Thermo fluids lab key

22 de May de 2016
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Thermo fluids lab key

• 1. 1 AKSUM UNIVERSITY COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING THERMO FLUIDS LAB MANUAL Compiled by A.Syed Bava Bakrudeen, Associate Professor, Mechanical Department, Aksum University, Ethiopia. LIST OF EXPERIMENTS FOR THERMO FLUIDS LAB 1. Forced vortex flow 2. Free vortex flow 3. Hydraulic ram pump 4. Orifice discharge 5. Rectangular weir 6. Triangular weir 7. Stability of the floating body 8. Flow visualization Apparatus 9. Darcy’s friction factor (Friction Loss – Major loss) 10. Velocity distribution in aerodynamic trainer kit 11. Drag model of aero foil section 12. Tubular heat exchanger (Parallel flow) 13. Tubular heat exchanger (Counter flow)
• 2. 2 1. FORCED VORTEX FLOW THEORY: Forced vortex flow: Flow of fluid along a curved path is called vortex flow, in forced vortex flow some external torque is required to rotate the fluid mass. AIM: To determine radius and height of the given forced vortex flow. APPARATUS REQUIRED: 1. Impeller vortex setup. 2. Collecting tank 3. Stop watch 4. Meter scale FORMULA: 1. Angular Velocity of Vane ωvane = 2 π n in 1/s When n= Revolution/Seconds. 2. The Angular vortex velocity ω = 0.8x ωvane 3. Height of the vortex h= ω2 r2 /(2xg) PROCEDURE: A. Set Up: 1. Position test set up on HM150 such that drain and overflow route water into drainage channel or volumetric tank. 2. Insert fittings in base plate in line with desired experiment (Forced vortex- Vane with cover plate and shaft). 3. Position selector ball cocks such that mark faces vertically upwards the two outlets are closed. 4. Make hose connection between HM150 and unit open drain of HM150. 5. Close main cock of HM 150 and switch on pump of HM 150. 6. Slowly turn selector ball cock to desired positions
• 3. 3 Anti clock wise : Tangential inlet for Forced vortex 7. The water volume can be exactly regulated with the slanted seat valve in the outlet. B. Height and Radius Measurement: 1. Insert combined radius and height gauge in mount from underneath and secure using star type nut. 2. For measurement, loosen the star type nut and set new height; then use knurled nut on cross member to move gauges to surface of vortex. Height: Shoulder of rotating rod; this indicates the height of the gauges above the bottom. Radius: Left and right edges of notch of slider; this indicates radius. C. Velocity Measurement: 1. For the forced vortex, a stop watch can be used to measure the time for the vane revolutions. As however the friction between liquid and vessel wall decelerates the vortex, the real vortex velocity is 20% lower than the vane velocity ( Vvane). GRAPHS: i Velocity (v) Vs Radius (r) (Ordinary Graph) RESULT: 1. The average value of height is = __________m. 2. The average value of radius is =___________m.
• 4. 4 TABULATION: S. No No of revolu tions (R) Time (S) Rev/ Sec (n) Radius (r) in mm Angular Velocity of vane (ω vane) in r/s Angular Velocity of vortex (ω) in r/s Height of vortex (h) (mm) 1 30 14.5 2.069 30 13 10.4 5 MODEL CALCULATIONS: 1. n=Revolutions/Second = 30/14.5 = 2.069 rev/s 2. ω = 2π n = 2xπx2.069 =13 r/s 3. ω = 0.8x ωvane = 0.8x8.6 =10.4 r/s 4. h= ω2 r2 /(2xg) = 10.42 x(30x10-3 )2 /(2x9.81) = 5 m m Fig 1 Forced Vortex Flow
• 5. 5 2. FREE VORTEX FLOW THEORY: Free vortex flow: Flow of fluid along a curved path is called vortex flow, in free vortex flow no external torque is required to rotate the fluid mass, it is imparted to the fluid previously. AIM: To determine radius and height of the given free vortex flow. APPARATUS REQUIRED: 1. Impeller vortex setup. 2. Collecting tank 3. Stop watch 4. Meter scale FORMULA: 1. Angular Velocity ω = 2 π n in 1/s When n= Revolution/Seconds. 2. The Velocity v=ω r in m/s When r= Radius of revolution in meter 3. Height of the vortex h= hmax - (k 2 /(2xg.r2 )) When hmax= height of the water level in meter k = v r PROCEDURE: A. Set Up: 1. Position test set up on HM150 such that drain and overflow route water into drainage channel or volumetric tank. 2. Insert fittings in base plate in line with desired experiment (Free vortex- Outlet of differing diameter and inlet ring for smoothing the flow).
• 6. 6 3. Position selector ball cocks such that mark faces vertically upwards the two outlets are closed. 4. Make hose connection between HM150 and unit open drain of HM150. 5. Close main cock of HM 150 and switch on pump of HM 150. 6. Slowly turn selector ball cock to desired positions Clock wise : Radial inlet for free vortex 7. The water volume can be exactly regulated with the slanted seat valve in the outlet. B. Height and Radius Measurement 1. Insert combined radius and height gauge in mount from underneath and secure using star type nut. 2. For measurement, loosen the star type nut and set new height; then use knurled nut on cross member to move gauges to surface of vortex. Height: Shoulder of rotating rod; this indicates the height of the gauges above the bottom. Radius: Left and right edges of notch of slider; this indicates radius. C. Velocity Measurement 1. To take small piece of paper or polystyrene to drop on to the water surface. Then stop watch use to determine the time taken for the piece of paper to perform 10 revolutions with the vortex. The radius on which the paper revolves must also be known. GRAPHS: i Velocity (v) Vs Radius (r) (Ordinary Graph) RESULT: 1. The average value of height is = __________m. 2. The average value of radius is =___________m.
