Heat Transfer Prof. Dr. Aloke Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati


 Denis Armstrong
 5 years ago
 Views:
Transcription
1 Heat Transfer Prof. Dr. Aloke Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module No. # 02 One Dimensional Steady State Heat Transfer Lecture No. # 05 Extended Surface Heat Transfer 2 Welcome to lecture 5 of module 2 on heat transfer. This lecture will continue our previous discussions on extended surface heat transfer, and the previous discussions what we have seen that why we need that extended surface heat transfer. And then for different case we have developed a generalized equation for temperature profile for a field with constant area of heat transfer, conduction heat transfer area. And then we seen that how the generalized temperature equation can be reduced to a specific form depending upon a see different boundary conditions that we have sorry people defined cases and in both the in all the 3 cases we have different boundary conditions that have been applied, and also you have seen how to find out the efficiency for the fins. Now, we will carry on that discussion and we will try to see that how are the efficiency expressions for the different cases that we have discussed in case of fins, and then later on we will take up the problem on this. (Refer Slide Time: 01:34)
2 So, to carry on the discussion then in the first case, case 1 we had as we have. So, Case 1 we have already discussed and this is the case 1 the field is in find at a long, and for that situation we have found out that temperature profile happens to be like this T naught T bar by T naught bar equals to e to the power minus m x, and for this kind of situations we know that if we also we have seen in case of find heat transfer that efficiency expression is like this it is minus k into A into dt by dx at x is equal to 0 and this is dt bar by dx at x equal to 0 divided by 8 into A T into T 0 bar. And this here. So, if we try to see that from this expression if we find out what is d T by d x. So, d T bar by d x at x equal to 0, this is equal to minus in T bar 0 and if we put this here, then it becomes minus minus cancels. So, k into a m into A into T bar 0 divided by 8 into. Now what is this A T, if we go back to our previous discussions, we will see that A T is that b this area plus the surface area of this. So, if this is L into this is the L into say B you can say this is the surface area and in the bottom also we will be having similar surface area and here also this. So, we can say that if this is the perimeter p into this length p into L will be the surface area of this side and the bottom side and this will be the surface area extra that is b into h, then this will be h into that 8 A T will be equal to the total perimeter total perimeter is P into L length plus B into H, that will be the total area into T 0 bar and we know that m is equal to we have seen we have already decided that m is equal to root over h P by k into A. So, then it becomes actually, if we substitute this value of n over here we will get root over of h P k into A divided by what we will do we will assume that and also it is very practical in case of long a field that B H this area will be much much less compared to P L. So, therefore, we will write at we will simplify it to be 8 into P L. So, if we write like that we can also write it as k A root over of k A by root over of P L into 1 by L sorry. So, this is equal to root over of k A by root over of h into p into 1 by L and you know that root over A 3 by k l is equal to m. So, we can write as one by m into l; that means, what for a long field the efficiency is 1 by m into L and is we understand as we understand that if l tends to infinity that indicates that eta efficiency is tending to 0. So, that is what I was telling that because of this reason infinitely long finishes got no practical relevance in the thing. In the point of view there is no efficiency if it is getting 0
3 and second point of this from the application of point of view for a long very long fins what are you going to fit it what are you going to accommodate it. So, that is not a practical solution. So, any way, but for case one that efficiency expression is becoming 1 by m L. (Refer Slide Time: 06:43) Now, if we go for case 2, in case of case 2 T bar already we have seen that T as we have seen in previously T bar by T 0 bar and that is equal to cos hyperbolic m into L minus x plus L by m k into sin hyperbolic m into L minus x divided by cos hyperbolic m L plus h by m k sin hyperbolic m into L. Now, d T by d x bar by d x at x is equal to 0 in this case is equal to minus m T 0 bar into sin hyperbolic m L plus h by m k cos hyperbolic m L divided by cos hyperbolic m L plus h by m k sin hyperbolic m L. Then eta efficiency is again as we have seen our previous case it will become root over of h p k A by h of 8 into p L plus B H into this sin hyperbolic m L plus h by m k cos hyperbolic m L divided by cos hyperbolic m L plus h by m k sin hyperbolic m L. Now, if we neglect again in certain situations we may neglect this and if we neglect B H if B H is less than p L then we can write it again as 1 by m L into as we have seen in the previous cases that this one is equal to 1 by m L h p L is equal to this equal to 1 by m L. So, it will be 1 by m L into this part. So, that is a factor that is coming for this case, which is sin hyperbolic m L plus 8 by m k cos hyperbolic m L divided by cos hyperbolic
4 m L plus h by m k sin hyperbolic m L. So, this we can see that with respect to the previous case here there is a factor that is come into picture when you have case of that a finite length of the field is considered. (Refer Slide Time: 10:11) Now coming to Case 3 when the tip is insulated in that case we have already find out a temperature profile as T bar by T 0 bar and that is equal to cos hyperbolic sorry m into L minus x divided by cos hyperbolic m into L. That is what we have got now if we again find out that eta if eta to find out. So, then we have to find out d T bar by d x at x is equal to 0 that for finding out the actual heat transfer at x equal to 0 then it is becoming minus n sin hyperbolic m L by cos hyperbolic m L into T bar. So, that says that if you put again that eta f this is equal to again root over of h p k A by h into p L plus B H and then tan hyperbolic m L and again in the same lump sum if you take that if B into h the product is less than p L, then we can write eta f is equal to tan hyperbolic m L by m L. So, that becomes the efficiency for the case 3 and in all the cases you can say that that the area of heat transfer by the field at the tip where which is equal to the breath of the fin into the thickness of the fin and in most of the situations, the thickness of the fins are in mille meters and then the total area with respect to the total area the p into L that main area of heat exchange by the fin happens to be much much higher compared to this area B into h. So it becomes a practical assumption engineering
5 assumption that B H is much much less compare to p L. So, B H is taken to B negligible or taken to be assumed to be 0. Under that situation, we have found out how what happens to the efficiency terms. Now whatever we have seen as of now is that 3 efficiency terms you have found out 3 efficiency expressions we have found out for t3 different cases and whatever we have found out in all the cases we have found, we have seen that heat conduction area for the fin that was constant. It is never changing, but as we have shown you in previously, in previous lecture there are different types of field we have shown and in many a situation we have seen there can be a variable area for the fins and when there is a variable area of the fin is involved then this derivation does not work out. Then we have because while deriving this we have taken area to be constant, but that area is a function of x. So, we have to take into consideration the function area as a function of x and then you have to go for the a derivation and then it looks to be a little bit more complicated. So, for the time being we are not discussing about that complications and not going for that derivations students interested in doing this they can he can always find out in that case he has to do that area is a function of x he has to find out it can be a linear function, but particularly when there is a trapezoidal kind of things when there is a conical end it can be linear function. But when there is a parabolic profile is there, then it need not be a linear function the area is need not be a linearly decreasing with respect to x and then resistant has to be little bit cost as above that. So, therefore, we just wanted to restrict ourselves with the simplest situation by that area of heat conduction through the fin is remaining more or less constant it is not changing as we go along the length of the length of the fin.
