Conduction

Heat transfer:Heat transfer is defined as the Science which deals with rate of heat exchange between hot and cold bodies. Hot and cold bodies are called source and receiver respectively.These are three different processes by which H.T occurs.Conduction.Convection.Radiation.

Conduction:If the temperature gradients exist with in a solid body the heat will be transferred from higher temperature to lower temperature. This phenomenon is called thermal conduction. The thermal energy may be transferred by means of electrons which are free to move through the lattice structure of the material. It may be transferred as vibrational energy in the lattice of structure.

The control of heat flow at the desired rate is one of the most important areas of Chemical Engineering It is a mode of heat transfer in which heat transfer occurs from source to receiver particle to particle or molecule to molecule .First we will discuss the Law of conduction. Secondly the steady state heat conduction where the temperature distribution within the solid does not change with time, and thirdly, we will discuss the unsteady state heat conduction where the temperature distribution does change with time.

Fourier’s Law:

Consider a flow of heat through a plane wall of thickness ‘dx’ as shown in the figure below.

The wall lower and upper surfaces are assume to be insulated so that direction of heat flow is at right angle (perpendicular ) to wall.Assume that left face is hot and right face is cold. Which

receives heat. Fourier’s Law states that rate of heat flow Q is directly proportional to temperature difference T between the two surfaces, directly proportional to area ‘A’ and inversely proportional to thickness ‘x’ of the wall.

T1 T2 Q d x Hot Surface Cold surface

Mathematically A T

Q -------- x

OrK A T

Q = ---------- x

In differential formdQ = K A ( -d T )

d x term d T is called the temperature gradient. d x where negative sign indicates that the temperature gradient is opposite to the heat flow. It shows that heat flow occur from hot to cold surface.The negative sign is also due to the decrease of temperature in the direction of flow. Where K is a constant called thermal conductivity. its units are.

K = Q x = B tu / hr x ft = Btu / hr . ft oFA T ft2 oF

M . K . S System .

Q x = Watt . m = Watt . A T m2 oC m oCIf K has higher values for a material, then material is called conductor and if the value of K is lower then the material is called insulator.

Q = ( T1 – T2) A x / k

Where x/k is known as the thermal resistance and k/x is the transfer co-efficient.

Thermal conductivity

thermal conductivity, k, is the property of a material that indicates its ability to conduct heat. It appears primarily in Fourier's Law for heat conduction. First, we define heat conduction by the formula:

• where is the rate of heat flow, k is the thermal conductivity, A is the total cross sectional area of conducting surface, ΔT is temperature difference and x is the thickness of conducting surface separating the 2 temperatures.

• Thus, rearranging the equation gives thermal conductivity,•

• (Note: is the temperature gradient)• In other words, it is defined as the quantity of heat, ΔQ,

transmitted during time Δt through a thickness x, in a direction normal to a surface of area A, due to a temperature difference ΔT, under steady state conditions and when the heat transfer is dependent only on the temperature gradient.

Alternately, it can be thought of as a flux of heat (energy per unit area per unit time) divided by a temperature gradient (temperature difference per unit length)

Typical units are SI: W/(m·K) and English units: Btu·ft/(h·ft²·°F). To convert between the two, use the relation 1 Btu·ft/(h·ft²·°F) = 1.730735 W/(m·K). [Perry's Chemical Engineers' Handbook, 7th Edition, Table 1-4]

List of thermal conductivitiesThis is a list of approximate values of thermal conductivity, k, for some common materials.

Material    

Thermal conductivityW/(m·K)      

Cement, Portland [1] 0.29

Concrete, stone [1] 1.7

Air 0.025

Wood 0.04 - 0.4

Alcohols and oils 0.1 - 0.21

Silica Aerogel 0.004-0.03

Soil 1.5

Rubber 0.16

Epoxy (unfilled) 0.19

(Nansulate) 0.018

LPG 0.23 - 0.26

Epoxy (silica-filled) 0.30

Water (liquid) 0.6

Thermal grease 0.7 - 3

Thermal epoxy 1 - 7

Glass 1.1

Diamond 900 - 2320

Ice 2

Sandstone 2.4

Stainless steel[2] 12.11 ~ 45.0

Lead 35.3

Aluminium 237

Gold 318

Copper 401

Silver 429

Problem 1:Find the heat loss per square meter of surface through a brick wall 0.5m thick when the inner surface is at 400K and the outside is at 300K. The thermal conductivity of the brick may be taken as 0.7 w/mK.Solution:

x = 0.5(T1 – T2) = T = (400 – 300)

