INDEX. Determination of overall heat transfer coefficient of Composite Wall. Determination of over all heat transfer coefficient of Lagged Pipe

Transcription

1 HEAT TRANSFER LAB

2 INDEX S.No. NAME OF EXPERIMENT PAGE No. 1 Determination of Thermal conductivity of insulation powder 2 Determination of overall heat transfer coefficient of Composite Wall 3 Determination of over all heat transfer coefficient of Lagged Pipe 4 Determination of Thermal Conductivity of given Metal Rod 5 Determination of heat transfer coefficient of Pin-Fin (Natural and Forced Convection) 6 Determination of heat transfer coefficient of Natural Convection 7 Determination of heat transfer coefficient of Forced Convection. 8 Determination of Stefan Boltzman Constant 9 Determination of Emissivity of test plate 10 Determination of effectiveness and overall heat transfer coefficient using Parallel and Counter flow Heat Exchanger 11 Determination of heat transfer coefficient in drop and film wise condensation 12 Determination of Critical Heat flux 13 Study of heat pipe and its demonstration

3 THERMAL CONDUCTIVITY OF INSULATING POWDER

4 AIM THERMAL CONDUCTIVITY OF INSULATING POWDER To determine the thermal conductivity of insulating powder at average temperature. INTRODUCTION Conduction of heat is flow of heat which occurs due to exchange of energy from one molecule to another without appreciable motion of molecules. In any heating process, heat is flowing outwards from heat generation point. In order to reduce losses of heat, various types of insulations are used in practice. Various powders e.g. asbestos powder, plaster of paris etc. are also used for heat insulation. In order to determine the appropriate thickness of insulation, knowledge of thermal conductivity of insulating material is essential. APPARATUS The apparatus consists of a smaller (inner) sphere, inside, which is fitted a mica electric heater. Smaller sphere is fitted at the center of outer sphere. The insulating powder, whose thermal conductivity is to be determined is filled in the gap between the two spheres. The heat generated by heater flows through the powder to the outer sphere. The outer sphere loses heat to atmosphere. The input to the heater is controlled by a dimmerstat and is measured on voltmeter and ammeter. Four thermocouples are provided on the outer surface of inner sphere and six thermocouples are on the inner surface of outer sphere, which are connected to a multi channel digital temperature indicator. SPECIFICATIONS 1. Inner sphere- 100mm O.D., halved construction 2. Outer sphere- 200mm I.D., halved construction

5 EXPERIMENTAL PROCEDURE 1. Keep dimmerstat knob at ZERO position and switch ON the equipment. 2. Slowly rotate the dimmerstat knob, so that voltage is applied across the heater. Let the temperatures rise. 3. Wait until steady state is reached. 4. Note down all the temperatures and input of heater in terms of volts and current. OBSERVAATIONS Sl. No. Temperatures 0 C Heat input T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 T 10 Volts Amps THEORY Consider the transfer of heat by heat conduction through the wall of a hallow sphere formed of insulating powder (Ref.fig.) Let, r i = raidus of inner sphere, m r 0 = radius of outer sphere, m T i = average inner sphere surface temp. O C T 0 = average outer sphere surface temp. O C Consider a thin spherical layer of thickness dr at radius r & temperature difference of dt across the layer. conduction, heat transfer rate, Q = -k. 4π. r 2. [dt/dr] Where k = thermal conductivity of insulating powder. Q dr Therefore x = dt 2 4πk r Integrating between r i to r o and T i to T o, we get Q 4πk r0 ri T 0 dr = dt 2 r Ti Applying Fourier law of heat

6 Q 1 x 4πk r or i 1 r 0 = (T T ) 4 kri ro (Ti To ) Q = π (r r ) From the measured values of Q, T i and T 0 thermal conductivity of insulating powder can be determined as Q (r0 ri ) k = 4 π.r r.(t T ) i. o i o 0 i i 0 CALCULATIONS 1. Heater input = Q = V x I Watts 2. Average inner sphere surface temperature, Ti T1 + T 2 + T 3 + T 4 = 4 0 C 3. Average outer sphere surface temperature T 0 T = 5 + T 6 + T T 6 4. Inner sphere radius = 50 mm = 0.05 m 5. Outer sphere radius = 100 mm = 0.1 m. 10 Q(r0 ri ) Ti + To o Now k = W / m K at C 4 π.r r.(t T ) 2 i. o i o PRECAUTIONS 1. Operate all the switches and controls gently. 2. If thermal conductivity of the powder other than supplied is to be determined then gently dismantle the outer sphere and remove the powder, taking care that heater connections and thermocouples are not disturbed. 3. Earthling is essential for the unit. RESULT Thermal conductivity of insulating powder is at temperature of

7 Fig. 1 Apparatus of thermal conductivity of insulating powder 1. Shell 2. Voltmeter 3. Ammeter 4. Temperature indicator 5. Selector switch 6. Main switch 7. Heater control Fig. 2 Location of thermocouples in spherical shell

8 COMPOSITE WALL APPARATUS

9 COMPOSITE WALL APPARATUS AIM To determine the thermal resistance, thermal conductivity of composite wall material and plot temperature gradient along composite wall structure. APPARATUS The apparatus consists of a plates of different materials sandwiched between two aluminum plates. Three types of slabs are provided on both sides of heater, which forms a composite structure. A small hand press frame is provided to ensure the perfect contact between the slabs. A dimmerstat is provided for varying the input to the heater and measurement of input is carried out by a Voltmeter and Ammeter. Thermocouples are embedded between interfaces of input slabs, to read the temperatures at the surface. SPECIFICATIONS Slab size: a. M.S. - ϕ 25 cm x 25 mm thick b. Bakelite - ϕ 25 cm x 10 mm thick c. Brass - ϕ 25 cm x 10 mm thick EXPERIMENTAL PROCEDURE 1. Start the supply of heater. By varying the dimmerstat adjust the input (range Watts) and start water supply. 2. Take readings of all the thermocouples at an interval of 10 minutes until steady state is reached. WALL THICKNESS CONDUCTIVITY a. M.S 2.5 cm 46 W / m K b. Bakelite 1.0 cm W / m K c. Brass 1.0 cm 110 W / m K

10 OBSERVATIONS Sl.No. Heat Supplied (Watts) Temperatures 0 C Voltmeter Ammeter T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 CALCULATIONS 1) Mean readings, ( T 2 a) ) 1 + T T A = 0 C 2 b) c) ( T ) 3 + T4 T B = 0 C 2 ( T ) 5 + T6 T C = 0 C 2 ( T 8 d) ) 7 + T T D = 0 C 2 2) Rate of heat supplied, Q = V x I Watts For calculating the thermal conductivity of composite walls, it is assumed that due to large diameter of the plates, heat flowing through central portion is unidirectional i.e. axial flow. Thus for calculations central half diameter area where unidirectional flow is assumed is considered. Accordingly thermocouples are fixed at close to center of the plates. Now, Heat flux, Q q = A 2 Watt / m 2 A = π x d ( d = half dia. of plates) 4 Total thermal resistance of composite slab: R TA TD 2 o total = m K / W q q. b i) Thermal conductivity of composite slab, K composite = W / m / k T T b = Total thickness of composite slab = m ii) Plot thickness of slab material against temperature gradient. A D

11 PRECAUTIONS 1. Keep the dimmerstat zero before start 2. Increase voltage slowly 3. Keep all the assembly undisturbed 4. Do not increase voltage above 200 V 5. Operate selector switch of temperature indicator slowly. GRAPH T A T B Temperature, 0 C Mild Steel Bakelite T C Brass T D Thickness of slab RESULT Thermal resistance of composite wall = Over all thermal conductivity of composite wall =