6 (Refer Slide Time: 14:51) Now, we will take of a problem on this fin heat transfer and we will see that how it work out. So, let us start the problem, problem 1 it says a carbon steel carbon steel pipe the outside diameter o d is equal to 60 mille meter has 8 longitudinal fins of thickness 1.6 mille meter the fin the fins extends sorry the fins extend the fins extend 4 centimeter from the pipe wall thermal conductivity of fin material of the fin material is 50 watt per meter per Kelvin the pipe surface temperature is 200 degree centigrade ambient temperature is 30 degree centigrade and heat transfer co efficient is and the heat transfer is 100. We have to calculate the percentage increase calculate the percentage increase, in the rate of heat transfer for the as well as the percentage fin effectiveness for the finned tube over the plain tube. So, this is the problem description it states it says that there is carbon steel and a carbon steel pipe and this pipe has got 8 longitudinal fins and they are of thickness 1.6 mille meter and the fin e extends 4 centimeter and that necessary information is like thermal conductivity the temperatures heat transfer coefficients are given. So, you have to find out what would be the speed efficiency and what would be the heat transfer and all this.
7 (Refer Slide Time: 19:12) So, if we just try to see that it is like this. So, we have a pipe and say this is the longitudinal fin that is being there and say this is the center line. So, there are like this way there are 8. So, if I take front sorry a top view of this it will look like this. So, there will be like this there will be some something like that. So, there will be sorry it will be like this it will be and 8. So, These e 8 fins are there at different sideways and this value has been given as diameter is this. So, 0.3 centimeter and this portion has been given as 4 centimeter. So, like this way there are 8 fins that is there here longitudinal fins are there fine. Now if you try to see that in how to tackle the problem or to find out the values of efficient first and to find out the efficiency, we need to know first which conditions we have to use this is a case straight forward inside case 2; that means, fin is of definite length fin is of definite length and that is too though tube is not insulated under that situation, if you try to calculate the efficiency of the fin the efficiency of the fin is eta is equal to 1 by m L sin hyperbolic m L plus h by m k cos hyperbolic m L then cos hyperbolic m L plus h by m k sin hyperbolic m L. So; that means, we have to calculate the value of m L and we have to calculate the value of m k h by m k now h by m k is nothing but we know h is equal to 100 and m is that we have to calculate. So, if you see that m is equal to root over of h p by k into area h p by k into area. So, this is that equal to h into root over of h into the p is equal to 2 into p is
8 equal actually b plus h as we have seen in the previous diagram we have seen that the weight blast thickness that is this is the p perimeter is this is the perimeter and this is the sorry this is B this is B this is H and this is H. So, 2 into B plus H is the perimeter and divided by k into area now if we say that h is relatively as you know is smaller than d, because we know that h is equal to 1.6 millemeter been given h is equal to. So, if we consider that perimeter like this is say B this is B this is H and this is H, B is that that breath like the length of the pipe and H is the thickness. So, we know that B plus B plus H plus H is the perimeter and in general we know that h is the thickness and which is much much less compared to the B values and therefore, we can write that 2 into B plus H we can write nearly equal to 2 into B. And therefore, what we can get is from here H into 2 into B divided by we have k into and area is equal to B into H. So, so this B B it cancels. So, we can get this is equal to root over of twice h by k into h now this is equal to m. So, m L is equal to sorry I will write here. So, m L is equal to then root over of twice h by k h into L and in this problem, this value is all this value has given h is equal to 100 k is 50 and H we know So, this is root over of 2 into 100 by k is equal to 50 into h is equal to into L is equal to we have 0.04 this is length this is L. So, then we can we will get the value of m L is equal to 2, you get the value of m L is equal to 2 and then and also we get the value of m is equal to value of m also we get if we put this square we get the value of m is equal to 50. So, then h by m k is becoming 50 into m is equal to 50 into k is equal to another 50. Then this value is becoming again So, h by m k is becoming Now if you put the value of all these here m L is equal to 2 h by m k is equal to So, eta value is becoming 1 by 2 into sin hyperbolic 2 plus 0.04 cos hyperbolic 2 by cos hyperbolic 2 plus 0.04 sin hyperbolic 2. And then this value is coming to be So, efficiency when we consider in these 2 situations the efficiency is becoming Incidentally also that if we consider even the case 3 here case 3 remains that T p is in; that means, there is no heat transfer from tube which is merely equal to assuming the heat transfer area negligible, because if we assume that B into H is very small if we assume that this heat transfer again is very small for the heat transfer which is; that means, the ratio heat transferred through the tube will be almost 0 which is equivalent to the case assuming that T is insulated.
