hmt

VELTECH VEL MULTIMEDIA VEL HIGHTECH

Department of Aeronautical engineeringSchool of Mechanical engineeringVel Tech Dr RR & SR Technical UniversityCourse MaterialU6AEA23- Heat Transfer

U6AEA23 HEAT TRANSFERUNIT I Heat Conduction9

Basic Modes of Heat Transfer One dimensional steady state heat conduction: Composite Medium Critical thickness Effect of variation of thermal Conductivity Extended Surfaces Unsteady state. Heat Conduction: Lumped System Analysis Heat Transfer in Semi infinite and infinite solids Use of Transient Temperature charts Application of numerical techniques.

UNIT II Convective Heat Transfer 9

Introduction Free convection in atmosphere free convection on a vertical flat plate Empirical relation in free convection Forced convection Laminar and turbulent convective heat transfer analysis in flows between parallel plates, over a flat plate and in a circular pipe. Empirical relations, application of numerical techniques in problem solving.

UNIT III Radiative Heat Transfer 9

Introduction to Physical mechanism Radiation properties Radiation shape factors Heat exchange between non black bodies Radiation shields.

UNIT IV Heat Exchangers 9

Classification Temperature Distribution Overall heat transfer coefficient, Heat Exchange Analysis LMTD Method and E-NTU Method, problems using LMTD and E-NTUmethds.

UNIT V Heat Transfer Problems In Aerospace Engineering9

High-Speed flow Heat Transfer, Heat Transfer problems in gas turbine combustion chambers Rocket thrust chambers Aerodynamic heating Ablative heat transfer, Heat transfer problems in nozzles.

TEXT BOOKS

1. Yunus A. Cengel., Heat Transfer A practical approach, Second Edition, Tata McGraw-Hill, 2002. 2. Incropera. F.P.and Dewitt.D.P. Introduction to Heat Transfer, John Wiley and Sons 2002. REFERENCE BOOKS

1. Lienhard, J.H., A Heat Transfer Text Book, Prentice Hall Inc., 1981.2. Holman, J.P. Heat Transfer, McGraw-Hill Book Co., Inc., New York, 6th Edn., 1991.3. Sachdeva, S.C., Fundamentals of Engineering Heat & Mass Transfer, Wiley Eastern Ltd., New Delhi, 1981.4.Mathur, M. and Sharma, R.P. Gas Turbine and Jet and Rocket Propulsion, Standard Publishers, New Delhi 1988.

UNIT IBasic Modes of Heat Transfer One dimensional steady state heat conduction: Composite Medium Critical thickness Effect of variation of thermal Conductivity Extended Surfaces Unsteady state. Heat Conduction: Lumped System Analysis Heat Transfer in Semi infinite and infinite solids Use of Transient Temperature charts Application of numerical techniques.

CONDUCTIONPART A

1. Define Heat Transfer.

Heat transfer can be defined as the transmission of energy from one region to another region due to temperature difference.

2. What are the modes of Heat Transfer?

Conduction Convection Radiation

3. Define Conduction.

Heat conduction is a mechanism of heat transfer from a region of high temperature to a region of low temperature within a medium (solid, liquid or gases) or between different medium in direct physical contact.

In condition energy exchange takes place by the kinematic motion or direct impact of molecules. Pure conduction is found only in solids.

4. Define Convection.

Convection is a process of heat transfer that will occur between a solid surface and a fluid medium when they are at different temperatures.

Convection is possible only in the presence of fluid medium.

5. Define Radiation.

The heat transfer from one body to another without any transmitting medium is known as radiation. It is an electromagnetic wave phenomenon. 6. State Fouriers Law of conduction.

The rate of heat conduction is proportional to the area measured normal to the direction of heat flow and to the temperature gradient in that direction.

Q - A

where A are in m2

- Temperature gradient in K/mK Thermal conductivity W/mK.

7. Define Thermal Conductivity.

Thermal conductivity is defined as the ability of a substance to conduct heat.

8. Write down the equation for conduction of heat through a slab or plane wall.

Heat transfer Where

T = T1 T2

- Thermal resistance of slab L = Thickness of slabK = Thermal conductivity of slabA = Area

9. Write down the equation for conduction of heat through a hollow cylinder.

Heat transfer Where

T = T1 T2

thermal resistance of slab L Length of cylinderK Thermal conductivityr2 Outer radius r1 inner radius

10. Write down equation for conduction of heat through hollow sphere.

Heat transfer Where

T = T1 T2

- Thermal resistance of hollow sphere.

11. State Newtons law of cooling or convection law.

Heat transfer by convection is given by Newtons law of cooling Q = hA (Ts - T)Where A Area exposed to heat transfer in m2h - heat transfer coefficient in W/m2KTs Temperature of the surface in KT - Temperature of the fluid in K.

12. Write down the equation for heat transfer through a composite plane wall.

Heat transfer Where

T = Ta Tb

L Thickness of slabha heat transfer coefficient at inner diameterhb heat transfer coefficient at outer side.

13. Write down the equation for heat transfer through composite pipes or cylinder.

Heat transfer Where

T = Ta Tb

14. Write down one dimensional, steady state conduction equation without internal heat generation.

15. Write down steady state, two dimensional conduction equation without heat generation.

16. Write down the general equation for one dimensional steady state heat transfer in slab or plane wall without heat generation.

17. Define overall heat transfer co-efficient.

The overall heat transfer by combined modes is usually expressed in terms of an overall conductance or overall heat transfer co-efficient U.

Heat transfer Q = UA T.

18. Write down the general equation for one dimensional steady state heat transfer in slab with heat generation.

19. What is critical radius of insulation (or) critical thickness.

Critical radius = rc Critical thickness = rc r1

Addition of insulating material on a surface does not reduce the amount of heat transfer rate always. In fact under certain circumstances it actually increases the heat loss up to certain thickness of insulation. The radius of insulation for which the heat transfer is maximum is called critical radius of insulation, and the corresponding thickness is called critical thickness.

20. Define fins (or) Extended surfaces.

It is possible to increase the heat transfer rate by increasing the surface of heat transfer. The surfaces used for increasing heat transfer are called extended surfaces or sometimes known as fins.

21. State the applications of fins.

The main application of fins are

1. Cooling of electronic components 2. Cooling of motor cycle engines.3. Cooling of transformers4. Cooling of small capacity compressors

22. Define Fin efficiency.

The efficiency of a fin is defined as the ratio of actual heat transfer by the fin to the maximum possible heat transferred by the fin.

23. Define Fin effectiveness.

Fin effectiveness is the ratio of heat transfer with fin to that without fin

Fin effectiveness =

24. What is meant by steady state heat conduction?

If the temperature of a body does not vary with time, it is said to be in a steady state and that type of conduction is known as steady state heat conduction.

25. What is meant by Transient heat conduction or unsteady state conduction?

If the temperature of a body varies with time, it is said to be in a transient state and that type of conduction is known as transient heat conduction or unsteady state conduction.

26. What is Periodic heat flow?

In periodic heat flow, the temperature varies on a regular basis.

Example:1. Cylinder of an IC engine.2. Surface of earth during a period of 24 hours.

27. What is non periodic heat flow?

In non periodic heat flow, the temperature at any point within the system varies non linearly with time.

Examples :

1. Heating of an ingot in a furnace.2. Cooling of bars.

28. What is meant by Newtonian heating or cooling process?

The process in which the internal resistance is assumed as negligible in comparison with its surface resistance is known as Newtonian heating or cooling process.

29. What is meant by Lumped heat analysis?

In a Newtonian heating or cooling process the temperature throughout the solid is considered to be uniform at a given time. Such an analysis is called Lumped heat capacity analysis.

30. What is meant by Semi-infinite solids?

In a semi infinite solid, at any instant of time, there is always a point where the effect of heating or cooling at one of its boundaries is not felt at all. At this point the temperature remains unchanged. In semi infinite solids, the biot number value is .

31. What is meant by infinite solid?

A solid which extends itself infinitely in all directions of space is known as infinite solid.

In semi infinite solids, the biot number value is in between 0.1 and 100.0.1 < Bi < 100.

32. Define Biot number.

It is defined as the ratio of internal conductive resistance to the surface convective resistance.

Bi =

Bi = .33. What is the significance of Biot number?

Biot number is used to find Lumped heat analysis, semi infinite solids and infinite solids

If Bi < 0.1 L Lumped heat analysis Bi = Semi infinite solids 0.1 < Bi < 100 Infinite solids.

34. Explain the significance of Fourier number.

It is defined as the ratio of characteristic body dimension to temperature wave penetration depth in time.

Fourier Number =

It signifies the degree of penetration of heating or cooling effect of a solid.

35. What are the factors affecting the thermal conductivity?

1. Moisture 2. Density of material 3. Pressure4. Temperature5. Structure of material

36. Explain the significance of thermal diffusivity.

The physical significance of thermal diffusivity is that it tells us how fast heat is propagated or it diffuses through a material during changes of temperature with time.

37. What are Heisler charts?

In Heisler chart, the solutions for temperature distributions and heat flows in plane walls, long cylinders and spheres with finite internal and surface resistance are presented. Heisler charts are nothing but a analytical solutions in the form of graphs. PART B

1. A wall of 0.6m thickness having thermal conductivity of 1.2 w/Mk. The wall is to be insulated with a material having an average thermal conductivity of 0.3 W/mK. Inner and outer surface temperatures are 1000 C and 10C. Heat transfer rate is 1400 W/m2 calculate the thickness of insulation.

Given Data

Thickness of wall L1 = 0.6 mThermal conductivity of wall K1 = 1.2 W/mK.Thermal conductivity of insulation K2 = 0.3 W/mK. Inner surface Temperature T1 = 1000C + 273 = 1273 KOuter surface Temperature T3 = 10C + 273 = 283 KHeat transfer per unit area Q/A = 1400 W/m2.

Solution:

Let the thickness of insulation be L2We know

[From equation (13)] (or) [HMT Data book page No. 34]Where

T = Ta Tb (or) T1 T3

Heat transfer coefficient ha, hb and thickness L3 are not given. So neglect that terms.

2. The wall of a cold room is composed of three layer. The outer layer is brick 30cm thick. The middle layer is cork 20 cm thick, the inside layer is cement 15 cm thick. The temperatures of the outside air is 25C and on the inside air is -20C. The film co-efficient for outside air and brick is 55.4 W/m2K. Film co-efficient for inside air and cement is 17 W/m2K. Find heat flow rate.

