Chap 4 Heat Transfer (PART 2)

BKF2422 HEAT TRANSFER Chapter 2 Principles of steady-state heat transfer in conduction

TOPIC OUTCOMES It is expected that students will be able to: • Solve problems using steady-state conduction principles for one-dimensional solid conduction heat transfer in parallel and series • Calculate overall heat transfer coefficient to solve problems related to combined conduction and convection heat transfer mechanism. • Solve the problem related to internal heat generation and determine the critical thickness and of insulation for a cylinder • Apply shape factor to estimate the multidimensional heat transfer

CONTENTS • One Dimensional Conduction Heat Transfer – – – – – –

Conduction Through a Plane Wall Conduction Through Solids In Series Conduction Through Solids In Parallel Conduction Through a Hollow Cylinder Conduction Through a Multilayer Cylinders Conduction Through a Hollow Sphere

• Combined Conduction and Convection and Overall Heat Transfer Coefficient

CONDUCTION: FOURIER’S LAW

• Flux of conduction heat transfer can be calculated by Fourier’s Law qx dT  k Fourier’s Law A dx qx : heat-transfer rate in the x direction (SI: W or J/s; cgs: cal/s; Eng.: btu/h) A : cross sectional are normal to the heat flow (m2) k : thermal conductivity ( SI: W/m. K; cgs: cal/s. cm. °C; Eng.: btu/h. °F. ft ) dT/dx : temperature different in the x direction • The minus sign is required in Fourier’s equation because the heat transfer is positive in the direction from initial point 1 to the final point 2. Since the T1 > T2 (heat is transport from high temperature to lower temperature region), minus sign is needed to make the value of heat rate positive.

HEAT TRANSFER – CONDUCTION • For steady state, the equation can be integrated, T2 q x x2 dx   k  dT  T1 A x1 qx k T1  T2   A x2  x1

This equation is basically a matter of putting in values to solve.

HEAT TRANSFER – CONDUCTION Conduction Through a Plane Wall. Temperature, (K) T1

T1 q

T2 Δx

T2

0

Δx

Distance,x (m)

T1  T2 T1  T2 q  x kA R The temperature various linearly with distance.

EXERCISE 1 Calculate the heat loss per m2 of surface area for an insulating wall composed of 25.4 mm thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1K. From Table A.3 (pg 599), thermal conductivity for fiber insulating board is 0.048 W/m.K. q k (T1  T2 ) 0.048   (352.7  297.1) A x2  x1 0.0254 q  105.1W / m 2 A

HEAT TRANSFER – CONDUCTION Conduction Through Solids In Series.

T1 A q

B T2

DxA DxB

C T3 DxC T4

T1  T2 T2  T3 T3  T4 T1  T4 • The rate q  of heat  transfer,   RA RB RC RA  RB  RC x A RA  where,k A A

xB RB  kB A

xC RC  kC A

EXERCISE 2 A cold storage room is constructed of an inner layer of 12.7 mm of pine, a middle layer of 101.6 mm of cork board and an outer layer of 76.2 concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of concrete. The conductivites for pine, 0.151; cork board, 0.0433; and concrete, 0.762 W/m.K. Calculate the heat loss in W for 1 m2 and the temperature at the interface between wood and cork board. Answer: (-16.48 W, 256.79 K)

SOLUTION T1  255.4 K , T4  297.1K , A  1m 2 k A  0.151, k B  0.043, kC  0.762 x A  0.0127 m, xB  0.1016m, xC  0.0762m x A 0.0127 K RA    0.0841 k A A 0.151(1) W xB 0.1016 K RB    2.346 k B A 0.043(1) W xC ? K RC   ? kC A ? W

HEAT TRANSFER – CONDUCTION • Conduction Through Solids In Parallel B A

C D

F E G

qT  qA  qB  qC  qD  qE  qF  qG kC AC k A AA kB AB kD AD qA  (T1  T2 )  (T2  T3 )  (T2  T3 )  (T2  T3 )  ....... xA xB xC xD

