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Heating In Agitated &
Non-Agitated VesselsIBD/BFBi Midland Section Engineering Symposium on
Heat Transfer and Refrigeration
Burton-on-Trent – Jan 2014
Mark Phillips – Briggs of Burton
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Contents
• Heat Energy
• Heat Transfer Mechanisms
• Heat Transfer Coefficients
• Vessel Heat Transfer
• Vessel Agitation
• MCV Heating (worked example)
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Heat Energy• Quantifying Energy
– SI Unit of Heat Energy - the joule
– 1 joule (J) = 1 newton force through 1 metre (Nm)
Heat Transfer – Rate of Change
– SI Unit of Power - the watt
– 1 watt (W) = 1 joule transferred in 1 second (J/s)
– 1 kW = 1,000 W
– 1 MW = 1,000,000 W
• Imperial units– 1 kW = 3,412 Btu/hr
1N
1mTHE JOULE
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Heat Energy
• Two Basic Forms
• Sensible Heat
− Heat associated with temperature (T)
− Also dependent on material property
Specific Heat Capacity (cp)
Heatenergy = mass × cp × T• Energy referenced to a datum (0oC)
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Heat Energy
• Typical cp values for common materials
Material Specific Heat CapacitykJ/kg.oC
Water 4.18
Wort 4.00
Beer 4.05
Stainless Steel 0.50
Copper 0.39
Air 1.01
Carbon Dioxide 0.83
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Heat Energy• Two Basic Forms
• Latent Heat
− Heat associated with change in state
− Solid to liquid (fusion)
− Liquid to vapour (vaporisation)
− Material property - latent heat (hfg)
Heatenergy = mass ×h��• Not dependent on temperature
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Heat Energy
• Typical latent heat values for water
• Vaporisation energy dependent on pressure –
steam tables
Material Type Latent HeatkJ/kg
Ice – Water Fusion 334
Water – Steam (0 barg) Vaporisation 2,257
Water – Steam (5 barg) Vaporisation 2,086
Water – Steam (10 barg) Vaporisation 1,999
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Heat Transfer Mechanisms
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Heat Transfer - Conduction
• Two Primary Mechanisms
• Molecular Interaction
– Higher temperature molecule imparts energy
via impact or vibration
• ‘Free’ Electron Drift
− Applicable for solids
• Pure Metals – High electron concentration
• Non Metallic – Low electron concentration
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Heat Transfer - Conduction
• Fourier’s Law of Heat Conduction
• (Heat Flux) is proportional to (Temperature
Gradient)
� = −�. ��• Proportionality constant – Thermal Conductivity
• k = Thermal Conductivity (W/ m.oC)
• ‘k’ is unique material property
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Heat Transfer - Conduction
• Typical k values for common materials
Material Thermal ConductivityW/m.oC
Water 0.58
Wort 0.52
Beer 0.55
Stainless Steel 16.0
Copper 401
Wood (oak) 0.17
Air 0.024
Carbon Dioxide 0.015
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Heat Transfer - Conduction
k
xQ
T1
T2
A• Thermal Potential = (T2 – T1)
• Thermal Resistance = x / k
Q = −kA T� − T�x
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Heat Transfer - Conduction
A
T2 T3T4
Q Q
ab
c
T1
• Resistance added in series
Q = −A T − T�x!k! "
x#k# "x$k$
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Heat Transfer - Conduction
r1T1
r2T2
r3T3
a
b
• Variable surface area
Q = − T% − T�&' (�/(�2+,�- " &' (%/(�2+,�.
