Ht Lab Manual-1.pdf

Experiment No:- 1

THERMAL CONDUCTIVITY OF METAL ROD AIM:To determine the thermal conductivity of given metal rod. APPARATUS:1. 2. 3. 4.

Measuring flasks Stop watch Thermocouples with temperature indicator Copper Rod

SPECIFICATIONS:1. Length of the metal rod =  Total length = 400mm  Test length = 250mm 2. Diameter of the bar = 25mm 3. Inner radius of insulating shell (𝑟 ) = 12.5 mm 4. Outer radius of insulating shell (𝑟 ) = 92.5 mm 5. Measuring flask capacity = 0 to 1000 ml 6. Specific heat of water = 4187 J/Kg k 7. Thermal conductivity of insulating powder 𝑲𝒊𝒏𝒔𝒖𝒍𝒂𝒕𝒊𝒏𝒈 =0.9304 W/mk PRACTICAL RELAVANCE:Thermal conductivity of physical property of a substance and is primarily a function of temperature and the nature of the material. BACKGROUND INFORMATION:The thermal conductivity of a metal rod (Isotropic Material) can be determined experimentally by measuring the rate of heat flow and the temperature gradient in the rod. The relevant equation for the experiment is Fourier’s law. 𝑄 =−KA

w

(Or) K =−𝑄

𝑑𝑇 w/mk 𝑑𝑥

Q=Heat transfer rate – watts A=Area normal to heat transfer - 𝑚 K = Thermal conductivity – w/mk

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝒅𝑻 𝒅𝒙

= Temperature gradient in the direction of heat flow.

The thermal conductivity for a given material depends on its state and it varies with direction, structure, humidity, pressure and temperature change. DESCRIPTION:The experimental setup consists of a metal bar, one end of which is heated by an electric heater while the other end projects in space a cooling water jacket. The middle portion is surrounded by a cylindrical shell filled with insulating powder and five thermocouples are placed on the bar for temperature measurement. For radical measurement of temperature thermocouples are placed at two sections in the insulating shell. The heater is provided with a dimmer stat for controlling the heat input and water tank under the constant head is provided for circulating the water through water jacket, the flow rate of a rise are measured and measuring jar and the thermocouples. SCHEMATIC:-

PRECAUTIONS:-

PROCEDURE:-

HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY OBSERVATION TABLE:S.NO

HEAT INPUT (W)

TEMP OF TEMP OF TIME FOR INSULATING WATER MEASURING SHELL JACKET FLASK 𝑇 𝑇 𝑇 𝑇

TEMP OF METAL ROD 𝑇

𝑇

𝑇

𝑇

𝑇

CALCULATIONS:For doing calculations we can approach two methods, 1. Experimental method 2. Sectional method 1st method:According to the “FIRST LAW OF THERMODYNAMICS” at any section, the rate of incoming energy must be equal to the rate of outgoing energy, Heat conducted of metal rod = heat absorbed by circulating water + heat absorbed by insulating powder.

𝑄 ̇ 𝑄 ̇

= 𝑄̇ + 𝑄 ̇ = −KA

Since K =

.

𝑄 ̇ = Heat carried away by water

=𝑚 𝑐 =𝑚 𝑐

(𝑇

∆𝑇 −𝑇 )

𝑚 = Mass flow rate of water kg/sec =Specific heat of water = 4187 J/Kg K

𝑐

HEAT TRANSFER LAB

𝑇

=Outlet water temperature - ℃

𝑇

=Inlet water temperature - ℃

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY (

𝑄 ̇ =

)

( / )

𝐾

=Heat conducted in insulating powder

=Thermal conductivity of insulating powder - w/mk

L = Length of shell (or) Length of test bar 𝑟 =Outer radius of shell – m 𝑟 =Inner radius of shell – m A = 𝜋DL =Surface area of metal bar - 𝑚 D = Diameter of bar – m L = Test length of the bar – m

=Temp gradient ↓ This is plotting of graph dT (vs) dx (or) RESULT:The thermal conductivity of the bar =___________ w/mk

HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Experiment no: 2

THERMAL CONDUCTIVITY OF INSULATING POWDER AIM:To determine the thermal conductivity of insulating powder.

