Question

To study the rate of heat flow through a metallic rod by conduction (Experiment)

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

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Explanation:

AIM OF THE EXPERIMENT:

a) To measure the temperature gradient along the length of the metal (copper) rod.

b) To determine the co-efficient of thermal conductivity of the metal (copper).

INTRODUCTION:

Conduction is a process of heat transfer through solids. When a temperature gradient exists in a body, experience has shown that there is a transfer of heat from the high temperature region to the low temperature region. The heat transfer rate per unit area is proportional to the temperature gradient given by:

q  T

A X ---Eq(1) Where, ‘q’ is the heat transfer rate (watts), A is the area of heat transfer (m2), T/ X is the temperature gradient in the direction of heat flow (C/m). When

the proportionality constant is inserted, we get,

q = -KT ---Eq(2)

A X

The positive constant ‘k’ is called the co-efficient of thermal Conductivity of the

material. The negative sign indicates that heat transfer takes place in the direction of decreasing temperature. Co- efficient of thermal conductivity has the units of watts/mC. Note that heat flow rate is involved and the numerical value of the co- efficient of thermal conductivity indicates how fast heat will flow in a given material.

Thermal conductivity co- efficient is a physical property of the material. Although it is fairly constant in a narrow temperature range, it varies over a wide temperature range. Metals which are good conductors of heat have high

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values of co-efficient of thermal conductivity; for example, 385 watts/mC for copper. Insulating materials have low values of co-efficient of thermal conductivity – for example 0.048 watt/mC for fiber insulating board.

In any conduction heat transfer problem, it is essential to have the knowledge of co-efficient of thermal conductivity of the material involved in the heat transfer process. This set-up has been designed to measure the temperature gradient along the length of the rod and to determine its co- efficient of the thermal conductivity.

APPARATUS:

It consists of a copper rod one end of which is heated by an electric heater and the other end projects inside the cooling water jacket. The middle portion of the rod is thermally insulated from the surroundings using asbestos rope. The temperature of the rod is measured at four different locations along its length. Following are the important features of the experimental setup.

a) Copper rod, Length Diameter

No. of thermocouples mounted

: 450mm : 20mm.

: 4 (at the interval of 58 mm) a long the length

b) Band heater used to heat up one end.

c) Thermal insulation covering the copper rod to reduce heat losses to the

surroundings.

d) Cooling water jacket at the other end with water supply connections

and thermocouples at both inlet T5 and outlet T6.

e) Heat controller or regulator to vary input power to the heater.

f) Measuring jar to measure water flow rate in the cooling water jacket.

g) Thermocouples to measure temperatures at 1, 2, 3 & 4 along the

length of the copper rod and 5 & 6 to measure temperatures at inlet & outlet of water jacket.

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h) Digital temperature indicator and channel selector.

PROCEDURE:

a) Switch ON the mains.

b) Open the valve at the inlet of the cooling water jacket and maintain constant

water flow rate.

c) Switch ON the heater.

d) Settheheatcontrolorregulatorandadjustthepowerinputtotheheater.

e) Wait for reasonable time till the temperatures T1 toT4 are fairly constant with time that is steady state is reached.

f) Read the temperatures T1 to T4 on the metal rod using channel selector and digital temperature indicator.

g) Read inlet and outlet water temperatures (T5 & T6) of the cooling water jacket.

h) Measurethecoolingwaterflowrateusingmeasuringjarandstopwatch.

i) Using the measured temperatures and water flow rate, the temperature gradient along the length of the brass rod and co- efficient of thermal

conductivity of copper are calculated using the procedure given below.

Formulae:

The heat balance equation is given by, qi = qo+q1 ---Eq (2)

Where,

qi = Input heat rate from the heater to the copper rod (Watts). qo = Output heat flow rate from the rod.

= Heat flow rate absorbed by water in the cooling water jacket (Watts). q1 = Heat loss from the rod to the surrounding s through thermal

insulation, watts (Watts), Assumed to be zero. 4

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