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Heat Transfer - Convection
• Energy exchange between a surface and a fluid
• Natural Convection
– Fluid next to solid boundary
causes circulation currents due
to density difference
• Forced Convection
– Fluid next to solid boundary
forced past its surface
Heat
HOT
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Heat Transfer - Convection
• Newton’s Law of Cooling
• (Heat Flux) is proportional to (Temperature Difference)
• Proportionality constant – Convective Heat Transfer Coefficient (Film Coefficient)
• h = Film Coefficient (W/ m2.oC)
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Heat Transfer - Convection
A
Tf
Tw
Fluid
Solid
Surface
A
h
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Overall Heat Transfer Coefficient
A
x
kFluid A
(Hot)
Fluid B
(Cooler)
TA
T1
T2
TB
h1 h2
Q
Wall
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Overall Heat Transfer Coefficient
A
U = T0 − T11h� " 345 " x
k " 346 " 1h�
Q
••TA (1/h1A)
•(x/kA)T1
•(1/h2A)T2
TB
U = 11h� "
xk "
1h�
= 1SystemResistance
Q = T0 − T11h�A "
xkA "
1h�A
= U. A. ∆T
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Overall Heat Transfer Coefficient
• Fouling Factors – ‘Clean’ surface altered that affects heat transfer capability
– Scale build up / Corrosion
U = 11h� " 345 " x
k " 346 " 1h�
– U experimentally determined in different conditions
R� = 1U;<=>? −
1U$@A!B
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Heat Transfer – Radiation
• Heat Transfer attributed to electromagnetic waves - No transfer medium required
• Heated bodies emit thermal radiation onto subsequent bodies
Reflected (ρ) Incident
Absorbed (α)
Transmitted (τ)
α " τ " ρ = 1Typically for solid objects
α " ρ = 1
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Heat Transfer – Radiation
• Pure black bodies are perfect absorbersα = 1
• Perfect absorbers are also perfect radiators
(emitters) – Defines ‘emissivity’, εα = ε = 1
• For non-perfect bodies, Kirchhoff's law applies: -
α = ε = GG.H-IJ
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Heat Transfer - Radiation
• Typical ε values for common materials
Surface Emissivity (new) Emissivity (typical)
True Black Body 1 1
Real object < 1 < 1
Copper 0.04 0.78
Stainless steel 0.08 0.85
Carbon steel 0.02 0.90
Aluminium 0.04 0.31
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Heat Transfer – Radiation• Heat transferred by radiation – Energy emission
Q#@!$K(W) = AσT Where σ = Stefan-Boltzmann constant, Wm-2K-4 (5.67x10-8)
T = Absolute temperature, K (K = oC + 273.15)
• For a non-black body in non-black surroundings
Q = σεA T� − T�
• Heat loss significant in large vessels
T1
Emitting
Body εQ
Surrounding
T2
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Q = Heat Transfer (W)
U = Overall Heat Transfer Coefficient (W/m².oC)
A = Heat Transfer Surface Area (m²)
∆T = Temperature Difference of System (oC)
• Summary For Steady State Heat Transfer
Heat Transfer Coefficients
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Heat Transfer Coefficients
• Typical U values for common systems
Example Material Typical Overall HTC (W/m². °C)
Radiator central heating Liquid Water 5-15
Steam radiator Vapour Water 5-20
Steam jackets - vessels with
stirrers
Condensing Water
Vapour
300-1000
Heat exchanger – water / water Liquid Water 900-2500
Condensers - steam / water Condensing Vapour
Water
1000-4000
Evaporators - steam / water Condensing Vapour
Water
1500 - 6000
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1. Heat transfer mechanism
– conduction, convection, radiation
2. Fluid dynamics
– e.g. velocity, turbulence, pressure
3. Media and surface properties
– e.g. composition, heat capacity, density,
absorptivity, fouling
4. Heat transfer geometry
– e.g. fluid paths, surface orientations
Heat Transfer Coefficients
What determines the heat transfer coefficient?
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Heat Transfer Coefficients • Usually one individual coefficient controls the
entire system
1U = 1
h�@P<; "1
hQ!@@" 1hR=S;
Product
TA
T1
T2
TB
hfluid hprod
Q
Wall
hwall
Heat Transfer
Fluid
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Heat Transfer Coefficients 1U = 1
h�@P<; "1
hQ!@@" 1hR=S;
• Real Case Study Values (W/m2.oC)
• hfluid = 4488, hwall = 3260, hprod = 290, U = 251
• 50% increase in hfluid, 2% increase in U
• 50% increase in hwall, 3% increase in U
• 50% increase in hprod, 40% increase in U
• Product coefficient is controlling
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Heat Transfer Coefficients
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• Product convection currents
are generally controlling
• Described by dimensionless
groups that link physical
and system properties
• Forced convection using
vessel agitation promotes
movement and turbulence
Heat Transfer Coefficients
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D
N
Heat Transfer Coefficients• Forced Convection is a function of Reynolds Number (Re)
and Prandtl (Pr) Number
TU = VWX�Y
Z( = [\Y�
] = �X 0.023TU`.aZ(`.