SPECIFICATIONS:1. Radius of inner sphere (𝑟 )=100 mm 2. Radius of outer sphere (𝑟 )=250 mm

PRACTICAL RELAVANCE:It is desirable to reduce the heat loss to the surroundings in much in heat exchanger equipment. Insulating materials have a very low value of thermal conductivity and are used in different shapes, sizes, and form. One important category of insulating material is the powder form. The powder can take any complicated shape between any two containing surfaces. In addition, its thermal conductivity is much lower than that of its basic solid form, because of the large number of air gaps present between particles. BACKGROUND INFORMATION:The heat transfer rate by conduction (Q) through a hollow sphere of a material of thermal conductivity (k) and maintaining at temperature (𝑇 ) at the inner surface (𝑟 ) and at (𝑇 ) the outer surface (𝑟 ) is given by 𝑄=

4𝜋𝑘𝑟0 𝑟̇ 𝑖 (𝑇 − 𝑇 ) −𝑤 (𝑟0 − 𝑟𝑖 )

(Or) 𝐾=

𝑄 (𝑟 0 − 𝑟 𝑖 ) − 𝑤/𝑚𝑘 4𝜋 𝑟0 𝑟𝑖 (𝑇 − 𝑇 ) Q= heat supplied – w

K= thermal conductivity of insulating powder – w/mk 𝑟 = outer radius of sphere – m 𝑟 = inner radius of sphere – m

𝑇 = average temperature of inner sphere - ℃ 𝑇 = average temperature of outer sphere - ℃

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝑇 =

…………

,𝑇 =

…………

DESCRIPTION:The apparatus consists of two thin walled concentric spheres. The inner sphere houses the heating coil. The insulating powder filled in the annular space between the two copper spheres; take the form of hollow sphere. The power supply to the heating coil is adjusted by dimmerstat and it is measured by wattmeter. The ten chromyl alum thermocouples are used to measure the temperature’s for analysis. The four thermocouples, numbered as 𝑻𝟏 to 𝑻𝟒 are embedded on outer surface of the inner sphere and four thermocouples numbered as 𝑻𝟓 to 𝑻𝟏𝟎 are embedded on the inner surface of outer sphere. SCHEMATIC:-

HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

D.SRINIVASA RAO

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY PROCEDURE:-

OBERVATION TABLE:S NO

HEAT TEMPERATURE OF INNER TEMPERATURE OF OUTER SPHERE INPUT SPHERE (w) T T𝟐 T T T T T T T T

CALCULATIONS:𝑄=

4𝜋𝑘𝑟0 𝑟̇ 𝑖 (𝑇 − 𝑇 ) −𝑤 (𝑟0 − 𝑟𝑖 )

(Or) 𝐾=

𝑄 (𝑟0 − 𝑟𝑖 ) − 𝑤/𝑚𝑘 4𝜋𝑟0 𝑟𝑖 (𝑇 − 𝑇 ) Q= heat supplied – w

K= thermal conductivity of insulating powder – w/mk 𝑟 = outer radius of sphere – m 𝑟 = inner radius of sphere – m

𝑇 = average temperature of inner sphere - ℃ 𝑇 = average temperature of outer sphere - ℃

𝑇 = 𝑇 =

-℃ -℃

PRECAUTIONS:-

RESULT:The thermal conductivity of insulating powder is =_________w/mk HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Experiment no: 3

LAGGED PIPE APPARATUS AIM:To determine the approximate thermal conductivity of lagging material. APPARATUS:  

Three concentric pipes mounted on suitable stand. Central heater. Thermocouples.

SPECIFICATIONS:1) Brass tube length=500mm Brass tube diameter=28mm 2) S.S tube length=500mm S.S tube diameter=50mm 3) M.S tube length=500mm M.S tube diameter=100mm. PRACTICLE RELAVANCE:Many practical situations in engineering practice involve heat transfer through a composite cylinder. In most of the multilayer cylindrical wall are frequently employed to reduce heat losses metallic pipes meant for handling a hot fluid. The pipe is generally wrapped in one or more layers of heat insulation.eg a steam pipe is used for conveying high pressure steam in a steam power plant may have cylindrical metal wall, a layer of insulating material and then a layer of protecting plaster. The arrangement is called lagging of the pipe system. BACKGROUND INFORMATION:In this experiment we use insulating powder instead of solid lagging between two layers of shell because insulating capacity in powder are very high compared to solid insulating, due to the air gap between each molecule space.

