Heat_mass_Transfer by naveen choudhary.pptx

heat mass and transfer notes , short notes of HM&T , Heat and Mass Transfer in Food Processing
3(2+1)
FT 236
Theory
Basic heat transfer processes, heat transfer coefficients, properties related to heat transfer, One-dimensional steady state conduction: Theory of heat conduction, Fourier's law and its derivation, Concept of electrical analogy and its application for thermal circuits, heat transfer through composite walls and insulated pipelines; One-dimensional steady state heat conduction with heat generation: Heat flow through slab, hollow sphere and cylinder with linear heat transfer, uniform/non-uniform heat generation, development of equations of temperature distribution with different boundary conditions; Steady-state heat conduction with heat dissipation to environment: Introduction to extended surfaces (fins) of uniform area of cross-section and with Equation of temperature distribution with different boundary conditions; Effectiveness and efficiency of the fins; Introduction to unsteady state heat conduction: System with negligible internal resistance and in various geometries; Convection: Forced and free convection, use of dimensional analysis for correlating variables affecting convection heat transfer, Concept of Nusselt number, Prandtl number, Reynolds number, Grashoff number, some important empirical relations used for determination of heat transfer coefficient; Heat exchangers: Principle and types; Heat transfer to flowing fluids; Radiation: Heat radiation, emissivity, absorptivity, transmissivity, radiation through black

1. Heat and Mass Transfer Course No: FT-236 Course Instructor: Er . Ashish Dhiman

2. Course Contents Heat Conduction ONE DIMENSIONAL STEADY STATE HEAT CONDUCTION HEAT TRANSFER THROUGH COMPOSITE WALL AND INSULATED PIPE Dimensionless numbers Introduction of condensation & Boiling Heat Transfer

3. Heat conduction and fourier’s law

4. Heat transfer Conduction Thermal energy is transferred by the direct contact of molecules ,not by the movement of material. Convection Thermal energy is transferred by the mass motion of group of molecule resulting in transport and mixing of properties Radiation Thermal energy is transferred by electromagnetic radiation [WAVES] Heat transfer is the process of transfer of heat from high temperature system to low temperature system . Mechanisms of heat transfer :

5. FOURIER’S LAW OF HEAT CONDUCTION It states that “the rate of heat transfer is directly proportional to the area normal to the area normal to the direction of heat flow and temperature gradient” Q ∝ A ∝ q=- where ,k= Thermal conductivity of material and negative sign indicate the drop in temperature Q=heat flux density  

6. Derivation of fourier’s law Rate of heat conduction ∝ [area][temperature difference]/thickness Suppose T1 and T2 are different temperature through a short distance of an area where the distance is Δ x the area is A and K is thermal conductivity of material. Therefore is one dimension equation can be represent as Q= =-KA q=- k∇T [Differential form] Q=- [One dimension form] [ Integral form]  

7. Assumptions in fourier law of heat conduction : The thermal conductivity of the material is constant throughout the material. There is no internal heat generation that occurs in the body . The temperature gradient is considered as constant. The heat flow is unidirectional and takes place under steady state conditions. The surfaces are isothermal. Essential features of fourier’s law of heat conduction : It is a vector expression. Negative sign indicates that heat flow in direction of decreasing temperature. It is applicable to heat conduction in all materials regardless of their state. It helps to determine the thermal conductivity ‘k’ of material through which heat transfer. It is not an expression derived from the first place ; instead, it is based on experimental evidence.

8. STEADY STATE HEAT TRANSFER VS UNSTEADY STATE HEAT TRANSFER STEADY STATE HEAT TRANSFER: The heat transfer process that does not affected by the time interval is know as steady state heat transfer . In this type of heat transfer , the temperature of the object does not change with respect of time . Unsteady state heat transfer: The heat transfer process that get affected by the time interval is know as unsteady state heat transfer . In this type of heat transfer , the temperature of the object increases with respect to time .

9. STEADY STATE Vs UNSTEADY STATE Steady state heat transfer unsteady heat transfer 1. The steady state heat transfer is denoted by =0 1. The unsteady state heat transfer is denoted by, ≠0 The temperature of the object doesn’t vary with respect to time. 2. The temperature of the object changes with respect to time. 3. No heat diffusion occurs in the object. 3. Heat diffusion occurs in the object. 4. For the steady state heat transfer, the energy entering the object is equal to the energy leaving from the , Ein = Eout . 4. For the unsteady state heat transfer, the energy entering the object is not equal to the energy leaving from the object, Ein≠Eout . 5. The change in the internal energy of the object is zero. 5. The change in the internal energy of the object is not equal to zero. 6. Example: heat generation because of electrical current. 6. Example: heating or cooling of water, heating of metal in a furnace. Steady state heat transfer unsteady heat transfer The temperature of the object doesn’t vary with respect to time. 2. The temperature of the object changes with respect to time. 3. No heat diffusion occurs in the object. 3. Heat diffusion occurs in the object. 4. For the steady state heat transfer, the energy entering the object is equal to the energy leaving from the , Ein = Eout . 4. For the unsteady state heat transfer, the energy entering the object is not equal to the energy leaving from the object, Ein≠Eout . 5. The change in the internal energy of the object is zero. 5. The change in the internal energy of the object is not equal to zero. 6. Example: heat generation because of electrical current. 6. Example: heating or cooling of water, heating of metal in a furnace.

10. FREE AND FORCED CONVECTION Free Convection : Free convection is also know as natural convection. It is the type of convection heat transfer in which the fluid molecules move due to density and temperature gradient to transfer the heat Natural convection is the convection occurring due to buoyant forces with the distinction in densities caused by distinction in temperature . The oceanic wind is an example of natural convection Forced convection Forced convection is the type of convective heat transfer in which the motion in molecules is generated by the use of an external sources for the transfer of heat. Example ; exhaust fan or a ceiling fan. The exhaust fan helps eject the heat inside the room by generating motion into air molecules

11. FREE VS FORCED CONVECTION FREE CONVECTION FORCED COVECTION In this, the molecules move due to density and temperature variation. In this, the fluid molecules are forced to move by an external source. The rate of heat transfer is lower. It has less overall heat transfer coefficient. The rate of heat transfer is higher. The overall heat transfer coefficient is higher. No external equipment is required. External equipment is necessary for convective heat transfer. The motion of molecules is comparatively slower. Molecules of fluid are forced to move faster. Equipment based on natural convection is larger in size. Equipment based on forced convection is compact in size. The flow of molecules cannot be controlled. E.g. movement of water molecules while boiling. The flow of molecules can be controlled by controlling the fan, pump, blower. E.g. movement of molecules due to fan or blower.

12. Fourier law and heat conduction Numerical Practice

13. Problem 1: A 10 cm thick block of ice with a temperature of 0 °C lies on the upper surface of 2400 cm 2 slab of stone. The slab is steam-exposed on the lower surface at a temperature of 100 °C. Find the heat conductivity of stone if 4000 g of ice is melted in one hour given that the latent heat of fusion of ice is 80 cal ⁄ gm. Solution: Given: Area of slab, A = 2400 cm 2 Thickness of ice, d = 10 cm Temperature difference, T h  – T c  = 100 °C – 0 °C  100 °C Time of heat transfer, t = 1 hr = 3600 s Amount of heat transfer, Q = m L = 4000 × 80  320000 cal Heat transfer rate, q = Q ⁄ t = 320000 cal ⁄ 3600 s  89 cal ⁄ s The formula for heat transfer rate is given as: q = K A ( T h  – T c ) ⁄ d Rearrange the above formula in terms of K . K = q d ⁄ A ( T h  – T c )  (89 × 10) ⁄ (2400 × 100) cal ⁄ cm s °C  3.7 × 10 -3   cal ⁄ cm s °C Hence, the thermal conductivity of stone is  3.7 × 10 -3   cal ⁄ cm s °C.