12 Fig. 1 Composite wall apparatus 1. Voltmeter 2. Ammeter 3. Temperature indicator 4 Main switch 5. Heater Control 6. Water connection Fig. 2 Thermocouple setting in composite wall apparatus A. Heater B & B1 M.S. Plate C & C1 Bakelite D & D1 Brass plate E & E1 Cooling plate

13 LAGGED PIPE

14 AIM LAGGED PIPE 1. To determine heat flow rate through the lagged pipe and compare it with the heater input for known value of thermal conductivity of lagging material. 2. To determine the approximate thermal conductivity of lagging material by assuming the heater input to be the heat flow rate through lagged pipe. 3. To plot the temperature distribution across the lagging material. APPARATUS The apparatus consists of three concentric pipes mounted on suitable stand. The hollow space of the innermost pipe consists of the heater. Between first two cylinders the insulating material with which lagging is to be done is filled compactly. Between second and third cylinders, another material used for lagging is filled. The third cylinder is concentric to other outer cylinder. Water flows between these two cylinders. The thermocouples are attached to the surface of cylinders appropriately to measure the temperatures. The input to the heater is varied through a dimmerstat and measured on voltmeter and ammeter. SPECIFICATIONS Pipes i) GI pipe inside Φ6 cm. (O.D) ii) GI pipe middle Φ8.5 cm. (Mean dia.) iii) GI pipe outer Φ10.7 cm. (I.D) iv) Length of pipes 1 meter. PROCEDURE 1) Start the supplies of heater and by varying dimmerstat adjust the input for desired value (range 60 to 120 Watts) by using voltmeter and ammeter, also start water supply.

15 2) Take readings of all the 6 thermocouples at an interval of 5 min until the steady state is reached. 3) Note down steady readings in observation table. OBSERVATIONS 1. Inner pipe O.D., D 1 = 0.06 m 2. Middle pipe mean dia., D 2 = m 3. Outer pipe I.D., D 3 = m Sl.No. Volt meter Ammeter Thermocouple Readings V I T 1 T 2 T 3 T 4 T 5 T 6 CALCULATIONS Mean readings T T T inside middle Outer T T = T = r i = Inner pipe radius = 0.03 m = T2 2 + T 2 + T r o = outer pipe radius = m r m = mean radius of middle pipe = m L = Length of pipe = 1m k = thermal conductivity W / m K Q= actual heat input = V x I Watts o Assumption: The pipe is so long as compared with diameter that heat flows in radial direction only middle half length. i) Now, first we find out the theoretical heat flow rate through the composite cylinder Q = 1 2 π L 1 k 1 T Log inside e o o C C C T rm + r k 1 = 0.22 W/m o K & k 2 = 0.13 W/m 0 K, where Actual heat input, Q act = V.I. i outside 1 k 2 Log e r r 0 m

16 ii) Now from known value of heat flow rate, value of combined thermal conductivity of lagging material can be calculated. Q act = 2 π Lk log ( Ti To ( r / r ) e 0 i ) Qact loge (r0 / ri ) k = W / mk 2πL(Ti To) The space between the pipes of φ6 cm and φ 8.5 cm contains commercial asbestos powder and the space between pipes of φ8.5 cm and φ10.5 cm contains saw dust. k k 1 2 = = Q act log 2 π L ( T Q act log 2 π L ( T e i e m ( r m T m / r ) ( r o / r m T ) o i ) W / m ) W / m iii) To plot the temperature distribution use formula o o K K. T T T o i T i V / S Log Log e e (r / r ) (r o i / r ) i Where r is the selected radius for corresponding to temperature T, between the two pipes of the same lagging material. Thus plot is made for different values of r. PRECAUTIONS 1) Keep dimmerstat to ZERO position before start. 2) Increase voltage gradually. 3) Keep the assembly undisturbed while testing. 4) While removing or changing the lagging materials do not disturb the thermocouples. 5) Do not increase voltage above 150V 6) Operate selector switch of temperate indicator gently. RESULTS Theoretical Heat flow rate = Thermal conductivity of lagged material = Thermal conductivity of asbestos powder = Thermal conductivity of saw dust =

17 Fig 1 Lagged pipe apparatus 1. Voltmenter 2. Ammeter 3. Temperature indicator 4. Selector switch 5. Main switch 6. Heater control 7. Assembly Fig. 2 Thermocouple settings in lagged pipe

18 Fig.3 Graph of temperature gradient

19 THERMAL CONDUCTIVITY OF METAL ROD

20 AIM THERMAL CONDUCTIVITY OF METAL ROD To determine the thermal conductivity of copper bar at various sections to study the variation of thermal conductivity with temperature. INTRODUCTION Thermal conductivity is the physical property of the material denoting the ease with which a particular substance can accomplish the transmission of thermal energy by molecular motion. Thermal conductivity of a material is found to depend on the chemical composition of the substance or substances of which it is a composed, the phase (i.e. gas, liquid or solid) in which it exists, its crystalline structure if a solid, the temperature and pressure to which it is subjected, and whether or not it is a homogeneous material. For pure copper thermal conductivity is 380 W/ m. K at 20 0 C. Thermal energy can be conducted in solids by free electrons and by lattice vibrations. Large number of free electrons moves about in the lattice structure of the material, in good conductors. These electrons carry thermal energy from higher temperature region to lower temperature region, in a similar way they transport electric charge. In fact, these electrons are frequently referred as electron gas. Energy may also be transferred as vibrational energy in the lattice structure of the material. In general, however, this mode of energy transfer is not as large as electron transport and hence, good electrical conductors are always good heat conductors, e.g. copper, silver etc. However, with increase in temperature, lattice vibrations come in the way of transport by free electrons and for most the metals thermal conductivity decreases with increase in temperature. APPARATUS The apparatus consists of a copper bar, one end of which is heated by an electric heater and the other end is cooled by a water-circulated heat sink. The middle portion, i.e. Test section of the bar is covered by a shell

21 containing insulation. The bar temperature is measured at 8 different section, while 2 thermocouples measure the temperature at the shell. Two thermometers are provided to measure water inlet and outlet temperatures. A dimmer is provided for the heater to control its input. Constant water flow is circulated through the heat sink. A gate valve provided controls the water flow. SPECIFICATIONS 1. Metal bar copper, 25mm O.D, approx. 430 mm long with insulation shell along the test length and water cooled heat sink at the outer end. 2. Test length of the bar 240 mm 3. Measuring flask to measure water flow. EXPERIMENTAL PROCEDURE 1. Start the electric supply. 2. Start heating the bar by adjusting the heater input to say 80 V or 100 V 3. Start cooling water supply through the heat sink and adjust it to around cc per minute. 4. Bar temperature will start rising. Go on checking the temperatures at time intervals of 5 minutes. 5. When all the temperatures remain steady, note down all the observations and complete the observation table. OBSERVATION TABLE Sl. No. TEST BAR TEMPERATURE O C Shell Temp. o C Water Temp. o C Water flow rate Lit/Sec. T 1 T 2 T 3 T 4 T 5 T 6 T 8 T 9 T 10 T 11 T 12 Using the temperatures of the bar at various points, plot the temperature distribution along the length of the bar and determine the slopes of the