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY

Temperature distribution for a composite cylindrical wall. DESCRIPTION:The apparatus consist of three concentric pipes mounted on suitable stand. The hallow space of the internal pipe consists of the heater, between the first two cylinders the insulating material with each lagging is to be filled compactly, between the second and third cylinder is filled by other insulating material. The thermocouples are attached to the surface of cylinders approximately to measure the temperatures. The input heater is varied through a dimmerstat. SCHEMATIC:-

HEAT TRANSFER LAB

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D.SRINIVASA RAO

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY PROCEDURE:-

PRECAUTIONS:OBSERVATION TABLE:S.NO

HEAT INPUT (W)

THERMOCOUPLE READING’S 𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

CALCULATIONS:-

𝑄=

(

𝑄=

(

𝑇 =

)

)

=

(

)

−℃

𝑇 =

𝑇 +𝑇 −℃ 2

𝑇 =

𝑇 +𝑇 −℃ 2

𝑟 =inner pipe radius in m 𝑟 =middle pipe radius in m 𝑟 =outer pipe radius in m 𝐾 =thermal conductivity of inner pipe powder in w/mk 𝐾 =thermal conductivity of outer pipe powder in w/mk RESULT:The obtained thermal conductivity of lagged pipe is 𝐾 [Asbestos lagging] = 𝐾 [Sawdust lagging] = HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Experiment No: 4

COMPOSITE WALL APPARATUS AIM:To determine the appropriate thermal conductivity and the overall heat transfer coefficient of a composite wall and temperature distribution across the width of the composite wall. APPARATUS:Composite slabs of different materials clamped in the center using screw rod at the center of the composite wall a heater are fitted. SPECIFICATIONS:1. Size of the plates(diameters)=M.S plate=Phylum plate=Wood=300mm 2. Width of the plates(thickness)=M.S plate=Phylum plate=Wood=20mm 3. Thermal conductive. PRACTICAL RELAVANCE:Many practical situations in engineering practice involve heat transfer through a medium of composing two or more materials of different thermal conductivity, e.g. the wall of buildings, refrigerator cold storage plants, hot water tanks, etc…therefore the thermal conductivity of such composite medium helps in better design of equipment’s. . The composite materials provides insulating properties as well as good strength. Therefore ,now a days its use is increasing

BACKGROUND INFORMATION:If we assume that the materials comprising a plane composite wall are in perfect thermal contact, and then the contact or inter phase resistance is negligible and then the temperature compatibility of two layers in contact exists. In practice we often encounter plane walls that consist of several layers of different materials. The thermal resistance concept can still be used to determine the rate of steady heat transfer through such composite walls

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY

Equivalent thermal circuit for a series of composite wall DESCRIPTION:The apparatus consists of three slabs of different materials of same thickness and sized clamped in the center using a screw rod at the center of the composite wall a heater is fitted. End losses from the composite wall are minimized by providing thick insulation all rounds to ensure unidirectional heat flow. Thermocouples are fitted at the interface of the plates at the different points so as to obtain the average temperature for each surface. Heat conducted through the composite wall is taken away the atmospheric air. SCHEMATIC:-

HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY PRECAUTIONS:-

PROCEDURE:-

OBSERVATION TABLE:S.NO

HEAT INPUT (W)

TEMPERATURE’S 𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

CALCULATIONS:Overall heat transfer coefficient: 𝑈=

1 𝐴𝜀𝑅

Where, A=heat transfer area = 𝑑 -𝑚 𝑑=diameter of composite slabs – m 𝜀𝑅

=Total thermal resistance

=

+

+

𝐿 =Length of slab 1 (dx) in m (or) thickness 𝐿 =Length of slab 2 (dx) in m (or) thickness 𝐿 =Length of slab 3 (dx) in m (or) thickness 𝐾 =Thermal conductivity of slab 1 in W/mk 𝐾 =Thermal conductivity of slab 2 in W/mk HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝐾 =Themal conductivity of slab 3 in W/mk