9 (Refer Slide Time: 27:43) So, that way if we apply case 3 to find out efficiency, we will find that efficiency expression,if we apply that is eta is equal to tan hyperbolic m L by m L, and we have seen the values that it is equal to tan hyperbolic m L is equal to 2 by 2 and this value is So, we say that previous case the efficiency is marginally more we have seen that in the previous case the efficiency was and now in present situation it is 0.48 only. So, there is a very small difference in the efficiency and which is only this much difference in the efficiencies obtained. So, this is almost negligible therefore, either of case 2 or of case 3 can be used for this kind of analysis, because that tube area this it depends upon the thickness of the fin where the tube area fin area is almost negligible and. So, heat transferred through this is almost negligible. So, case 2 and case 3 are almost coinciding with each other.
10 (Refer Slide Time: 29:04) Now, after we find out the efficiency then we have to find out the heat transfer we have to find out. So, for tube without fin. So, heat transfer rate if we try to calculate for that we have to find out that area of bear tube or according to the problem this is the pipe plus pipe. So, it area of the bear tube or pipe and that is equal to you know pi d 0 into in the into the L of the small l length of the here pipe. And if we this will become now this will become now pi into 0.06 is the outer diameter into say one is the per unit meter length. So, that we are assuming the length is equal to 1 meter this much meter square we are assuming L is equal to 1 meter length of the pipe. So, then it is becoming meter square and then heat transfer rate p is equal to h into area of the bear pipe into delta t and this is equal to h is 100 and area of the bear pipe is in is equal to into delta T is equal to 170. So, this becoming kilo watt.
11 (Refer Slide Time: 31:32) So, this is the heat transfer rate by the heat transfer rate by the pipe when there is no fin associated to it. Now if we try to find out the fin area of the fin fins transferring heat if we try to see here we will get that 2 into 1 into 0.04 this is the if we go back to our this thing we will see that we have taken this one to be 1 meter. You have taken this one to be 1 meter. So, 1 into 0.4 that is the area and it is in both sides heat is transferred from both the sides. So, then it becomes 2 into 1 into 0.04 this mille meter square this plus the area of tube and if we neglect the area of the tube. So, this is becoming actually 0.08 meter. And if we try to see that area of tube for each case is equal to into 1; that means, meter square. So, we can see that that area of the tube is very small compared to the total area therefore, we can simply neglect that area and then for 8 fins the area is 8 into 0.0 sorry 0.08 meter square and that is equal to 0.64 meter square this is the area of the fins that is transferring heat and then we have to find out the area which is not occupied by the fin and that area of their area of the pipe which is not occupied by the fin that is also responsible for heat transfer. So, that amount will be. So, they are area of pipe of finned pipe. This will be equal to 0 point I am sorry 0. Yeah. We have seen this is the area of the total pipe minus this is the area occupied by 1 fin there will be 8 fins. So, minus 8 into meter square and then it becomes meter square.
12 So, total area heat exchange area of the finned pipe and that is equal to 0.64 plus and that is equal to meter square. Now we have found out the area of this finned pipe. (Refer Slide Time: 35:10) So, heat transferred by the fin pipe heat transferred rate by v finned pipe p is now efficiency of the fin we have found out. So, 0.48 into the area of the fin is 0.64 into 170 is the driving force delta T into h is 100. So, this is actually we should write this will be heat transfer co efficient into area of the fin into driving force into eta A. So, this eta A efficiency of the fin is 0.48 then heat transferred area is equal to 0.64 then driving force is delta T is this and h is equal to 100 this plus this is the from the fin part and from the unoccupied part of the pipe and that is equal to say 8 again into a pi into delta T and that is plus will have h is equal to again 100 into area of the pipe here is just we have seen into 170 this area is the area of the pipe surface area of the pipe minus the portions which are occupied by the fins. So, then it becomes actually kilo watts. So, we have seen when the pipe was not finned, then that heat transfer was kilo watt and now we have found that heat transfer rate is kilo watt. Therefore. So, change percentage increase equals to minus by into 100. So, this is equal to percent. So, we can see that it is 156 times. This many percentage 156 percentage increase in the heat transfer rate is there, when the this kind of fin is there and
13 effectiveness is percentage effectiveness equals to that heat transfer just with fin divided by into 100 and this is may be called to percentage. So, effectiveness is 256 percent and percentage increase in heat transfer rate is percentage. So, this gives an idea that how fin helps us to increase the rate of heat transfer this problem will help us to understand that. Now one thing I should clear here, in case of circumferential fins or radial fins the area of conductive heat exchange that changes and that varies and in that situation, this kind of analysis will not be working to find out that values of m L and efficiency. So, what is being done in that case is there are certain charts are available to find out the values of efficiency for circumferential fins and or radial fins some charts are available. (Refer Slide Time: 39:27) And this charts are basically with respect to fin efficiency what says an expression that is h by k into A m whole to the power half into l to the power of 3 by 2 where A m is called profile area h is heat transfer coefficient k is thermal conductivity of the fin material and A m is called profile area this is equal to L into H the length of the fin into H thickness of the fin and L is the length of the fin. In some situations it may happen that that L is replaced by L c that is corrected length fin length. Then in that case the efficiency, if the plot is efficiency one says H by k into Am whole to the power half into L c to power 3 by 2 and L c is equal to L plus H by 2 and L
14 c is equal to L plus d by 4 this is for rectangular profile and this is for fin thin fins thins names this circular rod and d is the diameter of the rod fin rod. So, this way we find out the from the chart we can find out the values of efficiency for fins which has got a variable idea and then once we know the efficiency then rest of the part is the actually, heat transfer we can calculate that all that we have seen through this problem also actual heat transfer rate that is equal to that efficiency of the fin into a heat h A into T 0 bar. So, A is the total area of heat transfer h is the conductive heat transfer coefficient and T bar is the driving force delta T. From this, we can find out that actual heat transfer rate by the fins by the fins that has got variable conductive surface area. Now this way we try to explain to some extend on the extended surface area and how a heat transfer takes place through the fins and how the temperature profiles are there how are the efficiencies what is the effectiveness of the fin let us say that we have discussed. (Refer Slide Time: 43:12) Now, one specific topic I just wanted to tell here is that is called Thermal Conduct Resistances thermal conduct resistance now from the name itself we can understand like here.
