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UCL 2009/G017 Steady-state heat transfer II 1

Steady-state heat transfer II

Dr Farlan Veraitch

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Heat transfer between fluids - Recap

Consider heat transfer through the wall of a fermenter or througha pipe in a heat exchanger.

Hot and cold fluids flow on either side of thewall

Heat transfer across the wall is byconduction

Heat transfer between the fluid and the wallis by convection

Each layer represents a resistance to heattransfer

Rate of heat transfer:

But how to we calculate h?

Hot

fluid

Th

Tc

Cold

fluid

T1T2

Liquid

films

wall

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Estimation of heat transfer coefficients

Well now review empirical correlations to calculate individual heattransfer coefficients h

What factors need to be considered in the correlation to calculate h?Hot

fluid

Th

Tc

Cold

fluid

T1

T2

Liquid

films

wall

How do we correlate all these variables?

Why does this help? Simplifies problem as the number of groups < the number of variables a = f (b, c)

operating conditions: velocity u (pipe), stirrer speed N (vessel)

fluid properties: viscosity thermal conductivity specific heat capacity density

system geometry Characteristic length eg. diameter

h = f (u or N, , k,, Cp, d)

Dimensionless analysis to relate the most importantvariables together in the form ofdimensionless numbers

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Values ofh for flow in pipes or stirred vessels are usually evaluatedusing empirical correlations expressed in terms ofdimensionless

numbers

Dimensionless numbers that are used to calculate h are:Nusselt no. FlowReynolds no. ImpellerReynolds no. Prandtl no.

where

h individual heat transfer coefficient (W.m-2.K-1)

kfluid thermal conductivity of hot/cold fluid (W.m-1.K-1)

d Pipe or tank diameter (m)

Liquid density (kg.m-3)u Liquid velocity (m.s-1)

dp Pipe diameter (m) Liquid viscosity (N.s.m-2)N Impeller rotational speed (s-1)

di Impeller diameter (m)

Cp Specific heat capacity (J. kg-1.K-1)

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Estimation of heat transfer coefficients Significance of each dimensionless group:

Nusselt no. conductive resistance : convective resistance

FlowReynolds no. inertial forces : viscous forces

Ref< .LAMINAR FLOW

Ref> ....TURBULENT FLOW

ImpellerReynolds no. inertial forces : viscous forces

Rei < .. LAMINAR FLOW

Rei > .TURBULENT FLOW

Prandtl no. momentum transfer : heat transfer

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In most industrial applications, heat transfer occurs betweenturbulent fluids in pipes or in stirred vessels - forced convection in

these systems is more important than natural convection

For forced convection in pipes or stirred vessels, the general formof heat transfer correlations is:

Nu = c * Rea * Prb

The values of a, b and c depend on the heat-transfer equipmentand the flow regime We will look at correlations for:

turbulent flow in pipes turbulent flow in stirred tanks

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Turbulent flow in pipes

A widely accepted correlation for forced convection in pipes:Nu = 0.023 Ref

0.8

Pr0.4

This equation is valid for:

liquids with viscosity close to water

104 Ref 1.2 x 105

0.7 Pr 120L / d 60

When using these correlations to calculate h, it is important toalways check that Re and Prlie within the valid ranges!

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Turbulent flow in pipes

For very viscous liquids, there may be a marked difference betweenthe viscosity of fluid in bulk flow and the viscosity of fluid adjacent to

the wall

A modified form of the equation includes a viscosity correction term:

where b is the viscosity of bulk fluid

w is the viscosity of fluid adjacent to the wall

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Turbulent flow in pipes - Eg

A single-pass shell-and-tube heat exchanger is used to heat adilute salt solution for a large-scale protein chromatography

column. 25.5m3h-1 solution passes through 42 parallel tubes inside

the heat exchanger; the internal diameter is 1.5cm and the tube

length is 4m. The viscosity of the bulk salt solution is 10-3 kg m-1 s-1,

the density is 1010 kg m-3, the average heat capacity is 4 kJ kg-1 K-1

and the thermal conductivity is 0.64 W m-1 K-1.

a) Calculate the tube-side heat transfer coefficient (for the insideliquid film) and the resistance

b) Keeping all other variables constant, what is the expected %

change in h and Rif:

viscosity increases by 10%? velocity is doubled?Hint - spend some time working out the fluid velocity in the tubes

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RECAP: Single-pass shell-and-tube heat exchanger

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Solution

List known parametersN = 42

Qv = 25.5 m-3h-1 = ...