Thermal conductivity: 𝑄 = 𝐾𝐴 𝑄=

= 𝐾𝐴

𝐾 𝐴 (𝑇 − 𝑇 ) 𝐾 𝐴 (𝑇 − 𝑇 ) 𝐾 𝐴 (𝑇 − 𝑇 ) = = 𝐿 𝐿 𝐿

Therefore,

𝐾 =

(

)

𝐾 =

𝑄𝐿 𝐴 (𝑇 − 𝑇 )

𝐾 =

𝑄𝐿 𝐴 (𝑇 − 𝑇 )

Where, 𝑇 =Average temperature of central heater= 𝑇 =Average temperature of slab 1 = 𝑇 =Average temperature of slab 2 = 𝑇 =Average temperature of slab 3 = RESULT:The thermal conductivity of composite slabs : 𝐾 = ________W/mk 𝐾 = ________W/mk 𝐾 = ________W/mk The overall heat transfer coefficient : U=________W/mk HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Experiment No:- 5

NATURAL CONVECTION APPARATUS AIM:To determine the natural convection heat transfer coefficient heat transfer coefficient for the vertical tube exposed to atmospheric air. APPARATUS:Vertical metallic tube enclosed in a rectangular duct open from top to bottom with heating equipment. SPECIFICATIONS:1. 2. 3. 4.

Diameter of tube (D) = 28 mm Total length of tube (L) =500 mm Duct size = No. of thermocouples = 8

PRACTICLE RELAVANCE:Free (Or) natural convection is the principle mode of heat transfer from transmission lines, pipes, refrigerating coils, hot radiators, buildings and many other practice situations everyday life. BACKGROUND INFORMATION:Circulation of bulk fluid motion is caused by buoyancy effect by changes in fluid density resulting by the temperature gradient between the solid surface and the main mass of the fluid. The stagnant layer of fluid in the immediate vicinity of the hot body gets thermal energy by conduction later on by convection (due to density changes).

Fig.1

Fig.2

Fig.1-The cooling of a boiled egg in a coolerenvironment by natural convection.

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Fig.2-The warming up of a cold drink in a warmer environment by natural convection. Considering a situation in which the body force is gravitational and the change in the density is brought about by temperature gradient, the heat transfer rate in free convection is given by newton’s law of cooling i.e. 𝑄 = ℎ𝐴 (𝑇 − 𝑇∞ ) – Watts Q = rate of heat transfer – w 𝐴 = surface area of convection – 𝑚 𝑇 = mean surface temperature - ℃ 𝑇∞ = 𝑇 = 𝑇

=𝑇 surrounding fluid temperature ℃

h = average heat transfer coefficient –w/ 𝑚 𝑘 DESCRIPTION:The experimental setup consists of a brass tube fitted in a rectangular duct, vertically as shown in fig. the duct is open from top to bottom. An electrical heating element is kept in center of vertical tube, which in turn heats the tube surface longitudinally. The heat is lost from the tube to the surrounding air by natural convection. The temperature of a vertical tube is measured by six thermocouples at different locations and thermocouple 𝑇 and 𝑇 measures the duct temperatures. The heat energy input is measure by watt meter.

SCHEMATIC:-

HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

D.SRINIVASA RAO

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY PRECAUTIONS:PROCEDURE:OBSERVATION TABLE:-

S NO

HEAT INPUT (W)

THERMOCOUPLE READINGS (SURFACE TEMPERATURE) 𝑇

𝑇

𝑇

𝑇

𝑇

AMBIENT TEMPERATURE 𝑇

𝑇

𝑇

CALCULATIONS:1. Experimental Method:𝑄 = ℎ𝐴 (𝑇 − 𝑇∞ )

ℎ=

(

∞)

Q = rate of heat transfer – w 𝐴 = surface area of tube – 𝑚 = 𝜋𝐷𝐿 - 𝑚 D = diameter of tube – m L = length of the tube – m 𝑇 = mean surface temperatures - ℃ ⋯

=

n = no. of thermocouples on pipe

𝑇 = 𝑇∞ = 𝑇 = 𝑇 =

− ℃

2. Empirical relation:𝑁 =

HEAT TRANSFER LAB

ℎ𝐿 𝐾

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY ℎ=

𝑁 ∗𝐾 𝐿

For finding out 𝑵𝒖 [Nusselt number], we should use co relations 𝑮𝒓 &𝑷𝒓 , because 𝑁 = 𝐶(𝐺 ∗ 𝑃 ) 𝐺 =Grashof’s number =

