uid + Rcond,wall + Rf + Rconv,cold �uid
Deposits can occur by the precipitation of solid deposits (e.g.calcium in a kettle), corrosion or chemical fouling due tochemical reactions, and the growth of algae, biological fouling.
Can apply water treatment, coatings and chemical treatments.
Periodic cleaning of exchangers and the resulting down timeare additional penalties associated with fouling.
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Pumping of liquid �uids
Pumps are the devices used to add energy to a �uid in orderto maintain �ow.
Fluids �ow in the direction of decreasing pressure.
Pumps direct a �uid from one vessel to another or through along pipeline.
The energy added to the �uid compensates for the mechanicalenergy losses due to friction and provides an increase in thevelocity, the pressure, or the height of the �uid.
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Valves
Valves are used to control the �ow rate:
On/O� valves: gate (30% of all on/o� valves) and ball
Throttling valves: globe (50%), needle and diaphragm
Check valves: allow �ow only in one direction
Automatic valves
Relief valves (for safety): spring loaded, bursting disk.
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Pumping devices for liquids
Liquids used in the chemical industry di�er considerably inphysical and chemical properties so a variety of pump typesexists.
Most pumps fall into one of two major classi�cations:
Positive-displacement pumpsCentrifugal pumps
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Positive-displacement pumps
Reciprocating pumps involve a back-and-forth motion of apiston in a cylinder.
Rotary pumps depend upon a rotating motion.
The �ow from these pumps is pulsating. The higher the speedof the pump, the higher the �ow rate delivered.
Deliver a controlled amount of liquid for each stroke orrevolution.
Used when nearly constant delivery rates are required.
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Examples
Reciprocating pumps:
Diaphragm pumpPiston pump
Rotary pumps:
Gear pumpLobe pumpPeristaltic pumpScrew pump
For details on other types of pump refer to Coulson &Richardson.
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The Piston pump
In these pumps, the motion of a rotor is converted into aback-and-forward motion of a piston.
The rate of liquid delivery is a function of the volume sweptout by the piston in the cylinder and the number of strokesthe piston make per unit time.
For each stroke of the piston, a �xed volume of liquid isdischarged from the pump.
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Piston pumps
Piston pumps may be single-acting, with the liquid admittedonly to the portion of the cylinder in front of the piston.
When the piston moves towards the disk it creates a partialvacuum in the chamber. This allows atmospheric pressure topush the �uid from below.
On the return half of the cycle the top check valve opens andthe bottom one closes. The water is forced up through thepipe.
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Double-acting piston pumps
Piston pumps can also be double-acting, in which case thefeed is admitted to both sides of the piston.
As the piston moves downwards, it forces the �uid outthrough the bottom right valve and creates at the same time apartial vacuum in the upper chamber, pushing the �uidthrough the top left valve.
When the piston is moving upwards it forces the �uid outthrough the top right valve, while creates at the same time apartial vacuum in the lower chamber, pushing the �uid inthrough the bottom left valve.
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Piston pump applications
These pumps can deliver the highest pressure of any otherpump.
Piston reciprocating pumps have long been used in manyapplications, including pumping of oil, feed water and mud.
However, their capacities are relatively small compared tocentrifugal pumps.
Not used with liquid containing abrasive material as it candamage the machined surfaces of the cylinder and piston.
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Diaphragm pump
In one section a piston operates in a cylinderin which an inert liquid is displaced.
The movement of the �uid is transmitted bymeans of the �exible diaphragm.
They have been developed to handle corrosiveliquids or suspensions with abrasive solids.
They are used for example to move gasolinefrom the gas tank to the carburetor.
They are not used for high pressureapplications.
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Rotary pumps: the Gear pump
Has been developed to deal withviscous �uids.
Two gears operate within a casing.
Small packages of �uids are carriedbetween the teeth and the casingfrom the low pressure inlet side tothe high pressure delivery side.