Di = 0.015 m

L = 4 m = 10-3 kgm-1s-1 = 1010 kgm-3Cp = 4000 Jkg

-1K-1

kfluid = 0.64 Wm-1K-1

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Solution

What equation will you use to find h?Nu = 0.023 Re0.8 Pr0.4

What parameter values need to be calculated?

What order will you calculate parameters?hi d = 0.023 p u d

0.8 Cp0.4kfluid kfluid

? ?

u

Re

Pr

Nu

hR

u = Q/A

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Solution

Qv

u = Qv/A

A = d2 x number of tubes

4

= (0.015)2 x 42

4

A = 0.0074m2

u = Qv = 25.5/3600A 0.0074

u = 0.95ms-1

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Ref= ud= 1010 x 0.95 x 0.015

10-3

Ref= 1.44 x 104

Pr = Cpkfluid

= 4000x(1x10-3)

0.64

Pr = 6.25

L/d = 4 = 267

0.015

As 104 < Re < 1.25x105

0.7 < Pr < 120

L/d > 60

Eqn is VALID

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Nu = 0.023 Re0.8 Pr0.4

= 0.023(1.44 x 104)0.8(6.25)0.4

Nu = 101.6

Nu = hid

kfluid

hi = 101.6 x 0.64

0.015

hi = 4335 Wm-2K-1

R = 1 = 0.00023 m2KW-1

hi

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What happens if viscosity increases by 10%?

x 1.1hd = 0.023 ud 0.8 Cp 0.4k kh 1 0.80.4

-0.8.0.4h -0.4 [and therefore R 0.4]hnew = new -0.4hold oldhnew = (1.1)

-0.4 = 0.96

hold

4% in hi

Check Re + Pr still valid

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Rnew = new 0.4Rold old

= (1.1)0.4

= 1.04

4% in R as expected

R 1/h 0.4

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What happens if velocity doubles?h u0.8

hnew = unew0.8

hold uold

= 20.8

= 1.74

74% in h when u doubles

Rnew = 0.57

Rold

R u-0.8 2-0.843% in R when u x 2

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Turbulent flow in stirred tanks

Jacketed vessels

Inside film heat transfer coefficient hi Two correlations for heat transfer in jacketed vessel equipped with 6-turbine flat

blade stirrer:

1) Nu = 0.81 Rei0.68Pr0.33


105 Rei 7 x 105

liquids with physical properties close to water

Ref. Akse, H., et al Chem. Eng. Sci., 22, 135, 1967

2) Nu = 0.76 Rei0.67 Pr 0.33


4 x 103 Rei 2.8 x 105

1.9 Pr 1 x 105

Ref. Strek, F., et al Int. Chem. Eng., 7, 693, 1967

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Turbulent flow in stirred tanks

Helical cooling coils

Inside film heat transfer coefficient hi The value ofhican be found using the correlations for turbulent flow in pipes.

(eg. Nu = 0.023 Ref0.8Pr0.4)

These correlations have been obtained for straight tubes; with a coil somewhatgreater transfer is obtained for the same physical conditions:

hi(coil) = hi(straight pipe) ( 1 + 3.5 d/dc)

where d is the inside diameter of the tube and dc is the

diameter of the helix.

Outside film heat transfer coefficient ho The value ofho depends on the degree of agitation and the fluid properties.