∆

g = acceleration due to gravity = 9.81m/𝑠 𝛽 = Co-efficient of volumetric expansion of air =

𝑇 =

−𝑘 𝑇 −𝑇 −℃ 2

L = Length of vertical tube – m ∆𝑇 = (𝑇 − 𝑇∞ )℃ 𝜗 =Kinematic viscosity Taken from data book at ′𝑻𝒇 ′ 𝑃 =Prandtl number Taken from data book at ′𝑻𝒇 ′ 𝜇 = dynamic viscosity of air – Ns/𝑚 𝑐 = Specific heat of air – J/kgK 𝐾

= Thermal conductivity of air – w/mk

C = 0.56 and n = 0.25 → for 10 ≤(𝐺 ∗ 𝑃 ) ≤ 10 C = 0.13 and n = 1/3→ for 10 ≤(𝐺 ∗ 𝑃 ) ≤ 10

RESULT:The heat transfer coefficient for a vertical tube was found to be h = ___________𝒘/𝒎𝟐 k

HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Experiment No:- 6

FORCED CONVECTION APPARATUS AIM:To determine the heat transfer coefficient in forced convection of air in the given horizontal/vertical tube. APPARATUS:1. 2. 3. 4.

Blower unit fitted with the test pipe. Test section surrounded by nichrome band heater. Thermocouples embedded on the test section. Manometer setup for measurement of mass flow rate/volume flow rate of air.

PRACTICLE RELAVANCE:In many practical situations and equipment’s we invariably deal with flow of fluids in tubes ex-boiler super heater and condenser of a power plant; automobile radiator water and air heater or coolers etc…. the knowledge about the evaluation of forced convection heat transfer coefficient for fluid flow in tubes is essentially a prerequisite for an optimal design of all thermal systems. BACKGROUND INFORMATION:Convection is a process of energy transfer by the combined action of heat conduction, energy storage and mixing motion. When the mixing motion is induced by same external agency such as pump (or) a blower the process called forced convection. The intensity of the mixing motion is generally high in forced convection and consequently the heat transfer coefficients are higher than the fee convection.

Fig. 1 Fig. 2 Fig 1-Heat transfer from a hot surface to the surrounding fluid by forced convection Fig.2-We resort to forced convection whenever we need to increase the rate of heat transfer.

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY The heat transfer coefficient ‘h’ can be obtained from Newton Rock men’s law of convection [Newton’s Law of Cooling] i.e. 𝑄 = ℎ𝐴 ∆𝑇– Watts 𝑄 = ℎ𝐴 (𝑇 − 𝑇

) – Watts

Where, Q = rate of heat transfer – w 𝐴 = surface area of convection – 𝑚 𝑇 = mean surface temperature - ℃ 𝑇∞ = 𝑇 = 𝑇

= surrounding fluid temperature ℃

h = average heat transfer coefficient –w/ 𝑚 𝑘 DESCRIPTION:The experimental setup consist of a blower unit fitted with a test pipe as shown in fig. the pipe is surrounded by the heater; A part of the heat supplied through the pipe and wall of test section is flowing through the air by forced convection. Six thermocouples are attached on the test section and 𝑇 thermocouple measure of the atmospheric air. The test piece is connected with the orifice to measure the flow rate of air through the pipe. A lever valve is fitted in the delivery pipe in order to regulate the air flow rate. The energy input to the heater is measured by watt meter. Temperatures are displayed by digital temperature indicator with selector switch. SCHEMATIC:-

HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY SPECIFICATIONS:1. 2. 3. 4.