• 7. 7 TABULATION: S. N o No of revol utio ns Time (S) Rev/ Sec (n) Radi us (r) in mm Angul ar Veloci ty of Vane (ω) in r/s vortex velocity (v) in m/s height of the water level (hmax) K=vr Height of vortex (h) 1 40 34.57 1.157 55 7.27 0.4 144 0.022 144 MODEL CALCULATIONS: 1. n=Rev/sec = 40/34.5 =1.157 rev/Sec 2. ω = 2πn= 7.27 rad/Sec 3. v=ω r = 7.27x 0.055 =0.4 m/s 4. k = v r = 0.4x0.055 = 0.022 m2 /s 5. Height of the vortex h= hmax - (k 2 /(2xg.r2 )) h = 144- (0.0222 /(2x9.81x0.0552 ) )=144 Fig 2. Free vortex flow
• 8. 8 3. HYDRUALIC RAM PUMP THEORY: Using this type of pump it is possible to pump water to a higher level without provision of additional mechanical energy. In this process, the kinetic energy in the flowing water is converted into potential pressure energy by very sudden retardation. AIM: To determine mass flow ratio and efficiency of the given hydraulic ram pump. APPARATUS REQUIRED: 1. Hydraulic ram pump setup. 2. Collecting tank 3. Stop watch 4. Meter scale FORMULA: 1. Pump Head (H) = h1 – h2 (m) h1= First raised container (m) h2 = Second raised container (m) 2. Amount of water raised in the first raised level (m2)= m1 + mv (l/s) m1= Amount of water pumped (l/s) mv = Amount of water lost (l/s) 3. Flow ratio (µ) Practical = m2/m1 Theoretical= h1/h2 Percentage = (Practical/Theoretical)x100 4. Efficiency of the pump = P1/P2 P1= m1gh1; P1= Energy of the water pumped P2= m2gH; P2= Energy of the water used.
• 9. 9 PROCEDURE: 1. The unit is first to be connected to the water supply (HM 150). 2. Closed the bleed valve and out let valve. 3. Completely open the regulating valve. 4. Switch on HM 150 pump and open main valve. 5. Open inlet valve and fill fist raised vessel. 6. Once the overflow has been reached, reduce the feed a little to avoid the water level increasing further. 7. Normally the hydraulic ram pump now starts working. If does not work, step the action valve can be triggered by operating the valve spindle manually. 8. After pumping is finished the volumetric flow rate of the pump can be measured using a stop watch and a measuring container on the outlet with use of stop watch. GRAPHS: i Practical mass flow ratio Vs efficiency (Ordinary Graph) RESULT : 1. The average value of flow ratio is = ________. 2. The average value of efficiency =___________.
• 10. 10 TABULATION: S. N o First rai sed cont aine r h1 in m Sec o nd rai sed cont aine r h2 in m Pump Head (H = h1-h2) in m Amoun t of water pumpe d (liter /sec) (m2) Amout of water last (liter/ sec) (mv ) Amo unt of water in first raised level (m1= m2+m v) (L/s) Flow ratio (µ) Ener gy of the wate r pum ped (P1) (W) Ener gy of the wate r used (P2) (W) Efficie ncy of the pump = P1/P2 (W) Prac tical The orit cal In per cen tag e 1 0.88 1.15 0.27 1/40 =0.025 1/14.5 =0.069 0.094 .266 .765 35 0.81 .066 8.1 Average = Average = MODEL CAL CULATION: h1= 0.88m, h2= 1.15m, H= h1-h2 = 0.27m m2=1/40 = 0.025 kg/s; mv= 1/14.5= 0.069 kg/s m1=m2+mv=0.094 kg/s Practical flow ratio = m2/m1= 0.025/0.094 =0.266 Theoretical flow ratio = h1/h2 = 0.88/1.15 =0.765 % of flow ratio = 0.266/0.765 =0.347=35% Efficiency Pin = m1g h1 =0.094x9.81x0.88 = 0.81W Pout =m2gh2= 0.025x9.81x0.27 = 0.066 W Efficiency = Pout/Pin = (0.066/0.81)x100 =8.1%
• 11. 11 Fig 3 Principal of Hydraulic Ram ……………………………………………………………………………………… Fig 4 Orifice discharge
• 12. 12 4.ORIFICE DISCHARGE THEORY: Orifice is a small opening of any cross section (such as circular, rectangular or triangular etc) on the side or bottom of the tank, through which a fluid is flowing. AIM: To determine jet diameter and coefficient of discharge of the given orifice APPARATUS REQUIRED: 1. Orifice discharge setup. 2. Collecting tank fitted with piezo meter 3. Stop watch 4. Meter scale FORMULA: 1. Radius of jet = (10 –Value read on micrometer )x103 2. Jet contraction or contraction coefficient α = ( Ajet/Ainlet) = (d2 jet/ d2 inlet) djet - Dia of jet in „m‟ dinlet – Dia of inlet in „m‟ 3. Velocity coefficient φ = ( Wjet/Wtheor)=(√ 2 g hstat)/ (√ 2 g htotal) hstat = Height calculated in manometer in „ m‟. htotal = Height calculated in pitot tube in „ m‟. 4. Discharge coefficient µ = φ α PROCEDURE: 1. Screw the adjusting aid into the base of the tank from underneath. 2. Using the height adjusting screw, adjust the spindle so that the tip is at the height of the adjusting aid. 3. Loose the knurled nut. 4. Set the micrometer 4.00mm. 5. Hold the piece of paper between the tip of the spindle and the adjusting aid and using star screw, adjust the tip so that the sheet of paper is lightly clamped. 6. Hold the micrometer and securely tighten the knurled nut.