14. Problem 2: A metal rod 0.4 m long & 0.04 m in diameter has one end at 373 K & another end at 273 K. Calculate the total amount of heat conducted in 1 minute. (Given K = 385 J ⁄ m s °C) Solution: Given :
Thermal conductivity, K = 385 J ⁄ m s °C
Length of rod, d = 0.4 m
Diameter of rod, D = 0.04 m
Area of slab, A = π D2 ⁄ 4 = 0.001256 m2 note: cross-sectional inside pipe area= pi*(d/2) 2 Temperature difference, Th – Tc = 373 K – 273 K = 100 K
Time of heat transfer, t = 1 min = 60 s The formula for heat transfer rate is given as: Q ⁄ t = K A ( T h – T c ) ⁄ d Q = K A t ( T h – T c ) ⁄ d
= (385 × 0.001256 × 60 × 100) ⁄ 0.4 J  7.25 × 10 3 J Hence, the total amount of heat transfer is 7.25 × 10 3 J.

15. Problem 3: An aluminium rod and a copper rod of equal length 2.0 m and cross-sectional area 2 cm 2 are welded together in parallel. One end is kept at a temperature of 10 °C and the other at 30 °C . Calculate the amount of heat taken out per second from the hot end . (Thermal conductivity of aluminium is 200 W ⁄ m °C and of copper is 390 W ⁄ m °C). Solution: Given :
Thermal conductivity of aluminium, K al = 200 W ⁄ m °C
Thermal conductivity of copper, K cu = 390 W ⁄ m °C
Combined thermal conductivity for parallel combination, K = 200 W ⁄ m °C + 390 W ⁄ m °C = 590 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm 2 = 2 × 10-4 m2
Temperature difference, T h – T c = 30 °C – 10 °C = 20 °C The formula for heat transfer rate is given as: q = K A ( T h – T c ) ⁄ d = (590 × 2 × 10-4 × 20) ⁄ 2 W  1.18 W Hence, the total amount of heat transfer is 1.18 W.

16. Problem 4 : The average rate at which energy is conducted outward through the ground surface at a place is 50.0 mW ⁄ m 2 , and the average thermal conductivity of the near-surface rocks is 2.00 W ⁄ m K. Assuming surface temperature of 20.0 °C, find the temperature at a depth of 25.0 km. solution Given :
Average thermal conductivity, K = 2.00 W ⁄ m K
Depth, d = 25.0 km = 2.50 × 104 m
Surface temperature, Tc = 20.0 °C = (20 + 273) K = 293 K
Heat transfer rate per unit area, q ⁄ A = 50.0 mW ⁄ m 2 = 50.0 × 10 -3 W ⁄ m 2 The formula for heat transfer rate is given as: q = K A ( T h – T c ) ⁄ d Rearrange the above formula in terms of Th. Th = q d ⁄ KA + Tc  ((50.0 × 10-3 × 2.00 × 104) ⁄ 2.00) + 293  (500 + 293) K  893 – 273 K  520 °C Hence, the temperature at depth of 25.0 km is 520 °C.

17. Problem 5: The energy lost from a 10 cm thick slab of steel is 50 W. Assuming the temperature difference of 10.0 K, find the area of the slab. (Thermal conductivity of steel = 45 W ⁄ m K). solution Given :
Thermal conductivity, K = 45 W ⁄ m K
Thickness of slab, d = 10 cm = 0.1 m
Temperature difference, T h – T c = 10.0 K
Energy lost per sec, q = 50 W The formula for heat transfer rate is given as: q = K A ( T h – T c ) ⁄ d Rearrange the above formula in terms of A. A = q d ⁄ K ( T h – T c )  (50 × 0.1) ⁄ (45 × 10.0) m2  0.011 m2 Hence, the area of the slab is 0.011 m2.

18. Problem 6: One face of an aluminium cube of edge 5 meters is maintained at 60 ºC and the other end is maintained at 0 ºC. All other surfaces are covered by adiabatic walls. Find the amount of heat flowing through the cube in 2 seconds. (Thermal conductivity of aluminium is 209 W ⁄ m ºC). solution Given :
Edge length of cube, d = 5 m
Surface area of cube, A = d 2 = (5 m)2 = 25 m2
Temperature difference, T h – T c = 60 ºC – 0 ºC = 60 ºC
Thermal conductivity, K = 209 W ⁄ m ºC
Heat transfer time, t = 2 sec The formula for heat transfer rate is given as: q = K A ( T h – T c ) ⁄ d  (209 × 25 × 60) ⁄ 5 J  62700 J  62.7 KJ Hence, the amount of heat that flows through the cube is 62.7 KJ.

19. Problem 7: An aluminium rod and a copper rod of equal length 2.0 m and cross-sectional area 2 cm2 are welded together in series. One end is kept at a temperature of 10 °C and the other at 30 °C . Calculate the amount of heat taken out per second from the hot end . (Thermal conductivity of aluminium is 200 W ⁄ m °C and of copper is 390 W ⁄ m °C). solution Given :
Thermal conductivity of aluminium, Kal = 200 W ⁄ m °C
Thermal conductivity of aluminium, Kcu = 390 W ⁄ m °C
Combined thermal conductivity for series combination, 1 ⁄ K = 1 ⁄ 200 W ⁄ m °C + 1 ⁄ 390 W ⁄ m °C
K = (200 × 390) ⁄ (200 + 390) W ⁄ m °C
= 132.2 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm2 = 2 × 10-4 m2
Temperature difference, Th – Tc = 30 °C – 10 °C = 20 °C The formula for heat transfer rate is given as:p q = K A ( Th – Tc) ⁄ d  (132.2 × 2 × 10-4 × 20) ⁄ 2 W  0.2644 W Hence, the total amount of heat transfer is 0.2644 W .

20. ONE DIMENSIONAL STEADY STATE HEAT CONDUCTION

21. Steady State One Dimensional Conduction The formula for steady state one dimensional conduction describing heat transfer through a material is given by Fourier’s Law : Where : Q is the heat transfer rate, k is the material’s thermal conductivity, A is the cross sectional area through which the heat is conducted, is the temperature difference across the material, d is the thickness of material.  

22. Relation with Fourier’s Law : Steady-state one-dimensional conduction is a concept in heat transfer that deals with the steady flow of heat through a material along a single direction, typically in a straight line. The term "steady-state" implies a constant and unchanging condition over time. In this context, the temperature distribution within the material remains constant, and there is a continuous transfer of heat from one side to the other. The governing principle is Fourier's Law, which relates the heat transfer rate to the material's thermal conductivity, cross-sectional area, temperature difference, and thickness. This phenomenon is crucial in understanding how heat moves through solid objects under stable conditions.