Diameter of the tube (𝐷) =32 mm Length of test section (L) = 600 mm Orifice diameter (d) = 15 mm Diameter of the duct = 45 mm

PRECAUTIONS:PROCEDURE:-

OBSERVATION TABLE:-

S NO

HEAT INPUT (W)

MANOMETE R READING (∆𝐻)

AMBIENT TEMPERATURE 𝑇

THERMOCOUPLE READINGS SURFACE TEMPERATURES

𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

CALCULATIONS:1. Experimental method 𝑄 = ℎ𝐴 (𝑇 − 𝑇 ℎ=

(

) )

Where, Q = rate of heat transfer – w 𝐴 = surface area of convection – 𝑚 𝑇∞ = 𝑇 = 𝑇

= surrounding fluid temperature ℃

h = average heat transfer coefficient –w/ 𝑚 𝑘 𝑇 = mean surface temperatures - ℃ =

⋯

n = no. of thermocouples on pipe

HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝑇 =

𝑇 +𝑇 +𝑇 +𝑇 +𝑇 5

𝑇∞ = 𝑇 = 𝑇

=

2.Empirical relation:𝑁 = ℎ=

ℎ𝐿 𝐾

𝑁 𝐾 𝐿(𝑜𝑟)𝐿

Where, [𝐿 = Characteristic length] For, finding out 𝑁 , we should use co relations for this tube 𝑁 = 𝐶 ∗ (𝑅𝑒) ∗ (Pr ) → forced convection Where, 𝑅𝑒 =Reynolds’s number =

(Or)

(Or)

(Or)

𝜌 = Density of air – kg/ 𝑚 𝑣 = velocity of air – m/s D = diameter of duct L = Length of duct 𝜇 = dynamic viscosity – Ns/𝑚 𝜗 = kinematic viscosity - 𝑚 /𝑠 Pr = prandtl number = This can take directly from data book at 𝑻𝒇 . 𝑐 = Specific heat of air – J/kgK

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝐾

= Thermal conductivity of air – w/mk

Property values are taken from the data book at 𝑻𝒇 . 𝑇 =

𝑇 +𝑇 −℃ 2

Velocity of air can find out by; Q = A*V 𝑉= 𝑄

𝑄

= Actual discharge

=𝑐 ∗ 𝑑

2𝑔∆𝐻

𝑐 = Coefficient of discharge of orifice = 0.64 𝑑 = Diameter of orifice – m g = acceleration due to gravity = 9.81 m/𝑠 ∆𝐻 = Manometer head difference – m 𝜌 = Density of water = 1000 kg/ 𝑚 𝜌 = Density of air = 1.17774 kg/ 𝑚 A = Area of pipe =

(𝐷)

𝐷 = Diameter of tube – m RESULT:The heat transfer coefficient in forced convection ℎ =

HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Experiment No:- 7

HEAT TRANSFER THROUGH PIN FIN APPARTUS AIM:To determine the heat transfer coefficient, heat transfer efficiency and effectiveness of the fin by natural and forced convection using pin fin apparatus. APPARATUS:1. 2. 3. 4. 5. 6.

Delivery pipe Orifice plate Duct Blower Heater Fin rod 𝑇 to 𝑇 thermocouples position

SPECIFICATIONS:1. 2. 3. 4. 5. 6. 7. 8.

Length of the pin fin = 0.095 m Diameter of the pin fin = 0.0125 m Diameter of orifice = 15 mm Diameter of the pipe = 40 mm Co efficient of discharge (𝑐 ) = 0.64 Duct size = 150*100 mm Thermal conductivity of the fin = 110 w/mk Distance between each thermocouple on pin fin = 20 mm

PRACTICAL RELAVANCE:Heat transfer between a surface and the fluid surrounding it can be increased by attaching to the surface certain thin strip called fins (Or) extended surfaces. Pin fin (Or) spines are rod protruding from the surface. Fins are used in a wide range of practical application, Ex: cooling of motor cycle engine, electric motors, transformers, refrigerators etc… BACKGROUND INFORMATION:The term extended surface is commonly used in reference to a solid that experiences energy transfer by conduction and convection between its boundary and surroundings. A temp gradient in x-direction sustains heat transfer by conduction internally, at the same time there is energy transfer by convection into an ambient at 𝑻∞ from its surface temperature 𝑻𝒔 , given as

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝑄 = ℎ𝐴 (𝑇 − 𝑇∞ ) Where, h = Convective heat transfer co efficient 𝐴 = Heat transfer at area of surface When 𝑻𝒔 and 𝑻∞ are fixed by design consideration then there are only two ways to increase the heat transfer rate 1. To increase the convection co efficient ‘h’. (Or) 2. To increase the surface area ‘𝑨𝒔 ’. In situations, in which an increase in h is not practical (Or) economical, because h may requires the installation of a pump (Or) Fan (Or) replacing the existing one with larger one, but this approach may or may not be practical. The alternative is to increase the surface area by attaching to the surface extended is called fins made of highly conducting materials such as aluminum. Firmed surface are manufactured by extruding, welding, (Or) wrapping a thin sheet on a surface.