HEAT TRANSFER – CONDUCTION Conduction Through A Hollow Cylinder. T2 r2

L

q

r1 • The cross-sectional area normal to the heat flow is, A =2prL. q transfer,dT • The rate of heat  k A

dr

CONDUCTION THROUGH A HOLLOW CYLINDER q dT  k A dr T2 q r2 dr   k  dT  T1 2pL r1 r

2pL T1  T2  q  k ln r2 r1  or

T1  T2 q  kAlm r2  r1

Where:

Alm  

A2  A1 ln  A2 A1 

2pLr2   2pLr1  ln 2pLr2 2pLr1 

2pL(r2  r1 )  ln r2 r1  q

R

T1  T2 T  T2  1 r2  r1  kAlm R

r2  r1 ln r2 r1   kAlm 2pkL

EXERCISE 3 A thick wall cylindrical tubing of hard rubber having and inside radius of 5 mm and an outside radius of 20 mm is being used as a temporary cooling coil in a bath. Ice water is flowing rapidly inside, and the inside wall temperature is 2p L temperature is at 297.1 K. A 274.9 K. The outside qk (T1  T2 ) total of ln(r2 / r1 ) 14.65 W must be removed 2p Lfrom the bath by the q  0.151x (274.9  297.1) cooling 0.005) coil. How many mln(0.02 tubing /are needed?(k=1.15 q W/m.K)  15.2W / m L

14.65W length   0.964m 15.2W / m

HEAT TRANSFER – CONDUCTION Conduction Through a Multilayer Cylinders.

r1

r2

r3 r4

q A B C

T1 T2 T3 T4

Example, heat is being transferred through the walls of an insulated pipe.

HEAT TRANSFER – CONDUCTION • At steady-state, the heat-transfer rate q, be the same for each layer. • The rate of heat transfer,

T1  T2 T2  T3 T3  T4 T1  T4 q    RA RB RC RA  RB  RC where, ln r2 r1  RA  2pk A L

ln r3 r2  RB  2pk B L

ln r4 r3  RC  2pkC L

EXERCISE 4 A thick walled tube of stainless steel (A) having a k = 21.63 W/m.K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254 thick layer of insulation (B), k = 0.2423 W/m.K. The inside wall temperature of the pipe is 811 K and the outside is at 310.8 K. For a 0.305 m length pipe, calculate the heat loss and also the temperature at the interface between the metal and the insulation. Answer: (331.7 W, 805.5 K)

SOLUTION 0.0254 0.0508 r1   0.0127 m, r2   0.0254m...... 2 2 A1  2pLr1  2  p  0.305  0.0127  0.0243m 2 , A2 AAlm  RA 

A2  A1 0.0487  0.0243   0.0351m 2 ln  A2 / A1  ln 0.0487 / 0.0243

r2  r1 0.0127   0.01673K / W k A Alm 21.63(0.0351)

T1  T3 q  331.7W RA  RB T1  T2 q RA

HEAT TRANSFER – CONDUCTION Conduction Through a Hollow Sphere

T2 r2 q

r 1 T1

• The cross-sectional area normal to the heat flow is, A = 4pr2. q transfer,dT • The rate of heat  k A

dr

HEAT TRANSFER – CONDUCTION q  4p

T2 dr r1 r 2  k T1 dT 4pk T1  T2 T1  T2  q  T1  T2    1 1 R 1 1     4pk r1 r2  r1 r2  r2

HEAT TRANSFER – CONDUCTION Temperature Profile for Heat Transfer By Convection From One Fluid To Another. film

Metal wall film T2

T1

Cold fluid B T3

T4

Warm liquid A

T5

T6

q

HEAT TRANSFER – CONDUCTION Region, T1 – T2 : turbulent fluid flow. Mainly convective heat transfer. T2 – T3 : velocity gradient very steep. No turbulent flow, (i.e. only laminar). Mainly conductive heat transfer. T3 – T4 : conductive heat transfer. T4 – T5 : no turbulent in film, mainly conductive heat transfer T5 – T6 : turbulent flow, conductive heat transfer. T1 – T2 and T5 – T6 : different are small. Convective coefficient for heat transfer through a fluid: q = hA(T – Tw) where, h = convective heat transfer coefficient. T = average temperature in fluid. Tw = temperature of wall in contact.