23. For the following geometries, one-dimensional, steady-state heat conduction is taken into consideration : SLAB CYLINDER SPHERE

24. One Dimensional Steady State Heat Conduction through a Slab : One-dimensional steady-state heat conduction through a slab refers to the controlled and constant flow of heat in a straight path through a material with a uniform cross-section. Key points include : 1. Steady-state : The temperature distribution within the slab remains constant with time. There is no change in the temperature profile as time progresses. 2. One-dimensional : Heat transfer occurs primarily in one direction, usually along the thickness of the slab. This simplifies the analysis by focusing on a single dimension. 3. Slab : The material through which heat is conducted has a flat and often rectangular shape. This could represent various objects like walls, windows, or other planar structures.

25. Boundary Conditions for Heat Flow through Slab : Thickness = ‘ L’ , Area = ‘ A’ and Conductivity = ‘ k’ In the case of one-dimensional, steady heat conduction through a wall of uniform conductivity without heat generation : , and (1) Equation (1) reduces to : (2) Integrating equation (2) twice with respect to x : (3) where, and are constants of integration Now using the boundary conditions : At x = 0, Equation (3) is written as = (4) At x = L,  

26. Equation (3) can be written as OR (5) Now substituting the values of and in equation (3) (6) Equation (6) represents temperature distribution in the wall. Rate of heat transfer can be determined by using Fourier’s Law and can be expressed as : (7) Integrating equation (6) with respect to x to obtain the expression for temperature gradient  

27. Substituting the value of from above equation in equation (7) , we get (8) Equation (8) represents the heat transfer rate through the wall.  

28. One Dimensional Steady State Heat Conduction through a Cylinder : One-dimensional steady-state heat conduction through a cylinder involves the controlled and constant flow of heat along the radial direction of the cylindrical shape. Key points include : 1. Steady-state : Similar to the slab case, the temperature distribution within the cylinder remains constant with time. This implies that there is no change in the temperature profile as time progresses. 2. One-dimensional : Heat transfer primarily occurs in the radial direction, perpendicular to the length of the cylinder. This simplifies the analysis by focusing on a single dimension. 3. Cylinder : The material through which heat is conducted has a cylindrical shape. This could represent objects like pipes, rods, or other cylindrical structures.

29. Boundary Conditions for Heat Flow through a Cylinder : Inner Radii = a , Outer Radii = b, Length = H, Inner Surface Temp = , Outer Surface Temp = T = at r = a T = at r = b Upon integrating, we get i.e , (1)  

30. Upon integrating further, we get (2) By using the boundary conditions at r = a , b gives (3) (4) Subtracting equation (4) from equation (3) i.e , (5) Now using equation (5) in equation (1) , we get  

31. From the definition of heat flux, and from , we get = (6) where , Equation (6) represents the heat transfer rate through the cylinder.  

32. One Dimensional Steady State Heat Conduction through a Sphere : One-dimensional steady-state heat conduction through a sphere involves the controlled and constant flow of heat in the radial direction of a spherical object. Key Points Include : 1. Steady-state : Similar to the slab and cylinder cases, the temperature distribution within the sphere remains constant with time. There is no change in the temperature profile as time progresses. 2. One-dimensional : Heat transfer primarily occurs in the radial direction, from the center to the outer surface of the sphere. This simplifies the analysis by focusing on a single dimension. 3. Sphere : The material through which heat is conducted has a spherical shape. This could represent objects like a solid sphere or a spherical container.

33. Boundary Conditions for Heat Flow through a Sphere : Inner Radii = a , Outer Radii = b T = at r = a T = at r = b Upon integrating, we get i.e , (1)  

34. Upon integrating further, we get (2) Using the boundary conditions at r = a , b gives (3) (4) Subtracting equation (3) from equation (4) - = - = = i.e , =  

35. From the definition of heat flux, and from equation (1) , we get (5) Where , Equation (5) represents the heat transfer rate through the sphere.  

36. HEAT TRANSFER THROUGH COMPOSITE WALL AND INSULATED PIPE

37. HEAT TRANSFER THROUGH COMPOSITE WALL Consider a composite slab made of three different materials having conductivity k1, k2 and k3, length L1, L2 and L3 as shown in Figure 3. wall is exposed to a hot fluid having temperature T f and on the other side is atmospheric air at temperature T a . Convective heat transfer coefficient between the hot fluid and inside surface of wall is h i (inside convective heat transfer coefficient) and h o is the convective heat transfer coefficient between atmospheric air and outside surface of the wall (outside convective heat transfer coefficient). Temperatures at inner and outer surfaces of the composite wall are T 1 and T 4 whereas at the interface of the constituent materials of the slab are T 2 One side of the and T 3 respectively.

38. Heat is transferred from hot fluid to atmospheric air and involves following steps: (A) Heat transfer from hot fluid to inside surface of the composite wall by convection   (B) Heat transfer from inside surface to first interface by conduction   (C) Heat transfer from first interface to second interface by conduction

39. (D) Heat transfer from second interface to outer surface of the composite wall by conduction (E) Heat transfer from outer surface of composite wall to atmospheric air by convection Adding equations (1), (2), (3) and (5), we get  

40. If composite slab is made of ‘n’ number of materials, then equation (6) reduces to If inside and outside convective heat transfer coefficients are not to be considered, then equation (6) is expressed as

41. HEAT TRANSFRER THROUGH INSULATED PIPE Consider a composite cylinder consisting of inner and outer cylinders of radii r1, r2 and thermal conductivity k1, k2 respectively as shown in Figure 4. Length of the composite cylinder is L. Hot fluid at temperature T f is flowing inside the composite cylinder. Temperature at the inner surface of the composite cylinder exposed to hot fluid is T 1 and outer surface of the composite cylinder is at temperature T 3 and is exposed to atmospheric air at temperature T a . The interface temperature of the composite cylinder is T 2 . Convective heat transfer coefficient between the hot fluid and inside surface of composite cylinder is h i (inside convective heat transfer coefficient) and h o is the convective heat transfer coefficient between atmospheric air and outside surface of the composite cylinder (outside convective heat transfer coefficient). Heat is transferred from hot fluid to atmospheric air and involves following steps:

42. (A) Heat transfer from hot fluid to inside surface of the composite cylinder by convection (B) Heat transfer from inside surface to interface by conduction 1 2

43. (C) Heat transfer from interface to outer surface of the composite cylinder by conduction (D) Heat transfer from outer surface of composite wall to atmospheric air by convection 3 4

44. Adding both sides of equations (1), (2),(3) and (4), we get 5

45. If the composite cylinder consists of ‘n’ cylinders, then equation (5) can be expressed as: If inside and outside convective heat transfer coefficients are not to be considered, then equation (5) is expressed as   6 7

46. THERMAL RESISTANCE ANALOGY Consider heat flowing through a slab of thickness ‘L’ and area ‘A’ and T1 and T2 are the temperatures on the two faces of the slab as shown . Heat transfer from high temperature to low temperature side is expressed as -1 where (T1-T2) is the thermal potential kA/L is the thermal resistance. Now consider an electric circuit having resistance ‘R’ and electric potential E1 and E2 at the ends as shown . Current ‘I’ passing through the circuit can be expressed as -2  

47. Equations 1 and 2 are found to be symmetrical on comparison. ‘Q’ amount of heat flows through the slab having thermal resistance when a thermal potential (T1-T2) exits. Similarly, ‘I’ amount of current passes through the circuit having resistance ‘R’ when an electric potential (E1-E2) exists. Therefore, flow of heat through the slab can be represented by an electric circuit as shown below. If a hot gas at temperature Tg is in contact with one side of the slab and air at temperature Ta at the other side , then heat transfer from the hot gas to air through this slab of thickness can be represented by an electric circuit as shown.