Take K for brick = 2.5 W/mKK for cork= 0.05 W/mKK for cement= 0.28 W/mK

Given Data

Thickness of brick L3 = 30 cm = 0.3 mThickness of cork L2 = 20 cm = 0.2 mThickness of cement L1 = 15 cm = 0.15 mInside air temperature Ta = -20C + 273 = 253 KOutside air temperature Tb = 25C + 273 = 298 KFilm co-efficient for inner side ha = 17 W/m2KFilm co-efficient for outside hb = 55.4 W/m2KKbrick = K3 = 2.5 W/mKKcork = K2 = 0.05 W/mK.Kcement = K1 = 0.08 W/mK.

Solution:

Heat flow through composite wall is given by


T = Ta Tb

The negative sign indicates that the heat flows from the outside into the cold room.

3. A wall is constructed of several layers. The first layer consists of masonry brick 20 cm. thick of thermal conductivity 0.66 W/mK, the second layer consists of 3 cm thick mortar of thermal conductivity 0.6 W/mK, the third layer consists of 8 cm thick lime stone of thermal conductivity 0.58 W/mK and the outer layer consists of 1.2 cm thick plaster of thermal conductivity 0.6 W/mK. The heat transfer coefficient on the interior and exterior of the wall are 5.6 W/m2K and 11 W/m2K respectively. Interior room temperature is 22C and outside air temperature is -5C.

Calculate

a) Overall heat transfer coefficientb) Overall thermal resistancec) The rate of heat transferd) The temperature at the junction between the mortar and the limestone.

Given Data

Thickness of masonry L1 = 20cm = 0.20 mThermal conductivity K1 = 0.66 W/mKThickness of mortar L2 = 3cm = 0.03 mThermal conductivity of mortar K2 = 0.6 W/mKThickness of limestone L3 = 8 cm = 0.08 mThermal conductivity K3 = 0.58 W/mKThickness of Plaster L4 = 1.2 cm = 0.012 mThermal conductivity K4 = 0.6 W/mKInterior heat transfer coefficient ha = 5.6 W/m2KExterior heat transfer co-efficient hb = 11 W/m2KInterior room temperature Ta = 22C + 273 = 295 KOutside air temperature Tb = -5C + 273 = 268 K.

Solution:

Heat flow through composite wall is given by


T = Ta Tb

We knowHeat transfer Q = UA (Ta Tb) [From equation (14)]Where U overall heat transfer co-efficient

We know Overall Thermal resistance (R)

For unit Area

Interface temperature between mortar and the limestone T3

Interface temperatures relation

Temperature between Mortar and limestone (T3 is 276.5 K)

4. A steam to liquid heat exchanger area of 25.2 m2 is constructed with 0.5cm nickel and 0.1 cm plating of copper on the steam sides. The resistivity of a water-scale deposit on the steam side is 0.0015 K/W. The steam and liquid surface conductance are 5400 W/m2K ad 560 W/m2K respectively. The heated steam is at 110C and heated liquid is at 70C.

Calculate

1. Overall steam to liquid heat transfer co-efficient2. Temperature drop across the scale deposit

Take K(Copper) = 350 W/mK and K (Nickel) = 55 W/mK.Given

Area A = 25.2 m2Thickness of Nickel L1 = 0.5 cm = 0.5 10-2 mThickness of Copper L2 = 0.1 cm = 0.1 10-2 mResistivity of scale R3 = 0.0015 K/WLiquid surface conductance ha = 560 W/m2KSteam surface conductance hb = 5400 W/m2KSteam temperature Tb = 110C + 273 = 383 K Liquid temperature Ta = 70C + 273 = 343 K K2 (Copper) = 350 W/mK K1 (Nickel) = 55 W/mK

Solution:

Heat transfer through composite wall is given by


T = Ta Tb = 343 383 = -40 K

R3 value is given, R3 = 0.0015 K/W

[-ve sign indicates that the heat flows from, outside to inside]we know Heat transfer Q = UA (Ta Tb) [From equation No. (14)]

Temperature drop (T3 T4) across the scale is given by

5. A surface wall is made up of 3 layers one of fire brick, one of insulating brick and one of red brick. The inner and outer surface temperatures are 900C and 30C respectively. The respective co-efficient of thermal conductivity of the layers are 1.2, 0.14 and 0.9 W/mK and the thickness of 20cm, 8 cm and 11 cm. Assuming close bonding of the layers at the interfaces. Find the heat loss per square meter and interface temperatures.

Given

Inner temperature T1 = 900C + 273 = 1173 KOuter temperature T4 = 30C + 273 = 303 KThermal conductivity of fire brick K1 = 1.2 W/mKThermal conductivity of insulating brick K2 = 0.14 W/mKThermal conductivity of red brick K3 = 0.9 W/mKThickness of fire brick L1 = 20 cm = 0.2 mThickness of insulating brick L2 = 8 cm = 0.08 mThickness of red brick L3 = 11 cm = 0.11 m

Solution:

(i) Heat loss per square metre (Q/A)

Heat transfer [From equation (13)] (or) [HMT Data book page No. 34]Where

T = Ta Tb = T1 T4

[Convective heat transfer co-efficient ha, hb are not given. So neglect that terms]

(ii) Interface temperatures (T2 and T3)

We know that, interface temperatures relation

where

6. A furnace wall made up of 7.5 cm of fire plate and 0.65 cm of mild steel plate. Inside surface exposed to hot gas at 650C and outside air temperature 27C. The convective heat transfer co-efficient for inner side is 60 W/m2K. The convective heat transfer co-efficient for outer side is 8W/m2K. Calculate the heat lost per square meter area of the furnace wall and also find outside surface temperature.

Given Data

Thickness of fire plate L1 = 7.5 cm = 0.075 mThickness of mild steel L2 = 0.65 cm = 0.0065 mInside hot gas temperature Ta = 650C + 273 = 923 KOutside air temperature Tb = 27C + 273 = 300KConvective heat transfer co-efficient for Inner side ha = 60W/m2KConvective heat transfer co-efficient for Outer side hb = 8 W/m2K.

Solution:(i) Heat lost per square meter area (Q/A)

Thermal conductivity for fire plate K1 = 1035 10-3 W/mK [From HMT data book page No.11]

Thermal conductivity for mild steel plate

K2 = 53.6W/mK [From HMT data book page No.1]

Heat flow Where

T = Ta Tb

[The term L3 is not given so neglect that term]

(ii) Outside surface temperature T3

We know that, Interface temperatures relation

7. A mild steel tank of wall thickness 10mm contains water at 90C. Calculate the rate of heat loss per m2 of tank surface area when the atmospheric temperature is 15C. The thermal conductivity of mild steel is 50 W/mK and the heat transfer co-efficient for inside and outside the tank is 2800 and 11 W/m2K respectively. Calculate also the temperature of the outside surface of the tank. Given Data

Thickness of wall L1 = 10mm = 0.01 mInside temperature of water Ta = 90 + 273 = 363 KAtmospheric temperature Tb = 15C + 273 = 288 KHeat transfer co-efficient for inside ha = 2800 W/m2KHeat transfer co-efficient for outside hb = 11 W/m2KThermal conductivity of mild steel K = 50 W/mK

To find

(i) Rate of heat loss per m2 of tank surface area (Q/A)(ii) Tank outside surface temperature (T2)

Solution:

Heat loss

Where

T = Ta Tb

[L3, L2 not given so neglect L2 and L3 terms]

We know

8. A composite slab is made of three layers 15 cm, 10 cm and 12 cm thickness respectively. The first layer is made of material with K = 1.45 W/mK, for 60% of the area and the rest of material with K = 2.5 W/mK. The second layer is made of material with K = 12.5 W/mK for 50% of area and rest of material with K = 18.5 W/mK. The third layer is made of single material of K = 0.76 W/mK. The composite slab is exposed on one side to warn at 26C and cold air at -20C. The inside heat transfer co-efficient is 15 W/m2K. The outside heat transfer co-efficient is 20 W/m2K determine heat flow rate and interface temperatures.

Given Data

L1 = 15 cm = 0.15 mL2 = 10 cm = 0.1 mL3 = 12 cm = 0.12 mK1a = 1.45 W/mK, A1a = .60K1b = 2.5 W/mK A1b = .40K2a = 12.5 W/mK A2a = .50K2b = 18.5 W/mK A2b = .50K3 = 0.76 W/mKTa = 26C + 273 = 299 KTb = -20C + 273 = 253 Kha = 15 W/m2Khb = 20 W/m2K

Solution :

Heat flow Where

T = Ta Tb

Substitute R1a and R1b value in (1)

Similarly,

(ii) Interface temperatures (T1, T2, T3 and T4)

We know

9. An external wall of a house is made up of 10 cm common brick (K = 0.7 W/mK) followed by a 4 cm layer of zibsum plaster (K = 0.48 W/mK). What thickness of loosely packed insulation (K = 0.065 W/mK) should be added to reduce the heat loss through the wall by 80%. Given Data

Thickness of brick L1 = 10 cm = 0.1 mThermal conductivity of brick K1 = 0.7 W/mKThickness of zibsum L2 = 4 cm = 0.04 mThermal conductivity of zibsum K2 = 0.48 W/mKThermal conductivity of insulation K3 = 0.065 W/mKTo find

Thickness of insulation to reduce the heat loss through the wall by 80% (L3)

Solution:

Heat flow rate [From HMT data book Page No.34]Where

Heat loss is reduced by 80% due to insulation, so heat transfer is 20 W.

10. A furnace wall consists of steel plate of 20 mm thick, thermal conductivity 16.2 W/mK lined on inside with silica bricks 150 mm thick with conductivity 2.2 W/mK and on the outside with magnesia brick 200 mm thick, of conductivity 5.1 W/mK. The inside and outside surfaces of the wall are maintained at 650C and 150C respectively. Calculate the heat loss from the wall per unit area. If the heat loss is reduced to 2850W/m2 by providing an air gap between steel and silica bricks, find the necessary width of air gap if the thermal conductivity of air may be taken as 0.030 W/mK. Given Data

Steel plate thickness L1 = 20 mm = 0.02 mThermal conductivity of steel K1 = 16.2 W/mKThickness of the silica L2 = 150 mm = 0.150 mThermal conductivity of silica K2 = 2.2 W/mKThickness of the magnesia L3 = 200 mm = 0.2 mThermal conductivity of magnesia K3 = 5.1 W/mKInner surface temperature T1 = 650C + 273 = 923Outer surface temperature T4 = 150C + 273 = 423 KHeat loss reduced due to air gap is 2850 W/m2Thermal conductivity of the air gap Kair = 0.030 W/mK

Solution :

Heat transfer through composite wall is given by [without considering air gap]

Where

T = T1 T4

Neglecting unknown terms (ha and hb)

Heat loss is reduced to 2850 W/m2 due to air gap. So the new thermal resistance is

Thermal resistance of air gap Rair = Rnew R = 0.1754 0.1086

We know

Thickness of the air gap = 1.98 10-3 m11. A thick walled tube of stainless steel [K = 77.85 kJ/hr mC] 25 mm ID and 50 mm OD is covered with a 25 mm layer of asbestos [K = 0.88 kJ/hr mC]. If the inside wall temperature of the pipe is maintained at 550C and the outside of the insulator at 45C. Calculate the heat loss per meter length of the pipe.