15 (Refer Slide Time: 43:39) We can see that when 2 bodies come in physical contact A and B they are coming in physical contact, then what may happen there can be some extra resistance encountered due at this contact point. If you see that this is a just magnification of this is been shown here and once you see this that there is basically, some kind of roughness always they are in any surface there is some roughness when this 2 surface joins with each other then there will be some empty space that is been developed. And these empty spaces are being occupied by some gases it may be here it may be some other gases the empty spaces are occupied. So, what happens when the heat is being transferred say from A to B when the heat is being transferred in this direction from A to B what is going to happen is, that part fins is going through the direct contact of the solids in some places there is no contact directly by the solid the contact is by the gas phase inside these between the solids. So, part of this is by the direct contact between solids A and B and part of the transport is taking place through the gas or air which is entrapped in the wide space. Now, we understand or we know that thermal conductivity of the gas or air is very low compared to the solids and then therefore, what happens as because the as because it is expected that the whole things would have been the solid contact, but as because it is not the case then what happens there is a an extra resistance that is being developed and therefore, they we will find that there is a drop in temperature in the contact zone
16 between these 2 bodies and this is called this drop in temperature is happening due to the resistance being offered by the fluid entered into this place and therefore, it is called thermal contact resistant. And if you see that that if we draw a profile of the temperature we will find the suppose this is as we have seen that this is A and this is B and if we draw the profile here we will see heat transfer is taking place from A to B. So, here the heat transfer is taking place. So, now, this is these are the 2 temperature profile this is the temperature profile. So, they say this is T 1 say this is T 2 A this is T 2 B and this is say T 3. So, what is happening from T 1 to T 2 A there is a profile change there is a profile this is this is depending upon the thermal conductivity of a between T 2 B 2 T 3 again there is a another profile this is depending upon normal conductivity of the B, but at this junction there is a drop from T 2 A to T 2 B and this drop is basically, due to the resistance being offered due to the pressures of some gases entered in the wide space that is being created when they are in contact and that is being created due to the non smoothness of the surfaces or rather roughness of the surfaces. Now, this can be much easily understood if we increase the external pressure here, or in the other words other around, if we decrease the pressure here then what will happen, if we just decrease the pressure of the surrounding then the pressure of the fluid interrupt in this point then in this thermal contact region will be reduced. So, the number of molecules of the interrupt molecules will be reduced and when the numbers of molecules interrupt is reduced then what will happen thermal conductivity will be further reduced and that situation the drop will be further more. So, resistance will increase. So, resistance will increase with decrease in, we decrease in surrounding pressure there is one case another case is that, if we apply if we apply some kind of pressure at this junction at this junction. So, that some kind of deformation takes place here, at this junction this is the application of some kind of pressure then what will happen that give, if the deformations have to deformation takes place then thus if we can bring about more smoothness on this surface then that wideness will be reduced. So, amount of gas entered into the gas material will be reduced and then thermal contact resistance will be reduced.
17 That means what you say is that if we increase the joint pressure by some means resistances will we decrease with increased joint pressure this is what if we put some kind of pressure at the joint between this 2 body then what will happen the there will there will be more contact and, because of more contact less area for the air or gas has to be entered. So, less conduction by the airy material more conduction by this solid material and therefore, the thermal contradiction will decrease and in some in many situations these thermal contact resistance are reduced by putting some greasy material at this junction point if you put some greasy material we can reduce the thermal contact resistance. So, this way we have tried to see that in case of steady state heat conduction situation particularly in one dimension what happens point there is a cylindrical circulation when there is a plain wall and there is a spherical wall and there is a spherical body when there is an extension of the surface areas by extended surface resistant in the wire applying fin and then also we tried to see that what is the physical significance of thermal contact resistances in case of heat transfer. Now, in the next lecture onwards we will discuss on unsteady state heat transfer we will see that how or what will be the situation how are the performances when temperature changes with time what happens to the heat transfer rate if it changes if temperature changes with time all these things we will discuss from next lecture onwards. Thank you very much.
Heat Transfer Prof. Dr. Ale Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati
Heat Transfer Prof. Dr. Ale Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module No. # 04 Convective Heat Transfer Lecture No. # 03 Heat Transfer Correlation
More informationGas Dynamics Prof. T. M. Muruganandam Department of Aerospace Engineering Indian Institute of Technology, Madras. Module No  12 Lecture No  25
(Refer Slide Time: 00:22) Gas Dynamics Prof. T. M. Muruganandam Department of Aerospace Engineering Indian Institute of Technology, Madras Module No  12 Lecture No  25 PrandtlMeyer Function, Numerical
More informationModule 1 : Conduction. Lecture 5 : 1D conduction example problems. 2D conduction
Module 1 : Conduction Lecture 5 : 1D conduction example problems. 2D conduction Objectives In this class: An example of optimization for insulation thickness is solved. The 1D conduction is considered
More informationHydraulics Prof. A. K. Sarma Department of Civil Engineering Indian Institute of Technology, Guwahati. Module No. # 02 Uniform Flow Lecture No.