22 graph (i.e. temperature drop per unit length) dt/dx at the sections AA, BB and CC as shown in figure. (Note: As the value of temperature goes on decreasing along the length of the bar, the value of the slope dt/dx is negative) CALCULATIONS Heat is flowing through the bar from heater end to water heat sink. When steady state is reached, heat passing through the section CC of the bar is heat taken by water. 1) Heat passing through Section CC Where, Q cc = m. C P ΔT Watts. m = mass flow rate of cooling water, kg/s. C P = Specific heat of water= 4180 J / kg o C Δ T = (Water outlet temperature) (Water inlet temperature) dt Now, Q cc = -k cc dx cc π 2 A = Cross sectional area of the bar = 4 D K cc = W / m o C 2) Heat passing through section BB Q bb = Q cc + Radial heat loss between CC & BB. 2 k. L1 ( T6 T10 ) = Q cc + log ( r / ) e 0 r i.a Where k = Thermal conductivity of insulation = 0.35 W / m o C L 1 = Length of insulation cylinder = m r o = outer radius =0.105 m r i = inner radius = m dt Q bb = -k bb. dx.a bb K bb = W/m o C

23 3) Heat passing through section AA Q aa = Q bb + Radial heat loss between BB & AA. 2 k. L2 ( T3 T9 ) = Q bb + log ( r / ) Where,L 2 = m e 0 r i dt Q aa = -k aa. dx aa.a K aa = W/m o C RESULTS 1) Temperature of the bar decreases from hot end to cool end, which satisfies the Fourier law heat conduction. 2) Thermal conductivity of bar at three different sections. Thermal conductivity at section AA = k aa = Thermal conductivity at section BB = k bb = Thermal conductivity at section CC = k cc =

24 Fig.1 Apparatus of thermal conductivity of metal rod 1. Shell 2. Heater 3. Voltmeter 4. Ammeter 5. Temperature indicator 6. Main switch 7. Heater control Fig.3 Thermocouple settings in metal rod Distance between two thermocouple = m Fig.3 Graph for temperature gradient in metal rod

25 HEAT TRANSFER IN PIN FIN

26 AIM HEAT TRANSFER IN PIN FIN To study the temperature distribution, heat transfer coefficient and efficiency of a pin fin in natural and forced convection heat transfer. INTRODUCTION Extended surfaces or fins are used to increase the heat transfer rates from a surface to the surrounding fluid wherever it is not possible to increase the value of the surface heat transfer coefficient or the temperature difference between the surface and the fluid. Fins are fabricated in variety of forms. Fins around the air cooled engines are a common example. As the fins extend from primary heat transfer surface, the temperature difference with the surrounding fluid diminishes towards the tip of the fin. APPARATUS The apparatus consists of a simple pin fin which is fitted in a rectangular duct. The duct is attached to suction end of a blower. One end of fin is heated by an electrical heater. Thermocouples are mounted along the length of fin and a thermocouple notes the duct fluid temperature. When top cover over the fin is opened and heating started, performance of fin with natural convection can be evaluated and with top cover closed and blower started, fin can be tested in forced convection. SPECIFICATIONS 1) Fins 12 mm O. D., Effective length 102 mm with 5 nos of thermocouple positions along the length, made of brass, mild steel and aluminum - one each. Fin is screwed in heater block which is heated by a band heater. 2) Duct- 150 x 100mm cross-section, 1000mm long connected to suction side of blower 3) FHP centrifugal blower with orifice and flow control valve on discharge side. 4) Orifice dia 22mm, coefficient of discharge C d = ) Water manometer connected to orifice meter

27 THEORY Let A= Cross sectional area of the fin, m 2 P= Perimeter (circumference) of the fin, m L= Length of the fin = m T 1 = Base temperature of fin T f = Duct fluid temperature (Channel No. 6 of temperature indicator) θ = Temperature difference of fin and fluid temperature =T- T f h = Heat transfer coefficient, W / m 2 o C K f = Thermal conductivity of fin material = 110 W / m o C for brass = 46 W / m o C for mild steel = 232 W / m o C for aluminum Heat is conducted along the length of fin and also lost to surroundings. Applying first law of thermodynamics to a control volume along the length of fin at a station which is at length x from the base 2 d T 2 dx h. P θ = 0 k. A f (1) φ = ( 1 2 C. e mx ) + ( C. e mx ) (2) h. P Where m = (3) k A f. With the boundary conditions of θ = θ 1 at x = 0, θ 1 = T 1 -T f dθ Assuming tip is to be insulated, = 0 at x = L, dx Results in obtaining equation (2) in the form θ θ T T Cosh [ m( L X )] = Cosh [ m. ]. f = 1 T1 T f L (4) This is the equation for temperature distribution along the length of the fin. Temperatures T 1 and T f will be known for the given situation and the value of h depend upon mode of convection i.e. natural or forced.

28 EXPERIMENTAL PROCEDURE Sl. No. INPUT Manometer difference Fin Temperature o C Duct fluid temp. o C V I H (m of water) T 1 T 2 T 3 T 4 T 5 T 6 (T f ) A) NATURAL CONVECTION Open the duct cover over the fin. Ensure proper earthing to the unit and switch on the main supply. Adjust dimmerstat so that about 80 V are supplied to the heater. The fin will start heating. When the temperatures remain steady, note down the temperatures of the fin and duct fluid temperature. Sl. INPUT Fin temperatures o C NO. Duct fluid temperature V I T 1 T 2 T 3 T 4 T 5 T 6 (T f ) o C B) FORCED CONVECTION Close the duct cover over the fin. Start the blower. Adjust the dimmerstat so that about v are supplied to the heater. When the temperatures become steady, note down all the temperatures and manometer difference CALCULATIONS Nomenclature: T m = Average fin temperature = (T 1 + T 2 + T 3 + T 4 +T 5 ) /5 Δ T = T m T f T mf = Mean film temperature = (T m + T f ) / 2 ρ a = Density of air, kg / m 3 ρ w = Density of water, kg / m 3 = 1000 kg / m 3 D = Diameter of pin fin = 12 x 10-3 m d = Diameter of orifice = 22 x 10-3 m

29 C d = coefficient of discharge of orifice = 0.64 μ = Dynamic viscosity of air, N-s/m 2 C p = Specific heat of air, kj/kg.k Kinematic viscosity, m 2 /s =ט k air = Thermal conductivity of air, W/m K β = volume expansion coefficient = 1 / (T mf ) H = Manometer difference, m of water V = velocity of air in duct, m/s Q = volume flow rate of air, m 3 /s V tmf = velocity of air at mean film temperature All properties are to be evaluated at mean film temperature. NATURAL CONVECTION The fin under consideration is horizontal cylinder losing heat by natural convection. For horizontal cylinder, Nusselt number, from data book, page number 122. Nu= 1.02 (Gr.Pr) for 10-2 < Gr.Pr < 10 2 Nu = 0.85 (Gr.Pr) for 10 2 < Gr.Pr < Nu= 0.48 (Gr.Pr) for 10 4 < Gr.Pr < 10 7 Nu = (Gr.Pr) for 10 7 < Gr.Pr < g. β.d ΔT Where Gr = Grashof number. = 2 ν Cp. μ Pr = Prandtl number = k air Determine Nusselt number. Now, Nu = (hd)/k air Therefore, h = Nu. K air /D From h determine m from equation (3) (take from data book.) Using h and m, determine temperature distribution in the fin from equation (4) The rate of heat transfer from the fin and efficiency can be calculated as,