• 13. 13 7. Using the star screw, turn back the spindle so that the tip of the spindle is no longer touching the adjusting aid 8. Unscrew the adjusting aid. 9. Screw the relevant nozzle and sealing ring into the base of the tank from underneath. 10. Position test setup on HM 150 such that drain and overflow route of the later into outlet. 11. Make hose connection between HM 150 and unit, open drain of HM150 and unit and switch on pump of HM150. 12. Slowly open the main cock and adjust discharge height, the water level should be below the over flow. 13. Adjust the height of the inlet basket so that it is not immersed in the water. 14. Read off head of water column in tank on left pressure gauge. 15. Rotate pitot tube into centre of jet and read off height of water column on right pressure gauge. 16. Using the star screw, unscrew the spindle until the tip of the spindle just touches the water jet. Read off the value on the micrometer. GRAPHS i hstat Vs µ (Ordinary Graph) ii. htotal Vs µ (Ordinary Graph) OBSERVATION Dia of the Orifice (Dia of inlet) =……12x10-3 …………… m RESULT: 1. The average value of jet diameter is = ______________ m 2. The average value of velocity coefficient =___________.
• 14. 14 TABULATION: S. N o Static Head hstat in mm of w.c (x10-3 ) in m Total Head htotal in mm of w.c (x10-3 ) in m Value read in micro meter in mm Jet radiu s in m Jet diamet er in m Veloc ity of the jet Wjet in m/s Calcula ted velocit y Wtheo in m/s Contrac tion coeffici ent α Velocity coefficient φ Discharg e coefficie nt µ 1 250 249 5.1 4.9 9.8 2.21 2.215 0.66 0.997 0.65 Average = Average = MODEL CALCULATION: 1. Wjet = √(2ghtotal)=√(2x9.81x249x10-3 )=2.210 m/s 2. Wtheo = √(2ghstat)=√(2x9.81x250x10-3 )=2.215 m/s 3. φ = Wjet/Wtheo = 0.997 4. α= (djet)2 /(dinlet)2 = (9.8x10-3 )2 /( 12x10-3 )2 =0.667
• 15. 15 5.RECTANGULAR WEIR THEORY: A notch or weir is an opening in the side of a tank or a channel in such a way that the liquid surface in the tank or channel is below the top edge of the opening. A notch is smaller in size, it‟s made by metallic structure and commonly constructed in the tank. A weir is bigger in size, it‟s made of concrete or masonary structure and commonly constructed in channel. Nappe or Vein: The sheet of water flowing through a notch or over weir is called Nappe or Vein. Crest or Sill: The bottom edge of the notch or weir which the water flows is known a sill or crest. AIM: To determine the Coefficient of discharge for the given rectangular notch. APPARATUS REQUIRED: 1. Notch tank 2. Rectangular Notch 3. Hook Gauge 4. Collecting tank fitted with piezo meter 5. Stop watch 6. Meter scale FORMULA: 1. Head Over the Sill H = H1 – H2 (cm) H1 - Sill Level of the tank in „cm‟ H2 – Final Hook Gauge Reading in „cm‟ 2. Actual Discharge Qa = y / t (cm3 / s) Qa = Actual discharge in cm3 / s y = Discharge of water in liter t = time taken for y cm rise in the collection tank in „sec‟ (1 Liter = 1000 cm3 )
• 16. 16 3. Theoretical Discharge Qt = (2/3) x L x (√2g) H 3/2 (cm3 / s) Qt = Theoretical Discharge in cm3 /Sec L = Length of the notch in „cm‟ H = Head over the sill in „cm‟ 4. Co-efficient of discharge Cd = Qa /Qt Qa = Actual discharge in cm3 / s Qt = Theoretical Discharge in cm3 /s PROCEDURE: 1. The internal plan dimensions of the collecting tank and the length of the notch are measured. 2. The supply valve is opened and water is allowed to rise only up to the sill of the notch and the supply valve is closed tightly. 3. The tip of the points of the hook gauges is adjusted such that the tip coincides with free water surface. 4. The sill level of the notch is noted from the hook gauge reading (H1) 5. The supply valve is opened and water is allowed to flow through the notch. When the flow becomes steady, the reading in the hook gauge (H2) is noted. 6. The outlet valve of the collection tank is tightly closed and the time taken for „y‟ cm rise in the collection tank is noted. 7. The experiment is repeated by varying the discharge through the channel by adjusting the control valve and thus for various heads over the notch. 8. The observations are tabulated and the co-efficient of discharge is calculated. GRAPHS: i. H Vs Cd (Ordinary Graph) ii. H Vs Qa (Ordinary Graph) OBSERVATION: Breadth of the Notch b =………6………… cm RESULT: 1. The value of co-efficient of discharge of rectangular notch Cd =
• 17. 17 TABULATION: S. No Hook Gauge Reading in m Head over the sill H= H1-H2 in m Water lever rise in collect ing tank ‘y’in liter Time taken for ‘y’ liter rise in the tank in sec Actual discha rge Qa Theo retic al disch arge Qt Coefficie nt of discharg e Sill Level (H1) Final hook gauge readin g (H2) 1. 9.7 7.2 2.5 10 22 2200 2215 0.99 Average = MODEL CALCULATIONS: Qa = y / t = 22/10 =2.2 =2200 cm3 /s Qt = (2/3) x L x (√2g) H 3/2 = (2/3)x6x √(2x9810)x (2.5)3/2 =2215 cm3 /s Cd=Qa/Qt= 2200/2215=0.99