Hydraulics Prof. A. K. Sarma Department of Civil Engineering Indian Institute of Technology, Guwahati Module No. # 02 Uniform Flow Lecture No. # 04 Computation of Uniform Flow (Part 02) Welcome to this
More informationHeterogeneous Catalysis and Catalytic Processes Prof. K. K. Pant Department of Chemical Engineering Indian Institute of Technology, Delhi
Heterogeneous Catalysis and Catalytic Processes Prof. K. K. Pant Department of Chemical Engineering Indian Institute of Technology, Delhi Module  03 Lecture 10 Good morning. In my last lecture, I was
More informationSoil Dynamics Prof. Deepankar Choudhury Department of Civil Engineering Indian Institute of Technology, Bombay
Soil Dynamics Prof. Deepankar Choudhury Department of Civil Engineering Indian Institute of Technology, Bombay Module  2 Vibration Theory Lecture  8 Forced Vibrations, Dynamic Magnification Factor Let
More informationHigh Speed Aerodynamics Prof. K. P. Sinhamahapatra Department of Aerospace Engineering Indian Institute of Technology, Kharagpur
High Speed Aerodynamics Prof. K. P. Sinhamahapatra Department of Aerospace Engineering Indian Institute of Technology, Kharagpur Module No. # 01 Lecture No. # 06 Onedimensional Gas Dynamics (Contd.) We
More informationIntegration of a fin experiment into the undergraduate heat transfer laboratory
Integration of a fin experiment into the undergraduate heat transfer laboratory H. I. AbuMulaweh Mechanical Engineering Department, Purdue University at Fort Wayne, Fort Wayne, IN 46805, USA Email: mulaweh@engr.ipfw.edu
More informationRefrigeration and Airconditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Refrigeration and Airconditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture No. # 22 Refrigeration System Components: Compressor (Continued)
More informationSteady Heat Conduction
Steady Heat Conduction In thermodynamics, we considered the amount of heat transfer as a system undergoes a process from one equilibrium state to another. hermodynamics gives no indication of how long
More informationPower Electronics. Prof. K. Gopakumar. Centre for Electronics Design and Technology. Indian Institute of Science, Bangalore.
Power Electronics Prof. K. Gopakumar Centre for Electronics Design and Technology Indian Institute of Science, Bangalore Lecture  1 Electric Drive Today, we will start with the topic on industrial drive
More informationFluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  20 Conservation Equations in Fluid Flow Part VIII Good morning. I welcome you all
More informationDifferential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation
Differential Relations for Fluid Flow In this approach, we apply our four basic conservation laws to an infinitesimally small control volume. The differential approach provides point by point details of
More informationVISUAL PHYSICS School of Physics University of Sydney Australia. Why do cars need different oils in hot and cold countries?
VISUAL PHYSICS School of Physics University of Sydney Australia FLUID FLOW VISCOSITY POISEUILLE'S LAW? Why do cars need different oils in hot and cold countries? Why does the engine runs more freely as
More informationBasic Electrical Technology Dr. L. Umanand Department of Electrical Engineering Indian Institute of Science, Bangalore. Lecture  33 3 phase System 4
Basic Electrical Technology Dr. L. Umanand Department of Electrical Engineering Indian Institute of Science, Bangalore Lecture  33 3 phase System 4 Hello everybody. So, in the last class we have been
More informationThe soot and scale problems
Dr. Albrecht Kaupp Page 1 The soot and scale problems Issue Soot and scale do not only increase energy consumption but are as well a major cause of tube failure. Learning Objectives Understanding the implications
More informationCalculating Heat Loss by Mark Crombie, Chromalox
Calculating Heat Loss by Mark Crombie, Chromalox Posted: January 30, 2006 This article deals with the basic principles of heat transfer and the calculations used for pipes and vessels. By understanding
More informationTwoDimensional Conduction: Shape Factors and Dimensionless Conduction Heat Rates
TwoDimensional Conduction: Shape Factors and Dimensionless Conduction Heat Rates Chapter 4 Sections 4.1 and 4.3 make use of commercial FEA program to look at this. D Conduction General Considerations
More informationLecture  4 Diode Rectifier Circuits
Basic Electronics (Module 1 Semiconductor Diodes) Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati Lecture  4 Diode Rectifier Circuits
More informationChapter 2. Derivation of the Equations of Open Channel Flow. 2.1 General Considerations
Chapter 2. Derivation of the Equations of Open Channel Flow 2.1 General Considerations Of interest is water flowing in a channel with a free surface, which is usually referred to as open channel flow.
More informationHeat Transfer and Energy
What is Heat? Heat Transfer and Energy Heat is Energy in Transit. Recall the First law from Thermodynamics. U = Q  W What did we mean by all the terms? What is U? What is Q? What is W? What is Heat Transfer?
More informationNatural Convection. Buoyancy force
Natural Convection In natural convection, the fluid motion occurs by natural means such as buoyancy. Since the fluid velocity associated with natural convection is relatively low, the heat transfer coefficient
More informationCFD SIMULATION OF SDHW STORAGE TANK WITH AND WITHOUT HEATER
International Journal of Advancements in Research & Technology, Volume 1, Issue2, July2012 1 CFD SIMULATION OF SDHW STORAGE TANK WITH AND WITHOUT HEATER ABSTRACT (1) Mr. Mainak Bhaumik M.E. (Thermal Engg.)
More informationThe temperature of a body, in general, varies with time as well
cen2935_ch4.qxd 11/3/5 3: PM Page 217 TRANSIENT HEAT CONDUCTION CHAPTER 4 The temperature of a body, in general, varies with time as well as position. In rectangular coordinates, this variation is expressed
More information1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?
CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 33, with
More informationChapter 22: Electric Flux and Gauss s Law
22.1 ntroduction We have seen in chapter 21 that determining the electric field of a continuous charge distribution can become very complicated for some charge distributions. t would be desirable if we
More informationProbability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur
Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur Module No. #01 Lecture No. #15 Special DistributionsVI Today, I am going to introduce
More informationGeotechnical Measurements and Explorations Prof. Nihar Ranjan Patra Department of Civil Engineering Indian Institute of Technology, Kanpur
Geotechnical Measurements and Explorations Prof. Nihar Ranjan Patra Department of Civil Engineering Indian Institute of Technology, Kanpur Lecture No. # 13 (Refer Slide Time: 00:18) So last class, it was
More informationXI / PHYSICS FLUIDS IN MOTION 11/PA
Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A
More informationFREESTUDY HEAT TRANSFER TUTORIAL 3 ADVANCED STUDIES
FREESTUDY HEAT TRANSFER TUTORIAL ADVANCED STUDIES This is the third tutorial in the series on heat transfer and covers some of the advanced theory of convection. The tutorials are designed to bring the
More informationThe Fourth International DERIVETI92/89 Conference Liverpool, U.K., 1215 July 2000. Derive 5: The Easiest... Just Got Better!