48. Heat transfer from hot gas to air can be expressed as - 3 For a slab made of 3 material having thermal conductivities K1 , K2 and K3 respectively and is exposed to a hot gas on one side and atmospheric air on the other side as shown. Heat transfer from the hot gas to atmospheric air is expressed as - 4

49. Similarly for the composite slab shown below , an equivalent electric circuit has been shown here,

50. Plane and composite walls Numerical Practice

51. Problem 1 : Determine the loss of heat Q (watts) through a wall laid of lead brick of length 5 m, height 4 m and thickness 0.25 m, if the temperature of the surfaces of the wall are maintained at 110C & 40C. Assume thermal conductivity of brick = 0.7 watt/m-C. Solution λ thermal conductivity = 0.7 watt/m-C

δ = thickness = 0.25 m

∆t, temperature difference of surfaces = (110-40) = 70C

= 196 watt/m2

Q, loss of heat through 20 m 2 area of wall = q x A= 196 x 20 = 3920 watt

Q  3920 watt

52. Problem 2 : The inner surface of furnace wall is at 2000C and outer surface at 500C. Calculate the heat lost per m^2 area of the wall. If thermal conductivity of the brick is 0.5 W/m0C & the wall thickness is 200mm. solution Given T1 = 200°C , T2 = 50°C, L = 200mm = 0.2 , k = 0.5 w/m°c Heat loss = -ka dt/dx heat lost / m^2  -k(t2-t1)/L .  - 0.5(50-200)/0.2 .  375 W/m^2

53. Problem 3 : Heat transfer through a composite wall is shown in figure. Both the sections of the wall have equal thickness (l). The conductivity of one section is k and that of the other is 2k. The left face of the wall is at 600 K and the right face is at 300 K . The interface temperature Ti (in K) of the composite wall is ________ __? Solution When heat flows from a solid wall then total heat transfer from one face to another, Q =( t1 – t2)/(B1/k1a + B2/k2a) For interference temperature calculation, (t1 – t I)/(b1 /k1a) = ( Ti – t2)/(b2/k2a)........ eq 1 where, Ti = The interference temperature

Given:

k1 = k, k2 = 2k, T1 = 600 K, T2 = 300 K, b1 = b2, A1 = A2 From equation (1) (600 – ti )/ (1/k) =. ( ti – 300)/(1/2k)
⇒ Ti = 400 K

54. Problem 4 : Two surfaces of a plane wall of 15cm thickness and 5 m2 area are maintained at 240°C and 90°C respectively. Determine the heat transfer between the surfaces and temperature gradient across the wall if conductivity of the wall material is18.5 W/(m-K). Solution Given T1= 2 40°C, T2= 90°C, Thickness of wall, x = 15 cms .=0.15 m, Area, A= 5 m2 Thermal conducivity , k=18.5 W/(m-K) To determine: i ) Temperature gradient, , ii) Heat transfer rate, , i ) The temperature gradient in the direction of heat flow is (b) Heat flow across the wall is given by Fourier’s heat conduction equation =92500 W or 92.50 kW

55. Problem 5 : The bond between two plates, 2.5 cm and 15 cm thick, heat is uniformly applied through the thinner plate by a radiant heat source. The bonding epoxy must be held at 320 K for a short time. When the heat source is adjusted to have a steady value of 43.5 kW/m2, a thermocouple installed on the side of the thinner plate next to source indicates a temperature of 345 K. Calculate the temperature gradient for heat conduction through thinner plate and thermal conductivity of its material. Solution: t1 = 345 K ; t2 = 320 K δ = 2.5 cm = 0.025m # Temperature gradient, dt /dx = T2-T1/δ  320 – 345/0.025  - 1000°C/m

56. Hollow and composite cylinder

57. Problem 2 : A hollow cylinder 5 cm inner radius and 10 cm outer radius has inner surface temperature of 200°C and outer surface temperature of 100°C. If the thermal conductivity is 70 W/m K, find heat transfer per unit length. solution Given : inner radius, r1= 5 cm = 0.05 m Outer radius, r2= 10 cm = 0.1 m Inner surface temperature, T1= 200 + 273 = 473 K outer surface temperature, T2=100 + 273=373 K Thermal conductivity, k = 70 W/m K . to find:Heat flow per uniT length : solution : heat transfer through hollow cylinder is given by q= ∆t overall/r Where ∆t = t1-t2 and r= (1/2πlk) ln R2/r1  q= [t1-t2]/[(1/2πlk) Ln [ r2 / r1]]  q/l = 63453.04 w/m

58. Hollow and composite sphere

59. Problem 1 : If the inner and outer walls of a sphere having surface areas of A1 and A2 and inner and outer radii r1 and r2 are maintained at t1 and t2, then rate of heat flow will be ? Solution Heat conduction: The transfer of heat between two different bodies or two different locations of the same body through molecular vibrations is known as heat conduction.
The governing law for heat conduction is Fourier’s law of heat conduction. Fourier’s law of heat conduction: It states that ‘The rate of heat transfer is directly proportional to the temperature gradient’ and it is given as qk = -K × Dt/dx Where x = direction of heat flow, k = thermal conductivity, T = temperature
The heat conduction equation of the spherical wall is given as: Q= 4πk(t1-t2)R1r2/(R2-r1) So if we consider the surface areas at radius r1 and r2 then A1 = 4πr21 and A2 = 4πr22
q= K × (√4πR1^2)×(√4πr2^2 ) × T1 – t2/ R2 – R1
∴ The heat conduction in a sphere having surface areas of A1 and A2 and inner and outer radii r1 and r2 are maintained at temperatures t1 and t2 is given as: Q = K × √A1 × √a2 × T1-t2 / r2-r1

60. problem 2 : What will be the geometric radius of heat transfer for a hollow sphere of inner and outer radii r1 and r2? solution Heat transfer through the composite sphere is given by, Q = 4πkr1r2 (t1-t2)/ r2-r1 as Q = -K Am T2-t1/R2-r1 Am = 4πr^2M = 4πR1r2 . Therefore , Rm = √r1r2

61. Problem 3 : A spherical shaped vessel of 1.4 m outer diameter is 90 mm thick. Find the rate of heat leakage, if the temperature difference between the inner and outer surfaces is 220°C. Thermal conductivity of the material of the sphere is 0.083 W/m K. Solution R2 = 0.7 m
r1 = 0.7 – 0.09 = 0.61 m dT = 220°C, k = 0.083 W/ mK rate of heat leakage:  q= dt /r  r = r2-r1/ 4π(kr1r2)  0.09/4× 3.14× 0.083× 0.7× 0.61 . 0.20218  q= 220/2020.  1.088Kw