• 18. 18 6.TRAINGULAR WEIR THEORY: A notch or weir is an opening in the side of a tank or a channel in such a way that the liquid surface in the tank or channel is below the top edge of the opening. A notch is smaller in size, it‟s made by metallic structure and commonly constructed in the tank. A weir is bigger in size, it‟s made of concrete or masonary structure and commonly constructed in channel. Nappe or Vein: The sheet of water flowing through a notch or over weir is called Nappe or Vein. Crest or Sill: The bottom edge of the notch or weir which the water flows is known a sill or crest. AIM: To determine the Coefficient of discharge for the given triangular notch. APPARATUS REQUIRED : 1. Notch tank 2. Traingular Notch 3. Hook Gauge 4. Collecting tank fitted with piezo meter 5. Stop watch 6. Meter scale FORMULA: 1. Head Over the Sill H = H1 – H2 (m) H1 - Sill Level of the tank in „cm‟ H2 – Final Hook Gauge Reading in „cm‟ 2. 2. Actual Discharge Qa = y / t (cm3 / s) Qa = Actual discharge in cm3 / s y = Discharge of water in liter t = time taken for y cm rise in the collection tank in „sec‟ (1 Liter = 1000 cm3 )
• 19. 19 3. Theoretical Discharge Qt = (8/15) (√2g) H 5/2 tan (ϴ/2) (m3 / s) Qt = Theoretical Discharge in m3 /Sec H = Head over the sill in „m‟ ϴ = Angle of inclination of v notch in degrees 4. Co-efficient of discharge Cd = Qa /Qt Qa = Actual discharge in m3 / s Qt = Theoretical Discharge in m3 /s PROCEDURE: 1. The internal plan dimensions of the collecting tank and the length of the notch are measured. 2. The supply valve is opened and water is allowed to rise only up to the sill of the notch and the supply valve is closed tightly. 3. The tip of the points of the hook gauges is adjusted such that the tip coincides with free water surface. 4. The sill level of the notch is noted from the hook gauge reading (H1) 5. The supply valve is opened and water is allowed to flow through the notch. When the flow becomes steady, the reading in the hook gauge (H2) is noted. 6. The outlet valve of the collection tank is tightly closed and the time taken for „y‟ cm rise in the collection tank is noted. 7. The experiment is repeated by varying the discharge through the channel by adjusting the control valve and thus for various heads over the notch. 8. The observations are tabulated and the co-efficient of discharge is calculated. GRAPHS: i. H Vs Cd (Ordinary Graph) ii. H Vs Qa (Ordinary Graph) OBSERVATION: Angle of the Notch =……900 …… RESULT: 1. The value of co-efficient of discharge of triangular notch Cd =
• 20. 20 TABULATION: S. No Hook Gauge Reading in m Head over the sill H= H1-H2 in m Water lever rise in collect ing tank ‘y’ in liter Time taken for ‘y’liter rise in the tank in sec Actual discha rge Theor etical discha rge Coefficie nt of discharg e Sill Level (H1) Final hook gauge readin g (H2) 1 10.7 5.5 5.2 10 41 4,100 4,606 0.89 Average = MODEL CALCULATIONS: Qa = y / t = 41/10 =4.1L/S =4100 cm3 /s Qt = (8/15) x (√2g) H 5/2 x tan(ϴ/2)= (8/15)√(2x9810)x (5.2)5/2 =4606 cm3 /s Cd=Qa/Qt= 4100/4606=0.89.
• 21. 21 7.STABILITY OF THE FLOWTING BODY THEORY: a.Buoyancy: A body is immersed in the fluid, a upward force is exerted by the fluid on the body. This upward force is equal to weight of the fluid displaced by the body. This force is called buoyancy. The point at which the force of buoyancy act is called centre of buoyancy. b. Metacentre: It is defined as point at which the line of action of the force of the buoyancy will meet normal axis of the body, when the body is given a small angular displacement. c. Metacentric height: The distance between the meta centre of the body to the centre of gravity of the body is called metacentric height. AIM: To determine the stability of the floating body using metacentric height and also to find the buoyancy of it. APPARATUS REQUIRED: 1. Floating body apparatus. 2. Meter scale. FORMULA: 1. Flowting conditions A. If Zs-M is positive – Steady floating condition. B. If Zs-M is negative – Unsteady floating condition- It is submerged. Where Zs = Vertical position centroidal axis in meter M = Metacentre in meter. 2. Centre line of the floating body Xs= (MhxX) / (M+ Mh+ Mv) Where Mh = 194g, Mv = 576g, M=2636g. X= Distance of centre line of horizontal mass from bottom line. 3. The vertical position is referenced to the underside of the floating body Zs = (MvxZ)+ ((M+Mh)xZg) / (M+ Mh+ Mv) Zg = Centre gravithy of floating body without central weight = 7.13cm. 4. Stability gradient (dXs/dα) = Xs/α
• 22. 22 5. Buyonancy FA= FG= BxLxTxρxg B= Width of the floating body in m L= Length of the floating body in m T= Depth of immersion of a floating body in m PROCEDURE: 1. Set horizontal sliding weight to position X cm 2. Move vertical sliding weight at bottom position. 3. Fill the tank with necessary water and insert the whole body to float. 4. Gradually raise vertical sliding weight and read off angle on heel indicator. 5. Read off height of sliding weight at top edge of weight and enter in table together with angle. 6. A further experiment can be performed to determine buoyancy. As the floating body is block-shaped, the volume of water displaced can easily be calculated from the width, length and immersion depth. The immersion depth can be read off the vertical scale. A horizontal floating position is a prerequisite for this. GRAPHS: Vertical centre of gravity (Zs) Vs Stability gradient (dXs/dα) OBSERVATION: B= Width of the floating body = -----200 mm----- L= Length of the floating body = ------300mm----- T= Depth of immersion of a floating body =---80mm---- RESULT: 1. The value of metacentric height is in between ___ to ___cm. 2. The average buyonancy is =___________m.