A long circular fin of uniform cross section and the variation of temperature along the length

HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

D.SRINIVASA RAO

Page 25

LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY DESCRIPTION:A typical setup of pin fin is placed inside a duct open on one side. The other side of duct is connected to the suction side of blower. The delivery of air is through a gate valve and an orifice meter to the atmosphere. The air flow can be varied by the gate valve and can be measured by a U-Tube differential manometer connected to an orifice meter. A heating element operating through a dimmerstat is connected to end of the pin fin and five thermocouples are connected equidistant all along the length of the fin. Thermocouple is left in the duct. The panel of the apparatus is similar to the one described in experiments as forced convection apparatus, consisting of watt meter, temperature indicator, thermocouple selector switch, U-Tube manometer etc…. SCHEMATIC:-

PRECAUTIONS:-

PROCEDURE:-

HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

D.SRINIVASA RAO

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY OBSERVATION TABLE:1) NATURAL CONVECTION:S NO

HEAT INPUT (W)

DUCT FLUID TEMPERATURE 𝑇

FIN TEMPERATURE(℃) 𝑇

𝑇

𝑇

𝑇

𝑇

2) FORCED CONVECTON:S NO

HEAT MANOMETER INPUT DIFFERENCE (W)

FIN TEMPERATURE (℃) 𝑇

𝑇

𝑇

𝑇

𝑇

DUCT FLUID TEMPERATURE 𝑇

CALCULATIONS:Natural convection:Heat transfer coefficient (h) 𝑁 = ℎ=

ℎ𝑑 𝐾

𝑁 ∗𝐾 𝑑

𝑑 = Hydraulic diameter-m 𝑑 =

– m2

𝐴 = Cross sectional area of duct = 𝑑 for circular section’s = a*b for rectangular section’s P = perimeter = 𝜋𝑑 for circular section’s = 2(𝑎 + 𝑏) for rectangular section’s For finding out 𝑵𝒖 [Nusselt number], we should use co relations𝑮𝒓 &𝑷𝒓 , because 𝑁 = 𝐶(𝐺 ∗ 𝑃 ) HEAT TRANSFER LAB

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝐺 =Grashof’s number ∆

= Where,

g = acceleration due to gravity = 9.81m/𝑠 𝛽 = Co-efficient of volumetric expansion of air − 𝑘-1

=

𝑇 =

𝑇 −𝑇 −℃ 2

∆𝑇 = (𝑇 − 𝑇∞ ) ℃ 𝜗 =Kinematic viscosity Taken from data book at ′𝑻𝒇 ′ 𝑃 =Prandtl number Taken from data book at ′𝑻𝒇 ′ 𝜇 = dynamic viscosity of air – Ns/𝑚 𝑐 = Specific heat of air – J/kgK 𝐾

= Thermal conductivity of air – w/mk

C = 1.1 and n = 1.7 → for 0.1 ≤(𝐺 ∗ 𝑃 ) ≤ 10 C = 0.56 and n = 0.25 → for 10 ≤(𝐺 ∗ 𝑃 ) ≤ 10 C = 0.13 and n = 1/3 → for 10 ≤(𝐺 ∗ 𝑃 ) ≤ 10

Efficiency of fin ƞ=

tanh(𝑚𝐿 ) 𝑚𝐿

Heat transfer rate of pin fin 𝑄

= 𝜂 ℎ𝐴 (𝑇 − 𝑇∞ ) 𝐴 = 𝜋𝑑 𝐿

Effectiveness of fin 𝜀

HEAT TRANSFER LAB

=𝜂

∗

𝐴 𝐴

DEPARTMENT OF MECHANICAL ENGINEERING

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝐴

=

Where, h = heat transfer co efficient – w/𝑚 𝑘

𝜋 𝑑 4

p = perimeter of fin = 𝜋𝐷 – m D = diameter of fin – m K = thermal conductivity of fin = 110 w/m℃ (Or) 110w/mk [May vary with material] A = Surface area of fin = 𝜋𝐷𝐿 𝐿 = Corrected length of fin = L+ D = Diameter of the fin – m 𝑚=