62. Heat Loss of conduction

63. Problem 1 : A plane brick wall, 25 cm thick, is faced with 5 cm thick concrete layer. If the temperature of the exposed brick face is 70°C and that of the concrete is 25°C, find out the heat lost per hour through a wall of 15 m x10 m. Also, determine the interface temperature. Thermal conductivity of the brick and concrete are 0.7W/ m.K and 0.95 W/m. Solution Heat loss Q = (Ta- Tb) / r where Ta = 70°c; Tb = 25°c
and R = R(brick)+ R(concrete)
R(brick) = L(brick)/( Ak ( brick) R (concrete)= L(concrete) /( Ak (Concrete) A = 15 x 10  150 m2 R(brick ) = 0.25/(150 x 0.7)  2.381 10^-3 °c/w r(concrete) = 0.05/(150×0.95)  3.509 × 10^-4 °c/w R = 2.381 × 10^-3 + 3.509 × 10^-4  2.732 × 10^-3 °c/w q = ( 70-25) / (2.732 × 10^-3)  16471.4 J/sec  59.3 Mj /hr heat loss per hour = 59.3 Mj at steady state , this is the amount of Heat transferred through the Brick wall or concrete per hour . Q = A k(brick) Ta – ti /l(brick) = 16471.4W 150× 0.7× (70-ti) / 0.25  16471.4 70- Ti  39.2 Ti = 70- 39.2  30.8 °c interface temperature ti = 30.8°c

64. Problem 2 : The wall of a boiler is made up of 250mm of the brick, KFB= 1.05 W/m K; 120 mm of insulation brick KIB = 0.15W/ mK , and 200 mm of red brick, KRB= 0.85 W/m K. The inner and outer surface temperatures of the wall are 8500C and 650C respectively. Calculate the temperatures at the contact surfaces. solution :- Q/a = Q= (t1-t4) x1 + _ X2 + X3_ . K(fb) K(m) K(rb) Q/a = 1×(850 -65) ( 0.250 + 0.120 + 0.200 ) 1.05. 0.15 0.85  616.5 W/m^2 Q/a = T1-T2 = T2-t3 = t3-t4 . X1 . X2 . X3 k(fb). K(ib). K(Rb)  Q/a = T1-T2  616.5  850 - t2 X1 0.250 . K(fb) 1.05  t2 = 703°c Similarly t3 can be determined From the relation:- q/a = T2-T3  616.5 = 703 – t3 X2 . 0.120 K(ib). 0.15 . T3 = 209.8°c

65. Problem 3 : A steam pipe of inner diameter 200 mm is covered with 50mm thick high insulated material of thermal conductivity k = 0.01 W/m0C. The inner and outer surface temperatures maintained at 5000C and 1000C respectively. Calculate the total heat loss per meter length of pipe? Solution:- Given R1 = 200/2 = 100mm = 0.1m R2 = 200+100/ 2 = 150mm = 0.15m K = 0.01W/ M°c t1 = 500°c , t2 = 100°c heat transfer is given by :- q = t1-t2 . = 500 – 100 ln[ R2/r1] ln[0.15/0.1] 2πk l 2π × 0.01× 1  61.98W .

66. Problem 4 : A spherical shaped vessel of 1.2 m diameter is 100 mm thick. Find the rate of heat leakage, if the temperature difference between the inner and outer surface is 200oC. Thermal conductivity of the material of sphere is 0.3 kJ/m-h °c. Solution Given , outside diameter Of sphere = D2 = 1.2 m = R2 = 0.6m Inside diameter of sphere  1.2 – 2 × 0.100 = 1m . R1= 0.5m T1-T2 = 200°c, K = 0.3kJ/m°C  heat transfer in hollow sphere is given by :- q = 4πk(r1)(R2)(t1-t2) R2-R1 =. 4π × 0.3× 0.5×0.6 ×200 . 0.100  2262Kj/hr

67. Problem 5 : A cylinder having its diameter 30 cm and length 60 cm, has hemispherical ends, giving an overall length of 90 cm. The cylinder which is maintained at a steady temperature of 60oC is covered to a depth of 5 cm with lagging which has a coefficient of conductivity of 0.14 W/ mK. Calculate the rate of heat loss if the outer surface of lagging is at 30oC. solution Rate of heat loss From cylindrical portion Q = 2πLK(t1-t2) Ln[r2/r1]  2×π×0.6×0.14(60-30)  55w ln(20/15) Rate of heat loss from spherical portion q = 4πk(r1)(R2)(t1-t2) (r2-r1)  4π×0.14×0.15×0.20(60-30) 0.20-0.15  31.667w Hence total heat loss 55 + 31.66  86.667 W

68. Problem 6: The meat rolls of 25 mm diameter having k=1 W/ moC are heated up with the help of microwave heating for roasting. The centre temperature of the rolls in maintained at 100oC when the surrounding temperature is 30oC. The heat transfer coefficient on the surface of the meat roll is 20 W/ moC . Find the microwave heating capacity required in W/m3. Solution Given , R = 0.0125M , k =1 w/M°C; T(max) = 100°c , t(a) = 30°c , H = 20 W/m^2°c Microwave heating capacity ,q(g): The maximum temperature occurs at the centre and is given by :- T(max) = t(a) + q(g) × r + q(g) × r^2 . 2h. 4k 100. =. 30 + Q(g) ×0.0125 + q(g) × 0.0125^2 . 2×20 4×1 q(g) = (100-30)_______ 0.0003125 + 0.00003906) . 1.991 × 10^5 w/m^3

69. Combined convection and radiation

70. Problem 1 : A surface is at 200°C and is exposed to surroundings at 60°C and convects and radiates heat to the surroundings. The convection coefficient is 80W/m2K. The radiation factor is one. If the heat is conducted to the surface through a solid of conductivity 12 W/ mK , determine the temperature gradient at the surface in the solid. Solution  Heat convected + heat radiated = heat conducted considering,1m^2 , h(T1 – T2) + sigma(T1^2 – T2^4) = – kdT/dx Therefore, 80(200 – 60) + 5.67 {[(200 + 273)/100]^4 – [(60 + 273)/100]^4} = – 12 dT/dx Therefore dT/dx = – (11200 + 2140.9)/12  – 1111.7°C/m

71. Problem 2 : Heat is conducted through a material with a temperature gradient of – 9000 °C/m. The conductivity of the material is 25W/ mK. If this heat is convected to surroundings at 30°C with a convection coefficient of 345W/m2K, determine the surface temperature. If the heat is radiated to the surroundings at 30°C determine the surface temperature. Solution : In this case only convection and conduction are involved. – kAdT/dx = hA(T1 – T2). Considering unit area, – 25 × 1 × (– 9000) = 345 × 1 (T1 – 30) Therefore, T1 = 682.17°C in this case conduction and radiation are involved. Heat conducted = Heat radiated – 25 × 1 × (– 9000) = 5.67 [(T1/100)^4 – (303/100)^4] Therefore, T1 = 1412.14K = 1139°C

72. Dimensionless numbers  Reynolds number  Prandtl number . Grashof Number  NussElt number

73. Problem 1 : - Calculate Reynolds number, if a fluid having viscosity of 0.4 Ns/m2 and relative density of 900 Kg/m3 through a pipe of 20 mm with a velocity of 2.5 m. Solution Given that, viscosity of the fluid (mu) mu = 0.4 ns . M^2 Density of fluid (Rho) Rho = 900 Kg/m^2 diameter of the fluid l = 20 × 10^-3 M. So,  900×2.5× 20× 10^-3 . 0.4  112.5 From the above answer, we observe tha t the Reynolds number value is less than 2000. Therefore, the flow of liquid is laminar.