• 23. 23 TABULATION: Horizontal sliding weight distance X= _______8_______ cm Horizontal sliding weight distance Xs= ______0.0456________ cm S.No 1 Height of vertical sliding weight Z cm 3 Centre of gravity position Zs cm 6.431 Angle α 12.50 dXs/dα 0.0365 MODEL CALCUALTIONS: 1. Centre line of the floating body Xs= (MhxX) / (M+ Mh+ Mv) Where Mh = 194g, Mv = 576g, M=2636g. at X= 8cm Xs = 0.0456 cm. 2. The vertical position is referenced to the underside of the floating body Zs = (MvxZ)+ ((M+Mh)xZg) / (M+ Mh+ Mv) ; Zg = 7.13cm. Zs = 6.431cm 3. Stability gradient (dXs/dα) = Xs/α =0.0456/12.5 =0.0365 cm/0 4. Buyonancy FA= FG= BxLxTxρxg=0.2x0.3x0.08x9810 = 47.088 N M –Meta centre G – Centre of gravity M- Metacentre FB – Buoyancy force MG- Metacentric height
• 24. 24 8. FLOW VISUVALISATION APPARATUS THEORY: Path lines are the trajectories that individual fluid particles follow. These can be thought of as "recording" the path of a fluid element in the flow over a certain period. AIM: To visualize the flow path of a liquid with subjected to various shapes of obstacles. APPARATUS REQUIRED: 1. Flow visualization apparatus setup 2. Collecting tank 3. Ink PROCEDURE: A. Venting of Apparatus: 1. Before tests or experiments can be performed , all the hoses and valves must be vented. 2. Install and position of flat gasket and install glass pane and fasten it using 10 knurled head screw. 3. Connect the reservoir tank to the flow chamber, supply water and open all the valves of the apparatus. The hoses and valves will be rinsed out by a strong water flow. 4. When no more air bubbles can be seen, close all the valves again and maintain the water supply. 5. Carefully remove the glass pane. If necessary push it out to the side in order to prevent adhesion. B. Assembly of the glass pane: 1. Position the flat gasket exactly onto the bore holes on the flow chamber. 2. If desired, install obstacle models and press firmly on to the plastic plate. 3. Lay the glass pane onto the gasket on one side at the supply side. Ensure the water supply via the supply valve and slowly lower the glass pane down on
• 25. 25 the other side. Make sure that there are no air bubbles. If necessary, repeat this procedure. 4. Install all of the knurled-head screws and tighten them by hand. C. Filling and injecting the ink: 1. For good results, it is sufficient to dilute the provided ink in a ratio of 1:5. 2. Make sure that valve (B) is closed and that the pump of the HM 150 is switched on. 3. Slightly open valve (9a). Water will flow through the flow chamber. 4. Fill the reservoir tank to the 2/3 with diluted ink. 5. Open valve (C) and slightly open valve (B). Ink will now flow through the thin hose to the injection bore holes. 6. When the water at valve (C) has a bluish colour, close valves (B) and (C). 7. The apparatus is now ready to perform experiments. 8. In order to avoid the formation of air bubbles, make sure that there is always enough ink in the reservoir tank! RESULT: Thus the fluid flow path of the liquid is shown according to the obstacle. Fig 8. Flow Visualization
• 26. 26 9. DARCY’S FRICTION FACTOR (FRICTION LOSS –MAJOR LOSS) THEORY: When a liquid is flowing through a pipe with high velocity a shear stress is produced in the whole liquid due to viscosity. The viscous action causes loss of energy which usually known as friction loss. This is major loss of energy when a liquid flow through pipe. The minor losses are loss due to sudden contraction, loss due to bending and loss due to sudden enlargement etc. AIM: To determine the Darcy‟s friction factor for a material of the given pipe and to study the variation of Reynolds number under varying flow conditions (laminar or turbulent flow). APPARATUS REQUIRED: 1. Darcy‟s friction apparatus 2. Collecting tank 3. Stop watch 4. Meter scale FORMULA: 1. Difference in Head H = H1 – H2 (m) H1 - Sill Level of the tank in „m‟ H2 – Final Hook Gauge Reading in „m‟ 3. Actual Discharge Qa = y / t (m3 /s) y = Depth of changes of water in collecting tank in “liter” t = time taken for rise of water in the collecting tank in „sec‟ 4. Velocity v= Qa/A Qa is in m3 /sec (1 liter/Sec = 0.001 m3 /Sec) A= Area of pipe at diameter = 3mm 5. Actual friction factor fa = 2 g d H/(4xlxv2 ) g- Specific gravity in m/s2 . d- Dia of pipe = 3 mm l = Length of pipe = 400 mm.