ℎ𝑃 𝐾𝐴

𝐴 = Cross section of the fin 𝜋 = 𝑑 4

Forced convection:Heat transfer coefficient (h) 𝑁 = ℎ=

ℎ𝑑 𝐾

𝑁 ∗𝐾 𝑑

𝑑 = Hydraulic diameter-m 𝑑 =

– m2

𝐴 = Cross sectional area of duct = 𝑑 for circular section’s = a*b for rectangular section’s P = perimeter HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

D.SRINIVASA RAO

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY = 𝜋𝑑 for circular section’s = 2(𝑎 + 𝑏) for rectangular section’s

For, finding out 𝑵𝒖 , we should use co relations for this tube 𝑁 = 𝐶 ∗ (𝑅𝑒) ∗ (Pr ) → forced convection 𝑅𝑒 = Reynolds’s number =

(Or)

(Or)

(Or)

𝜌 = Density of air – kg/ 𝑚 𝑣 = velocity of air – m/s D = diameter of duct L = Length of duct 𝜇 = dynamic viscosity – Ns/𝑚 𝜗 = kinematic viscosity - 𝑚 /𝑠 Pr = prandtl number = This can take directly from data book at 𝑇 . 𝑐 = Specific heat of air – J/kgK 𝐾

= Thermal conductivity of air – w/mk

Property values are taken from the data book at 𝑇 =

𝑇 +𝑇 −℃ 2

Velocity of air can find out by; Q = A*V 𝑉= 𝑄 Since 𝑄 HEAT TRANSFER LAB

𝑄 𝐴

= Actual discharge =𝑐 ∗ 𝑑

2𝑔∆𝐻

DEPARTMENT OF MECHANICAL ENGINEERING

D.SRINIVASA RAO

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY 𝑐 = Coefficient of discharge of orifice = 0.64 𝑑 = Diameter of orifice – m g = acceleration due to gravity = 9.81 m/𝑠 ∆𝐻 = Manometer head difference – m 𝜌 = Density of water = 1000 kg/ 𝑚 𝜌 = Density of air = 1.17774 kg/ 𝑚 A = Area of duct = (𝐷 ) − 𝑚 𝐷 = Inner diameter of tube – m = L*B - 𝑚 if rectangular duct C 0.989 0.911 0.683 0.193 0.0266

𝑅 0.4 - 4 4 – 40 40 – 4000 4000 – 40000 40000 400000

m 0.330 0.385 0.466 0.618 0.805

Efficiency of fin ƞ=

tanh(𝑚𝐿 ) 𝑚𝐿

Heat transfer rate of pin fin 𝑄

= 𝜂 ℎ𝐴 (𝑇 − 𝑇∞ ) 𝐴 = 𝜋𝑑 𝐿

Effectiveness of fin 𝜀

=𝜂 𝐴

Where, h = heat transfer co efficient – w/𝑚 𝑘

=

∗

𝐴 𝐴

𝜋 𝑑 4

p = perimeter of fin = 𝜋𝐷 – m D = diameter of fin – m K = thermal conductivity of fin = 110 w/m℃ (Or) 110w/mk [May vary with material] HEAT TRANSFER LAB

DEPARTMENT OF MECHANICAL ENGINEERING

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LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY A = Surface area of fin = 𝜋𝐷𝐿 𝐿 = Corrected length of fin = L+ D = Diameter of the fin – m 𝑚=

ℎ𝑃 𝐾𝐴

𝐴 = Cross section of the fin 𝜋 = 𝑑 4

RESULT:Free convection H =______ w/𝑚 𝑘 Q = _______ watts Ƞ = _______ Ɛ = _______

HEAT TRANSFER LAB

Forced convection H = ______ w/𝑚 𝑘 Q = _______ watts Ƞ = _______ Ɛ = _______

DEPARTMENT OF MECHANICAL ENGINEERING

D.SRINIVASA RAO

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