A = 1 m2

K = 0.7 w/mK = - K A T = - K A (T2 – T1)

x x

Q/A =K ( T1 – T2 ) = 0.7 x (400-300) x 0.5 m

Q/A = 140 w/m2

Problem 2:One face of a copper slab is maintained at 1000 oF and the other face is at 200 oF. How much heat is conducted through the slab per unit area if the slab is 3 inches thick? The thermal conductivity of copper may be taken as K 215 Btu/hr.ft oF. Solution:

q = - K A d T d x

q /A = - K ∫TiT2dT / ∫x2

x1dx .

q / A = - K T2 - T1 x2 - x1

T1 = 1000 oF x = x2 – x1 = 3/12 ft .T2 = 200 oFq/A = - 215 (200 - 1000) = 6.88x105 Btu/hr. ft2

3/12

Thermal Resistances in series:

A composite wall made up of three materials with thermal conductivities K1, K2, and K3 with thicknesses x1, x2, x3 and with temperature T1, T2, T3, and T4 at the faces as shown in figure given below.

Conduction of heat through an attached wall. Apply the Fourier's law to each section assuming that quantity of heat Q must pass through each area A.

Q = - K1A (T2 – T1) = K1A(T1 – T2) x1 x1

Then for first section:T1 – T2 = x1 Q I

K1A

Q

T1

T2

T3

T4

K1 K2 K3

x1 x2 x3

Second section:T2 – T3 = x2 Q II

K2AFor third section:T3 – T4 = x3 Q III

K3AAdd all these eq. We will get: T1 – T4 = x1 Q + x2 Q + x3 Q

K1A K2A K3A

T1 – T4 = Q/A (x1/K1 + x2/K2 + x3/K3 )

Be written as. Q = (T1 – T4)A = Total driving force ∑ x/K (Total Thermal resistance/area)

Problem:3A furnace is constructed with 0.20m of fire brick, 0.10m of insulating brick and 0.20m of building brick. The inside temperature is 1200K and the outside temperature is 330K. If the thermal conductivities are

Kf = 1.4, KI = 0.21 and K

b = 0.7 w/m

oK

Find the heat loss o1200K

per unit area Insulating Building

and the temperature Fire brick brick brick

at the Junction of the fire

0.20m 0.10m 0.20m

brick and insulating

brick

Kf = 1.4 K

I = 0.21 K

b = 0.7

330 oK A = 1m

2

q = T1 - T4 x1/k1A1 + x2 + x3 K2

A1 K3

A1

q = 1200 - 330 0.20 + 0.10 + 0.20 1.4A 1 0.21A1 0.7A1

q/A = 870 = 870 0.143 + 0.476 + 0.286 0.905 = 961 w/m2

x1 q = T1 – T2 K A 1

T2 = (1200 - 0.2 x 961) = 1062 oK 1.4 1

Problem:4An Industrial furnace wall is constructed of 0.7 ft thick fireclay brick having K = 0.6 Btu/hr. ft oF. This is covered on the outer surface with a layer of insulating material having K = 0.04 Btu/hr. ft oF. The inner most surface is at 1800 oF and the outer most is at 100 oF. The maximum allow able heat transfer rate for the furnace is 300 Btu/hr. ft2. Calculate the thickness of the insulating material for these conditions.Solution:

K1 = 0.6 K2 = 0.04 K1=0.6 K= 0.042 D x1 = 0.7 ft

1800 oF 100 oF

q = T1 – T3 Dx1/K1 + Dx2//K2

T1 T2 T3

Q /A = 300Btu/hr.ft2 300 = 1800 – 100 0.7/0.6 D x + 2

300 = 1

+ D x 1700 (0.7/.6 0.04) 2 /

0.7 + D x 2 = 17/3 0.6 0.04 D x 2 = (17/3 – 0.7/0.6) 0.04 D x 2 = 0.18 ft.

/ 0.04

Problem:5A flat wall is constructed of a 4.5 inches layer of refractory brick with a thermal conductivity of 0.08 Btu/h.ft.oF. Backed by a 9 inch layer of common brick of conductivity 0.8Btu/hrft 0F. The temperature of the inner face of the wall is 1400 oF and that of the outer face is 1700F. What is the heat loss through the wall? What is the temperature of the inner face between the refractory brick and the common brick.

A = 1ft2 2

K 1

K 2

T1

1 = 0.08 = 0.8

1400 o F x

9 170 o F = 4.5 inch x2 = 9 inch T 2 T 3

x 1 4.5/12 , x 2 = 9/12 K1 = 0.08 K 2 = 0.8 4.5 + 9 12 x 0.08 12 x 0.8

Q = T1 – T3

x1 + x2 /K2 1

K1

Q /A = (1400 – 170)/ (4.5/12x0.08 +9/12x0.8)

Q /A = 218.666 = 219 Btu/hr ft 2.