Given Data

Inner diameter of steel d1 = 25 mmInner radius r1 = 12.5 mm 0.0125 mOuter diameter D2 = 50 mmOuter radius r2 = 25 mm 0.025 mRadius r3 = r2 + 25 mm = 50 mm 0.05 mThermal conductivity of stainless steel

K1 = 77.85 kJ/hr mC = kJ/sec mC = 0.0216 kJ/sec mC 0.0216 kW/mC

Similarly,

Thermal conductivity of asbestos K2 = 0.88 kJ/hr mC

Solution :

Heat flow through composite cylinder is given by

[From equation No.(19) or HMT data book Page No.35]Where

T = Ta Tb

Convective heat transfer co-efficient are not given so neglect ha and hb terms.

12. A steel tube (K = 43.26 W/mK) of 5.08 cm inner diameter and 7.62 cm outer diameter is covered with 2.5 cm layer of insulation (K = 0.208 W/mK) the inside surface of the tube receivers heat from a hot gas at the temperature of 316C with heat transfer co-efficient of 28 W/m2K. While the outer surface exposed to the ambient air at 30C with heat transfer co-efficient of 17 W/m2K. Calculate heat loss for 3 m length of the tube. Given

Steel tube thermal conductivity K1 = 43.26 W/mKInner diameter of steel d1 = 5.08 cm = 0.0508 mInner radius r1 = 0.0254 mOuter diameter of steel d2 = 7.62 cm = 0.0762 mOuter radius r2 = 0.0381 mRadius r3 = r2 + thickness of insulation Radius r3 = 0.0381 + 0.025 m r3 = 0.0631 m

Thermal conductivity of insulation K2 = 0.208 W/mKHot gas temperature Ta = 316C + 273 = 589 KAmbient air temperature Tb = 30C + 273 = 303 KHeat transfer co-efficient at inner side ha = 28 W/m2KHeat transfer co-efficient at outer side hb = 17 W/m2KLength L = 3 m

Solution :

Heat flow [From equation No.(19) or HMT data book Page No.35]Where

T = Ta Tb

[The terms K3 and r4 are not given, so neglect that terms]

Heat loss Q = 1129.42 W.

13. A hollow sphere (K = 65 W/mK) of 120 mm inner diameter and 350 mm outer diameter is covered 10 mm layer of insulation (K = 10 W/mK). The inside and outside temperatures are 500C and 50C respectively. Calculate the rate of heat flow through this sphere.

Given

Thermal conductivity of sphere K1 = 65 W/mKInner diameter of sphere d1 = 120 mmRadius r1 = 60 mm = 0.060 mOuter diameter of sphere d2 = 350 mmRadius r2 = 175 mm = 0.175 mRadius r3 = r2 + thickness of insulation R3 = 0.175 + 0.010

Thermal conductivity of insulation K2 = 10 W/mKInside temperature Ta = 500C + 273 = 773 KOutside temperature Tb = 50C + 273 = 323 K

Solution:

Heat loss through hollow sphere is given by

[From equation No.(19) or HMT data book Page No.34 & 35]Where

T = Ta Tb

ha, hb not given so neglect that terms.

Heat transfer = Q = 28361 W

Radius r3 = r2 + thickness of insulation = 0.0455 + 90 10-3 m r3 = 0.1355 m

Radius r4 = r3 + thickness of insulation = 0.1355 + 40 10-3 m r4 = 0.1755 m

Thermal conductivity of pipe K1 = 47 W/mKThermal conductivity of insulation (I) K2 = 0.5 W/mKThermal conductivity of insulation (II) K3 = 0.25 W/mKOutside temperature T4 = 20C + 273 = 293 KSolution :

Heat flow through composite cylinder is given by

[From equation No.(19) or HMT data book Page No.35]Where


Heat transfer Q/L = 448.8 W/m.

14. A hollow sphere has inside surface temperature of 300C and then outside surface temperature of 30C. If K = 18 W/mK. Calculate (i) heat lost by conduction for inside diameter of 5 cm and outside diameter of 15 cm (ii) heat lost by conduction, if equation for a plain wall area is equal to sphere area. Given Data :

T1 = 300C + 273 = 573 KT2 = 30C + 273 = 303 KK1 = 18 W/mKd1 = 5 cm = 0.05 mr1 = 0.025 md2 = 15 cm = 0.15 mr2 = 0.075 m

Solution:

(i) Heat lost (Q)

Heat flow [From HMT data book Page No.34 & 35]Where


[The terms ha, hb not given so neglect that terms].

(ii) Heat loss (If the area is equal to the plain wall area) Q1

L = r2 r1 = 0.075 0.025

Derive an expression of Critical Radius of Insulation For A Cylinder.

Consider a cylinder having thermal conductivity K. Let r1 and r0 inner and outer radii of insulation.

Heat transfer [From equation No.(3)]

Considering h be the outside heat transfer co-efficient.

To find the critical radius of insulation, differentiate Q with respect to r0 and equate it to zero.

15. A wire of 6 mm diameter with 2 mm thick insulation (K = 0.11 W/mK). If the convective heat transfer co-efficient between the insulating surface and air is 25 W/m2L, find the critical thickness of insulation. And also find the percentage of change in the heat transfer rate if the critical radius is used.

Given Data

d1= 6 mmr1 = 3 mm = 0.003 mr2 = r1 + 2 = 3 + 2 = 5 mm = 0.005 mK = 0.11 W/mKhb = 25 W/m2KSolution:

1. Critical radius [From equation No.(21)]

Critical thickness = rc r1

2. Heat transfer through an insulated wire is given by

Heat flow through an insulated wire when critical radius is used is given by

Percentage of increase in heat flow by using

Critical radius =

Internal Heat Generation Formulae used For plane wall :

1. Surface temperature

2. Maximum temperature whereT - Fluid temperature, Kq - Heat generation, W/m3L Thickness, mh - Heat transfer co-efficient, W/m2KK Thermal conductivity, W/mK.For Cylinder 1. Heat generation 2. Maximum temperature 3. Surface temperature Where V Volume - r2 L r radius m

For sphere 1. Temperature at the centre

16. A current of 200 A is passed through a stainless steel wire (K = 19 W/mK) 3 mm in diameter. The resistivity of the steel may be taken as 70 cm and the length of the wire is submerged in a liquid at 110C with heat transfer co-efficient h = 4 kW/m2C. Calculate the centre temperature of the wire.

Given

Current A = 200 AThermal conductivity K = 19 W/mKDiameter d = 3 mm = 3 10-3 mResistivity = 70 - cm Liquid temperature Tw = 110C + 273 = 383 KHeat transfer co-efficient h = 4 kW/m2C = 4 10-3 W/m2C

Solution:

The maximum temperature in the wire occurs at the centre.

We know that Q = I2R = (200)2 (0.099)

Heat generated

Substituting q value in Equation (A)

17. A sphere of 100 mm diameter, having thermal conductivity of 0.18 W/mK. The outer surface temperature is 8C and 250 W/m2 of energy is released due to heat source. Calculate 1. Heat generated 2. Temperature at the centre of the sphere.

Given

Diameter of sphere d = 100 mmr = 50 mm = 0.050 mThermal conductivity K = 0.18 W/mKSurface temperature Tw = 8C + 273 = 281 KEnergy released Q = 250 W/m2

Solution:

Heat generated

Temperature at the centre of the sphere

18. One end of the long solid rod of 50 mm diameter is inserted into a furnace with the other end is projecting the atmosphere at 25C. Once the steady state is reached, the temperature of the rod is measured at two points 20 cm apart are found to be 150C and 100C. The convective heat transfer co-efficient between the rod and the surrounding air is 30 W/m2K. Calculate the thermal conductivity of the rod material.

Given Data:

Atmospheric Temperature T = 25C + 273 = 298 KDistance x = 20 cm = 0.20 mBase temperature Tb = 150C + 273 = 423 KIntermediate temperature T = 100C + 273 = 373 KHeat transfer co-efficient h = 30 W/m2K.

Solution:

Since the rod is long, it is treated as long fin. So, temperature distribution

[From HMT data book (CPK) Page No.41]

We know that,

h heat transfer co-efficient = 30 W/m2KP Perimeter = d = 0.050

19. An aluminium alloy fin of 7 mm thick and 50 mm long protrudes from a wall, which is maintained at 120C. The ambient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mK respectively. Determine

1. Temperature at the end of the fin.2. Temperature at the middle of the fin.3. Total heat dissipated by the fin.

Given

Thickness t = 7mm = 0.007 mLength L= 50 mm = 0.050 mBase temperature Tb = 120C + 273 = 393 KAmbient temperature T = 22 + 273 = 295 KHeat transfer co-efficient h = 140 W/m2KThermal conductivity K = 55 W/mK.

Solution:

Length of the fin is 50 mm. So, this is short fin type problem. Assume end is insulated. We know Temperature distribution [Short fin, end insulated]

[From HMT data book Page No.41]

(i) Temperature at the end of the fin, Put x = L

A Area = Length thickness = 0.050 0.007

(ii) Temperature of the middle of the fin,

Put x = L/2 in Equation (A)

Temperature at the middle of the fin

(iii) Total heat dissipated


20. Ten thin brass fins (K = 100 W/mK), 0.75 mm thick are placed axially on a 1 m long and 60 mm diameter engine cylinder which is surrounded by 35C. The fins are extended 1.5 cm from the cylinder surface and the heat transfer co-efficient between cylinder and atmospheric air is 15 W/m2K. Calculate the rate of heat transfer and the temperature at the end of fins when the cylinder surface is at 160C.

Given

Number of fins = 10Thermal conductivity K = 100 W/mKThickness of the fin t = 0.75 mm = 0.75 10-3 mLength of engine cylinder = 1mDiameter of the cylinder d = 60 mm = 0.060 mAtmosphere temperature T = 35C + 273 = 300 KLength of the fin L = 1.5 cm = 1.5 10-2 mHeat transfer co-efficient h = 15 W/m2KCylinder surface temperaturei.e. Base temperature Tb = 160C + 273 = 433 K

Solution

Assuming that the fin end is insulated and length of the fin is 1.5 cm. So this is short fin end insulated type problem.