HEAT TRANSFER – CONDUCTION Combined Convection and Conduction and Overall Coefficients. • Heat flow with convective boundaries: plane q T wall h T i

o

1

T2

hi

xA

To

Ti  To Ti  To q  1 hi A  x A k A A  1 ho A Ri  RA  Ro q

ATi  To   UATi  To  1 hi   x A k A   1 ho 

HEAT TRANSFER – CONDUCTION Heat flow with convective boundaries: cylindrical wall with insulation. rO T1

r1

ho T2 T3

ri T4 hi A B

Similarly,

T1  T4 T1  T4 q  1 hi Ai   ro  ri  k A AAlm  1 ho Ao   R

Overall heat transfer coefficient,

T1  T4 q  U i Ai T1  T4   U o Ao T1  T4   R

where, Ui 

1 1 hi  ro  ri  Ai k A AAlm  Ai Ao ho

and 1 Uo  Ao Ai hi  ro  ri Ao k A AAlm  1 ho

HEAT TRANSFER • Other way we can used,

T1  T4 q  U i Ai T1  T4   U o Ao T1  T4  Ri  RA  RB  Ro

where, 1

1 Ri   2pLri hi  hi Ai

1 1 Ro   2pLro ho  ho Ao

ln r1 ri  RA  2pk A L

ln ro r1  RB  2pk B L

A thick-walled tube of stainless steel (A) having a k = 21.63 W/m.k with dimensions of 0.0254m ID and 0.0508m OD is covered with a 0.0254m layer of asbestos (B) insulation, k = 0.2423 W/m.k. The inside wall temperature of the pipe is 811K and the outside surface of the insulation is at 310.8K. For a 0.305m length of pipe, calculate the heat loss and also the temperature at the interface between the

T1  T3 q R A  RB The resistances are

ln( r2 / r1 ) ln( d 2 / d1 ) RA   2pk A L 2pk A L 0.0254 ln( ) 0.0127  2p (21.63)(0.305)  0.01673 K/W The heat transfer rate is

ln( r2 / r1 ) ln( d 2 / d1 ) RB   2pk B L 2pk B L 0.1016 ) 0.0508  2p (0.2423)(0.0.305)  1.493 K/W ln(

T1  T3 q R A  RB

T1  T2 q RA

811  310.8 q 0.01673  1.493  331.7 W

811  T2 331.7  0.01673 T2  805.5 K

CRITICAL THICKNESS OF INSULATION FOR A CYLINDER q  h0 A(T2  T1 )

k (r2 ) cr  h • If outer radius < rcr: adding more insulation will increase heat transfer rate • If outer radius > rcr: adding more insulation will decrease heat transfer rate

With insulation:

2pL(T1  T0 ) q ln r2 / r1  1  k r2 h0

EXAMPLE • An electric wire having a diameter of 1.5 mm covered with a plastic insulation (thickness = 2.5mm) is exposed to the air at 300K and ho = 20 W/m2.K. It is assumed that the wire surface temperature is constant at 400K and is not affected by the covering. • A) calculate the value of the critical radius • B) calculate the heat loss per (m) of wire length with no insulation • C) repeat (b) for insulation being present a) 20 mm b) 9.42W c) 32.98 W

CONVECTION HEAT TRANSFER Convection: Heat transfer using movement of fluids. Heat transfer is considered as convection with the presence of bulk fluid motion. Fluid motion enhances heat transfer where the higher the fluid velocity, the higher the rate of heat transfer.  2 main classification of convective heat transfer; 1. Forced Convection : fluid flow by pressure differences, a pump, a fan and so on 2. Natural Convection: motion of fluid results from the density changes in heat transfer

Tw = 80 oC

q

To = 30 oC

Fluid flow The rate of heat transfer :

q  Ah(Tw  To )



h = heat transfer coefficient (W/m2.K) A= surface area (m2)

The convection coefficient is a measure of how effective a fluid is at carrying heat to and away from the surface.