Ref. Chilton, T. H., et al Ind. Eng. Chem., 36, 510-516, 1944

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Logarithmic and arithmetic mean temp differences

Thi

Tho

Tci

Tco

Use Ta when one fluid is maintained at a constant temperature

E.g. In a fermenter

T1 = Tci - Tferm

T2 = Tco - Tferm

LM is used if temperature varies in both fluids in eithercounter-currentorco-currentflow

T1 = Thi - Tci

T2 = Tho - Tco

Thi

Tho

Tci

Tco

T1 = Thi - Tco

T2 = Tho - Tci

co-current counter-current

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RECAP: Single-pass shell-and-tube heat exchanger

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Turbulent flow in stirred tanks Case studyHeat transfer coefficient for stirred vessel and cooling coil length

A fermenter used for antibiotic production must be kept at 27C. After considering theoxygen demand of the organism and the heat dissipation from the stirrer, themaximum heat-transfer rate required is estimated at 550 kW.

Cooling water is available at 10C and the exit temperature of the cooling water is

calculated as 25C. It is proposed to install a helical cooling coil inside the fermenter;

the outer diameter of the coil pipe is 8 cm, the pipe thickness is 5 mm and the

thermal conductivity of the steel is 60 W m-1 C-1. The heat transfer coefficient for the

cooling water is 14,000 W m-2 C-1. The internal fouling factor of the coil is 8500 W

m-2 C-1; the fermenter side of the coil is kept relatively clean.

The fermenter has diameter 5m and is mixed using a turbine impeller 1.8 m indiameter operated at 60 rpm. The fermentation broth has the following properties: b

= 5 x 10-3 Pa s; = 1000 kg m-3; Cp = 4.2 kJ kg-1 C-1; k = 0.7 W m-1 C-1.

Neglecting viscosity changes at the wall of the coil, Calculate the heat transfer coefficient for the broth Calculate the length of cooling coil required. Comment on your answer.

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Turbulent flow in stirred tanks Case study

List parametersLet broth be hot fluid denoted as h

Let broth be cold fluid denoted as c

Broth Cold water Coil

Th = 27C Tci = 10C qmax = 550,000w

= 5 x 10-3 PaS Tco = 25C do = 0.08m

= 1000kg/m3 x = 0.005m

Cp = 4200J/kgC Ksteel = 60 W/mC

kh = 0.7W/mC hc = 14,000W/m2C

1/hfh = 0 hfc = 8500W/m2C

Fermenter: dv = 5m, di = 1.8m, N = 60rpm = 1s-1

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CW

hc

hfc

hh

BROTH

x or roln(ro/ri)

k k

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Calculate the heat transfer coefficient for the broth1) Calculate hh?

Nu = 0.87 Rei0.62Pr0.33

Re Ndi2 1000 x 60/60 x 1.82 6.48x105 5x10-3Pr Cp 4.2x103 x 5x10-3 30

Kh 0.7

Nu = 0.87(6.48x105)0.62 (30)0.33

= 1.07x104

hh = Nu,kh 1.07x104 x 0.7 1501W/m2C

dv 5

===

= = =

==

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Calculate the length of cooling coil required. What equation will you use to find L?

What parameter values need to be calculated?

What order will you calculate parameters?

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Calculate L

q = Uo Ao TaCalculate Ta

UoAo

L

Ta T1 + T2 (27-10)+(27-25)2 2

Ta = 9.5C

= =

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1 1 + 1 + x + 1 + 1U hh hfh k hfc hc

1 + 0 + 5x10-3 + 1 + 1

1501 60 8500 14,000

6.67x10-4 + 8.33x10-5 + 1.17x10-4 + 7.14x10-5

small small

= 9.386 x 10-4

U = 1065W/m2C

or U = 1113W/m2C if use

roln(ro/ri)

k0

=

=

=

roln(ro/ri)

k

Wall resistance +

cw resistance - minor

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q = Uo Ao TaAo= q

Uo Ta= 550000

1065 x 9.5

Ao = 54.36m2

AO = do L

L = 54.36 X 0.08

L = 216m

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hlL3 -SSHT II 201213