We knowHeat transferred Q = (hPKA)1/2 (Tb - T) tan h (mL).(A)[From HMT data book Page No.41]WhereP Perimeter = 2 Length of the cylinder = 2 1

A = Area = length of the cylinder thickness = 1 0.75 10-3 m

Heat transferred per fin = 58.1 WThe heat transfer for 10 fins = 58.1 10

Heat transfer from unfinned surface due to convection is

So, Total heat transfer Q = Q1 + Q2 Q = 581 + 375.8

We know that,

Temperature distribution [short fin, end insulated]


Temperature at the end of fin, so put x = L 21. Aluminium fins 1.5 cm wide and 10 mm thick are placed on a 2.5 cm diameter tube to dissipate the heat. The tube surface temperature is 170C ambient temperature is 20C. Calculate the heat loss per fin. Take h = 130 W/m2 C and K = 200 W/m2 C for aluminium.

Given

Wide of the fin b = 1.5 cm = 1.5 10-2 mThickness t = 10 mm = 10 10-3 mDiameter of the tube d = 2.5 cm = 2.5 10-2 mSurface temperature Tb = 170C + 273 = 443 KAmbient temperature T = 20C + 273 = 293 KHeat transfer co-efficient h = 130 W/m2 CThermal conductivity K = 200 W/mC

Solution

Assume fin end is insulated, so this is short fin end insulated type problem.Heat transfer [short fin, end insulated]

Q = (hPKA)1/2 (Tb - T) tan h (mL) ..(1) [From HMT data book Page No.41]Where A Area = Breadth thickness

22. A straight rectangular fin has a length of 35 mm, thickness of 1.4 mm. The thermal conductivity is 55W/mC. The fin is exposed to a convection environment at 20C and h = 500 W/m2C. Calculate the heat loss for a base temperature of 150C. Given

Length L = 35 mm = 0.035 mThickness t = 1.4 mm = 0.0014 mThermal conductivity K = 55 W/mCFluid temperature T = 20C + 273 = 293 KBase temperature Tb = 150C + 273 = 423 KHeat transfer co-efficient h = 500 W/m2K.

Solution

Length of the fin is 35 mm, so this is short fin type problem. Assume end is insulated.

Heat transferred [Short fin, end insulated]Q = (hPKA)1/2 (Tb - T) tan h (mL) .(1)[From HMT data book Page No.41] Where P Perimeter = 2 Length (Approximately) = 2 0.035

A Area = Length thickness = 0.035 0.0014

Substituting h, p, K, A, Tb, T, m, L values in equation (1)

23. A heating unit made in the form of a cylinder is 6 cm diameter and 1.2 m long. It is provided with 20 longitudinal fins 3 mm thick which protrude 50 mm from the surface of the cylinder. The temperature at the base of the fin is 80C. The ambient temperatures is 25C. The film heat transfer co-efficient from the cylinder and fins to the surrounding air is 10 W/m2K. Calculate the rate of heat transfer from the finned wall to the surrounding. Take K = 90 W/mK.

Given

Diameter of the cylinder d = 6 cm = 0.06 mLength of the cylinder = 1.2 mNumber of fins = 20Thickness of fin (t) = 3 mm = 0.003 mLength of fin L = 50 mm = 0.050 mBase temperature Tb = 80C + 273 = 353 KAmbient temperature T = 25C + 273 = 298 KFilm heat transfer co-efficient h = 10 W/m2KThermal conductivity K = 90 W/mK.

Solution

Length of the fins is 50 mm. Assume end is insulated. So this is short fin end insulated type problem.

We know Heat transferred [short fin, end insulated]

Q = (hPKA)1/2 (Tb - T) tan h (mL) ..(1) [From HMT data book Page No. 41]Where P Perimeter = 2 Length of the cylinder = 2 1.2

A Area = Length of the cylinder thickness of fin

= 1.2 0.003

Heat transferred per fin = 62.16 WNumber of fins = 20So, Total heat transferred Q1 = 62.16 20 Q1 = 1243.28 W

Heat transfer from unfinned surface due to convection is Q2 = h A T

So, Total heat transfer Q = Q1 + Q2 Q = 1243.28 + 122.75

Total heat transfer

24. An aluminium cube 6 cm on a side is originally at a temperature of 500C. It is suddenly immersed in a liquid at 10C for which h is 120 W/m2K. Estimate the time required for the cube to reach a temperature of 250C. For aluminium = 2700 kg/m3, C = 900 J/kg K, K = 204 W/mK.

Given

Thickness of cube L = 6 cm = 0.06 mInitial temperature T0 = 500C + 273 = 773 KFinal temperature T = 10C + 273 = 283 KIntermediate temperature T = 250C + 273 = 523 KHeat transfer co-efficient h = 120 w/m2K

Density = 2700 kg/m3Specific heat C = 900 J/Kg kThermal conductivity K = 204 W/mK

Solution

For Cube,

Characteristic length

We know

Biot number

Bi =

Biot number value is less than 0.1. So this is lumped heat analysis type problem

For lumped parameter system,

[From HMT data book Page No.48]We know,

Characteristics length

Time required for the cube to reach 250C is 144.86 s.

25. A copper plate 2 mm thick is heated up to 400C and quenched into water at 30C. Find the time required for the plate to reach the temperature of 50C. Heat transfer co-efficient is 100 W/m2K. Density of copper is 8800 kg/m3. Specific heat of copper = 0.36 kJ/kg K.Plate dimensions = 30 30 cm.

Given

Thickness of plate L = 2 mm = 0.002 mInitial temperature T0 = 400C + 273 = 673 KFinal temperature T = 30C + 273 = 303 KIntermediate temperature T = 50C + 273 = 323 KHeat transfer co-efficient h = 100 W/m2KDensity = 8800 kg/m3Specific heat C= 360 J/kg kPlate dimensions = 30 30 cm

To find Time required for the plate to reach 50C. [From HMT data book Page No.2]

Solution:

Thermal conductivity of the copper K = 386 W/mKFor slab,

Characteristic length

We know,

Biot number

Bi =

Biot number value is less than 0.1. So this is lumped heat analysis type problem.


.(1)[From HMT data book Page No.48]We know,

Characteristics length Lc =

Time required for the plate to reach 50C is 92.43 s.26. A 12 cm diameter long bar initially at a uniform temperature of 40 C is placed in a medium at 650C with a convective co-efficient of 22 W/m2K. Determine the time required for the center to reach 255C. For the material of the bar, K = 20 W/mK, Density = 580 kg/m3, specific heat = 1050 J/kg K.

Given:

Diameter of bar, D = 12 cm = 0.12 mRadius of bar, R = 6 cm = 0.06 mInitial temperature T0 = 40C + 273 = 313 KFinal temperature T = 650C + 273 = 923 KIntermediate temperature T = 255C + 273 = 528 KHeat transfer co-efficient h = 22 W/m2KThermal conductivity K = 20 W/mKDensity = 580 kg/m3Specific heat C = 1050 J/kg k

Solution

For cylinder,

Characteristic Length

We know,

Biot number

Bi = 0.033 < 0.1

Biot number value is less than 0.1. So this is lumped heat analysis type problem.For lumped parameter system,



Time required for the cube to reach 255C is 360.8 s.

27. A steel ball (specific heat = 0.46 kJ/kgK. and thermal conductivity = 35 W/mK) having 5 cm diameter and initially at a uniform temperature of 450C is suddenly placed in a control environment in which the temperature is maintained at 100C. Calculate the time required for the balls to attained a temperature of 150C. Take h = 10W/m2K.

Given

Specific heat C = 0.46 kJ/kg K = 460 J/kg KThermal conductivity K = 35 W/mKDiameter of the sphere D = 5 cm = 0.05 mRadius of the sphere R = 0.025 mInitial temperature T0 = 450C + 273 = 723 KFinal temperature T = 100C + 273 = 373 KIntermediate temperature T = 150C + 273 = 423 KHeat transfer co-efficient h = 10 W/m2K

To find

Time required for the ball to reach 150C[From HMT data book Page No.1]Solution

Density of steel is 7833 kg/m3

For sphere,


We know,

Biot number

Bi = 2.38 10-3 < 0.1



.(1)



Time required for the ball to reach 150C is 5840.54 s.28. An aluminium sphere mass 5.5 kg and initially at a temperature of 290o is suddenly immersed in a fluid at 15C with heat transfer co-efficient 58 W/m3K. Estimate the time required to cool the aluminium to 95C. For aluminium take = 2700 kg/m3, C = 900 J/kg K, K = 205 W/mK. Given

Mass, m = 5.5 kgInitial temperature T0 = 290C + 273 = 563 KFinal temperature T = 15C + 273 = 288 KIntermediate temperature T = 95C + 273 =368 KHeat transfer co-efficient h = 58 W/m2KThermal conductivity K = 205 W/mKDensity = 2700 kg/m3Specific heat C = 900 J/kg K.

Solution

We know,

Density =

We know,

Volume of sphere

For sphere,


We know,

Biot number

Bi = 7.41 10-3 < 0.1



.(1)



Time required to cool the aluminium to 95C is 1355.6 s.

29. Alloy steel ball of 2 mm diameter heated to 800C is quenched in a bath at 100C. The material properties of the ball are K = 205 kJ/m hr K, = 7860 kg/m3, C = 0.45 kJ/kg K, h = 150 KJ/ hr m2 K. Determine (i) Temperature of ball after 10 second and (ii) Time for ball to cool to 400C. Given

Diameter of the ball D = 12 mm = 0.012 mRadius of the ball R = 0.006mInitial temperature T0 = 800C + 273 = 1073 KFinal temperature T = 100C + 273 = 373 KThermal conductivity K = 205 kJ/m hr K

Density = 7860 kg/m3Specific heat C = 0.45 kJ/kg K = 450 J/kg K

Heat transfer co-efficient h = 150 kJ/hr m2 K

Solution

Case (i) Temperature of ball after 10 sec.

For sphere,


We know,

Biot number

Bi = 1.46 10-3 < 0.1





Case (ii) Time for ball to cool to 400C

T = 400C + 273 = 673 K

30. A large wall 2 cm thick has uniform temperature 30C initially and the wall temperature is suddenly raised and maintained at 400C. Find

1. The temperature at a depth of 0.8 cm from the surface of the wall after 10 s.2. Instantaneous heat flow rate through that surface per m2 per hour.

Take = 0.008 m2/hr, K = 6 W/mC.

Given

Thickness L = 2 cm = 0.02 mInitial temperature Ti = 30C + 273 = 303 KSurface temperature T0 = 400C + 273 = 673 KThermal diffusivity = 0.008 m2/h= 2.22 10-6 m2/sThermal conductivity K = 6 W/mC.

Case (i)Depth 0.8 cm = 0.8 10-2 m = 0.008 m Time t = 10 sCase (ii)

Time t = 1 h = 3600 sSolution

In this problem heat transfer co-efficient h is not given. So take it as . i.e. h .We know that,

Biot number Bi = h =

Bi value is . So this is semi infinite solid type problem. Case (i)

For semi infinite solid.