30 Q fin tanh [ ml ] η = = h P. k. A ( T T ) and ml. f 1 f FORCED CONVECTION For flow across Horizontal cylinder loosing heat by forced convection, from data book, page number 100. Nu = (Re) Pr for 4 < Re < 40 Nu = (Re) Pr for 40 < Re < 4000 Nu= (Re) Pr for 4000 < Re < 40,000 Vtmf. D Where, Re = ν V. ( Tmf + 273) V tmf = ( T + 273) f Velocity of air is determined from air volume flow. π 2 3 Q = Cd d 2. g. H ( ρ w / ρ a ) m / s 4 V = Q / Duct cross sectional area = Q /(0.15 X 0.1) m / s From Nusselt Number, find out h and from h, find out m Now temperature distribution, heat transfer rate and effectiveness of the fin can be calculated using equations 4, 5 and 6 respectively. CONCLUSION 1. Comment on the observed temperature distribution and calculation by theory, it is expected that observed temperatures should be slightly less than their calculated values because of radiation and non- insulated tip. 2. Plot the graphs of temperature distribution in both natural and forced convection. PRECAUTIONS 1. Operate all the switches and controls gently 2. Do not obstruct the suction of the duct or discharge pipe 3. Open the duct cover over the fin for natural convection experiment

31 4. Fill up water in the manometer and close duct cover for forced convection experiment 5. Proper earthing to the unit is necessary 6. While replacing the fins, be careful for fixing the thermocouples. Incorrectly fixed thermocouples may show erratic readings T 1 T 2 T 3 T 4 T Fig.1: Thermocouple position on fin GRAPH Fin Temperature, 0 C Natural Convection Forced Convection Thermo couple distance, x Fig 2: Variation of fin temperature along the length of fin with natural convection and forced convection. RESULTS Natural convection: Heat transfer coefficient = Efficiency of pin fin = Forced convection: Heat transfer coefficient =

32 Efficiency of pin fin = Fig. 3 Pin fin apparatus 1. Manometer 2. Ammeter 3. Voltmeter 4. Temperature indicator 5. Selector switch 6. Blower switch 7. Heater control 8. Main switch 9. Suction duct 10. Orifice meter

33 HEAT TRANSFER IN PIN FIN

34 AIM HEAT TRANSFER IN NATURAL CONVECTION To determine the experimental and theoretical heat transfer coefficient for vertical tube losing heat by natural convection. INTRODUCTION In contrast to the forced convection, natural convection phenomenon is due to the temperature difference between the surface and the fluid and is not created by any external agency. The present experimental set up is designed and fabricated to study the natural convection phenomenon from a vertical cylinder in terms of the variation of local heat transfer coefficient along the length and also the average heat transfer coefficient and its comparison with the value obtained by using and appropriate correlation. APPARATUS The apparatus consists of a brass tube fitted in a rectangular vertical duct. The duct is open at the top and bottom and forms an enclosure and serves the purpose of undisturbed surrounding. One side of the duct is made up of perspex for visualization. An electric heating element is kept in the vertical tube which in turn heats the tube surface. The heat is lost from the tube to the surrounding air by natural convection. The temperature of the vertical tube is measured by seven thermocouples. The heat input to the heater is measured by an ammeter and a voltmeter and is varied by a dimmerstat. The vertical cylinder with the thermocouple positions is shown in figure. The tube surface is polished to minimize the radiation losses. SPECIFICATIONS 1. Diameter of the tube (d)= 38mm 2. Length of tube (L) = 500mm 3. Duct size 200mm x 200mm x 800mm Length

35 THEORY When a hot body is kept in still atmosphere, heat is transferred to the surrounding fluid by natural convection. The fluid layer in contact with the hot body gets heated, rises up due to the decrease in its density and the cold fluid rushes in to take place. The process is continuous and the heat transfer takes place due to the relative motion of hot and cold fluid particles. The heat transfer coefficient is given by: Q QR h = (1) A (T T ) s s a Where h = average surface heat transfer coefficient (W/m 2 o C) Q = Heat transfer rate V. I (watts) A s = Area of the heat transferring surface =.d L (m 2 ) ( T1 + T2 + T3 + T4 + T5 + T6 + T7 ) T s = Average surface temperature = 0 C 7 T a = Ambient temperature in the duct = T 0 8 C 4 4 Q R = Heat loss by radiation = σ. A. ( T s T a ) Where σ. = Stefan Boltzmann constant = 5.667x 10-8 W/m 2 K 4 = Emissivity of pipe material = 0.06 T s & T a = Surface and ambient temperatures in o K respectively. The surface heat transfer coefficient, of a system transferring heat by natural convection depends on the shape, dimensions and orientation of the fluid and the temperature difference between heat transferring surface and the fluid. The dependence of h on all the above mentioned parameters is generally expressed in terms of non-dimensional groups as follows: 3 hxl g. L β. ΔT = A 2 k υ x Cρμ K n (2) Where hxl is called the Nusselt number. k g. L 3. βδt = is called the Grashof Number and 2 v

36 Cρ. μ = is the Prandtl Number. k A and n are constants depending on the shape and orientation of the heat transferring surface. Where L = A characteristic dimension of the surface. K= Thermal conductivity of fluid Kinematic viscosity of fluid =ט μ = Dynamic viscosity of fluid C p = Specific heat of fluid β = Coefficient of volumetric expansion for the fluid g = Acceleration due to gravity. Δ T = [T s T a ] 1 For gases β = K -1, T f = (T s + T a )/2 ( T f + 273) For a vertical cylinder losing heat by natural convection, the constants A and n of equation (2) have been determined and the following empirical correlation s obtained from data book, page number 120. hl = 0.59 (Gr.Pr.) 0.25 for 10 4 < Gr.Pr. < 10 9 (3) k hl = 0.10 (Gr.Pr.) 1/3 for 10 9 < Gr.Pr. < (4) k L = Length of the cylinder. All the properties of the fluid are determined at the mean film temperature (T f ) PROCEDURE 1. Put ON the supply and adjust the dimmerstat to obtain the required heat input (say 40 W, 60 W, 70 W etc). 2. Wait till the steady state is reached, which is confirmed from temperature readings ( T 1 - T 7 ) 3. Measure surface temperature at the various points i.e. T 1 to T 7 4. Note the ambient temperature i.e. T 8

37 OBSERVATIONS 1) O.D. Cylinder = 38 mm 2) Length of cylinder = 500 mm 3) Input to heater = V. I Watts Temperature, 0 C Sl.No. Volt Amp T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 CALCULATIONS 1) Calculate the value of average surface heat transfer coefficient neglecting end losses using equation (1) 2) Calculate and plot (Fig.4) the variation of local heat transfer coefficient along the length of the tube using: T = T 1 to T 7 and h = Q/[A s (T-T a )] 3) Compare the experimentally obtained value with the predictions of the correlation equation (3) or (4). PRECAUTIONS 1. Proper earthing is necessary for the equipment. 2. Keep dimmerstat to ZERO volt position before putting on main switch and increase it slowly. 3. Keep at least 200 mm space behind the equipment. 4. Operate the change-over switch of temperature indicator gently from one position to other, i.e. from 1 to 8 positions. 5. Never exceed input above 80 Watts. RESULTS AND DISCUSSIONS The heat transfer coefficient is having a maximum value at the beginning as expected because of the just staring of the boundary layer and it decreases as expected in the upward direction due to thickening of layer and which is laminar one. This trend is maintained up to half of the lengths (approx) and beyond that there is little variation in the value of local heat transfer coefficient because of the transition and turbulent boundary layers. The last point shows some what increase in the value

38 of heat transfer coefficient which is attributed to end loss causing a temperature drop. The comparison of average heat transfer coefficient is also made with predicted values are some what less than experimental values due to the heat loss by radiation. RESULTS Experimental heat transfer coefficient = Theoretical heat transfer coefficient = Fig. 1 Natural convection apparatus