(b) Q /A = T1- T2

x1/K1

219 = 1400 – T2

4.5/12 x 0.08

219 = 1400 – T2

4.687

219 x 4.687 = 1400 – T2

T2 = 1400 – 219 x 4.687 = 373.547 = 374 oF

The temperature of the interface = 374 oF

Problem:6The inside temperature of a furnace is 15000F and its walls are lined with a 4 inch thick refractory bricks. The ambient temperature is 1000F. Compute the rate of heat flow through the walls and the temperature profile across the wall. If the variation of thermal conductivity of the refractory is given by K =0.10 + 5 x 10-5 T.

Solution:q/A=?Temperature Profile =?T1 = 15000 F, T2 = 1000F, K = 0.10 + 5x10-5Tx = 4inch =1/3 ft Q/A = -K dT/dx qdx = -KdT

2

1

5

0

0.10 5 10Tx

T

q dx TdT

5 2 5 22 2 1 1

1 1[0.10 (5 10 ) 0.1 (5 10 ) ]

2 2qx T T T T

2

1

5 210.10 (5 10 )

2

T

T

qx T T

5 2 5 22 2 1 1

1 1 10.10 5 10 0.1 (5 10 )

3 2 2q T T T T

5 2 213[0.1 1500 100 5 10 1500 100 ]

2q

2

3 140 56

588 / .

q

q Btu hr ft

Temperature profile across the wall.For obtaining the temperature profile.

5

0 1500

5 21500

0.1 5 10

1[0.1 5 10 ]

2

x T

T

q dx T dT

qx T T

25 2 51 1588 [0.1 5 10 0.1 1500 5 10 1500 ]

2 2x T T

5 2

5 2

4 8 2

588 0.1 2.5 10 150 56.25

588 206.25 0.1 2.5 10

0.350 1.7 10 4.25 10

x T T

x T T

x T T

This equation shows the non-linear characteristic of temperature Variation across the wall.

Conduction through a Cylinder:

Consider the hollow cylinder of inside radius is r1 the outside radius is r2 and length of cylinder is L. the thermal conductivity of the cylindrical material is K. The temperature of the out side surface is at T2 and the inside surface is at T1. It is desired to calculate the rate of heat flow out ward for this case.Consider a very thin cylinder of Wall thickness is dr.The heat flow at the thickness is given by

Q = - K 2 r L dT

drSeparate the Variable and Integrate the Eq. between the limits.

Q r1∫r2dr = - 2KLT1∫T2dT r

r1

r2

r

dr

T1

T2

T+dT

Q ( lnr)r1r2 = -2KL(T)T1

T2

Q(lnr2 – lnr1) = - 2 KL (T2 – T1)

Q ln (r2/r1) = 2 KL (T1 – T2)

Q = 2 K L (T1 – T2) ln r2/r1

The Eq can be written in the form of Fourier’s Eq we can use the log mean radius

rm = r2 – r1 /ln r2/r1

Then ln r2/r1 = r2 – r1 rm

Put the value of ln r2/r1 in the above Eq.

Then Q = K 2 L (T1 – T2) r2 – r1/rm

Q = 2 K L rm (T1 – T2)r2 – r1

This relation is used for heat flow through a thick-walled tube.For thin walled tubes it is sufficient to use the arithmetic mean radius ra instead of rm

Problem :A tube 60mm OD(Outer Dia) is insulated with a 50 mm layer of silica foam for which the conductivity is 0.055 w/m oC. followed with a 40mm layer of cork with a conductivity of cork is 0.05 w/m oC. If a temperature of the outer surface of the pipe is150 oC and the temp. of the outer surface of the cork is 30 oC, calculate theQ heat loss in watts per meter of pipe. Solution:The layer are too thick to use the arithmetic mean radius. Therefore the logarithmic mean radius should be used.

Then Q = 2 p K L rm (T 1 – T 2) r2– r1 160mm

for silica layer 80mm 60 rm = 80 – 30 = 50.97 mm 120mm ln 80/30 240mm

for cork layer 120 80 mm rm = 120 - 80 ln 120/80 B Tx 30mm Silica aA o Ts pipe Cork Tc Tc 50mm 60mm 50mm Tx Ts 160 -40- D = -40- mm mm 240mm.

For the silica layers. Ts = 150

oC, Te = 30

oC

KA = 0.055 w/m

oC , x

A = 50mm = 0.05m

rma

= r2 – r

1 /lnr

2/r

1 = 80 – 30 = 50.97mm.

ln 80/30

and for the cork layer = 0.05097m.