[From HMT data book Page No. 50]

Where,

Put x = 0.008 m, t = 10 s, = 2.22 10-6 m2/s.

X = 0.848, corresponding erf (X) is 0.7706

[Refer HMT data book Page No.52]

Case (ii)

Instantaneous heat flow

[From HMT data book Page No.50]t = 3600 s (Given)

Intermediate temperature Tx = 387.85 K

Heat flux qx = 13982.37 W/m2.

31. A large cast iron at 750C is taken out from a furnace and its one of its surface is suddenly lowered and maintained at 45C. Calculate the following:

1. The time required to reach the temperature 350C at a depth of 45 mm from the surface. 2. Instantaneous heat flow rate at a depth of 45 mm and on surface after 30 minutes.3. Total heat energy after 2 hr for ingot, Take = 0.06 m2/hr, K = 48.5 W/mK.

Given

Initial temperature Ti = 750C + 273 = 1023 KSurface temperature T0 = 45C + 273 = 318 KIntermediate temperature Tx = 350C + 273 = 623 KDepth x = 45 mm = 0.045 mThermal diffusivity = 0.06 m2/hr = 1.66 10-5 m2/sThermal conductivity K = 48.5 W/mK.Solution

In this problem heat transfer co-efficient h is not given. So take it as , i.e. h .

We know that,

Biot number Bi = h =

Bi value is . So this is semi infinite solid type problem.

1. For semi infinite solid.

[From HMT data book Page No. 50]

0.432 = erf (X)

erf (X) = 0.432

erf (X) = 0.432, corresponding X is 0.41

We know

Time required to reach 350C is 181.42 s.

2. Instantaneous heat flow

[From HMT data book Page No.50]t = 30 minutes (Given)t = 1800 s

[Negative sign shows that heat lost from the ingot].

3. Total heat energy

[Time is given, 2 hr = 7200 s]

[Negative sign shows that heat lost from the ingot]32. A large steel plate 5 cm thick is initially at a uniform temperature of 400C. It is suddenly exposed on both sides to a surrounding at 60C with convective heat transfer co-efficient of 285 W/m2K. Calculate the centre line temperature and the temperature inside the plate 1.25 cm from themed plane after 3 minutes.

Take K for steel = 42.5 W/mK, for steel = 0.043 m2/hr.Given

Thickness L = 5 cm = 0.05 mInitial temperature Ti = 400C + 273 = 673 KFinal temperature T = 60C + 273 = 333 KDistance x = 1.25 mm = 0.0125 mTime t = 3 minutes = 180 sHeat transfer co-efficient h = 285 W/m2KThermal diffusivity = 0.043 m2/hr = 1.19 10-5 m2/s.Thermal conductivity K = 42.5 W/mK.

Solution

For Plate :


We know,

Biot number

0.1 < Bi < 100, So this is infinite solid type problem. Infinite Solids

Case (i)

[To calculate centre line temperature (or) Mid plane temperature for infinite plate, refer HMT data book Page No.59 Heisler chart].

X axis value is 3.42, curve value is 0.167, corresponding Y axis value is 0.64

Case (ii)

Temperature (Tx) at a distance of 0.0125 m from mid plane

[Refer HMT data book Page No.60, Heisler chart]

X axis value is 0.167, curve value is 0.5, corresponding Y axis value is 0.97.

Temperature inside the plate 1.25 cm from the mid plane is 544 K.

33. A 10 cm diameter apple approximately spherical in shape is taken from a 20C environment and placed in a refrigerator where temperature is 5C and average heat transfer coefficient is 6 W/m2K. Calculate the temperature at the centre of the apple after a period of 1 hour. The physical properties of apple are density = 998 kg/m3. Specific heat = 4180 J/kg K, Thermal conductivity = 0.6 W/mK.

Given:

Diameter of sphere D = 10 cm = 0.10 mRadius of sphere R = 5 cm = 0.05 mInitial temperature Ti = 20C + 273 = 293 KFinal temperature T = 5C + 273 = 278 KTime t = 1 hour = 3600 sDensity = 998 kg/m3Heat transfer co-efficient h = 6 W/m2KSpecific heat C = 4180 J/kg KThermal conductivity K = 0.6 W/mK

Thermal diffusivity =

Solution

For Sphere,


We know,

Biot number

0.1 < Bi < 100, So this is infinite solid type problem. Infinite Solids

[To calculate centre line temperature for sphere, refer HMT data book Page No.63].

X axis value is 0.20, curve value is 0.5, corresponding Y axis value is 0.86.

Center line temperature T0 = 290.9 K.

34. A long steel cylinder 12 cm diameter and initially at 20C is placed into furnace at 820C with h = 140 W/m2K. Calculate the time required for the axis temperature to reach 800C. Also calculate the corresponding temperature at a radius of 5.4 cm at that time. Physical properties of steel are K = 21 W/mK, = 6.11 10-6 m2/s.

Given:

Diameter of cylinder D = 12 cm = 0.12 mRadius of sphere R = 6 cm = 0.06 mInitial temperature Ti = 20C + 273 = 293 KFinal temperature T = 820C + 273 = 1093 KHeat transfer co-efficient h = 140 W/m2K

Intermediate radius r = 5.4 cm = 0.054 mThermal diffusivity = 6.11 10-6 m2/s.Thermal conductivity K = 21W/mK

To find

1. Time (t) required for the axis temperature to reach 800C.2. Corresponding temperature (Tt) at a radius of 5.4 cm.

Solution

For Cylinder,


We know,

Biot number Bi =

0.1 < Bi 2300, flow is turbulent

For turbulent flow, general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

This is heating process so, n = 0.4

20. Air at 30C, 6 m/s flows over a rectangular section of size 300 800 mm. Calculate the heat leakage per meter length per unit temperature difference.

Given : Air temperature Tm = 30nnnn C Velocity U = 6 m/s Area A = 300 800 mm2A = 0.24 m2

To find: 1. Heat leakage per metre length per unit temperature difference.

Solution:

Equivalent diameter for 300 800 mm2 cross section is given by

We know

Since Re > 2300, flow is turbulent.

For turbulent flow general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

Assuming the pipe wall temperature to be higher than a temperature. So heating process n = 0.4

We know

Heat transfer coefficient h = 18.09 W/m2KHeat leakage per unit per length per unit temperature difference Q = h P

=Q = 39.79 W

21. Air at 333K, 1.5 bar pressure, flow through 12 cm diameter tube. The surface temperature of the tube is maintained at 400K and mass flow rate is 75 kg/hr. Calculate the heat transfer rate for 1.5 m length of the tube.

Given : Air temperature Tm = 333 K = 60C Diameter D= 12 cm = 0.12 mSurface temperature Tw= 400 K = 127C

Mass flow rate m= 75 kg/hr = m = 0.020 Kg/sLength L = 1.5 m

To find:1. Heat transfer rate (Q)

Solution:

Since the pressure is not much above atmospheric, physical properties of air may be taken at atmospheric condition

We know

Since Re > 2300, so flow is turbulent For turbulent flow, general equation is (Re>10000)

Nu = 32.9

22. 250 Kg/hr of air are cooled from 100C to 30C by flowing through a 3.5 cm inner diameter pipe coil bent in to a helix of 0.6 m diameter. Calculate the value of air side heat transfer coefficient if the properties of air at 65C are

K = 0.0298 W/mK = 0.003 Kg/hr mPr = 0.7 = 1.044 Kg/m3

Given : Mass flow rate in = 205 kg/hr

Inlet temperature of air Tmi = 100COutlet temperature of air Tmo = 30CDiameter D = 3.5 cm = 0.035 m

Mean temperature

To find: Heat transfer coefficient (h)

Solution:

Reynolds Number Re =

Kinematic viscosity

Since Re > 2300, flow is turbulent For turbulent flow, general equation is (Re > 10000)

We know that,

Heat transfer coefficient h = 2266.2 W/m2K

23. In a long annulus (3.125 cm ID and 5 cm OD) the air is heated by maintaining the temperature of the outer surface of inner tube at 50C. The air enters at 16C and leaves at 32C. Its flow rate is 30 m/s. Estimate the heat transfer coefficient between air and the inner tube.

Given : Inner diameter Di = 3.125 cm = 0.03125 m Outer diameter Do = 5 cm = 0.05 mTube wall temperature Tw = 50CInner temperature of air Tmi = 16COuter temperature of air tmo = 32CFlow rate U = 30 m/s

To find: Heat transfer coefficient (h)

Solution:

Mean temperature Tm =

We know,

Hydraulic or equivalent diameter

= 0.05 0.03125Dh = 0.01875 m

Re = 35.3 10-6

Since Re > 2300, flow is turbulent

For turbulent flow, general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

This is heating process. So n = 0.4

24. Engine oil flows through a 50 mm diameter tube at an average temperature of 147C. The flow velocity is 80 cm/s. Calculate the average heat transfer coefficient if the tube wall is maintained at a temperature of 200C and it is 2 m long.

Given : Diameter D = 50 mm = 0.050 mAverage temperature Tm= 147C Velocity U= 80 cm/s = 0.80 m/sTube wall temperature Tw = 200C Length L= 2m

To find: Average heat transfer coefficient (h)

Solution : Properties of engine oil at 147C

We know

Since Re < 2300 flow is turbulent

For turbulent flow, (Re < 10000)

25. A system for heating water from an inlet temperature of 20C to an outlet temperature of 40C involves passing the water through a 2.5cm diameter steel pipe. The pipe surface temperature is maintained at 110C by condensing steam on its surface. For a water mass flow rate of 0.5 kg/min, find the length of the tube desired.

Given : Inlet temperature Tmi = 20C Outlet temperature Tmo = 40C Diameter D = 2.5 cm = 0.025 mPiper surface temperature Tw = 110CMass flow rate m = 0.5 Kg/min = 8.33 10-3 Kg/s

To find: Length of the tube (L)

Solution: We know

We know

For laminar flow,

Nusselt number Nu = 3.66


Formulae used for free convection

1. Film temperature where Tw Surface temperature in C T - Fluid temperature in C

2. Coefficient of thermal expansion

3. Nusselt Number Nu =

Where h Heat transfer coefficient W/m2K L Length, m K Thermal conductivity, W/mK

4. Grashof number for vertical plate

5. If GrPr value is less than 109, flow is laminar. If GrPr value is greater than 109, flow is turbulent.

i.e., GrPr > 109, Laminar flow GrPr > 109, Turbulent flow

6. For laminar flow (Vertical plate):

Nusselt number Nu = 0.59 (GrPr)0.25

This expression is valid for,104 < Gr Pr < 109

7. For turbulent flow (Vertical plate):Nusselt Number Nu = 0.10 [Gr Pr]0.333

8. Heat transfer (vertical plate):Q = h A (Tw - T)

9. Grashof number for horizontal plate:

Where Lc Characteristic length = W Width of the plate.