74. Problem 2 : Determine the flow of fluid having a relative density of 100 kg/m3, the viscosity of 0.5 Ns/m2 with a velocity of 5 m/s through a pipe of 0.2 m. Solution Given :

Velocity of fluid, V=5 m/s
Diameter of pipe, D= 0.2 m
Relative density of fluid, p=100 kg/m^3
Viscosity of fluid=0.5 Ns/m^2 The formula of Reynolds number is given as : Re = pVD/u

Substitute all the values in the formula to calculate the Reynolds number . Re  (100 kg/m3)(5 m/s)(0.2 m)/(0.5 Ns/m2)  200 Since, the Reynolds number is less than 2000, the flow of liquid is laminar.

75. Problem 3 :Calculate the Reynolds number, Re, for oil flow in a circular pipe. The diameter of the pipe is 60 mm, the density of the oil is 910 kg/m3, the volumetric oil flow rate is 60 L/min, and the dynamic viscosity of the oil is 50 m Pa s. Given: Diameter of the pipe, D = 60 mm = 0.06 m.
Density of the oil, p = 910 kg/m3
Volume flow rate, Q = 60 L/min = 0.01 m3/s.
Dynamic viscosity, u=50 m Pa s =0.05 Pa s The area of the pipe is given as: A = π(D/2)2
= π(0.06/2)2
= 0.0283 m2 The formula for volume flow rate is given as: Q = Av 0.01 = 0.0283 × V V = 0.353 m/s
So, Re = pVD /u Re = (910 kg/m3)(0.353 m/s)(0.06 m)/(0.05 Pa s)  386 Thus, the Reynolds number of the flow is 386.

76. Problem 4: The Reynold’s number for the flow of a fluid in a horizontal circular tube of constant diameter is 1200. If the diameter of the tube and the kinematic viscosity of the fluid are doubled and that discharged at the pipe exit is unchanged, then the new Reynold’s number for the flow in the tube will be ? Given: Initially Reynold’s number, Re1 = 1200, D1 = D, D2 = 2 × D, (kinematic viscosity)2 = 2 × (kinematic viscosity)1 ⇒ v2 = 2×v1 Also, discharge at the pipe exit is unchanged. Q1=q2 So, A1×v1 = A2×v2 D1^2 × v1 = d2^2×v2 putting the value of D1 and D2 in the above equation, we get V2 = v1/4 Re = Vd/v re1/re2 = d1×v1 × . V2 = D×v1 ×. 2×v1 . V1. D2×v2. V1. 2×d×v1 . 4 Re(2) = Re(1)/4  300 So, New Reynold’s number for the flow in the tube will be 300.

77. Problem 5 : In a circular tube of diameter 100 mm and length 13 m with laminar flow, the friction factor is estimated to be 0.05. Calculate the Reynolds number? Solution Laminar flow Friction factor ⇒ F = 64/re Where Re is the Reynolds number. For laminar flow friction factor depends only upon the Reynolds number of the flow . Calculation :- Given , f = 0.05 0.05 = 64  Re = 64  1280 . Re. 0.05 .

78. Prandtl number In the case of boundary layer flow, the Prandtl number is given by, Pr =

79. Problem 1 : The water is flowing over the heated plate. The Prandtl number of water is 6. Find the relation between velocity boundary layer thickness and thermal boundary layer thickness. Solution Given : Pr = 6 The Prandtl number is given by, Therefore the velocity boundary layer thickness is 1.817 times the thermal boundary layer thickness.

80. Grashof number Problem 1 : Check the flow is laminar or turbulent. The properties of the fluid are given below, Characteristics length of body, L=6 m. Surface temperature = 650 K Fluid temperature= 450 K Kinematic viscosity= 0.2 * 10^-3 Solution ẞ = 2 = _. 2_____ Ts + t∞ 650 + 450 ẞ =. 1.81 × 10^-3 The Grashof number is given by, gr = GL^3ẞ(t s - t ∞ ) V 2 gr = 9.81 × 6 3. × (1.81 × 10^-3) (650-450) (0.2 × 10^-3)^2 Gr =. 1.917 × 10^10

81. Problem 2: The sphere at 600 K with a characteristic length of 5.5 m is cooled by the gas at 300 K by free convection. If the cooling gas is considered as ideal and if the kinematic viscosity is 0.25 × 10^-3, find the type of flow over the surface. Solution l = 5.5 m, ts = 600k, t∞ = 300k, v= 0.25 × 10^-3M^2/s ß = 2. =. 2.____ ts + T∞. 600 + 300 ß = 2.22 × 10^-3 For free convection, the Grashof number is given by, Gr = Gl^3ẞ(ts – t∞) V^2 Gr = 9.81 × 5.5^3 × ( 2.22× 10^-3) (600 – 300) (0.25 × 10^-3)^2 gr = 1.73 × 10^10

82. Nusselt number Problem 1 : at 1 atmospheric Pressure And 27°c blows accross a 12mm Diameter sphere At a small heater inside The sphere maintains The surface temperature at 77°c. With k = 0.026 w/mk And with nu = 31.4 , the heat Loss by the sphere would be ? Solution Given , nu = 31.4 ,K =0.026 w/mk , D = 12Mm  r= 6 × 10^-3 M , Ts = 27°c , t = 77°c nu = Hlc/k H = Nu × k/ Lc h = 31.4 × 0.026. . 68.03 w/m^2k . 0.012 surface area of sphere a = 4×π ×r^2 = 4×3.14 × (6×10^-3)^2 q(loss) = 68.03× 4× 3.14× (6×10^-3)^2 × (77-27)  1.54 j/s

83. Problem 2 :For a hydrodynamically and thermally fully developed laminar flow through a circular pipe of constant cross-section, the Nusselt number at constant wall heat flux ( Nuq ) and that at constant wall temperature ( NuT ) are related as Solution :- Part 1) For constant surface heat flux ( qs = constant) hydronamically and Thermally Fully developed laminar flow Through a circular pipe Of constant cross section nuq = 4.36  eq (1)

84. Part 2) For constant wall temperature ( tw = constant) for this case nut = 3.66  eq (2) Comparing (1) and (2) Nuq > nut

85. Introduction of condensation & Boiling Heat Transfer Introduction Boiling: Liquid – to – vapors phase change Like Evaporation Condensation: Gas phase into liquid phase Water vapors to liquid water when in contact with a liquid or solid surface Reverse of evaporation

86. BOILING Boiling occurs at the solid–liquid interface when a liquid is brought into contact with a surface maintained at a temperature sufficiently above the saturation temperature of the liquid . Evaporation occurs at the liquid–vapor interface when the vapor pressure is less than the saturation pressure of the liquid at a given temperature. Types- Nucleate Critical heat flux Transition Film

87. Boiling Curve

88. Classification Pool Boiling:- Motion of bubbles under the influence of buoyancy. Boiling is called pool boiling in the absence of bulk fluid flow. Flow Boiling:- Continuous flow of bulk fluid. Boiling is called flow boiling in the presence of bulk fluid flow .