• 27. 27 6. Reynolds number Re = v x d/υ . v = Velocity of water in m/s υ = Viscosity of water =0.0107x10-4 m2 /Sec. If Re < 2300 Laminar flow Re > 2300 Turbulent flow 7. Theoretical Friction factor fth fth = 16/Re for laminar flow fth = 0.0791/(Re)0.25 for turbulent flow 8. Percentage of deviation % = (fa- fth)x100/fa PROCEDURE: 1. The supply valve is opened and water is allowed to flow in the pipe. 2. Different in monometer readings are noted (H1, H2) 3. The outlet valve of the collection tank is tightly closed and the time taken for „y‟ liter rise in the collection tank is noted. 4. The experiment is repeated by varying the discharge through the channel by adjusting the control valve. GRAPHS i. Velocity Vs Head loss (Ordinary Graph) OBSERVATION Diameter of the pipe =………………… m Length of the pipe = …………………...m Kinematic viscosity =…………………. m2 /Sec RESULT The average coefficient of friction (practical) f = The average coefficient of friction (Theortical) f =
• 28. 28 TABULATION: S. No Manomete r Reading in m Loss of head H= H1-H2 in x10-2 m Water lever rise in collect ing tank ‘y’lite r Time taken for ‘y’ liter rise in the tank in sec Actual discharge Qa m3 /Sec Velocity (V) in m/sec Actual friction factor (fa) Renol ds numbe r (Re) Theoriti cal friction factor (fth) % of Devia tion H1 H2 1 30 28 2 0.2 230 0.00087 0.123 0.0486 345 0.0463 4.7 Average = MODEL CALCULATION: 1. H= 30 -28 = 2 cm. 2. Qa = 0.2/230 = 0.00087l/s = 0.00087x10-3 m3 /s. 3. V = Qa/A = 0.00087x10-3 /7.07x10-6 = 0.123 m/s. 4. A= (π/4)xd2 = (π/4)x(3x10-3 )2 = 7.07x10-6 m2 . 5. fa = 2 g d H/(4xlxv2 ) = 2x9.81x0.003x0.02/(4x0.4x0.1232 ) = 0.0486 6. Re = v x d/υ = 0.123x 0.003/ 0.0107x10-4 = 344.85 7. fth = 16/Re = 16/ 344.85 = 0.0463 8. % = (fa- fth) x100/fa = (0.0486- 0.0463)x100/0.0486 Fig 9. Friction loss through pipe
• 29. 29 10. VELOCITY DISTRIBUTION IN AERODYNAMIC TRAINER THEORY: Aerodynamics Trainer can be used to carry out extensive experiments on the subject of aerodynamics. The experiments from the field of flow around bodies are 1.velocity measurement of flows with Pitot tube. 2. Boundary layer analysis on a flat plate with incident flow 3.Visualisation of streamlines AIM: To determine the velocity distribution on a plate using aerodynamic trainer kit. APPARATUS REQUIRED 1. Aerodynamic trainer kit 2. Plate. 3. Vaporization chamber. FORMULA: 1. Pdyn = Ptotal- Pstat. Pdyn = Dynmic pressure in Pa. Ptotal = Total pressure in Pa. Pstat = Static pressure in Pa. 2. Velocity (v) = √(2xPdyn/ρ) ρ=1.204 kg/m3 . PROCEDURE: 1. Install necessary arrangements 2. Measure ambient temperature and air pressure and use these to determine the current air density. 3. Push the plate to the vertical position x = 0 mm (distance from the leading edge of the plate) and secure with knurled screw. 4. Read the atmospheric pressure off a manometer tube not connected to a measuring point. 5. Read the measured values for Ptotal and Pstat off the manometer tubes. 6. Use the micrometer screw to increase the horizontal position y (horizontal distance from the plate) and read the measured values for Ptotal and Pstat off the manometer tubes.
• 30. 30 7. Repeat this step several times to obtain a series of measurements. GRAPHS i Vertical position (x) Vs Flow velocity n (Ordinary Graph) ii. Horizontal position (y) Vs Flow velocity (Ordinary Graph) OBSERVATION Density of vapour ρ=1.204 kg/m3 . RESULT: Thus the velocities are calculated and velocity gradients are generated.
• 31. 31 TABULATION: S.No Horizontal y Axis=10mm Horizontal y Axis Ptotal in Pa Pstat in Pa Velocity in m/s Ptotal in Pa Pstat in Pa Velocity in m/s Verti cal X- axis 0.35 142 202 31.6 MODEL CALCULATIONS: 1. Pdyn = Ptotal- Pstat. = 1420-2020 =600 Pa. 2. Velocity (v) = √2xPdyn/ρ. = √(2x600/1.204) =31.6 m/s. Fig 10.Velocity distribution in plate
• 32. 32 11. DRAG MODEL OF AERO FOIL SECTION THEORY: A. Coefficient of lift (CA) is a dimensionless coefficient that relates the lift generated by a lifting body to the fluid density around the body, the fluid velocity and an associated reference area. A lifting body is a foil or a complete foil-bearing body such as a fixed-wing aircraft. CA is a function of the angle of the body to the flow, its Reynold number and its Mach number. The lift coefficient cl refers to the dynamic lift characteristics of a two-dimensional foil section, with the reference area of the foil. B. Coefficient of drag (Cw) is a dimensionless quantity that is used to quantify the drag or resistance of an object in a fluid environment, such as air or water. AIM: To determine the coefficient of drag and coefficient of lift for the given aero foil section with use of wind tunnel. APPARATUS REQUIRED 1. Wind tunnel. 2. Aerofoil section. 3. Computer system. FORMULA: 1. Holder resistance FH =0.5 x ρ CwAHOLDER x V2 . ρ= Density of air = 1.225 Kg/m3 . CwAHOLDER = 0.000125 m2 . V = Velocity of air in m/s. 2. Area of the aero foil A = C x L C – Chord length in mm L – Chord width in mm 3. Drag force Fw= Fwgem – FH. Fwgem - Drag force in Newton. 4. Coefficient of lift (CA) = 2FA / (ρ A V2 ). FA = Lift force in newton.
• 33. 33 5. Coefficient of drag (CW) = 2FW / (ρ A V2 ). FW = Drag force in Newton. 6. Re = Vx Lx ρ/μ μ = Dynamic viscosity = 18.1x10-6 kg/ms. PROCEDURE: 1. Install the aero foil section in the wind tunnel and close the tunnel 2. Apply the necessary speed to the fan. 3. Note down the velocity of air, lift and drag force. 4. Repeat the process with higher speeds. 5. Note down the value and tabulate the record. GRAPHS: 1. Coefficient of lift Vs Coefficient of Drag 2. Coefficient of Drag Vs Angle of inclination 3. Coefficient of lift Vs Angle of inclination. OBSERVATION: i. Chord Length L = 100 mm ii. Chord Width W= 7.5mm RESULT: Thus the average coefficient of lift= ____________. Thus the average coefficient of Drag= ____________.