10. For horizontal plate, upper surface heated,

Nusselt number Nu = 0.54 [Gr Pr]0.25This expression is valid for

11. For horizontal plate, lower surface heated

Nusselt Number Nu = 0.27 [Gr Pr]0.25This expression is valid for 105 < Gr Pr < 1011

12. Heat transfer (Horizontal plate)

Q = (hu + hj) A (Tw - T)Where hu Upper surface heated, heat transfer coefficient W/m2 KHi Lower surface heated, heat transfer coefficient, W/m2K

13. For horizontal cylinder

Nusselt number Nu = C [Gr Pr]m

14. For horizontal cylinder,

Heat transfer Q = h A (Tw - T)Where A - DL15.For sphere,

Nusselt number Nu = 2 + 0.43 [Gr Pr]0.25Heat transfer Q = h A (Tw - T)Where A - 4r2

16. Boundary layer thickness

26. A vertical plate of 0.75 m height is at 170 C and is exposed to air at a temperature of 105C and one atmosphere calculate:

1. Mean heat transfer coefficient,2. Rate of heat transfer per unit width of the plate

Given : Length L= 0.75 mWall temperature Tw= 170CFluid temperature T= 105C

To find:

1. Heat transfer coefficient (h)2. Heat transfer (Q) per unit width

Solution: Velocity (U) is not given. So this is natural convection type problem.

We know that

We know

Grahsof number Gr =

Since Gr Pr < 109, flow is laminar

Gr Pr value is in between 104 and 109 i.e., 104 < Gr Pr < 109So, Nusselt Number

Nu = 0.59 (Gr Pr)0.25

We know

Nusselt number Nu =

Heat transfer coefficient h = 4.24 W/m2KWe know

27. A large vertical plate 4 m height is maintained at 606C and exposed to atmospheric air at 106C. Calculate the heat transfer is the plate is 10 m wide.

Given : Vertical plate length (or) Height L = 4 mWall temperature Tw = 606CAir temperature T = 106CWide W = 10 m

To find: Heat transfer (Q)

Solution:

Gr = 1.61 1011Gr Pr = 1.61 1011 0.676Gr Pr = 1.08 1011 Since Gr Pr > 109, flow is turbulent For turbulent flow, Nusselt number Nu = 0.10 [Gr Pr]0.333

We know that,

Nusselt number


Heat transfer Q = h A T

28. A thin 100 cm long and 10 cm wide horizontal plate is maintained at a uniform temperature of 150C in a large tank full of water at 75C. Estimate the rate of heat to be supplied to the plate to maintain constant plate temperature as heat is dissipated from either side of plate.

Given :

Length of horizontal plate L = 100 cm = 1mWide W = 10 cm = 0.10 mPlate temperature Tw = 150CFluid temperature T = 75C

To find: Heat loss (Q) from either side of plate

Solution:

Lc = 0.05 m

Gr Pr = 5.29 109Gr Pr value is in between 8 106 and 1011i.e., 8 106 < Gr Pr < 1011

For horizontal plate, upper surface heated:

Nusselt number Nu = 0.15 (Gr Pr)0.333

We know that,

Upper surface heated, heat transfer coefficient hu = 3543.6 W/m2K

For horizontal plate, lower surface heated:

Nusselt number Nu = 0.27 [Gr Pr]0.25

Lower surface heated, heat transfer coefficient h1 = 994.6 W/m2K

Total heat transfer Q = (hu + h1) A T

= (hu + h1) W L (Tw - T)= (3543.6 + 994.6) 0.10 (150 75)Q = 34036.5 W

29. A hot plate 20 cm in height and 60 cm wide is exposed to the ambient air at 30C. Assuming the temperature of the plate is maintained at 110C. Find the beat loss from both surface of the plate. Assume horizontal plate.

Given:

Height (or) Length of the Plate L = 20 cm = 0.20 mWide W= 60 cm = 0.60 mFluid temperature T= 30CPlate surface temperature Tw= 110C

To find:

Heat loss from both the surface of the plate (Q)

Solution:

We know

Coefficient of thermal expansion}

We know

Gr Pr value is in between 8 106 and 1011i.e., 8 106 < Gr Pr < 1011

For horizontal plate, Upper surface heated,

Nusselt number Nu = 0.15 (Gr Pr)0.333

Upper surface heated, heat transfer coefficient hu = 6.99 W/m2K

For horizontal plate, lower surface heated:

Nusselt number Nu = 0.27 (Gr Pr)0.25= 0.277 [1.06 108]0.25Nu = 28.15We know that,

Lower surface heated, heat transfer coefficient h = 2.78 W/m2K

30. A vertical pipe 80 mm diameter and 2 m height is maintained at a consent temperature of 120C. The pipe is surrounded by still atmospheric air at 30C. Find heat loss by natural convection.

Given : Vertical pipe diameter D = 80 mm = 0.080 mHeight (or) Length L= 2 mSurface temperature Tw= 120CAir temperature T= 30C

To find: Heat loss (Q)

Solution: We know

We know

For turbulent flow,

Nu = 0.10 [Gr Pr]0.333= 0.10 [3.32 1010]0.333Nu = 318.8

We know that,

UNIT IIIIntroduction to Physical mechanism Radiation properties Radiation shape factors Heat exchange between non- black bodies Radiation shields.

RADIATIONPART A

1. Define Radiation.

The heat transfer from one body to another without any transmitting medium is known as radiation. It is an electromagnetic wave phenomenon.

2. Define emissive power [E] The emissive power is defined as the total amount of radiation emitted by a body per unit time and unit area. It is expressed in W/m2.

3. Define monochromatic emissive power. [Eb]

The energy emitted by the surface at a given length per unit time per unit area in all directions is known as monochromatic emissive power.

4. What is meant by absorptivity?

Absorptivity is defined as the ratio between radiation absorbed and incident radiation.

Absorptivity

5. What is meant by reflectivity?

Reflectivity is defined as the ratio of radiation reflected to the incident radiation.

Reflectivity

Absorptivity

6. What is meant by transmissivity?

Transmissivity is defined as the ratio of radiation transmitted to the incident radiation.

Transmissivity

7. What is black body?

Black body is an ideal surface having the following properties.

1. A black body absorbs all incident radiation, regardless of wave length and direction.

2. For a prescribed temperature and wave length, no surface can emit more energy than black body.

8. State Plancks distribution law.

The relationship between the monochromatic emissive power of a black body and wave length of a radiation at a particular temperature is given by the following expression, by Planck.

Where Eb = Monochromatic emissive power W/m2

= Wave length mc1 = 0.374 10-15 W m2c2 = 14.4 10-3 mK9. State Wiens displacement law.

The Wiens law gives the relationship between temperature and wave length corresponding to the maximum spectral emissive power of the black body at that temperature.

Where c3 = 2.9 10-3 [Radiation constant]

10. State Stefan Boltzmann law.

The emissive power of a black body is proportional to the fourth power of absolute temperature.

11. Define Emissivity. It is defined as the ability of the surface of a body to radiate heat. It is also defined as the ratio of emissive power of any body to the emissive power of a black body of equal temperature.

Emissivity

12. What is meant by gray body? If a body absorbs a definite percentage of incident radiation irrespective of their wave length, the body is known as gray body. The emissive power of a gray body is always less than that of the black body.

13. State Kirchoffs law of radiation.

This law states that the ratio of total emissive power to the absorbtivity is constant for all surfaces which are in thermal equilibrium with the surroundings. This can be written as

It also states that the emissivity of the body is always equal to its absorptivity when the body remains in thermal equilibrium with its surroundings. 1 = E1; 2 = E2 and so on.

14. Define intensity of radiation (Ib).

It is defined as the rate of energy leaving a space in a given direction per unit solid angle per unit area of the emitting surface normal to the mean direction in space.

15. State Lamberts cosine law.

It states that the total emissive power Eb from a radiating plane surface in any direction proportional to the cosine of the angle of emission

Eb cos

16. What is the purpose of radiation shield?

Radiation shields constructed from low emissivity (high reflective) materials. It is used to reduce the net radiation transfer between two surfaces.

17. Define irradiation (G)It is defined as the total radiation incident upon a surface per unit time per unit area. It is expressed in W/m2.

18. What is radiosity (J) It is used to indicate the total radiation leaving a surface per unit time per unit area. It is expressed in W/m2.

19. What are the assumptions made to calculate radiation exchange between the surfaces?

1. All surfaces are considered to be either black or gray 2. Radiation and reflection process are assumed to be diffuse. 3. The absorptivity of a surface is taken equal to its emissivity and independent of temperature of the source of the incident radiation.

20. What is meant by shape factor?

The shape factor is defined as the fraction of the radiative energy that is diffused from on surface element and strikes the other surface directly with no intervening reflections. It is represented by Fij. Other names for radiation shape factor are view factor, angle factor and configuration factor.

PART B

1. A black body at 3000 K emits radiation. Calculate the following:

i) Monochromatic emissive power at 7 m wave length.ii) Wave length at which emission is maximum.iii) Maximum emissive power.iv) Total emissive power,v) Calculate the total emissive of the furnace if it is assumed as a real surface having emissivity equal to 0.85. Given: Surface temperature T = 3000K

Solution:

1. Monochromatic Emissive Power :

From Plancks distribution law, we know

[From HMT data book, Page No.71]Where c1 = 0.374 10-15 W m2c2 = 14.4 10-3 mK = 1 10-6 m[Given]

2. Maximum wave length (max)

From Wiens law, we know

3. Maximum emissive power (Eb) max:

Maximum emissive power (Eb)max = 1.307 10-5 T5 = 1.307 10-5 (3000)5 (Eb)max = 3.17 1012 W/m2

4. Total emissive power (Eb):

From Stefan Boltzmann law, we know that Eb= T4[From HMT data book Page No.71]

Where = Stefan Boltzmann constant= 5.67 10-8 W/m2K4 Eb = (5.67 10-8) (3000)4 Eb = 4.59 106 W/m2

5. Total emissive power of a real surface:

(Eb)real = T4Where = Emissivity = 0.85

(Eb)real =

2. A black body of 1200 cm2 emits radiation at 1000 K. Calculate the following:

1. Total rate of energy emission2. Intensity of normal radiation3. Wave length of maximum monochromatic emissive power. 4. Intensity of radiation along a direction at 60 to the normal.

Solution:

From Stefan Boltzmann law.