89. Classification Subcooled Boiling:- When the temperature of the main body of the liquid is below the saturation temperature. Saturated Boiling:- When the temperature of the liquid is equal to the saturation temperature

90. POOL BOILING In pool boiling, the fluid is not forced to flow by a mover such as a pump. Any motion of the fluid is due to natural convection currents and the motion of the bubbles under the influence of buoyancy. Boiling Regimes and the Boiling Curve Boiling takes different forms, depending on the D t excess = T s - T sat

91. Pool Boiling Curve

92. Flow Boiling In flow boiling , the fluid is forced to move by an external source such as a pump as it undergoes a phase-change process. External flow boiling , over a plate or cylinder is similar to pool boiling, but the added motion increased both the nucleate boiling heat flux and the maximum heat flux considerably. Internal flow boiling , commonly referred to as two-phase flow, is much more complicated in natural because there is no free surface for the vapour to escape, and thus both the liquid and the vapour are forced to flow together.

93. Flow boiling curve

94. Film Boiling (beyond Point D) Beyond point D the heater surface is completely covered by a continuous stable vapor film . Point D, where the heat flux reaches a minimum is called the Leidenfrost point . The presence of a vapor film between the heater surface and the liquid is responsible for the low heat transfer rates in the film boiling region. The heat transfer rate increases with increasing excess temperature due to radiation to the liquid.

95. Condensation Condensation of vapor is the reverse phenomenon of the evaporation of liquid . Condensation (Vapor to Liquid): Heat energy is released Boiling (Liquid to Vapor): Heat energy is absorbed When a saturated vapor is brought in contact with a surface at a lower temperature, heat is received by the surface from the vapor and condensation occurs. The heat exchange is equivalent to the latent heat.

96. Condensation When does condensation occurs on a surface ? Consider a vertical flat plate which is exposed to a condensable vapour . If the temperature of the plate is below the saturation temperature of the vapour , condensate will form on the surface and flows down the plate due to gravity. It is to be noted that a liquid at its boiling point is a saturated liquid and the vapour in equilibrium with the saturated liquid is saturated vapour . A liquid or vapour above the saturation temperature is called superheated. If the non-condensable gases will present in the vapour the rate of condensation of the vapour will reduce significantly.

97. Types of condensation: Film-wise Condensation Drop wise Condensation

98. Film wise vs Drop-wise condensation Film wise condensation Drop-wise condensation The surface over which the steam condenses is wet-able and hence, as the steam condenses, a film of condensate is formed. The surface over which condensation takes place is non wet-able. In this mode, when steam condenses, the droplets are formed. Low heat transfer rates as the film of condensate impedes the heat transfer. High heat transfer rates are achieved and hence, many times, chemicals are used to ensure that condensation takes place drop wise. The thickness of the film formed depends on many parameters including orientation of the surface, viscosity, rate of condensation etc. When the drops become bigger, they simply fall under the gravity.

99. RADIATION

100. WHAT IS RADIATION Heat is transferred between two bodies even though they are not in direct physical contact and vacuum exists between these bodies. Such a heat transfer is termed as radiation. It involves the transfer of heat energy via electromagnetic waves. The electromagnetic radiation that can be detected as heat is termed as thermal radiation. Thermal radiation emitted by a body or substance is directly proportional to its temperature. Heat transfer by radiation occurs through open space and is most significant when temperature differences are substantial. For example: In food processing, heat is transferred by: Radiation: From the oven walls to the product surface.

101. Absorptivity, Transmissivity and Emissivity Absorptivity( α ) : refers to the ability of a material to absorb electromagnetic radiations. It is defined as the ratio of the absorbed radiant energy to the incident radiant energy. Materials with high absorptivity are efficient at absorbing radiation across a range of wavelengths. Transmissivity(T) : it is the measure of how much electromagnetic radiation can pass through a material. It is defined as the ratio of transmitted radiant energy to incident radiant energy. Transparent materials have high transmissivity whereas opaque materials have low transmissivity .

102. Emissivity( ε ) : describes the ability of a material to emit electromagnetic radiation. It is defined as the ratio of the radiant emittance of a particular surface to that of a particular surface to that of a blackbody at the same temperature. Materials with high emissivity radiate thermal energy more effectively than those with low emissivity TYPE ABSORPTIVITY TRANSMISSIVITY EMISSIVITY Black Body 1 1 Transparent body 1 LESS THAN 1 Opaque body LESS THAN 1 BETWEEN 0-1

103. BLACK BODY A perfect black body is one that absorbs all the thermal radiation, irrespective of wavelength, received by it. It does not reflect or transmit incident thermal radiation; therefore, absorptivity of such a body is 100%. At a given temperature and wavelength, a black body emits more energy as compared to any other body. A black body absorbs all incident radiation irrespective of their wavelength and direction. The radiation emitted by a black body depends upon wavelength and temperature, but it is independent of direction .

104. Net radiation heat exchange between black bodies Total emissive power for black body 1 per unit area: = W/m 2 K 4 Total emissive power for black body 1: ε 1 = A 1 T 1 4 For black body 2: ε 2 A 2 T 2 4 Total radiation leaving from surface 1 and falling on surface 2: Q 12 = A 1 T 1 4 F 12 α From surface 2-1: Q 21 = A 2 T 2 4 F 21 α As we are dealing with black bodies so whatever radiations coming on surface will absorb completely: ( α =1) [ Q 12 -Q 21 ]: NET RADIATION HEAT EXCHANGE Q net = A 1 T 1 4 F 12 - A 2 T 2 4 F 21 Q net = A 1 F 12 [T 1 4 - T 2 4 ]  

105. GRAY BODY A Gray body is defined as a surface whose emissivity is constant at all temperatures and throughout the entire range of wavelength. For a Gray body emissivity and absorptivity are independent of wavelength. Therefore, a Gray body, like black body, is an ideal body and values of its absorptivity and emissivity are less than unity. NET RADIATION HEAT EXCHANGE BETWEEN GREY BODIES- Radiation emitted by S1 in the form of electromagnetic waves. Radiation emitted by grey surface having emissivity ε 1 . = ε = ε 1 = ε × ε b ( ε b = A 1 T 1 4 ) = ε 1 = ε × A 1 T 1 4 Fraction of radiation emitted by surface 1 and absorbed factor α E 1 = ε 1 A 1 T 1 4 F 12 α 2 Similarly for surface 2, fraction of radiation is E 2 = ε 2 A 2 T 2 4 F 21 α 1  

106. Net radiation heat exchange between two surfaces : Q net = E 1 - E 2 ε 1 A 1 T 1 4 F 12 α 2 - ε 2 A 2 T 2 4 F 21 α 1 Q net = A 1 F 12 [ ε 1 T 1 4 α 2 = ε 2 T 2 4 α 1 ] According to kirchoff’s law of thermal equilibrium- α = ε Considering α 1 = ε 1 And α 2 = ε 2 Q net = A 1 F 12 [ ε 1 ε 2 T 1 4 = ε 2 ε 1 T 2 4 ] Q net = A 1 F 12 ε 1 ε 2 { T 1 4 - T 2 4 }  

107. BASIC LAWS OF THERMAL RADIATION STEFAN-BOLTZMAN LAW: It states that total energy emitted by black body per unit area and per unit time is directly proportional to the fourth power of its absolute temperature and is expressed as: b = 4 Eb- Energy emitted from a black body per unit area per unit time, W/m 2 b - Stefan- Boltzman constant =5.67 10 -8 W/(m 2 -K 4 ) T= Absolute temperature of the emitting surface, K. Total emissive power ‘E’ of a body is defined as total energy radiated by the body in all directions over entire range of wavelength per unit surface area per unit time. The total emissive power ‘E’ of a black body can be determined easily by using Stefan- Boltzman law if absolute temperature of the black body is known .  