• 34. 34 TABULATION: S. No Angle of inclina tion in Degree Veloc ity (S) Drag force in Newton (Fwgem) Lift force in Newton (FA) Holder resistance (FH) in Newton Drag force (Fw) in Newton Coeffici ent of lift (CA) Coefficie nt of drag in (CW) Renol d‟s Numb er 1 10 25 2.23 2 0.05 2.18 MODEL CALCULATIONS: 1. FH =0.5 x ρ CwAHOLDER x V2 = 0.5 x 1.225x0.000125x252 = 0.04785=0.05 N. 2. A = C x L = 0.1x0.0075 = 7.5x10-4 m2 . 3. Fw= Fwgem – FH.= 2.23 – 0.05 =2.18 N. 4. CA = 2FA / (ρ A V2 ) = 2x2/(1.225x 7.5x10-4 x252 ) = 6.96 5. CW = 2FW / (ρ A V2 ) = 2.18x2/(1.225x 7.5x10-4 x252 ) = 7.6. 6. Re = Vx Lx ρ/μ = 25x0.0075x1.225/1.8x10-6 = 127604 s/m2 . Fig 11. Drag model of aerofoil section
• 35. 35 12. TUBULAR HEAT EXCHANGER (PARRALLEL FLOW) THEORY: A heat exchanger is a device used to transfer heat between one or more fluids. The fluids may be separated by a solid wall to prevent mixing or they may be in direct contact.[1] They are widely used in space heating, refrigeration, air conditioning, power stations, chemical plants, petrochemical plants, petroleum refineries, natural-gas processing, and sewage treatment. The classic example of a heat exchanger is found in an internal combustion engine in which a circulating fluid known as engine coolant flows through radiator coils and air flows past the coils, which cools the coolant and heats the incoming air. There are several types heat exchanger shell and tube heat exchanger and plate type heat exchanger etc. AIM: Parameter determination of the tubular heat exchanger in parallel flow of water. APPARATUS REQUIRED: 1. Tubular heat exchanger setup. 2. Water tank. FORMULA: 1. LMTD = Logarithmic Mean Temperature Difference. LMTD = [Thi –Tci]- [Tho – Tco] / {ln [(Thi- Tci)/(Tho- Tco)]} Where Tci = T6 = Entry temperature of cold fluid in Kelvin. Thi = T1 = Entry temperature of hot fluid in Kelvin. Tco = T4 = Exit temperature of cold fluid in Kelvin. Tho = T3 = Exit temperature of hot fluid in Kelvin. 2. Qh = Heat transfer rate from hot water in KJ = mh x Cph [Thi – Tho] Where mh = Mass flow rate of hot water [Kg/s] Cph = Specific heat of hot water [KJ/KgK] = 4.187 KJ/KgK 3. Qc = Heat Transfer rate to the cold water = mc x Cpc [Tco- Tci] Where mc = Mass flow rate of cold water [Kg/s] Cpc = Specific heat of cold water [KJ/KgK] =4.187 KJ/KgK 4. Q = Heat transfer rate in Watts = [Qh + Qc] / 2 5. U = Overall Heat transfer co-efficient W/m2K = Q/(A x[ΔT]M) Where [ΔT]M = LMTD
• 36. 36 A = Area = πdl 6. Cr = Cmin/ Cmax Ch = Cph x mh Cc = Cpc x mc In Cc , Ch which is minimum called Cmin and which is maximum called Cmax. 7. NTU = No of transfer units = Ux A/ Cmin 1- exp [ - NTU x (1+ Cr)] 8. Effectiveness E = ------------------------------------ 1+ Cr PROCEDURE: 1. Give the necessary connection to the set up. 2. Heat the water in the setup using heater. 3. Give the flow of hot water and cold water using the valve according to the diagram. Note down the flow rate of hot and cold water. 4. Now the change in temperatures take place, note down the temperatures after the change in temperatures reaches a steady value 5. Repeat the process of for other flow rates. 6. Tabulate the value and plot the graph. GRAPHS: Heat transfer rate Vs Effectiveness. Observation: Overall length (L) = ……..560…….. mm. Diameter (D) = ………..7…….. mm. RESULT: 1. Heat transfer rate (Q)= _______________ W 2. Overall heat transfer coefficient (U)= _____________ W/m2 k. 3. Effectiveness (E) = ___________ .
• 37. 37 TABULATION: S . N o Flow rate of Hot water Flow rate of Cold water Inlet temp of hot water (Thi) (T1) outlet temp of hot water (Tho) (T3) Inlet temp of cold water (Tci) (T6) Outlet temp of hot water (Tco) (T4) LMT D Hea t tran sfer rate (Q) Ove r all heat tran sfer coef ficie nt (U) Effe ctiv enes s (E) L/hr Kg /s L/mi n Kg /s 0 C K 0 C K 0 C K 0 C K 1 144 .04 120 .0333 56 329 45 318 34 307 39 312 12.31 1.27 8.4 .403 MODEL CALCULATIONS: LMTD = [Thi – Tci] - [Tho – Tco] / ln [Thi – Tci/Tho – Tco] = [329 – 307] – [318 – 312] / ln [(329 – 307) / (318 – 312)]= 12.31 K. Qh = mh x Cph [Thi – Tho] = 0.04 x 4.187 x [329 – 318] = 1.842 KJ/sec. Qc = mc x cpc [Tco –Tci]= 0.0333 x 4.187 [312 – 307] = 0.691 KJ/sec. Q = [Qh + Qc] / 2 = [1.842 + 0.691] / 2 = 1.27 KJ/sec. A = π x D x L= π x 0.007 x 0.56= 0.0123 m2 . U = Q/(A x[ΔT]M)= 1.27 / (0.0123 x 12.31)= 8.4 W/m2 K. Cr = Cmin/ Cmax = 0.14/0.167= 0.8383 Ch = Cph x mh = 4.187 x 0.04= 0.167 = Cmax Cc = Cpc x mc = 4.187 x 0,0333= 0.140 = Cmin NTU = Ux A/ Cmin = 8.4x0.0123/ 0.14 = 0.737.