• HTC dependent on physical properties density (ρ), viscosity
(µ), specific heat capacity (cp) and conductivity (k)
• HTC dependent on system properties agitator diameter (D)
and agitator speed (N), and agitator type
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• Natural convection even more complicated!
• For turbulent and laminar flow
] = bJ 0.14. d(`.%e Z(`.�fg − 0.55 d( ≥ 10j
] = bJ 0.68. d(`.�gZ(`.�fg mn
`.ae�omn`.�g d( < 10j
• Grashof Number (Gr) a function of buoyancy
force as this creates the fluid velocity
Heat Transfer Coefficients
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Advantages Disadvantages
Fluids in contact with vessel wall Low heat transfer performance
Low contamination potential Relatively high flows needed
Easy to clean Limited surface area
Vessel Heat Transfer• Vessel heating commonly performed using
jackets, coils and external recirculation loops
• Relative merits of heating jackets
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Full Jackets
• Viscous fluids
• Hygienic applications
• Clean-in-place
• Zoned heating
• External surface area maximised
• High cost
Vessel Heat Transfer
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Vessel Heat Transfer
Dimple Jackets
• Welds located in a regular pattern
• Maintains strength using thin shell material
• Dimples impart turbulence
• Flow guiding system can be installed for liquid heating to increase contact time
• Lower capital cost
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Vessel Heat Transfer
Limpet Jackets
• Half pipe configuration around vessel shell
• Uniform fluid velocity
• Good distribution and contact time with vessel wall
• High pressure capability (pipe)
• High cost
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Vessel Heat Transfer
Vessel Coils
• Most commonly used for batch processes
• Full helical or small ringlet coils used
• Larger surface areas can be created
• Poor hygiene and CIP capability
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Vessel Agitation
• Agitators promote mixing
and facilitate heat transfer
• Five common types of
agitator:-
− Propeller
− Turbine
− Paddle
− Anchor
− Helical Ribbon
• Many proprietary design
agitators exist
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Propeller Agitator
• Simple three blade device
• Creates axial flow patterns within the vessel fluid
• Smaller diameter –operated at high speed to compensate
• Relatively low cost
Vessel Agitation
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Turbine Agitator
• Simple design to facilitate low capital and cleaning
• Operated at high speed in low viscosity liquids
• Larger diameter than propeller type
• Blades can be flat, curved or pitched
• Commonly used within solid-liquid applications
Vessel Agitation
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Vessel Agitation
Paddle Blade Agitator
• Can be used at high and low speeds of rotation
• Low speed creates axial flow patterns
• High speed and pitched blades generate radial patterns
• Generally larger diameter and cost
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Vessel Agitation
Anchor Agitator
• Used for viscous applications - Laminar
• Low speed and large diameter
• Disturbance close to vessel wall
• Suitable for jacket heating
• Higher complexity and cost
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Helical Ribbon Agitator
• Similar applications and concept as anchor type
• Imparts intimate mixing at vessel wall and core of the product
• Highest complexity and cost
Vessel Agitation
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Propeller Turbine Paddle Ribbon Anchor
Viscosity Low Low High &
Low
High High
Speed High High High &
Low
Low Low
Diameter Small Small Large &
Small
Large Large
Typical
Diameter (%)
25-35 35-45 50-70 75-95 80-95
Vessel Agitation
• Mixer type summary
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Effective Mixing for Low Viscosity Applications
• Vortex formation for centrally driven mixers
• Poor mixing and heat transfer
Vessel Agitation
• Vortex breaking
mechanisms includes
– Vessel baffles
– Offset angle agitators
– Offset vertical agitators
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Vessel Agitation
Vessel Baffles
• Typically 3-4 baffles create localised
mixing zones
• Increased turbulence aids heat
transfer and uniformity
• Baffle width 8-10% of diameter
• Mounted off the vessel wall
• Increased power consumption
• Low hygiene / CIP capability
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Vessel Agitation
Offset Angle Agitators
• Typically 10-15% from
vertical
• Unbalanced forces can
become severe
• Limited power delivery
• Complicated installation
in larger vessels
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Vessel Agitation
Offset Vertical Agitators
• Mounted 15-20% off
vessel centre
• Typically used with
hygienic beverage
industries
− Effective mixing
− Good CIP capability
− Low capital cost
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Mash Conversion Vessel • Offset configuration
• Large diameter / HTC
• Low speed / shear
• Solids mixing / suspension
Vessel AgitationProprietary Agitation
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MCV Heating ExampleA stainless steel MCV contains 48 tonne of mash with a specific heat
capacity of 3.9 kJ/kg.oC. The vessel is heated using 3.0 barg saturated
steam. The mash requires to be heated from 65oC to 76oC.