39 Fig. 2 Thermocouple locations Fig.3 Variation of local heat transfer coefficient

40 HEAT TRANSFER IN FORCED CONVECTION

41 HEAT TRANSFER IN FORCED CONVECTION AIM To determine the experimental and theoretical heat transfer coefficient in forced convection heat transfer for internal flow. APPARATUS The apparatus consists of a circular pipe, through which cold fluid, i.e. air is being forced. Pipe is heated by a band heater outside the pipe. Temperature of pipe is measured with thermocouples attached to pipe surface. Heater input is measured by a voltmeter and ammeter. Thus, heat transfer rate and heat transfer coefficient can be calculated. SPECIFICATIONS Test pipe 33mm I.D. 500 mm long Band heater for pipe. Blower to force the air through test pipe Orifice meter with water manometer. EXPERIMENTAL PROCEDURE 1. Put ON main supply. 2. Adjust the heater input with the help of dimmerstat. 3. Start the blower and adjust the air flow with valve 4. Wait till steady state is reached and note down the reading in the observation table Sl. No Vol t Amp. Temperatures O C Manomete r Difference V I T 1 T 2 T 3 T 4 T 5 T 6 T 7 h w

42 CALCULATIONS 1. Air inlet temp T 1 = 0 C 2. Air outlet temp. T 7 = 0 C 1.293x Density of air, ρ a = T kg/m 3 4. Diameter of orifice = 22 mm Manometer difference = Water head = h w meters Air head, h a = h w (ρ w / ρ a ) Where ρ w = density of water = 1000 Kg/m 3 Air volume flow rate, i.e., discharge Q = C a 2 m 3 /sec. d 0 gh a Where C d = 0.64 a o = cross section area of orifice 5. Mass flow rate of air, m a = Q x ρ a kg/sec. Velocity of air, Q V = m/sec. a ρ 1 Where, a ρ = Cross Sectional area of pipe = m 2 6. Heat gained by air, Q air = m a x C pa x (T 7- T 1 ) Where C pa = Specific heat of air = 1KJ/Kg.K. or 10 3 J/Kg K ( T 2 + T3 + T4 + T5 + T 7. Average inside surface temperature, T s = 6) 0 C 5 ( T 1 + T 7) 8. Bulk mean temperature of air, T m = 0 C 2 9. Average heat transfer coefficient Actual Heat Loss Due to Forced Convection= Q air Q R Heat Loss due to radiation, Q R = σ 0.4 x A x (T s4 - T a4 ) T a = atmospheric temperature = T 1 h expt Qair QR = A( T T ) W/m 2 K m s Where A = Inside surface are of the pipe =π x d i x L VxD 10. Reynolds number, R ed = υ υ = Kinematic Viscosity at T m

43 D = m If R ed < 2300, flow is laminar. h.d For laminar flow, N ud = = 4.36, from data book, page number 109. k air If Reynolds number exceeds 2300, flow is turbulent. For turbulent flow, N ud = (R ed ) 0.8 (Pr.) n from data book, page number 110. Where n = 0.4 when fluid is being heated. n = 0.3 When fluid is being cooled. Determine h theo from Nu D. Note: The calculated values and actual values may differ appreciably because of heat losses. The heat loss through natural convection, conduction and heat loss through insulation over the heater is not considered, but they are present. Also, the heat flux is not uniform practically, as assumed in theory, which gives difference between actual and theoretical value. PRECAUTIONS 1. While putting ON the supply, keep dimmerstat at zero position and blower switch OFF 2. Operate all the switches and controls gently. 3. Do not obstruct the flow of air while experiment is going on. RESULT Experimental heat transfer coefficient, h expt = Theoretical heat transfer coefficient = h theo =

44 Fig.1 Forced convection apparatus 1. Manometer 2. Voltmeter 3. Ammeter 4. Temperature indicator 5. Selector switch 6. Blower switch 7. Heater control 8. Main switch 9. Blower 10. Orifice meter 11. Test section 12. Thermocouple setting

45 STEFAN BOLTZMANN APPARATUS

46 STEFAN BOLTZMANN APPARATUS AIM To determine the Stefan- Boltzmann s constant in the radiation heat transfer. INTRODUCTION All the substances emit thermal radiation. When heat radiation is incident over a body, part of radiation is absorbed, transmitted through and reflected by the body. A surface which absorbs all thermal radiation incidents over it is called black surface. For black surface, transmissivity and reflectivity are zero and absorptivity is unity. Stefan Boltzmann Law states that emissivity of a surface is proportional to fourth power of absolute surface temperature i.e. E α T 4 or E = σ ε T 4 Where E = emissive power of surface, W / m 2 T = absolute temperature σ = Stefan Boltzmann constant ε = emissivity of the surface Value of Stefan Boltzmann constant is taken as σ = x 10-8 W / m 2 K 4 For black surface ε = 1, hence above equation reduces to E = σ. T 4 APPARATUS The Apparatus consists of a water-heated jacket of hemispherical shape. A copper test disc is fitted at the center of jacket. The hot water is obtained from a hot water tank, fitted to the panel, in which water is heated by an electric immersion heater. The hot water is taken around the hemisphere, so that hemisphere temperature rises. The test disc is then inserted at the center. Thermocouples are fitted inside hemisphere to average out hemisphere temperature. Another thermocouple fitted at the center of test disc measures the

47 temperature of test disc. A timer with a small buzzer is provided to note down the disc temperatures at the time intervals of 5 seconds. EXPERIMENTAL PROCEDURE 1. See that water inlet cock of water jacket is closed and fill up sufficient water in the heater tank. 2. Put ON the heater. 3. Blacken the test disc with the help of lamp black and let it cool. 4. Put the thermometer and check water temperature. 5. Boil the water and switch OFF the heater 6. See that drain cock of water jacket is closed and open water inlet cock. 7. See that there is sufficient water above the top of hemisphere (A piezometer tube is fitted to indicate water level) 8. Note down the hemisphere temperatures (up to channel 1 to 4) 9. Note down the test disc temperature (i.e.. channel 5) 10. Start the timer. Buzzer will start ringing. At the start of timer cycle, insert test disc into the hole at the bottom of hemisphere. 11. Note down the temperatures of disc, every five times of the buzzer rings. Take at least 8-10 readings OBSERVATIONS Hemisphere Temperature ( o C) T 1 = T 2 = T 3 = T 4 =

48 Time Interval (Sec) Test disc Temperature ( o C) CALCULATIONS 1) Area of test disc, A = (π/4)d 2 = m 2 (d = 20 mm) 2) mass of test disc, m = 7.6 gr = 7.6 x 10-3 kg. 3) Plot a graph of temperature rise of test disc with time as base and dt find out its slope at origin. i.e. dt K / sec T Hemisphere temperature, T H = 4) Initial Test disc temperature T D = T k 1 + T 2 + T 4 3 att=0 + T As area of hemisphere is very large as compared to that disc, we can put Q = σ є.a (T H 4 T D4 ) Where Q = heat gained by disc/sec. Q = m. c P. (dt/dt) t=0 σ = Stefan Boltzmann Constant m = Mass of test disk = 7.6 x 10-3 kg. є = Emissivity of test disc = 1 A = Area of Test disc c P = Specific heat of copper = 381 J/Kg 0 C K

49 σ = m. c P A.( T.( dt 4 H / dt ) T t = 0 4 D ) W/ m 2 K 4 Theoretical value of σ is 5.667x 10-8 W/m 2 K 4. In the experiment this value may deviate due to reasons like convection, temperature drop of hemisphere, heat losses etc. PRECAUTIONS 1) Never put ON the heater before putting water in the tank. 2) Put OFF the heater before draining the water from heater tank. 3) Drain the water after completion of experiment. 4) Operate all the switches and controls gently RESULT Stefan Boltzmann s constant, σ = W / m 2 K 4 Fig.1 Stefan Boltzmann apparatus 1. Water tank 2. Main switch 3. Temperature indicator 4. Temperature selector switch 5. Buzzer switch 6. Heater switch 7. Shell