• 38. 38 1- exp [ - NTU x (1+ Cr)] 1- exp [- 0.737x 1.8383] 9. E = ---------------------------------- = --------------------------------- = 0.4036 1+ Cr 1.8383 Fig 12.Parrallel Flow Tubular Heat Exchanger
• 39. 39 13. TUBULAR HEAT EXCHANGER (COUNTER FLOW) THEORY: A heat exchangers are classified parallel flow and counter flow based on the direction of both the fluids flow. In parallel flow heat exchanger both the fluids (hot and cold) are flowing in the same direction. But in counter flow heat exchanger both the fluids are flowing in the opposite direction. AIM: Parameter determination of the tubular heat exchanger in counter flow of water. APPARATUS REQUIRED: 1. Tubular heat exchanger setup. 2. Water tank. FORMULA: 1. LMTD = Logarithmic Mean Temperature Difference. LMTD = [Thi –Tci]- [Tho – Tco] / {ln [(Thi- Tci)/(Tho- Tco)]} Where Tci = T4 = Entry temperature of cold fluid in Kelvin. Thi = T1 = Entry temperature of hot fluid in Kelvin. Tco = T6 = Exit temperature of cold fluid in Kelvin. Tho = T3 = Exit temperature of hot fluid in Kelvin. 2. Qh = Heat transfer rate from hot water in KJ = mh x Cph [Thi – Tho] Where mh = Mass flow rate of hot water [Kg/s] Cph = Specific heat of hot water [KJ/KgK] = 4.187 KJ/KgK 3. Qc = Heat Transfer rate to the cold water = mc x Cpc [Tco- Tci] Where mc = Mass flow rate of cold water [Kg/s] Cpc = Specific heat of cold water [KJ/KgK] =4.187 KJ/KgK 4. Q = Heat transfer rate in Watts = [Qh + Qc] / 2 5. U = Overall Heat transfer co-efficient W/m2K = Q/(A x[ΔT]M) Where [ΔT]M = LMTD A = Area = πdl 6. Cr = Cmin/ Cmax Ch = Cph x mh Cc = Cpc x mc
• 40. 40 In Cc , Ch which is minimum called Cmin and which is maximum called Cmax. 7. NTU = No of transfer units = Ux A/ Cmin 1- exp [ - NTU x (1- Cr)] 8. Effectiveness E = ------------------------------------------- 1- {Cr xexp [ - NTU x (1- Cr)]} PROCEDURE: 1. Give the necessary connection to the set up. 2. Heat the water in the setup using heater. 3. Give the flow of hot water and cold water using the valve according to the diagram. Note down the flow rate of hot and cold water. 4. Now the change in temperatures take place, note down the temperatures after the change in temperatures reaches a steady value 5. Repeat the process of for other flow rates. 6. Tabulate the value and plot the graph. GRAPHS: Heat transfer rate Vs Effectiveness. Observation: Overall length (L) = ……..560…….. mm. Diameter (D) = ………..7…….. mm. RESULT: 1. Heat transfer rate (Q)= _______________ W 2. Overall heat transfer coefficient (U)= _____________ W/m2 k. 3. Effectiveness (E) = ___________ .
• 41. 41 TABULATION: S. N o Flow rate of Hot water Flow rate of Cold water Inlet temp of hot water (Thi) (T1) outlet temp of hot water (Tho) (T3) Inlet temp of cold water (Tci) (T4) Outlet temp of hot water (Tco) (T6) LMT D Hea t tran sfer rate (Q) Over all heat trans fer coeff icien t (U) Effe ctiv enes s L/h r Kg /s L/hr Kg /s 0 C K 0 C K 0 C K 0 C K 1 288 .08 191 .053 82 355 57 330 35 308 51 324 26.24 5.95 18.43 0.44 MODEL CALCULATIONS: LMTD = [Thi – Tci] - [Tho – Tco] / ln [Thi – Tci/Tho – Tco] = [355 – 324] – [330 – 308] / ln [(355 – 324) / (330 – 308)]= 26.24 K. Qh = mh x Cph [Thi – Tho] = 800 x 10-4 x 4.187 x [355 – 330] = 8.347 KJ/sec. Qc = mc x cpc [Tco –Tci]= 0.053 x 10-4 x 4.187 [32 4 – 308] = 3.551 KJ/sec. Q = [Qh + Qc] / 2 = [8.347 + 3.551] / 2 = 5.95 KJ/sec. A = π x D x L= π x 0.007 x 0.56= 0.0123 m2 . U = Q/(A x[ΔT]M)= 5.95 / (0.0123 x 26.24)= 18.4352 W/m2 K. Cr = Cmin/ Cmax = 0.335/0.222 = 1.51 Ch = Cph x mh = 4.187 x 0.08 = 0.335 = Cmax Cc = Cpc x mc = 4.187 x 0.053 = 0.222 = Cmin NTU = Ux A/ Cmin = 18.4352x0.0123/ 0.222 = 1.02.
• 42. 42 1- exp [ - NTU x (1- Cr)] 1- exp [- 1.02 x -0.51] E = ------------------------------------ = --------------------------------------- = 0.44 1- {Cr xexp [ - NTU x (1- Cr)]} 1- {1.51xexp [- 1.02 x -0.51]} Fig 13. Counter Flow Tubular Heat Exchanger