The Fourth International DERIVETI9/89 Conference Liverpool, U.K., 5 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de technologie supérieure 00, rue NotreDame Ouest Montréal
More informationArea & Volume. 1. Surface Area to Volume Ratio
1 1. Surface Area to Volume Ratio Area & Volume For most cells, passage of all materials gases, food molecules, water, waste products, etc. in and out of the cell must occur through the plasma membrane.
More information4 Microscopic dynamics
4 Microscopic dynamics In this section we will look at the first model that people came up with when they started to model polymers from the microscopic level. It s called the Oldroyd B model. We will
More informationB = 1 14 12 = 84 in2. Since h = 20 in then the total volume is. V = 84 20 = 1680 in 3
45 Volume Surface area measures the area of the twodimensional boundary of a threedimensional figure; it is the area of the outside surface of a solid. Volume, on the other hand, is a measure of the space
More informationCHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON SOLIDS
CHAPTER 9 VOLUMES AND SURFACE AREAS OF COMMON EXERCISE 14 Page 9 SOLIDS 1. Change a volume of 1 00 000 cm to cubic metres. 1m = 10 cm or 1cm = 10 6m 6 Hence, 1 00 000 cm = 1 00 000 10 6m = 1. m. Change
More informationLecture 6  Boundary Conditions. Applied Computational Fluid Dynamics
Lecture 6  Boundary Conditions Applied Computational Fluid Dynamics Instructor: André Bakker http://www.bakker.org André Bakker (20022006) Fluent Inc. (2002) 1 Outline Overview. Inlet and outlet boundaries.
More informationAmpere's Law. Introduction. times the current enclosed in that loop: Ampere's Law states that the line integral of B and dl over a closed path is 0
1 Ampere's Law Purpose: To investigate Ampere's Law by measuring how magnetic field varies over a closed path; to examine how magnetic field depends upon current. Apparatus: Solenoid and path integral
More informationApplied Fluid Mechanics
Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and
More informationMeasuring Optical and Thermal Properties of High Temperature Receivers
www.dlr.de Folie 1 Measuring Optical and Thermal Properties of High Temperature Receivers Johannes Pernpeintner, Thomas Fend 4 th SFERA Summerschool, May 1516, 2013, Burg Hornberg www.dlr.de Folie 2 Part
More informationExergy Analysis of a Water Heat Storage Tank
Exergy Analysis of a Water Heat Storage Tank F. Dammel *1, J. Winterling 1, K.J. Langeheinecke 3, and P. Stephan 1,2 1 Institute of Technical Thermodynamics, Technische Universität Darmstadt, 2 Center
More informationBasic Electronics Prof. Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati
Basic Electronics Prof. Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati Module: 2 Bipolar Junction Transistors Lecture2 Transistor
More informationHEAT TRANSFER IM0245 3 LECTURE HOURS PER WEEK THERMODYNAMICS  IM0237 2014_1
COURSE CODE INTENSITY PREREQUISITE COREQUISITE CREDITS ACTUALIZATION DATE HEAT TRANSFER IM05 LECTURE HOURS PER WEEK 8 HOURS CLASSROOM ON 6 WEEKS, HOURS LABORATORY, HOURS OF INDEPENDENT WORK THERMODYNAMICS
More informationDerive 5: The Easiest... Just Got Better!
Liverpool John Moores University, 115 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de Technologie Supérieure, Canada Email; mbeaudin@seg.etsmtl.ca 1. Introduction Engineering
More informationChapter 8: Flow in Pipes
Objectives 1. Have a deeper understanding of laminar and turbulent flow in pipes and the analysis of fully developed flow 2. Calculate the major and minor losses associated with pipe flow in piping networks
More information1 DESCRIPTION OF THE APPLIANCE
1 DESCRIPTION OF THE APPLIANCE 1.1 INTRODUCTION The cast iron SF boilers are a valid solution for the present energetic problems, since they can run with solid fuels: wood and coal. These series of boilers
More informationOpen channel flow Basic principle
Open channel flow Basic principle INTRODUCTION Flow in rivers, irrigation canals, drainage ditches and aqueducts are some examples for open channel flow. These flows occur with a free surface and the pressure
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME 2 ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS
ENGINEERING COMPONENTS EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS Structural members: struts and ties; direct stress and strain,
More informationCarbon Cable. Sergio Rubio Carles Paul Albert Monte
Carbon Cable Sergio Rubio Carles Paul Albert Monte Carbon, Copper and Manganine PhYsical PropERTieS CARBON PROPERTIES Carbon physical Properties Temperature Coefficient α 0,0005 ºC1 Density D 2260 kg/m3
More informationChapter 15 Collision Theory
Chapter 15 Collision Theory 151 Introduction 1 15 Reference Frames Relative and Velocities 1 151 Center of Mass Reference Frame 15 Relative Velocities 3 153 Characterizing Collisions 5 154 OneDimensional
More informationIntroduction to Solid Modeling Using SolidWorks 2012 SolidWorks Simulation Tutorial Page 1
Introduction to Solid Modeling Using SolidWorks 2012 SolidWorks Simulation Tutorial Page 1 In this tutorial, we will use the SolidWorks Simulation finite element analysis (FEA) program to analyze the response
More informationNotes on Polymer Rheology Outline
1 Why is rheology important? Examples of its importance Summary of important variables Description of the flow equations Flow regimes  laminar vs. turbulent  Reynolds number  definition of viscosity
More informationFeature Commercial codes Inhouse codes
A simple finite element solver for thermomechanical problems Keywords: Scilab, Open source software, thermoelasticity Introduction In this paper we would like to show how it is possible to develop a
More informationThe two dimensional heat equation
The two dimensional heat equation Ryan C. Trinity University Partial Differential Equations March 6, 2012 Physical motivation Consider a thin rectangular plate made of some thermally conductive material.