108. PLANCK’S LAW: In 1900, German physicist Max Planck proposed Planck’s law. Planck’s constant h=6.625×10 -34 Joule-Sec. It describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T, when there is no net flow of matter or energy between the body and its environment. The higher the temperature of a body, the more radiation it emits at every wavelength. It relates the energy of electromagnetic radiation to its frequency. In the context of food processing, Planck’s law can be applied to various aspects, such as thermal processing, heating and cooling, which are essential in preserving food quality and safety.

109. Planck’s law of distribution equation: Where C 1 =2 c 2 h=37.404 10 17 J-m 2 /K C 2 =ch/k=1.4387 10 ⁻ 2 m-K An increase in temperature results in decrease in value of wavelength for which emissive power is maximum. In this figure, area under each curve represents energy emitted by the black body at a particular temperature for the range of wavelength considered. With increase in temperature area under the curve increases as energy emitted increases.  

110. Kirchoff’s Law: According to kirchoff’s law ratio of total emissive power to absorptivity is constant for all bodies which are in thermal equilibrium with their surroundings. For three bodies which are in thermal equilibrium with each other: = = equation(1) Assuming third body to be black body, equation(1) can be written as: = = [ ] → equation(2) = = → equation(3) However, according to definition of emissivity, = = → equation(4) Comparing equation(3) and equation(4) , we can write: = , = → equation(5) Therefore, kirchoff’s law states that for a body in thermal equilibrium with its surroundings, its absorptivity is equal to its emissivity.  

111. WEIN’S DISPLACEMENT LAW: It states that product of absolute temperature and wavelength at which emissive power is maximum, is constant. It has been established that monochromatic emissive power of a black body depends upon its temperature and wavelength of emitted radiations. For a given temperature, emissive power initially increases with increase in wavelength, attains a maximum value corresponding to particular wavelength of emitted radiations. With increase in temperature, maximum emissive power occurs at smaller wavelengths. Wein’s displacement law gives the value of wavelength at which emissive power of a body is maximum for a given temperature and is expressed as : = wavelength at which monochromatic power is maximum corresponding to temperature “T”.  

112. SHAPE FACTOR Shape factor, also known as form factor is a dimensionless quantity that characterizes the shape and size of a heat source, and it is used to determine the heat radiation emitted from the source. Heat radiation is the transfer of heat energy from one object to another through electromagnetic waves, which are mainly in the form of infrared radiation. The amount of heat radiation emitted from a surface is determined by its temperature, emissivity, and view factor.  The view factor is a dimensionless quantity that accounts for the geometric configuration of surfaces and their orientations, while emissivity is a measure of a surface’s ability to emit radiant energy. The shape factor is related to the view factor, and it is used to determine the heat radiation from non-uniform heat sources.  The shape factor, denoted as σ, is defined as the ratio of the heat transfer rate through a surface to the heat transfer rate through a reference surface of unit heat transfer coefficient and the same surface area. In other words, the shape factor is a measure of how efficiently a heat source transfers heat to its surroundings.

113. DETERMINATION OF SHAPE FACTOR There are three rules for the determination of the shape factor: 1.Summation 2.Reciprocity 3.Superposition The shape factor is purely a function of geometrical parameters. When two bodies radiating energy with each other only, the shape factor relation is expressed as : A 1 F 12 = A 2 F 21 F 12 = Heat emitted from Body 1 reached Body 2. The shape factor of convex surface or flat surface with the other surface enclosing the first is always unity. This is because all the radiation coming out from the convex surface is intercepted by the enclosing surface but not vice versa. (F 11 = 0) A concave surface has a shape factor with itself because the radiation energy coming out from one part of the surface is intercepted by another part of the same surface. The shape factor of a surface with respect to itself is denoted by F 11. If a surface of area A 1 is completely enclosed by a second surface of area A 2 and if A 1 does not see itself(F 11 =0) Then F 12 = 1.

114. Shape factor for spherical surface F 21 = 1 (inner to outer) F 11 + F 21 = 1 Reciprocity theorem: A 1 F 12 = A 2 F 21 F 12 = × F 21 F 11 + × F 21 = 1 × F 21 = 1 - F 11 F 11 = 1 - × F 21 (F 21 = 1) F 11 = 1-  

115. NUMERICALS QUESTION 1. A thin metal plate of 4 cm diameter is suspended in atmospheric air whose temperature is 290 K. the plate attains a temperature of 295 K when one of its face receives radiant energy from a heat source at the rate of 2 W. If heat transfer coefficient on both surfaces of the plate is stated to be 87.5 W/m 2 -deg, workout the reflectivity of the plates. Solution:         Heat lost by convection from both sides of the plate = 2hA t The factors 2 accounts for two sides of the plate =2 87.5 (295 – 290) = 1.1 W For most of solids, the transmissivity is zero. Energy lost be reflection = 2.0 – 1.1 = 0.9 W Reflectivity = = 0.45  

116. QUESTION 2. A black body of total area 0.045 m 2  is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m 2  and thermal conductivity 1.07 W/m-deg. If the inner surface of the enveloping wall is to be maintained at 215°C and the outer wall surface is at 30°C, calculate the temperature of the black body. Neglect the difference between inner and outer surfaces areas of enveloping material. Solution:    Net heat radiated by the black body to the enclosing wall,    Q r = b A(T b 4 – T w 4 ) = 5.67 -8 Where is the temperature of the black body is kelvin Heat conducted through the wall , Q c = = = 1979.5 W Under steady state conditions, the heat conducted through the wall must equal the net radiation loss from the black body. Thus = 5.67 Tb4 = + 488 4 = 8349.47 temperature of the black body, Tb = 955.9 K  

117. QUESTION 3.  A furnace radiation at 2000K. Treating it as a black body radiation, calculate the (i) Monochromatic radiant flux density at 1 μm wavelength. (ii) Wavelength at which emission is maximum and the corresponding radiant flux density (iii) Total emissive power Solution:   (a) From Planck’s law of distribution, (E = = = 2.81 Wavelength (b) From Wein’s displacement law: = 1.449   Maximum radiant flux density, 1.285×10⁻𝟓 T5 = 1.285×10⁻𝟓×(2000)5 =4.11×1011 𝑊/𝑚2 (c) From Stefan- Boltzman law, E= 𝝈𝒃 𝑻4 = 5.67×10⁻⁸×(2000)⁴=907200 𝑊/𝑚2

118. Maximum radiant flux density, 1.285 10 T5 = 1.285 5 (c) From Stefan- Boltzman law, E= 4 = 5.67 2  

119. QUESTION 4. A gray surface has an emissivity  at a temperature of 550 K source. If the surface is opaque, calculate its reflectivity for a black body radiation coming from a 550 K source. (b)         A small 25 mm square hole is made in the thin-walled door of a furnace whose inside walls are at 920 K. if the emissivity of the walls is 0.72, calculate the rate at which radiant energy escapes from the furnace through the hole to the room. Solution:      =1 Here, (i) (ii) This is in accordance with Kirchoff’s law which states that absorptivity equals emissivity under the same temperature conditions. Reflectivity Thus the surface reflects 65 percent of incident energy coming from a source at 550K. (b) The small hole acts as a black body and accordingly the rate at which radiant energy leaves the hole is E= 4 = = 25.38 watts  

120. Thank you