50 Fig.2 Thermocouple setting Fig.3 Variation of temperature of disc with time

51 EMISSIVITY MEASUREMENT

52 AIM EMISSIVITY MEASUREMENT To determine the emissivity of the test plate. INTRODUCTION All the bodies emit and absorb the thermal radiation to and from surroundings. The rate of thermal radiation depends upon the temperature of body. Thermal radiations are electromagnetic waves and they do not require any medium for propagation. When thermal radiation strikes a body, part of it is reflected, part of it is absorbed and part of it is transmitted through body. The fraction of incident energy, reflected by the surface is called reflectivity (ρ). The fraction of incident energy, absorbed by the surface is called absorptivity (α) and the fraction of incident energy transmitted through body is called transmissivity (τ). The surface which absorbs all the incident radiation is called a black surface. For a black surface, ρ+α+τ = 1. The radiant flux, emitted from the surface is called emissive power (E). The emissivity of a surface is ratio of emissive power of a surface to that of black surface at the same temperature. Thus, ε = E / E b APPARATUS The apparatus uses comparator method for determining the emissivity of test plate. It consists of two aluminum plates, of equal physical dimensions. Mica heaters are provided inside the plates. The plates are mounted in an enclosure to provide undisturbed surroundings. One of the plates is blackened outside for use as a comparator (because black surface has ε = 1). Another plate is having natural surface finish. Input to heaters can be controlled by separator dimmer stats. Heater input is measured on common ammeter and voltmeter. One thermocouple is fitted on surface of each plate to measure the surface temperature with digital temperature indicator. By adjusting input to the heaters, both the plates are brought to same temperature, so that conduction and

53 convection losses form both the plates are equal and difference in input is due to different emissivities. Holes are provided at backside bottom and at the top of enclosure for natural circulation of air over the plates. The plate enclosure is provided with Perspex acrylic sheet at the front. EXPERIMENTAL PROCEDURE 1. Blacken one of the plates with the help of lamp black (Normally this is blackened at the works, but if blackening is wiped out, then blackening is necessary) 2. Keep both the dimmer knobs at ZERO position. 3. Insert the supply pin-top in the socket (which is properly earthed) and switch ON the mains supply. 4. Switch ON the mains switch on the panel. 5. Keep the meter selector switch (toggle switch) at the black plate side position. 6. Adjust dimmer of black plate, so that around volts are supplied to black plate. 7. Now, switch the meter selector switch on other side. 8. Adjust test plate voltage slightly less than that of black plate (say volts) 9. Check the temperatures (after, say 10 min) and adjust the dimmers so that temperatures of both the plates are equal and steady. Normally, very minor adjustments are required for this. 10. Note down the readings after the plate temperatures reach steady state. OBSERVATIONS Plate Input Surface temperature, V I Test plate T 1 = Black plate T 2 = 0 C Enclose temperature, T 3 = 0 C

54 CALCULATIONS 1. Enclose temperature: T E =T 0 3 = C = (T ) 0 K 2. Plate surface temp. T = T 1 =T 2 = 0 C T S = (T ) 0 K 3. Heat input to black plate, W b = V x I Watts 4. Heat input to test plate, W T = V x I Watts 5. Surface area of test plates, A = 2 x (π/4) D 2 + (π D t) = Where, D = dia. Of plates = 0.16 m. And t = thickness of plates = m. 6. For black plate, W b = W CVb + W Cdb + W Rb (i) Where, W Cvb = Convection losses W Cdb = Conduction losses W Rb = Radiation losses Similarly, for test plate, W T = W CvT + W CdT + W RT (ii) As both plates are of same physical dimensions, same material and at same temperatures, W Cvb = W CvT and W Cdb = W CdT Subtracting equation (ii) from (i) we get, W b - W T = W Rb - W RT = [σ A (T 4 s T E4 )] - [σ A ε T (T 4 s T E4 )] = σ A (T 4 s T E4 ) (ε b ε) As emissivity of black plate is 1, W b - W T = σ A (T 4 s T E4 ) (1 ε) Where, ε = Emissivity of test plate σ = Stefan Boltzman constant = x 10-8 W/m 2 K 4 PRECAUTIONS 1. Black plate should be perfectly blackened. 2. Never put your hand or papers over the holes provided at the top of enclosure.

55 3. Keep at least 200 mm distance between the backside of unit and the wall. 4. Operate all the switches and knobs gently. Note: Emissivity of oxidized aluminum plate i.e. test plate is normally with in the range of 0.3 to 0.7. RESULT Emisvity of the test plate surface = at temperature of Fig.1 Emissivity measurement apparatus 1. Voltmeter 2. Ammeter 3. Temperature indicator 4. Meter selector switch 5. Heater control 6. Heater control 7. Black plate 8 Test plate

56 CONCENTRIC TUBE HEAT EXCHANGER

57 CONCENTRIC TUBE HEAT EXCHANGER AIM To determine the heat transfer rate, LMTD, over all heat transfer coefficient and effectiveness of heat exchangers in parallel flow and counter flow concentric tube heat exchanger. INTRODUCTION Heat exchangers are the devices in which the heat is transferred from one fluid to another. Exchange of heat is required at many industrial operations as well as chemical process Common examples of heat exchangers are radiator of a car, condenser of a refrigeration unit or cooling coil of an air conditioner. Heat exchanger are of basically three types i) Transfer type- in which both fluids pass through the exchanger and heat gets transferred through the separating walls between the fluids, ii) Storage type in this, firstly the hot fluid passes through a medium having high heat capacity and then cold fluid is passed through the medium to collect the heat. Thus hot and cold fluids are alternately passed through the medium. iii) Direct contact type in this type, the fluids are not separated but they mix with each other and heat passes directly from one fluid to the other. Transfer type heat exchangers are the type most widely used. In transfer type heat exchangers, three types of flow arrangements are used, viz. parallel, counter or cross flow. In parallel flow, the fluids flow in the same direction while in counter flow, they flow in the opposite direction. In cross flow, they flow at right angles to each other. APPARATUS The apparatus consists of two concentric tubes in which fluids pass. The hot fluid is hot water, which is obtained from an electric geyser. Hot water flows through the inner tube in one direction. Cold fluid is cold water, which flows through annulus. Control valves are provided so that direction of cold water can be kept parallel or opposite to that of hot water. Thus, the heat exchanger can be operated either as parallel or

58 counter flow heat exchanger. The temperatures are measured with thermometers. Thus, the heat transfer rate, heat transfer coefficient, LMTD and effectiveness of heat exchanger can be calculated for both parallel and counter flow. SPECIFICATIONS 1. Heat exchanger- a) Inner tube - ϕ 12.7 mm O.D., ϕ 11.7 mm I.D. copper tube b) Outer tube ϕ 25 mm G.I. Pipe. c) Length of heat exchanger 1 m. 2. Valves for flow and direction control 5nos 3. Thermometers to measure temperatures 10 to C 4nos 4. Measuring flask and stop clock for flow measurement. EXPERIMENTAL PROCEDURE 1. Start the water supply. Adjust the water supply on hot and cold sides. Firstly, keep the valves V 2 and V 3 closed and V 1 and V 4 opened so that arrangement is parallel flow. 2. Put few drops of oil in thermometer pockets. Put the thermometer in the thermometer pockets. 3. Switch ON the geyser. Temperature of water will start rising. After temperatures become steady, note down the readings and fill up the observation table. 4. Repeat the experiment by changing the flow. 5. Now open the valvesv 2 and V 3 and then close the valves V 1 and V 4. The arrangement is in now counter flow. 6. Wait until the steady state is reached and note down the readings.