CONVECTION HEAT TRANSFER

Metal wall

T1 T2

Cold fluid B T3

Turbulence region

Warm fluid A q

q = hA (T-

Turbulence absent

FORCED CONVECTION INSIDE PIPES  Forced convection – fluid forced to flow by pressure differences

 Types of fluid, laminar or turbulent – great effect on heat-transfer coefficient More turbulent– greater heat-transfer coefficient

 Reynolds number, NRe

 NRe  D where 

v = velocity of fluid (m/s)  = viscosity of fluid (Pa.s) = density of fluid (kg/m3) D = diameter of pipe (m)

FKKSA

FORCED CONVECTION Dimensionless numbers:  Prandtl number, NPr μ cμ NPr  ρ  P k ρc k P

 Nusselt number, NNu

where = viscosity of fluid (Pa.s) = density of fluid (kg/m3) k = thermal conductivity of fluid (W/m.K) cP = heat capacity of fluid (J/kg.K) h = heat transfer coefficient (W/m2.K) D = diameter of pipe (m)

FKKSA

NNu  hD k

LAMINAR FLOW INSIDE HORIZONTAL PIPE NRe  2100 & NReNP r D  100 : L 1   3 0.14 h D D     b  N   a 1.86N N   Nua Re Pr k L  w   where

D = inside diameter of pipe (m)  L = length of pipe (m) b = viscosity of fluid at bulk temperature (Pa.s) w = viscosity of fluid at wall temperature (Pa.s) ha = average heat transfer coefficient (W/m2.K)

 T  T   bo bi  T  All physical properties at b mean 2

q = haA∆Ta where FKKSA

Limitations N Re  2100 N Re N Pr

except  w

 T  T    T  T  bi   w bo  ΔT   w a 2

D  100 L

TURBULENT FLOW INSIDE HORIZONTAL PIPE Rate of heat transfer is greater Many industrial heat transfer processes in the turbulent region L Limitations NRe  6000 , 0.7 ≤ NP r ≤ 16000 &D  60: N Re  6000 NNu 

hLD  0.027 NRe 0.8 N k

 

1  0.14 3  b  Pr    w 

where

0.7  N Pr

D  16000 L

L  60 D

 b =viscosity of fluid at bulk average temperature (Pa.s)  w = viscosity of fluid at wall temperature (Pa.s) k = thermal conductivity of fluid (W/m.K) cP = heat capacity of fluid (J/kg.K) hL = heat transfer coefficient based on the log mean driving force ∆Tlm (W/m2.K) D = inside diameter of pipe (m) FKKSA

EXAMPLE 4.5-1 Page 262: Heating of Air in Turbulent Flow Air at 206.8 kPa and an average of 477.6 K is being heated as it flows through a tube of 25.4mm inside diameter at velocity of 7.62 m/s. The heating medium of 488.7 K steam condensing on the outside of the tube. Since the heat-transfer coefficient of condensing steam is several thousand W/m2.K and the resistance of the metal wall is very small, it will be assumed that the surface wall temperature of the metal in contact with the air is 488.7 K. Calculate the heat-transfer coefficient for an L/D > 60 and also the heat-transfer flux q/A.