1. Energy emission Eb = T4

[From HMT data book, Page No.71]

Eb =

Here Area = 1200 10-4 m2, Eb = 5.67103 1200 10-4 Eb = 6804 W

2. Intensity of normal radiation

3. From Wiens law, we know that

3. Assuming sun to be black body emitting radiation at 6000 K at a mean distance of 12 1010 m from the earth. The diameter of the sun is 1.5 109 m and that of the earth is 13.2 106 m. Calculation the following.

1. Total energy emitted by the sun.2. The emission received per m2 just outside the earths atmosphere.3. The total energy received by the earth if no radiation is blocked by the earths atmosphere. 4. The energy received by a 2 2 m solar collector whose normal is inclined at 45 to the sun. The energy loss through the atmosphere is 50% and the diffuse radiation is 20% of direct radiation.

Given: Surface temperature T = 6000 KDistance between earth and sun R = 12 1010 m Diameter on the sun D1 = 1.5 109 m Diameter of the earth D2 = 13.2 106 m

Solution:

1. Energy emitted by sun Eb = T4

2. The emission received per m2 just outside the earths atmosphere:

The distance between earth and sun R = 12 1010 m

3. Energy received by the earth:

Energy received by the earth

4. The energy received by a 2 2 m solar collector;

Energy loss through the atmosphere is 50%. So energy reaching the earth.

Energy received by the earth

Diffuse radiation is 20%

Total radiation reaching the collection

Energy received by the collector

4. A large enclosure is maintained at a uniform temperature of 3000 K. Calculate the following:

1. Emissive power2. The wave length 1 below which 20 percent of the emission is concentrated and the wave length 2 above which 20 percent of the emission is concentrated. 3. The maximum wave length.4. Spectral emissive power.5. The irradiation incident.

Given : Surface temperature T = 3000 K

1. Emissive power Eb = T4

2. The wave length 1 corresponds to the upper limit, containing 20% of emitted radiation.

, corresponding

1T = 2666 K[From HMT data book, Page No.72]

The wave length 2 corresponds to the lower limit, containing 20% of emitted radiation.

3. Maximum wave length (max):max T = 2.9 10-3 mK

4. Spectral Emissive Power:

From Plancks distribution law, we know


5. Irradiation:

The irradiation incident on a small object placed within the enclosure may be treated as equal to emission from a black body at the enclosure surface temperature.

So, G = Eb = .

5. The sun emits maximum radiation at = 0.52. Assuming the sun to be a black body, calculate the surface temperature of the sun. Also calculate the monochromatic emissive power of the suns surface.

Given :

To find : 1. Surface temperature T. 2. Monochromatic emissive power Eb.Solution:

1. From Wiens law, we know

2. Monochromatic emissive power (Eb):

From Plancks law, we know


6. A furnace wall emits radiation at 2000 K. Treating it as black body radiation, calculate 1. Monochromatic radiant flux density at 1m wave length.2. Wave length at which emission is maximum and the corresponding emissive power. 3. Total emissive power

Given: Temperature T = 2000 K; = 1 m = 1 10-6

Solution:

1. Monochromatic emissive power (Eb):


2. Maximum Wave Length (max):

From Wiens Law, we know that

max T = [From HMT data book, Page No.71]

Corresponding emissive power

3. Total emissive power (Eb):

From Stefan Boltzmann law, we know

Eb = T4Where - Stefan Boltzmann constant

7. The temperature of a black surface 0.25 m2 of area is 650C. Calculate,

1. The total rate of energy emission2. The intensity of normal radiation.

The wavelength of maximum monochromatic emissive power. Given :A = 0.25 m2T = 650 + 273 = 923 KTo find : 1. Eb ; 2. In ; 3. max

Solution:

1. We know Emissive power Eb = T4

Here Area = 0.25 m2

2. We know

3. From Wiens law,

8. Calculate the heat exchange by radiation between the surfaces of two long cylinders having radii 120mm and 60mm respectively. The axis of the cylinder are parallel to each other. The inner cylinder is maintained at a temperature of 130C and emissivity of 0.6. Outer cylinder is maintained at a temperature of 30C and emissivity of 0.5.

Given : r1 = 60 mm = 0.060 m r2 = 120 mm = 0.12 T1 = 130C + 273 = 403 1 = 0.6 T2 = 30C + 273 = 303 K 2 = 0.5To find : Heat exchange (Q)

Solution: Heat exchange between two large concentric cylinder is given by

9. Two concentric spheres 30 cm and 40 cm in diameter with the space between them evacuated are used to store liquid air at - 130C in a room at 25C. The surfaces of the spheres are flushed with aluminium of emissivity = 0.05. Calculate the rate of evaporation of liquid air if the latent heat of vaporization of liquid air is 220 kJ/kg.

Given: Inner diameter D1 = 30 cm = 0.30 mInner radius r1 = 0.15 mOuter diameter D2 = 40 cm = 0.40 mOuter radiusr2 = 0.20 mT1 = - 130C + 273 = 143 KT2 = 25C + 273 = 298 K = 0.05Latent heat of vapourisation = 220kJ /kg = 220 103 J / kg

To find: Rate of evaporation

Solution: This is heat exchange between large concentric sphere problem.

Heat transfer

Where

10. A pipe of outside diameter 30 cm having emissivity 0.6 and at a temperature of 600 K runs centrally in a brick of 40 cm side square section having emissivity 0.8 and at a temperature of 300K. Calculate the following:

1. Heat exchange per metre length.2. Convective heat transfer coefficient when surrounding of duct is 280 K.

Given:

Pipe diameter D1 = 30 cm D1 = 0.30 mSurface area A1 = D1L = 0.30 1 A1 = 0.942 m2

1 = 0.6 T1 = 600 K

Brick duct side = 40 cm = 0.40 mSurface area A2 = (0.4 1) 4[length L = 1m; No. of sides = 4]

To find: 1. Heat exchange (Q) 2. Convective heat transfer coefficient (h) when T = 280 KSolution:

Case 1: We know that

Heat exchange

where

Case (ii) :

Heat transfer by convection Q = hA (T - T) Q12 = h A (T2 - T) Q12 = h 1 (300 280)

Equating (2) and (3),3569.2 = 20h

11. Emissivities of two large parallel plates maintained at 800C and 300C are 0.5 respectively. Find net radiant hat exchange per square metre for these plates. Find the percentage reduction in heat transfer when a polished aluminium radiation shield of emissivity 0.06 is placed between them. Also find the temperature of the shield.

Given : T1= 800C + 273 = 1073 K T2= 300C + 273= 573 K 1= 0.3 2 = 0.5Shield emissivity 3 = 0.06

To find:

1. Net radiant heat exchange per square metre. (Q/A)2. Percentage reduction in heat loss due to radiation shield.3. Temperature of the shield (T3).

Solution: Heat exchange between two large parallel plates without radiation shield is given by

Heat exchange between plate 1 and radiation shield 3 is given by

Heat exchange between radiation shield 3 and plate 2 is given by

We know Q13 = Q32

Substituting T3 value in equation (A) (or) equation (B), Heat transfer with radiation shield

Heat transfer with radiation shield

Reduction in heat loss due to radiation shield

12. A pipe of diameter 30 cm, carrying steam runs in a large room and is exposed to air at a temperature of 25C. The surface temperature of the pipe is 300C. Calculate the loss of heat of surrounding per meter length of pipe due to thermal radiation. The emissivity of the pipe surface is 0.8.

What would be the loss of heat due to radiation of the pipe is enclosed in a 55 cm diameter brick of emissivity 0.91?

Given :

Case 1: Diameter of pipe D1 = 30 cm = 0.30 mSurface temperature T1 = 300C + 273 = 573 KAir temperature T2 = 25C + 273 = 298 KEmissivity of the pipe 1 = 0.8Case 2: Outer diameterD2 = 55 cm = 0.55m Emissivity 2 = 0.91

To find: 1. Loss of heat per metre length (Q/L). 2. Reduction in heat loss.

Solution:

Case 1:

Heat transfer

Heat loss per metre length = 4271.3 W/m

Case 2: When the 30 cm dia pipe is enclosed in a 55 cm diameter pipe, heat exchange between two large concentric cylinder is given by

Reduction in heat loss = 4271.3 4057.8= 21.3.4

13. Emissivities of two large parallel plates maintained at T1 K and T2 K are 0.6 and 0.6 respectively. Heat transfer is reduced 75 times when a polished aluminium radiation shields of emissivity 0.04 are placed in between them. Calculate the number of shields required.

Given: 1 = 0.6 2 = 0.6

Heat transfer reduced = 75 times Emissivity of radiation shield, s = 3 = 0.04

To find: Number of screens require.

Solution: Heat transfer with n shield is given by

14. Find the relative heat transfer between two large plane at temperature 1000 K and 500 K when they are

1. Black bodies2. Gray bodies with emissivities of each surface is 0.7.

Given: T1 = 1000 K T2 = 500 K 1 = 0.7 2 = 0.7

Solution :

Case 1: Heat exchange between two large parallel plate is given by

Case 2:

15. The inner sphere of liquid oxygen container is 40 cm diameter and outer sphere is 50 cm diameter. Both have emissivities 0.05. Determine the rate at which the liquid oxygen would evaporate at -183C when the outer sphere at 20C. Latent heat of oxygen is 210 kJ/kg.

Given : Inner diameter D1 = 40 cm = 0.40 m Inner radius r1 = 0.20 m Outer diameter D2 = 50 cm = 0.50 m Outer radiusr2 = 0.25 m Emissivity1 = 0.052 = 0.05 Inner temperature T1 = -183C + 273 = 90K Outer temperature T2 = 20C + 273 = 293 K Latent heat of oxygen = 210 kJ / kg = 210 103 J/kgTo find : Rate of evaporation

Solution :

This is heat exchange between two large concentric spheres problem.

Heat transfer [From equation No.27]

[Negative sign indicates heat is transferred from outer surface to inner surface.]

Rate of evaporation =

16. Emissivities of two large parallel plates maintained at 800C are 0.3 and 0.5 respectively. Find the net radiant heat exchange per square metre of the plates. If a polished aluminium shield ( = 0.05) is placed between them. Find the percentage of reduction in heat transfer.

Given : T1 = 800C + 273 = 1073 K T2 = 300C + 273 = 573 K 1 = 0.3 2 = 0.5

Radiation shield emissivity 3 = 0.05

To find: 1. Net radiant heat exchange per square metre 2. Percentage of reduction in heat loss due to radiation shield.

Solution:

Case 1 : Heat transfer without radiation shield:

Heat exchange between two large parallel plats without radiation shield is given by

Case 2: Heat transfer with radiation shield:

Heat exchange between plate 1 and radiation shield 3 is given by

Heat exchange between radiation shield 3 and plate 2 is given by

We know Q13 = Q32

Substitute T3 value in equation (A) or (B).