59 OBSERVATION TYPE OF FLOW Parallel Flow Counter Flow HOT WATER Time for 1 Temperatures Lit. Water Inlet Outlet 0 C 0 C x h sec COLD WATER Time for 1 Temperatures Lit. Water Inlet Outlet 0 C 0 C x c sec CALCULATIONS 1. Hot water inlet temperature, t hi = 0 C Hot water outlet temperature, t ho = 2. Hot water flow rate, m h Let time required for 1lit of water be x h sec Mass of 1lit water = 1kg There fore, m h = 1/ x h kg/s 3. Heat given by hot water (inside heat transfer rate) Q h = m h c p (t hi t ho ) Watts where c p =specific heat of water = 4200 J/kgK 4. Similarly, for cold water Heat collected by cold water (out side heat transfer rate) Q c = m c c p (t co t ci ) Watts 5. Logarithmic mean temperature difference (LMTD) LMTD = ΔT m = (T i T o ) / ln (T i / T o ) Where for parallel flow, for counter flow 0 C T i = t hi - t ci T o = t ho - t co T i = t hi - t co T o = t ho - t ci 6. Overall heat transfer coefficient, U a) Inside overall heat transfer coefficient, U i Inside diameter of tube = m

60 Inside surface area of the tube, A i = Π x x L Now Q h = U i ΔT m A i Therefore U i = Q h / (ΔT m A i ) W/m 2 0 C b) Outside overall heat transfer coefficient, U o Outside diameter of tube = m Outside surface area of the tube, A o = Π x x L Similarly Q c = U o ΔT m A o Therefore U o = Q c / (ΔT m A o ) W/m 2 0 C 7. Effectiveness of heat exchanger transfer rate Є = Rate of heat transfer in heat exchanger / Max. possible heat ε = mhc p ( thi tho ) [ mc ] ( t t ) p min hi ci Where [mc p ] min is smaller of two capacity rates of m h or m c PRECAUTIONS 1. Never switch on the geyser unless there is water supply through it. 2. If the red indicator on geyser goes off during operation, increase the water supply, because it indicates that water temperature exceeds the set limit. 3. Ensure steady water flow rate and temperatures before noting down the readings, as fluctuating water supply can give erratic results. TYPE OF HEAT TRANSFER RATE LMTD Heat Transfer Coefficient Effective - ness FLOW Inside (Watts) Outside (Watts) 0 C U i W/m 2 k U O W/m 2 k ε Parallel Flow Counter Flow

61 RESULTS Fig.1 Concentric tube heat exchanger (plain tube type) 1. Tci 2. Tho 3. V4 4. V3 5. Thi 6. V1 7. Tco 8. V2 Parallel flow V1 & V4 open V2 & V3 close Cross flow V2 & V3 open V1 & V4 close

62 HEAT TRANSFER IN DROP AND FILM WISE CONDENSATION

63 HEAT TRANSFER IN DROP AND FILM WISE CONDENSATION AIM To determine the experimental and theoretical heat transfer coefficient for drop wise and film wise condensation. INTRODUCTION Condensation of vapor is needed in many of the processes, like steam condensers, refrigeration etc. When vapor comes in contact with surface having temperature lower than saturation temperature, condensation occurs. When the condensate formed wets the surface, a film is formed over surface and the condensation is film wise condensation. When condensate does not wet the surface, drops are formed over the surface and condensation is drop wise condensation APPARATUS The apparatus consists of two condensers, which are fitted inside a glass cylinder, which is clamped between two flanges. Steam from steam generator enters the cylinder through a separator. Water is circulated through the condensers. One of the condensers is with natural surface finish to promote film wise condensation and the other is chrome plated to create drop wise condensation. Water flow is measured by a Rota meter. A digital temperature indicator measures various temperatures. Steam pressure is measured by a pressure gauge. Thus heat transfer coefficients in drop wise and film wise condensation cab be calculated. SPECIFICATION 1. Condensers: Made of copper, 19 mm O.D., 150 mm long, one with natural surface and one with chrome-plated surface. 2. Pressure gauge to measure steam pressure 3. Necessary valves for water and steam control.

64 EXPERIMENTAL PROCEDURE Fill up the water in the steam generator and close the water-filling valve. Start water supply through the condensers. Close the steam control valve, switch on the supply and start the heater. After some time, steam will be generated. Close water flow through one of the condensers. Open steam control valve and allow steam to enter the cylinder and pressure gauge will show some reading. Open drain valve and ensure that air in the cylinder is expelled out. Close the drain valve and observe the condensers. Depending upon the condenser in operation, drop wise or film wise condensation will be observed. Wait for some time for steady state, and note down all the readings. Repeat the procedure for the other condenser. OBSERVATIONS Drop wise Condensation Steam Pressure, kg/cm 2 Water flow rate, LPH Steam temperature, T 1, 0 C Drop wise condensation surface temperature, T 2, 0 C Water inlet temperature, T 4, 0 C Water outlet temperature, T 5, 0 C

65 Film wise Condensation Steam Pressure, kg/cm 2 Water flow rate, LPH Steam temperature, T 1, 0 C Film wise condensation surface temperature, T 3, 0 C Water inlet temperature, T 4, 0 C Water outlet temperature, T 6, 0 C CALCULATIONS (Film wise condensation) Water flow m w = LPH = kg/sec Water inlet temperature T 4 = Water outlet temperature = o C o C (T 5 for drop-wise condensation and T 6 for film-wise condensation) Heat transfer rate at the condenser wall, Q = m w.c P. (T 6 -T 4 ) Watts Where c p = Specific heat of water = 4.2 x 10 3 J / Kg K Surface area of the condenser, A =πdl m 3 Experimental heat transfer coefficient, wise and drop wise condensation) Where T s = Temperature of steam (T 1 ) T W = Condenser wall temperature (T 2 or T 3 ) Theoretically, for film wise condensation h h = ( T fg s 2 3. ρ. g. k Tw ). μ. L 0.25 Q h = 2 o W / m C A( T T ) (for both film Where h fg = Latent heat of steam at T S J/kg (take from temperature tables in steam tables) ρ = Density of water, Kg / m 3 g = Gravitational acceleration, m / sec 2 s w

66 k = Thermal conductivity of water W / m o C μ = Viscosity of water, N.s/m 2 and, L = Length of condenser = 0.15 m Above values at mean temperature, T m ( Ts + TW ) o = C (from data book) 2 (For drop wise condensation, determine experimental heat transfer coefficient only) In film wise condensation, film of water acts as barrier to heat transfer where as, in case of drop formation, there is no barrier to heat transfer, Hence heat transfer coefficient in drop wise condensation is much greater than film wise condensation, and is preferred for condensation. But practically, it is difficult to prolong the drop wise condensation and after a period of condensation the surface becomes wetted by the liquid. Hence slowly film wise condensation starts. PRECAUTIONS 1. Operate all the switches and controls gently 2. Never allow steam to enter the cylinder unless the water is flowing through condenser. 3. Always ensure that the equipment is earthed properly before switching on the supply. RESULTS Film wise condensation Experimental average heat transfer coefficient = Theoretical average heat transfer coefficient = Drop wise condensation Experimental average heat transfer coefficient =

67 Fig. 1 Condensation in drop and film forms 1. Steam generator 2. Water level 3. Rota meter 4. Steam pressure 5. Condensers 6. Temperature indicator 7. Selector switch 8. Heater control 9. Main switch