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The 3 lobe pump
Works on the same principle of thegear pump but the gear teeth arereplaced by two or three lobes.
A small clearance between thelobes can be maintained and wearis reduced.
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The Peristaltic pump
Delivers �ow precisely controlled by thespeed of a rotor.
An elastic tube is compressed in stages bya rotor.
As the rollers rotate, they �atten the tubeagainst the track at the points of contact.
These �ats move the �uid along the tube.
They are mainly used in labs.
They are particularly good in handlingbiological �uids, where all forms ofcontacts must be avoided.
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The screw pump
One of the oldest pumps: its usage goes back 2000 years. Itconsist of a helical screw. Suitable for very viscous �uids (e.g.polymers) and also for sewage �uids. The �uid is sheared inthe screw channel and so is raised to the delivery side.
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Figure: Example of a screw pump used in a waste water treatmentplant.
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Centrifugal pumps
Pumps that cause the pumped �uid to rotate are calledcentrifugal pumps.
Centrifugal pumps are the most widely applied in the chemicaland petroleum industry.
They are applied for large capacity applications.
They pump liquids with very wide-ranging properties andsuspensions with high solids content.
They cannot handle highly viscous �uids.
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A Centrifugal pump has an impeller enclosed by a casing orvolute. The impeller consists of a series of curved vanes. Thegreater the number of vanes, the greater is the control overthe direction of motion of the �uid.
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The �uid is fed to the centre of therotating impeller and is thrownfrom the impeller vanes into thecasing by centrifugal force.
As a result, the liquid acquires ahigh kinetic energy. This velocityenergy is converted in pressure asthe �uid leaves the impeller andenters the casing.
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Turbine pump
In the turbine pump, the liquid�ows from the impeller into a seriesof �xed vanes, called di�users.
These give more gradual change indirection to the �uid and moree�cient conversion of kineticenergy into pressure energy.
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Summary of types
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Pump characteristics
Centrifugal Reciprocating Rotary(piston) (gear or screw)
Head High, single stage Highest available Intermediateup to 600 ft 100,000 psi up to 600 psiMultistageup to 6000 psi
Capacity Low (100 gal/min) Intermediate Low (1 gal/min)to very high (500 gal/min) to intermediate(200,000 gal/min) (500 gal/min)
Liquids Clear or dirty, Clean High viscosity,non viscous no solids non abrasive
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Summary of characteristics
Pump Pressure Flow Capacity Fluiddelivered delivered
Piston very high pulsating not high non-corrosiveDiaphragm not high pulsating not high corrosiveRotating not high pulsating not high highly viscousCentrifugal not high continuous very high not viscous
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Operating characteristics of
centrifugal pumps
Centrifugal pumps are the most widely applied in the chemicaland petroleum industry.
Centrifugal pumps operate at constant speed and the capacitydepends upon the total head, H, the design and the suctionconditions.
Pumps usually achieve maximum e�ciency at one particular�ow rate.
Operating characteristics are described through use ofcharacteristic curves.
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Characteristic curves
For a pump at a particular speed, the characteristic curvesshow the inter-relation between:
Total head and capacity, the H − Q curve.Power input and pump capacity, the P − Q curve.Pump e�ciency and capacity, the η − Q curve.
The duty point shows the optimum conditions for operation.
This is the point where the head curve cuts the ordinatethrough the point of maximum e�ciency.
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Characteristic curve graph
For a pump having the characteristics shown above: maximume�ciency would occur at a capacity of 2500 gal/min and atotal head of 80ft.
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Variable speeds
When a pump is capable of being operated at variable speeds,then, at higher speed of rotation gives higher capacity andrequires more horsepower to supply the increased supply ofliquid.
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Pump design
The energy required for a pump will depend on
the height through which the �uid raises,
the pressure required at the delivery point,
the length and diameter of the pipe,
the rate of �ow, and
the physical properties of the �uid, density and viscosity.