Substituting T3 value in equation (A) (or) equation (B),

17. The amount of radiant energy falling on a 50 cm 50 cm horizontal thin metal plate insulated to the bottom is 3600kJ /m2 hr. If the emissivity of the plate surface is 0.8 and the ambient air temperature is 30C, find the equilibrium temperature of the plate.Given : Area A = 50 cm 50 cm = 0.5 0.5 m

Emissivity = 0.8

Ambient air temperature T2 = 30C + 273 = 303 K

To find : Plate temperature T1

Solution : We know

Heat transfer

18. Calculate the shape factors for the configuration shown in fig.

1. A black body inside a black enclosure.

2. A tube with cross section of an equilateral triangle.

3. Hemispherical surface and a plane surface Solution:Case 1:[All radiation emitted from the black surface 2 is absorbed by the enclosing surface 1.]We know F1-1 + F1 2 = 1 .(1)

By reciprocity theorem A1F12 = A2F21

Now considering radiation from surface 2,

By reciprocity theorem, we know

A1F12 = A2F21

Result: F1 1 = 0, F21 = F12 = 0.5 F22 = 0 F1 2 = 0.5, F2 3 = 0.5 F1 3 = 0.5

Case 3: We know F1 1 + F1 2 = 1

By reciprocity theorem, A1 F1 2 = A2 F2 1

[Since all radiation emitting from the black surface 2 are absorbed by the enclosing surface 1]

19. Two black square plates of size 2 by 2 m are placed parallel to each other at a distance of 0.5 m. One plate is maintained at a temperature of 1000C and the other at 500C. Find the heat exchange between the plates.

Given: Area A = 2 2 = 4 m2T1 = 1000C + 273 = 1273 KT2 = 500C + 273 = 773 KDistance = 0.5 m

To find : Heat transfer (Q)

Solution : We know

Heat transfer general equation is

where [From equation No.(6)]

For black body

Where F12 Shape factor for square plates

In order to find shape factor F12, refer HMT data book, Page No.76.

Curve 2 [Since given is square plates]

X axis value is 4, curve is 2. So corresponding Y axis value is 0.62.

i.e.,

20. Two circular discs of diameter 0.3 m each are placed parallel to each other at a distance of 0.2 m. one is disc is maintained at a temperature of 750C and the other at 350C and their corresponding emissivities are 0.3 and 0.6. Calculate heat exchange between the discs.

Given : D1 = 0.3 m D2 = 0.3 m

T1 = 750C + 273 = 1023 KT2 = 350C + 273 = 623 K1 = 0.3 2 = 0.6

Distance between discs = 0.2 m.

To find : Heat exchange between discs (Q),

Solution:

Heat transfer by radiation general equation is

Where F12 Shape factor for disc


X axis value is 1.5, curve is 1. So, corresponding Y axis value is 0.28.

21. Two parallel rectangular surfaces 1 m 2m are opposite to each other at a distance of 4m. The surfaces are black and at 300C and 200C. Calculate the heat exchange by radiation between two surfaces.

Given: Area A = 2 2 = 2 m2 Distance = 4 m T1 = 300C + 273 = 573 K T2 = 200C + 273 = 473 K

To find: Heat exchange (Q12)

Solution : We know, Heat transfer general equation is

Where F12 Shape factor for parallel rectangles

In order to find shape factor refer HMT data book, Page No.77 and 78.

From graph, we know,

22. Two parallel plates of size 3 m 2 m are placed parallel to each other at a distance of 1 m. One plate is maintained at a temperature of 550C and the other at 250C and the emissivities are 0.35 and 0.55 respectively. The plates are located in a large room whose walls are at 35C. If the plates located exchange heat with each other and with the room, calculate.

1. Heat lost by the plates.2. Heat received by the room.

Given: Size of the plates = 3 m 2 mDistance between plates= 1 m

First plate temperature T1 = 550C + 273 = 823 KSecond plate temperatureT2= 250C + 273 = 523 KEmissivity of first plate 1= 0.35Emissivity of second plate2= 0.55Room temperature T3 = 35C + 273 = 308 K

To find: 1. Heat lost by the plates 2. Heat received by the room.

Solution: In this problem, heat exchange take place between two plates and the room. So this is three surface problem and the corresponding radiation network is given below.

Area A1 = 3 2 = 6 m2

Since the room is large

From electrical network diagram.

Apply values in electrical network diagram.

To find shape factor F12 refer HMT data book, Page No.78.

X value is 3, Y value is 2, corresponding shape factor [From table]F12 = 0.47

We know that, F11 + F12 + F13 = 1But, F11 = 0

Similarly, F21 + F22 + F23 = 1We know F22 = 0

From electrical network diagram,

From Stefan Boltzmann law, we know

[From diagram]

The radiosities, J1 and J2 can be calculated by using Kirchoffs law.

The sum of current entering the node J1 is zero.At Node J1:

At node j2

-+*

Solving equation (7) and (8),

Heat lost by plate (1) is given by

Heat lost by plate 2 is given by

Total heat lost by the plates Q = Q1 + Q2 = 49.36 103 3.59 103

Heat received by the room

From equation (9), (10), we came to know heat lost by the plates is equal to heat received by the room.

23. Two black square plates of size 1 by 1 m are placed parallel to each other at a distance of 0.4 m. One plate is maintained at a temperature of 900C and the other at 400. Find the net heat exchange of energy due to radiation between the two plates. Given: Area A = 1 1 = 1 m2Distance = 0.4 m T1 = 900C + 273 = 1173 KT2 = 400C + 273 = 673 K

To find: Heat exchange (Q)

Solution: Heat transfer by radiation general equation is

Where F12 shape factor for square plates.


Curve 2 [since given is square plate]

X axis value is 2.5, curve is 2, so corresponding Y axis value is 0.42.i.e., F12 = 0.42

24. Two circular discs of diameter 20 cm each are placed 2 m apart. Calculate the radiant heat exchange for these discs if there are maintained at 800C and 300C respectively and the corresponding emissivities are 0.3 and 0.5.

Given : D1 = 20 cm = 0.2 mD2 = 0.2 mT1 = 800C + 273 = 1073 KT2 = 300C + 273 = 573 K1 = 0.32 = 0.5

To find: Heat exchange (Q)

Solution: Area = A1 = 0.031 m2

A2 = 0.031 m2 Heat transfer by radiation generation equation is

Where F12 = Shape factor for disc.

In order to find shape factor, F12 refer HMT data book, Page No.76.

Curve 1 [since given is disc]

X axis value is 0.1, curve is 1, so corresponding Y axis value is 0.01.

F12 = 0.01

F12 = 0.01

(1) Q12 = 20.7 Watts.

25. A long cylindrical heater 30 in diameter is maintained at 700C. It has surface emissivity of 0.8. The heater is located in a large room whose wall are 35C. Find the radiant heat transfer. Find the percentage of reduction in heat transfer if the heater is completely covered by radiation shield ( = 0.05) and diameter 40 mm.

Given : Diameter of cylinder D1=30mm=0.030 mm Temperature T1=700C + 273 = 973 K Emissivity 1 = 0.8 Room temperature T2 = 35C + 273 = 308 K

Radiation Shield :

Emissivity 3 = 0.05Diameter D3 = 40 mm = 0.040 m

Solution:

Case 1 : Heat transfer without shield:

Heat transfer by radiation general equation is

Since room is large F12 = Shape factor

Small body enclosed by large body F12 = 1

[Refer HMT data book, Page No.73]

Heat transfer without shield

Case 2: Heat transfer with shield:

Heat transfer between heater (1) and radiation shield (3) is given by

Shape factor for concentric long cylinder F13 = 1[Refer HMT data book, Page No. 73]

Heat exchange between radiation shield (3) and Room (2) is given by

Since room is large, A2 =

Shape factor for small body enclosed by large body F32 = 1[Refer HMT data book, Page No.73]

Substitute T3 value in (3) or (4).Heat transfer with radiation shield

26. A gas is enclosed in a body at a temperature of 727C. The mean beam length of the gas body is 3 m. The partial pressure of water vapour is 0.2 atm and the total pressure is 2 atm. Calculate the emissivity of water vapour.

Given : Temperature T = 727C + 273 = 1000KMean beam length Lm = 3m

Partial pressure of water vapour Total pressure P = 2 atm.

To find : Emissivity of water vapour

Solution:

From HMT data book, Page No.92, we can find emissivity of H2o.

From graph,

Emissivity of H2o = 0.3

To find correction factor for H20:

From HMT data book, Page No.94, we can find correction factor for H2o

From graph,

Correction factor for H2o = 1.36

So, Emissivity of H2o,

27. A gas mixture contains 20% CO2 and 10% H2o by volume. The total pressure is 2 atm. The temperature of the gas is 927C. The mean beam length is 0.3 m. Calculate the emissivity of the mixture.

Given : Partial pressure of CO2, = 20% = 0.20 atm

Partial pressure of H2o, = 10% = 0.10 atm. Total pressure P = 2 atm Temperature T = 927C + 273 = 1200 KMean beam length Lm = 0.3 m

To find: Emissivity of mixture (mix).

Solution : To find emissivity of CO2

From HMT data book, Page No.90, we can find emissivity of CO2.

From graph, Emissivity of CO2 = 0.09

To find correction factor for CO2

Total pressure, P = 2 atm

From HMT data book, Page No.91, we can find correction factor for CO2

From graph, correction factor for CO2 is 1.25

To find emissivity of :

From HMT data book, Page No.92, we can find emissivity of

From graph Emissivity of = 0.048

To find correction factor for :

From HMT data book, Page No.92 we can find emission of H20

From graph,

Correction factor for = 1.39

Correction factor for mixture of CO2 and H2O:

From HMT data book, Page No.95, we can find correction factor for mixture of CO2 and

From graph,

Total emissivity of gascous mixture

28. A furnace of 25 m2 area and 12 m2 volume is maintained at a temperature of 925C over is entire volume. The total pressure of the combustion gases is 3 atm, the partial pressure of water vapour is 0.1 atm and that of CO2 is 0.25 atm.Calculate the emissivity of the gaseous mixture.

Given : Area A = 25 m2Volume V = 12 m3 Temperature T= 925 + 273= 1198 K Total pressure P= 3 atm

Partial pressure of water vapour,

Partial pressure of CO2

To find: Emissivity of mixture

Solution : We know Mean beam length for gaseous mixture.

To find emissivity of CO2

From HMT data book, Page No.90, we can find emissivity of CO2.

From graph, Emissivity of CO2 = 0.15

To find correction factor for CO2:

Total pressure P = 3 atm.

From HMT data book, Page No.91, we can find correction factor for CO2.

From graph, we find

To find emissivity of H2O:

From HMT data book, Page No.92, we can find emissivity of H2O.

From graph, Emissivity of H2O = 0.15

To find correction factor for H2O:

From HMT data bo

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