a) Calculate the mass of steam required to heat the vessel contents assuming a fully insulated system and an isothermal process.
b) If the fully insulated system has a heating jacket surface area of 42m2 and an overall heat transfer coefficient of 1100 W/m2.oC, calculate the heating time required to achieve 76oC.
c) Calculate the radiated heat loss from the vessel to its 20oC surroundings if a surface area of 50m2 with emissivity 0.5 is fully exposed when at 76oC.
Steam
Water Grist CIP
Condensate
Mash To Mash Separator
Agitator
Mash
Conversion
Vessel
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MCV Heating a) Calculate the mass of steam required to heat the vessel contents
assuming a fully insulated system and an isothermal process.
From steam tables, Tsteam (3.0 barg) = 144oC,
latent heat (hfg) = 2,133 kJ/kg.oC
For an isothermal process
Tsteam (In) = Tsteam (Out)
Therefore steam condensation is the heat source to the mash
System Energy Balance (0oC datum)
Energy required to heat mash;
Er!st = mr!stcR∆TEr!st = (48 × 1000) × 3.9 × 76 − 65Er!st = 2,059,200kJ
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MCV Heating a) Calculate the mass of steam required to heat the vessel contents
assuming a fully insulated system and an isothermal process.
Energy provided by steam Es>A!r= Er!stEs>A!r = ms>A!r]yz (isothermal process)
ms>A!r = 2,059,2002,133
{|}~�{= �����
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MCV Heating b) If the fully insulated system has a heating jacket surface area of 42m2
and an overall heat transfer coefficient of 1100 W/m2.oC, calculate the
heating time required to achieve 76oC.
For steady state heat transfer
Q = mcRT = UA∆TBatch heating is an unsteady steady process due to a continuously changing
product temperature. Differential equations are required to describe this rate of
change;
Q = mr!stcR dTdt = UA Ts>A!r − Tr!stIntegration of the this equation generates;
� dTTs>A!r = UA
mr!stcR� dt>
S
����
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MCV Heating b) If the fully insulated system has a heating jacket surface area of 42m2
and an overall heat transfer coefficient of 1100 W/m2.oC, calculate the
heating time required to achieve 76oC.
� dTTs>A!r = UA
mr!stcR� dt>
S
����
ln Ts>A!r − Tr!st<Ts>A!r − Tr!st�
= UAmr!stcR t
t = mr!stcRUA . ln Ts>A!r − Tr!st<
Ts>A!r − Tr!st�
t = 48 × 1000 × 3.9 × 10001100 × 42 × ln 144 − 65
144 − 76} = ���|~� = 5�. 5{��
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MCV Heating b) If the fully insulated system has a heating jacket surface area of 42m2
and an overall heat transfer coefficient of 1100 W/m2.oC, calculate the
heating time required to achieve 76oC.
Note; for non-isothermal processes (such as using hot water instead of steam as
the heating medium), the log mean temperature difference (LMTD) is used to
describe temperature differential and the mass flowrate of the heating medium is required: -
ln ������������������������
= �������r����$� .
A��
���������A
���������
.t
where M is the mass flowrate of the heating medium with a specific heat capacity
Cpheat and inlet temperature of Theat.
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MCV Heating b) Calculate the radiated heat loss from the vessel to its 20oC
surroundings if a surface area of 50m2 with emissivity 0.5 is fully
exposed when at 76oC.
Heat transfer equation for radiation (non-black bodies)
Q = σεA(T� − T� )Q = 5.67 × 10�a × 0.5 × 50 × 76 " 273 − 20 " 273
Q = 10,582W� = 5�. ���
Note: if oC is instead of K used then
Q = 5.67 × 10�a × 0.5 × 50 × 76 −20 � = �. ����Very different answers!
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Heating In Agitated &
Non-Agitated Vessels