KB = 0.05 w/ moC , xb = 40mm 0.04m

rmb

= r2 – r

1 = 120 – 80 = 98.64mm.

ln r2/r

1 ln 120/80

=0.09864m

Ts – Tx = 2.838 qA/L Tx – Toc = 1.291 qB/L Add these two Eq. after rearrangingthen

Ts – Toc = 2.838 q/L + 1.291 q/L 150-30 = q/L (4.1290) q/L = 120 4.129 q/L = 29.06 w/m Ans.

Temperature distribution in concentric cylindrical metallic shells.

Problem:Two Concentric cylindrical metallic shells are separated by a solid material. If

the two metal surfaces are maintained at different constant temperatures what is the steady state temperature distribution within the separating

material?

Solution:

There are many problems where the properties of the system are functions of position instead of time, in this case, the temperature T and heat flow rate

per unit area Q are both functions of the radius r.

Figure

T0

T1a

b

r

r+∂r

T

The heat transfer must be related to a space interval b/w r and r + ∂r

Properties of system at r r + ∂ r.

Temperature T T + dT ∂ r dr

Heat transfer area/unit length 2r 2(r + ∂ r)Heat flow rate per unit area Q Q + dQ ∂ r.or Radial heat flow dr

Total radial heat flow 2rQ 2(r + ∂ r) (Q + dQ ∂ r)dr

Heat input to inner surface = 2rQ.Heat output from outer surface = 2(r + ∂ r) (Q + dQ ∂ r)

drAccumulation of heat = 0Heat input – Heat output = accumulation of heat

2prQ - 2p(r + ∂ r) (Q + dQ ∂ r) = 0 dr 2prQ - (2pr + 2p ∂ r) (Q + d Q ) = 0 dr 2prQ - (2prQ + 2pr dQ ∂ r + 2pQ ∂ r = 0. dr + 2pdQ ∂ r

2/dr 2prQ - 2prQ - 2prdQ ∂rr - 2pQ∂ 8r - 2p dQ ∂r

2 = 0. dr dr -(2pr dQ ∂ r + 2pQ∂ r + 2pdQ ∂ r2) = 0 dr dr 2pr dQ ∂ r + 2pQ∂ r + 2pdQ ∂ r2 = 0. dr dr

∂ r

divide both side by 2 p ∂rwe will get. r dQ + Q + dQ ∂r = 0 dr dr If ∂r 0. Then r dQ + Q = 0, r dQ = - Q dr dr separate the variable and intgrate. ò dQ = -ò dr Q r lnQ = -ln r + lnA ln Q + ln r = ln A Q r = A Q = A/r we know that Q = - K dT dr put the value of Q in the above Eq. then A/r = - K dT dr

Again separate the variable and integrate the Eq. ò dr = - ò K d T r A lnr = -KT/A +B In this Eq there are two constant of integration A and B these can be eliminated by applying the boundary condition

A t r = a T = T o And A t r = b T = T1

Apply the Boundary condition in the above Eq.

ln a = - K T0 /A + B ln b = - K T1 /A + B

- + - ln a – lnb = (K T1 – KT0)/A A = K T1 - K T0/ lna – lnb

Put the value of A in any of the above two Eq.

Then. ln a = - K T0 +B. ( K T1 – K T0) / (lna – lnb) ln a = - K T0 (lna – lnb) + B. K (T1 – T0) ln a = - T0 (lna – lnb) + B T1 – T0 B = lna + T0 (lna –lnb) T1 – T0

B = lna (T1 –T0) + T0 (lna – lnb) T1 –T0 B = T1 lna – T0 lna + T0 lna – T0 lnb T1 –T0 B = (T1 lna – T0 lnb) / (T1 – T0)

Put the value of A & B in the Eq. ln r = - KT + B A ln r = - KT + T1 lna – T0 lnb K(T1 – T0) / (ln a – lnb) (T1 –T0) ln r = - K T(lna – lnb) + T1 lna – To lnb K (T1 – T0) (T1 – T0) (T1 –T0) lnr = (ln b – lna) T + T1 ln a – T0 lnb (T1 –T0) lnr = (lnb – lna) T + T1 ln a – T0 lnb + T0 lna – T0 lna (T1 –T0) lnr = (lnb – lna) T – T0 lnb + T0 lna + T1 lna – T0 lna (T1 –T0) lnr = (lnb – lna) T – T0 (lnb – lna ) + T1 lna – T0 lna

Unsteady state heat conduction:

The unsteady state heat conduction is that where the temperature distribution does change with time.

A derivation of the partial differential Eq. for one dimensional heat flow will be discussed here.

(ln r) (T1 –T0) = (lnb – lna) (T – T0) + (T1 – T0) lna (ln r)(T 1 –T0)-(T1-T0 ) lna = (lnb – lna) (T – T0) (T1 – T0) (lnr – lna) = (T – T0) (lnb – lna) T – T0 = lnr – lna T1 – T0 lnb – lna Ans.