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General scenario for pumping
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Energy balanceThe work done on a �uid by a pump is expressed as head, H(m), and is given by the mechanical energy balance (per unitmass):
H = ∆
(p
ρg+
u2
2g+ z
)+ ∆hfT
=P2 − P1
ρg+
u22− u2
1
2g+ (z2 − z1) + ∆hfT (7)
where ∆hfT represents the total friction loss in the system andis the sum of the losses in the whole pipe length and otherlosses due to �ttings:
∆hft = ∆hf + ∆h�tting loss
where hf is the frictional head loss.
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Frictional losses
∆hf is expressed as a frictional pressure drop:
∆hf = 2cf1
d
u2
g=
∆pf
ρg
where the Fanning friction factor, cf , is given as a function ofthe Reynolds number, Re (eq. 3):
Laminar �ow cf = 16
Re
Turbulent �ow cf = 0.079Re−0.25
Rough pipes use Moody diagram to estimate the value
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Fitting losses
Loss due to �ttings, such as bends and valves, can generatelarge-scale turbulence in which energy is dissipated as heat.
For turbulent �ow, these losses are proportional to the squareof the �uid velocity and can be expressed as the frictional lossdue to an equivalent length of straight pipe, le , estimatedas a multiple of the pipe diameter:
le = nd
so that
∆h�tting loss = 2cfle
d
u2
g
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Total friction losses
Total friction loss is therefore equal to
∆hfT = 2cfl + le
d
u2
g(8)
which in terms of pressure losses is
∆pfT = ρg∆hfT = 2cfl + le
dρu2
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Power requirements
Thus, from eq. 7, the head, H, that must be supplied by thepump is:
H =P2 − P1
ρg+
u22− u2
1
2g+ (z2 − z1) + 2cf
l + le
d
u2
g
The power required by the pump to deliver H is given by:
Power = ∆PpQ = H ρg Q
where Q is the volumetric �owrate of the �uid that the pumpmoves and is known as the capacity of the pump.
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Example
2.32m3
hof water is pumped in a 35 mm internal diameter pipe
through a distance of 125 m in a horizontal direction and thenup through a vertical height of 12 m. The friction loss in the90◦ square elbow may be taken as equivalent to 60 pipediameters. Also in the line there is a control valve fully openand frictional losses may be taken equivalent to 200 pipediameters.Calculate the total head ∆hfT to be developed to overcomethe total frictional losses in the pipeline. You may assume thatfor this pipe f = 0.079Re−0.25. Assume the water to �ow inturbulent regime through the pipe. Density and viscosity ofwater in the pipe are 1000 kg
m3 and 0.65mNm2 s respectively.
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Solution: data collection
Quantity Value Units
Volumetric Flow rate Q 2.32 m3
h
Density of water ρ 1000 kgm3
Viscosity of water µ 0.65 mNm2 s
0.65× 10−3 Nm2 s
Internal pipe diameter d 35 mm
0.035 m
Horizontal pipe length lh 125 mVertical pipe length lv 12 m90◦ square elbow friction loss 60d mControl valve friction loss 200d mFriction coe�cient cf 0.079Re−0.25
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SolutionThe total friction losses, eq. 8, in the pipeline for turbulent�ow regime is expressed as:
∆hfT =2cf (l + le)u2
gd
with l = lh + lv = 125m + 12m = 137m and le from thede�nition of the equivalent pipe length:
le = 200d + 60d = 200x0.035m + 60x0.035m = 9.1m
The velocity of the water in the pipe is given by the volumetric�ow rate divided by the pipe cross-section area:
u =Q
π(d2
)2 =2.32m
3
h× 1h
3600s
π ×(0.035m
2
)2 = 0.67m
s
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Solution
The friction coe�cient depends on Re, the Reynolds number(eq. 3):
Re =duρ
µ=
0.035m × 0.67ms× 1000 kg
m3
0.65× 10−3 Nm2 s
= 36076
⇒ cf = 0.079Re−0.25 = 0.079× 36076−0.25 = 0.0057
⇒ ∆hfT =2cf (l + le)u2
gd
=2× 0.0057× (137 + 9.1)m × (0.67m
s)2
9.81ms2× 0.035m
= 2.17m
⇒ ∆pfT = ∆hfTρg = 2.17m × 9.81m
s2× 1000
kg
m3= 21287
N
m2
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Cavitation in centrifugal pumps
Cavitation is the formation of cavities or bubbles in a pumped�uid when the pressure on the �uid falls below the vapourpressure of the liquid, Pv (or p∗).
