Lecture Notes for CO2 (Part 3)Lecture Notes for CO2 (Part 3)1-D STEADY STATE HEAT

CONDUCTION

Rizalman Mamat

Obed Majeed AliUniversiti Malaysia Pahang

Week – 5

Course Outcome 2 (CO2)

Students should be able to understand and evaluate one-dimensional heat flow and in different geometries

2

Lesson Outcomes from CO2 (Part 3) To derive the equation for temperature

distribution in various geometries Thermal Resistance concept – to derive

expression for various geometries To evaluate the heat transfer using thermal

resistance in various geometries To evaluate the critical radius of insulation To evaluate heat transfer from the rectangular fins

3

CRITICAL RADIUS OF INSULATIONAdding more insulation to a wall or to the attic results:

a) decrease heat transfer and,

b) increase the thermal resistance of the wall.

In a cylindrical pipe or a spherical shell, the additional insulation results :

a) increase the conduction resistance but,

b) decreases the convection resistance because of the surface for convection become larger .

An insulated cylindrical pipe exposed to convection from the outer surface and the thermal resistance network associated with it.

• Cylindrical pipe of outer radius r1 whose outer surface temperature T1 is maintained constant.

• The pipe is covered with an insulator (k and r2).

• Convection heat transfer at T∞ and h.

• The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as

1 1

2 1

2

ln / 12 2

ins conv

T T T TQ

r rR R

Lk h r L

5

The critical radius of insulation for a cylindrical body:

The critical radius of insulation for a spherical shell:

The variation of heat transfer rate with the outer radius of the insulation r2 when r1 < rcr.

• The rate of heat transfer from the cylinder increases with the addition of insulation for r2 < rcr

• Reaches maximum, when r2 = rcr

• Decreases when r2 > rcr

• Thus, insulating the pipe may actually increase the rate of heat transfer instead of decreasing it when r2 < rcr.

A 3 mm diameter and 5 m long electrical wire is tightly wrapped with a 2 mm thick plastic cover whose thermal conductivity is k = 0.15 W/m°C. Electrical measurements indicate that a current of 10A passes through the wire and there is a voltage drop of 8V along the wire.

If the insulating wire is exposed to a medium at T∞ = 30°C with a heat transfer coefficient of h = 24 W/m2 °C, find the temperature at the interface of the wire and the plastic cover in steady state operation. Also, determine the effect of doubling the thickness of the plastic cover on the interface temperature.

6

Problem

7

W80108 VIWQ e

Solution

• Outer surface Area

• Thermal resistances

• The interface temperature can be determined

• Critical radius

• Doubling thickness?

8

22 m 11.050035.022 LrA

C/W 0.940180.76R

C/W 18.0515.02

5.15.3ln

2

ln

C/W 76.011.012

11

0plasticconvtotal

012plastic

0

2conv

RR

kL

rrR

hAR

C 103T94.0

03-T80

T-T 01

1

total

1

RQ

mm 12.5m 0012512

15.0rcr

h

k

cr

crnew2,

r reaches radiusouter theuntilfer heat trans theincreasing Thus

r than less is which mm, 6r

A 5-mm-diameter spherical ball at 50 0C is covered by a 1-mm-thick plastic insulation (k=0.13 W/m ·0C). The ball is exposed to a medium at 15 0C, with a combined convection and radiation heat transfer coefficient of 20 W/m2·0C.

Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball.

9

Problem

10

2

cr

ran greater th iswhich

mm 13m 013.020

13.022

h

kr

Solution

11

HEAT TRANSFER FROM FINNED SURFACES

When Ts and T are fixed, there are two

ways to increase the rate of heat transfer:

• Increase the convection heat transfer coefficient h. This may require the installation of a pump or fan or add larger size of fan. Not adequate.

• Increase the surface area As by attaching to the surface extended surfaces called fins made of highly conductive materials such as aluminum.

Newton’s law of cooling: The rate of heat transfer from a surface to the surrounding medium

12

The thin plate fins of a car radiator greatly increase the

rate of heat transfer to the air.

13

Fin Equation

Volume element of a fin at location x having a length of x, cross-sectional area of Ac, and perimeter of p.

14

Fin Equation

Volume element of a fin at location x having a length of x, cross-sectional area of Ac, and perimeter of p.

where

Temperature excess

and

15

The general solution of the differential equation

Boundary condition at fin base

Boundary conditions at the fin base and the fin tip.

16

Under steady conditions, heat transfer from the exposed surfaces of the fin is equal to heat conduction to the fin at the base.

1 Infinitely Long Fin (Tfin tip = T)

Boundary condition at fin tip

The variation of temperature along the fin

The steady rate of heat transfer from the entire fin

17

2 Negligible Heat Loss from the Fin Tip(Adiabatic fin tip, Qfin tip = 0)

Boundary condition at fin tip

The variation of temperature along the fin

Heat transfer from the entire fin

Fins are not likely to be so long that their temperature approaches the surrounding temperature at the tip. A more realistic assumption is for heat transfer from the fin tip to be negligible since the surface area of the fin tip is usually a negligible fraction of the total fin area.

18

3 Specified Temperature (Tfin,tip = TL)

In this case the temperature at the end of the fin (the fin tip) is fixed at a specified temperature TL.

This case could be considered as a generalization of the case of Infinitely Long Fin where the fin tip temperature was fixed at T.

19

4 Convection from Fin Tip

The fin tips, in practice, are exposed to the surroundings, and thus the properboundary condition for the fin tip is convection that may also include the effectsof radiation. Consider the case of convection only at the tip. The conditionat the fin tip can be obtained from an energy balance at the fin tip.

20

Replace the fin length L in the relation for the insulated tip (slide 49) by a corrected length defined as

Corrected fin length Lc is defined such that heat transfer from a fin of length Lc with insulated tip is equal to heat transfer from the actual fin of length L with convection at the fin tip.

t is the thickness of the rectangular fins.D is the diameter of the cylindrical fins.

4 Convection from Fin Tip - cont

21

Fin Efficiency

Zero thermal resistance or infinite thermal conductivity (Tfin = Tb)

22

mL

23

Efficiency of straight fins of rectangular, triangular, and parabolic profiles.

Triangular and parabolic are more efficient than rectangular, contain less material and more suitable for applications requiring less weight and less space

24

25

• The fin efficiency decreases with increasing fin length because of decrease in fin temperature with length.

• Fin lengths that cause the fin efficiency to drop below 60 percent usually cannot be justified economically.

• The efficiency of most fins used in practice is above 90 percent.

26

Fin Effectiveness

Effectiveness of a fin

• The thermal conductivity k of the fin material should be as high as possible. Use aluminum, copper, iron.

• The ratio of the perimeter to the cross-sectional area of the fin p/Ac should be as high as possible. Use slender or thin pin fins.

• Low convection heat transfer coefficient h. Place fins on the gas (air) side.

27

Proper Length of a Fin

Because of the gradual temperature drop along the fin, the region near the fin tip makes little or no contribution to heat transfer.

mL = 5 an infinitely long fin mL = 1 offers a good compromise between heat transfer performance and the fin size.

28

• Heat sinks: Specially designed finned surfaces which are commonly used in the cooling of electronic equipment, and involve one-of-a-kind complex geometries.

• The heat transfer performance of heat sinks is usually expressed in terms of their thermal resistances R.

• A small value of thermal resistance indicates a small temperature drop across the heat sink, and thus a high fin efficiency.

29

30

31