68 CRITICAL HEAT FLUX APPARATUS

69 CRITICAL HEAT FLUX APPARATUS AIM To determine the experimental and theoretical value of critical heat flux in pool boiling of water. INTRODUCTION When heat is added to a liquid from a submerged solid surface which is at a temperature higher than the saturation temperature of the liquid, it is usual for a part of the liquid to change phase. This change of phase is called boiling. Boiling is of various types, the type depending upon the temperature difference between the surface and the liquid. The different types are indicated in figure, in which a typical experimental boiling curve obtained in a saturated pool of liquid is drawn. The heat flux supplied to the surface is plotted against (T w T s ) the difference between the temperature of the surface and the saturation temperature of the liquid. It is seen that the boiling curve can be divided into three regions Heat flux, W/m 2 Nucleate Film boiling boiling I II III II a II b Bubbles rise Bubbles condense A III a Unstable film B III b Radiation coming in to play Stable film C Excess temperature (T w - T s ) I) Natural convection region II) Nucleate boiling region and III) Film boiling region

70 The region of natural convection occurs at low temperature differences (of the order of 10 0 C or less). Heat transfer from the heated surface to the liquid in its vicinity causes the liquid to be superheated. This superheated liquid rises to the free liquid surface by natural convection, where vapour is produced by evaporation. As the temperature difference (T w T s ) is increased, nucleate boiling starts. In this region, it is observed that bubbles start to form at certain locations on the heated surface region (II) consists of two parts. In the first part (II-a) the bubbles formed are very few in number. They condense in the liquid and do not reach the free surface. In the second part (II-b) the rate of bubble formation as well as the number of locations where they are formed increase. Some of the bubbles now rise all the way to the free surface. With increasing temperature difference, a stage is finally reached when the rate of formation of bubbles is so high, that they start to coalesce and blanket the surface with a vapour film. This is the beginning of region (III) viz, film boiling. In the first part of this region (III-a) the vapour film is unstable, so that film boiling may be occurring on a portion of the heated surface area, while nucleate boiling may be occurring on the remaining area. In the second part (III-b) a stable film covers the entire surface. At the end of region (II) the boiling curve reaches a peak (point A). Beyond this, in region (III-A) in spite of increasing temperature difference, the heat flow increases with the formation of a vapour film. The heat flux passes through a minimum (point B) at the end of region (III-a). It starts to increase again with (T w T s ) only when stable film boiling begins and radiation becomes increasingly important. It is of interest to note how the temperature of the heating surface changes as the heat flux is steadily increased from zero. Up to the point A, natural convection boiling and then nucleate boiling occur and the temperature of the heating surface is obtained by reading off the value of (T w T s ) from the boiling curve and adding to it the value of T s. If the heat flux is increased even a little beyond the value of A, the temperature of the surface will shoot up to the value corresponding to the point C. It is

71 obvious from figure that the surface temperature corresponding to point C is high. For most surfaces it is high enough to cause the material to melt. Thus in most practical situation, it is undesirable to exceed the value of heat flux corresponding to point A. This value is therefore of considerable engineering significance and is called the critical or peak heat flux. The discussions so far has been concerned with the various type of boiling which occurring saturated pool boiling. If the liquid is below the saturation temperature we say that sub-cooled pool boiling is taking place. Also in many practical situations, e.g. steam generators, one is interested in boiling in a liquid flowing through tubes. This is called forced convection boiling may also be saturated or sub cooled and of the nucleate or film type. Thus in order to completely specify billing occurring in any process, one must state that (i) whether it is forced convection boiling or pool boiling, (ii) whether the liquid is saturated or sub cooled and (iii) whether is in the natural convection nucleate of film region. APPARATUS The apparatus consists of a cylindrical glass housing the test heater and heater coil for heating of the water. This heater coil is direct connected to the mains (Heater R1) and the test wire is also connected to mains via. variac. An ammeter is connected in series while a voltmeter across it to read the current and voltage respectively. The glass container is kept on a stand. There is provision of observing the test heater wire with the help of a lamp light from back and the heater wire can be view a lens. SPECEIFICATIONS 1) Glass container Diameter 250 mm Height 100 cm 2) Heater for initial heating, Nichrome Heater (R 1) 1 kw 3) Test Heater (R-2), Nichrome wire size - Φ mm (To be calculated according to wire used say 36 SWG to 40 SWG.) 3) Length of test Heater (R-2) = 100 mm 4) Thermometer 0 to 100 o C

72 EXPERIMENTS This experimental set up is designed to study the pool boiling phenomenon up to critical heat flux point. The pool boiling over the heater wire can be visualized in the different regions up to the critical heat flux point at which the wire melts. The heat flux from the wire is slowly increased by gradually increasing the applied voltage across the test wire the change over from natural convection to nucleate boiling can be seen. The formation of bubbles and their growth in size and number can be visualized followed by the vigorous bubbles formation and their immediate carrying over to surface and ending this in the breaking of wire indicating the occurrence of critical heat flux point. PROCEDURE 1) Take sufficient amount of distilled water in the container. 2) See that both the heaters are completely submerged. 3) Connect the heater coil R-1 (1KW Nichrome coil) and test heater wire across the studs and make the necessary electrical connections. 4) Switch on the heater R-1(Let variac be at O position.) 5) Keep it ON till you get the required bulk temperature of water in the container say 50 O C, 60 O C, 70 O C temperature. 6) Switch off the heater R-1. 7) Very gradually increase the voltage across test heater by slowly changing the variac position and stop a while at each position to observe the boiling phenomenon on wire. 8) Go on increasing the voltage till wire brakes and carefully note the voltage and current at this point. PRECAUTIONS 1) Keep the variac to zero voltage position before starting the experiments. 2) Take sufficient amount of distilled water in the container so that both the heaters are completely immersed. 3) Connect the test heater wire across the stud.

73 4) Do not touch the water or terminal points when the main switch ON. 5) Operate the variac gently in steps and sufficient time in between. 6) After the attainment of critical heat flux decrease slowly the voltage and bring it to zero position. OBSERVATIONS 1) Diameter of test heater wire, d = m. 2) Length of the test heater, L = 0.1 m 3) Surface area A = π.d.l m 2 = Bulk Temperature of water O C Ammeter Reading (I Amps) Voltmeter Reading (V volts) Note: - The ammeter and voltmeter readings are to be note down when wire melts. CALCULATIONS The critical heat flux at various bulk temperatures water can be calculated by the following procedure. 1. Heat input Q = V. l Watts 2. Critical Heat flux Q q exp t = = A V.I A W / m 3. Zubfer has given following equation for calculating peak heat flux in saturated pool boiling. 2 q theor = Q A = σ. ( LV g ρ L ρ v 0.18 h fg ρ v 2 ρ v 1 / 4 Where Q/A = Heat Flux, W/m 2 h fg = Latent Heat of vaporization J/kg (from steam table) σ LV = Liquid vapour surface tension N/m (from Chart) ρ L = Density of Liquid

74 ρ ν = Density of vapour = 1/v g kg/m 3 (from steam table) h fg, σ LV, ρ L and ρ ν are evaluated at the water temperature. The experimental value of critical heat flux at the sat temperature is comparable to that obtained by Zuber s correlation. RESULT 1. Experimental critical heat flux = 2. Theoretical critical heat flux = Fig.1 Critical heat flux apparatus 1. Voltmeter 2. Ammeter 3. Heater switch 4. Lamp switch 5. Main switch 6. Heater control 7. Class container 8. Heater fitting