ho  hsteam

25.4 mm

air

Tbi

Tave  477.6 K v  7.62 m/s P  206.8 kPa Steam, Tw  488.7 K

L

From Appendix A.3, at P  101.32 kPa, Tave  Tbm  477.6 K

b  2.6 10 5 Pa.s N Pr  0.686 k  0.03894 W/m

  0.74 kg/m

3

From Appendix A.3, at Tw  488.7 K

 w  2.64 105 Pa.s

Tbo

For  ,  m PV  nRT  PV    RT M  m PM    RT  PM  RT V  PM  is dep end on P & T    RT  P  T   2  1  2  2   P1  T1  T1  T2  206.8  3   206.8 kPa  0.74   1.509 kg/m  101.35 

N Re 

D



25.4 10 3 (7.62)(1.509)  2.6 10 5  1.122 10 4 ( 6000)

N Nu 

hL D 0.8  0.027 N Re N Pr k

1 3

 b     w 

q  hL Tw  Tbm  A  63.2488.7  477.6  701.1 W/m 2 0.14

1 0.0260 hL (25.4 10 3 )   4 0.8 3      0.027 1.122 10 0.686   0.03894  0.0264 

hL  63.2 W/m 2 .K

0.14

Thi

Tho

T1

Tci

Tco

Thi

Tho

T2

Tco Tci

countercurrent flow Thi Thi

Tho

Tci

Tco

Tho

T1 Tci

parallel flow

Tco

T2

EXAMPLE 4.5-4 Page 268: Heat Transfer Area and Log Mean Temperature Difference A heavy hydrocarbon oil which has a cpm = 2.30kJ/kg is being cooled in a heat exchanger from 371.9 K to 349.7 K and flows inside the tube at a rate of 3630 kg/h. A flow of 1450kg water/h enters at 288.6K for cooling and lows outside the tube. a) Calculate the water outlet temperature and heat-transfer area if the overall Ui = 340 W/m2.K and the streams are countercurrent b) Repeat for parallel flow

(a) countercurrent flow 397.1 K  Thi

T1

  3630 kg/h oil, m

Tho  349.7K

Tco

T2

c  c 

p oil

 2.3 kJ/kg.K

p water

 4.187 kJ/kg.K

U i  340 W/m 2 .K Ai  ?

Tci  288.6 K

  1450 kg/h water, m

q  m c p h Tho  Thi 

 3630(2.3)(371.9  349.7) / 3600  51490 W

q  m c p c Tco  Tci  51490  1450(4.187)(Tco  288.6) Tco  319.1 K

T1  371.9  319.1  52.8 K T2  349.7  288.6  61.1 K

T1  T2 61.1  52.8 Tlm    56.9 K  T1   61.1  ln    ln   52.8   T2  q  U i Ai Tlm 51490  340 Ai (56.9) Ai  2.66 m 2

(b) parallel flow T1  349.7  319.1  30.6 K T2  371.9  288.6  83.3 K

T1  T2 83.3  30.6 Tlm    52.7 K 83 . 3  T    ln   ln  1  30 . 6    T2  q  U i Ai Tlm 51490  340 Ai (52.7) Ai  2.87 m 2

This is a larger area than for counterflow. This occurs because counterflows gives larger tem perature driving forces.

•Radiation heat transfer is the transfer of heat by electromagnetic radiation •Occur in solid, liquid and gas •Not require heat transfer medium •Fastest energy transfer •Example: microwave, radar, cordless telephones

Absorptivity • When thermal radiation (light waves) falls upon a body, part is absorbed, part is reflected into space and part is transmitted through the body. •

  absorptivity/fraction absorbed  1.0 BLACK BODY – one that absorb all radiant   reflectivity/fraction reflected  0

energy and reflect none.

E total emissive power of a surface Emissivity,    E B total emissive power of black body

• Kirchoff’s Law states at the same  1  1 temperature T1 Black body,  1

 1

• For • For a perfect black body with Gray body,   1

q  A T 4

q  AT 4

:

• Substances that have emissivity < than 1.0 are called gray bodies