When a centrifugal pump is operated at high capacity, lowpressure may develop at the impeller eye or vane tips and sovapourisation may occur.
Cavitation leads to:
A reduction in pump capacity.A reduction in the head of the pump.A noise that can be heard when the pump is running.Mechanical damage that can be seen on the pump impellerand volute.
To avoid cavitation, the pressure at the pump inlet mustexceed the vapour pressure of the liquid.
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Pump con�gurations
Negative suction head
Pump is placed above thereservoir of �uid to be pumped,drawing �uid up with suction:
Positive suction head
Pump is placed below thereservoir of �uid to be pumpedand is fed by gravity action:
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Net positive suction head (NPSH)
To avoid cavitation, the pressure at the pump inlet mustexceed the vapour pressure of the liquid. There are two cases:
NPSH is the amount by which the pressure at the suctionpoint of the pump must exceed the vapour pressure of theliquid and is expressed as a head of the liquid to be pumped.
For any pump, the manufacturers specify the minimum valueof the NPSH required at the impeller eye to avoid cavitation.
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Typical con�guration
P1, pressure at the tankliquid surface.
h1, height of the liquidsurface above the pumpcentre-line at the suctioninlet.
u1, liquid velocity on thesurface.
u2, velocity at the pumpinlet.
hf , total piping friction lossbetween (1) and (2).
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Available NPSHStart with energy balance (per unit mass):
P1
ρg+
u21
2g+ h1 − hf =
P2
ρg+
u22
2g+ h2 (9)
If the reference plane is taken at h2, and the liquid velocity inthe reservoir, u1, is negligible compared with u2:
P1
ρg+ h1 − hf =
P2
ρg+
u22
2g(10)
The available NPSH is the di�erence between the static headand the head corresponding to the vapour pressure of theliquid at the suction inlet.
NPSHA =
(P2
ρg+
u22
2g
)− Pv
ρg=
P1 − Pv
ρg+ h1 − hf (11)
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Required NPSH - I
From eq. 10, the total head at the suction inlet is:
P2
ρg=
P1
ρg+ h1 − hf −
u22
2g(12)
Cavitation usually occurs at the impeller eye where thepressure will be less than at the suction inlet by
∆P = φu23
2g(13)
Where φ is a pressure drop coe�cient characteristic of pumpgeometry and u3 is the �uid velocity at the eye.
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Required NPSH � IICavitation is probable if the total head at the impeller eye (rhsof eq. 12 minus eq. 13) is equal to or less than the vapourpressure:
P1
ρg+ h1 − hf −
u22
2g− φ u
2
3
2g≤ Pv
ρg
Limiting case therefore is
P1 − Pv
ρg+ h1 − hf =
u22
2g+ φ
u23
2g
where the right hand side is the NPSH required at the impellereye:
NPSHR =u22
2g+ φ
u23
2g
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Avoiding cavitation
In order to avoid cavitation, the NPSHA available has to begreater than the NPSHR required at impeller eye:
NPSHA > NPSHR
The value of the required NPSHR for the particular pumpbeing used may be obtained from the pump manufacturer.
If NPSHA is too low, then cavitation can be avoided byincreasing h1.
This is why pumping �uids that are close to saturatedconditions require that the vessel upstream be elevated . . .
. . . or, more generally, that the pump be lowered.
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