More informationStability of Evaporating Polymer Films. For: Dr. Roger Bonnecaze Surface Phenomena (ChE 385M)
Stability of Evaporating Polymer Films For: Dr. Roger Bonnecaze Surface Phenomena (ChE 385M) Submitted by: Ted Moore 4 May 2000 Motivation This problem was selected because the writer observed a dependence
More informationRValue and Thermal Conductivity of PEX and PERT TR48/2014
RValue and of PEX and PERT TR48/2014 Foreword RVALUE AND THERMAL CONDUCTIVITY OF PEX AND PERT TR48/2014 This technical report was developed and published with the technical help and financial support
More informationEffect of design parameters on temperature rise of windings of dry type electrical transformer
Effect of design parameters on temperature rise of windings of dry type electrical transformer Vikas Kumar a, *, T. Vijay Kumar b, K.B. Dora c a Centre for Development of Advanced Computing, Pune University
More informationCalculating Viscous Flow: Velocity Profiles in Rivers and Pipes
previous inex next Calculating Viscous Flow: Velocity Profiles in Rivers an Pipes Michael Fowler, UVa 9/8/1 Introuction In this lecture, we ll erive the velocity istribution for two examples of laminar
More information1. Fluids Mechanics and Fluid Properties. 1.1 Objectives of this section. 1.2 Fluids
1. Fluids Mechanics and Fluid Properties What is fluid mechanics? As its name suggests it is the branch of applied mechanics concerned with the statics and dynamics of fluids  both liquids and gases.
More informationModule 2  GEARS Lecture 7  SPUR GEAR DESIGN
Module 2  GEARS Lecture 7  SPUR GEAR DESIGN Contents 7.1 Spur gear tooth force analysis 7.2 Spur gear  tooth stresses 7.3 Tooth bending stress Lewis equation 7.4 Tooth bending stress AGMA procedure
More informationCO 2 41.2 MPa (abs) 20 C
comp_02 A CO 2 cartridge is used to propel a small rocket cart. Compressed CO 2, stored at a pressure of 41.2 MPa (abs) and a temperature of 20 C, is expanded through a smoothly contoured converging nozzle
More informationStack Contents. Pressure Vessels: 1. A Vertical Cut Plane. Pressure Filled Cylinder
Pressure Vessels: 1 Stack Contents Longitudinal Stress in Cylinders Hoop Stress in Cylinders Hoop Stress in Spheres Vanishingly Small Element Radial Stress End Conditions 1 2 Pressure Filled Cylinder A
More informationFREE CONVECTION FROM OPTIMUM SINUSOIDAL SURFACE EXPOSED TO VERTICAL VIBRATIONS
International Journal of Mechanical Engineering and Technology (IJMET) Volume 7, Issue 1, JanFeb 2016, pp. 214224, Article ID: IJMET_07_01_022 Available online at http://www.iaeme.com/ijmet/issues.asp?jtype=ijmet&vtype=7&itype=1
More informationElectromagnetism Laws and Equations
Electromagnetism Laws and Equations Andrew McHutchon Michaelmas 203 Contents Electrostatics. Electric E and Dfields............................................. Electrostatic Force............................................2
More information10.1 Powder mechanics
Fluid and Particulate systems 424514 /2014 POWDER MECHANICS & POWDER FLOW TESTING 10 Ron Zevenhoven ÅA Thermal and Flow Engineering ron.zevenhoven@abo.fi 10.1 Powder mechanics RoNz 2/38 Types of flow of
More informationExam 1 Practice Problems Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8 Spring 13 Exam 1 Practice Problems Solutions Part I: Short Questions and Concept Questions Problem 1: Spark Plug Pictured at right is a typical
More informationExercises on Voltage, Capacitance and Circuits. A d = (8.85 10 12 ) π(0.05)2 = 6.95 10 11 F
Exercises on Voltage, Capacitance and Circuits Exercise 1.1 Instead of buying a capacitor, you decide to make one. Your capacitor consists of two circular metal plates, each with a radius of 5 cm. The
More informationCryptography and Network Security. Prof. D. Mukhopadhyay. Department of Computer Science and Engineering. Indian Institute of Technology, Kharagpur
Cryptography and Network Security Prof. D. Mukhopadhyay Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur Module No. # 01 Lecture No. # 12 Block Cipher Standards
More informationVERTICAL STRESS INCREASES IN SOIL TYPES OF LOADING. Point Loads (P) Line Loads (q/unit length) Examples:  Posts. Examples:  Railroad track
VERTICAL STRESS INCREASES IN SOIL Point Loads (P) TYPES OF LOADING Line Loads (q/unit length) Revised 0/015 Figure 6.11. Das FGE (005). Examples:  Posts Figure 6.1. Das FGE (005). Examples:  Railroad
More informationIf A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?
Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question
More informationPractice Problems on Boundary Layers. Answer(s): D = 107 N D = 152 N. C. Wassgren, Purdue University Page 1 of 17 Last Updated: 2010 Nov 22
BL_01 A thin flat plate 55 by 110 cm is immersed in a 6 m/s stream of SAE 10 oil at 20 C. Compute the total skin friction drag if the stream is parallel to (a) the long side and (b) the short side. D =
More informationMath 1B, lecture 5: area and volume
Math B, lecture 5: area and volume Nathan Pflueger 6 September 2 Introduction This lecture and the next will be concerned with the computation of areas of regions in the plane, and volumes of regions in
More information(Refer Slide Time: 1:42)
Introduction to Computer Graphics Dr. Prem Kalra Department of Computer Science and Engineering Indian Institute of Technology, Delhi Lecture  10 Curves So today we are going to have a new topic. So far
More informationDimensional analysis is a method for reducing the number and complexity of experimental variables that affect a given physical phenomena.
Dimensional Analysis and Similarity Dimensional analysis is very useful for planning, presentation, and interpretation of experimental data. As discussed previously, most practical fluid mechanics problems
More informationPHYSICAL QUANTITIES AND UNITS
1 PHYSICAL QUANTITIES AND UNITS Introduction Physics is the study of matter, its motion and the interaction between matter. Physics involves analysis of physical quantities, the interaction between them
More informationMath 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers.
Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used
More informationManagement Information System Prof. B. Mahanty Department of Industrial Engineering & Management Indian Institute of Technology, Kharagpur
(Refer Slide Time: 00:54) Management Information System Prof. B. Mahanty Department of Industrial Engineering & Management Indian Institute of Technology, Kharagpur Lecture  18 Data Flow Diagrams  III
More informationCryptography and Network Security Prof. D. Mukhopadhyay Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur
Cryptography and Network Security Prof. D. Mukhopadhyay Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur Lecture No. # 11 Block Cipher Standards (DES) (Refer Slide
More informationLearning Module 4  Thermal Fluid Analysis Note: LM4 is still in progress. This version contains only 3 tutorials.
Learning Module 4  Thermal Fluid Analysis Note: LM4 is still in progress. This version contains only 3 tutorials. Attachment C1. SolidWorksSpecific FEM Tutorial 1... 2 Attachment C2. SolidWorksSpecific
More information7.2 Quadratic Equations
476 CHAPTER 7 Graphs, Equations, and Inequalities 7. Quadratic Equations Now Work the Are You Prepared? problems on page 48. OBJECTIVES 1 Solve Quadratic Equations by Factoring (p. 476) Solve Quadratic
More informationGeometry Unit 6 Areas and Perimeters
Geometry Unit 6 Areas and Perimeters Name Lesson 8.1: Areas of Rectangle (and Square) and Parallelograms How do we measure areas? Area is measured in square units. The type of the square unit you choose
More informationGeneral Allowances for Insulation & Cladding
TRADE OF Industrial Insulation PHASE 2 Module 1 Sheet Metal and Insulation Fundamentals UNIT: 4 General Allowances for Insulation & Cladding Produced by In cooperation with subject matter expert: Michael
More informationTHE MODIFICATION OF WINDTUNNEL RESULTS BY THE WINDTUNNEL DIMENSIONS
THE MODIFICATION OF WINDTUNNEL RESULTS BY THE WINDTUNNEL DIMENSIONS 13\' MAX M. MONK, Ph.D., Dr.Eng. Technical Assistant, National Advisory Committee for Aeronautics RIlPRINTED FROM THII JOURNAL OF THE
More information= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm
Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by
More informationChapter 13 OPENCHANNEL FLOW
Fluid Mechanics: Fundamentals and Applications, 2nd Edition Yunus A. Cengel, John M. Cimbala McGrawHill, 2010 Lecture slides by Mehmet Kanoglu Copyright The McGrawHill Companies, Inc. Permission required
More informationLecture 24  Surface tension, viscous flow, thermodynamics
Lecture 24  Surface tension, viscous flow, thermodynamics Surface tension, surface energy The atoms at the surface of a solid or liquid are not happy. Their bonding is less ideal than the bonding of atoms
More informationModule 7 (Lecture 24 to 28) RETAINING WALLS
Module 7 (Lecture 24 to 28) RETAINING WALLS Topics 24.1 INTRODUCTION 24.2 GRAVITY AND CANTILEVER WALLS 24.3 PROPORTIONING RETAINING WALLS 24.4 APPLICATION OF LATERAL EARTH PRESSURE THEORIES TO DESIGN 24.5
More informationdu u U 0 U dy y b 0 b
BASIC CONCEPTS/DEFINITIONS OF FLUID MECHANICS (by Marios M. Fyrillas) 1. Density (πυκνότητα) Symbol: 3 Units of measure: kg / m Equation: m ( m mass, V volume) V. Pressure (πίεση) Alternative definition:
More informationp atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh
IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: P..Sleigh@leeds.ac.uk r J Noakes:.J.Noakes@leeds.ac.uk January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1
More informationInformation Theory and Coding Prof. S. N. Merchant Department of Electrical Engineering Indian Institute of Technology, Bombay
Information Theory and Coding Prof. S. N. Merchant Department of Electrical Engineering Indian Institute of Technology, Bombay Lecture  17 ShannonFanoElias Coding and Introduction to Arithmetic Coding
More informationMachine Design II Prof. K.Gopinath & Prof. M.M.Mayuram. Module 2  GEARS. Lecture 17 DESIGN OF GEARBOX
Module 2  GEARS Lecture 17 DESIGN OF GEARBOX Contents 17.1 Commercial gearboxes 17.2 Gearbox design. 17.1 COMMERCIAL GEARBOXES Various commercial gearbox designs are depicted in Fig. 17.1 to 17.10. These
More informationPerimeter. 14ft. 5ft. 11ft.
Perimeter The perimeter of a geometric figure is the distance around the figure. The perimeter could be thought of as walking around the figure while keeping track of the distance traveled. To determine
More informationSURFACE AREA AND VOLUME
SURFACE AREA AND VOLUME In this unit, we will learn to find the surface area and volume of the following threedimensional solids:. Prisms. Pyramids 3. Cylinders 4. Cones It is assumed that the reader has
More informationExpress Introductory Training in ANSYS Fluent Lecture 1 Introduction to the CFD Methodology
Express Introductory Training in ANSYS Fluent Lecture 1 Introduction to the CFD Methodology Dimitrios Sofialidis Technical Manager, SimTec Ltd. Mechanical Engineer, PhD PRACE Autumn School 2013  Industry
More informationFundamentals Of. Heat Transfer. Theory and Applications. (Class Notes for ME 371) Michael C. Wendl
Fundamentals Of Heat Transfer Theory and Applications (Class Notes for ME 371) Michael C. Wendl Department of Mechanical Engineering and School of Medicine Washington University Saint Louis, U.S.A. (25)
More informationRate of Heating Analysis of Data Centers during Power Shutdown
2011. American Society of Heating, Refrigerating and AirConditioning Engineers, Inc. (www.ashrae.org). Published in ASHRAE Transactions, Volume 117, Part 1. For personal use only. Additional reproduction,
More informationMath 115 Extra Problems for 5.5
Math 115 Extra Problems for 5.5 1. The sum of two positive numbers is 48. What is the smallest possible value of the sum of their squares? Solution. Let x and y denote the two numbers, so that x + y 48.
More informationManagerial Accounting Prof. Dr. Vardaraj Bapat Department of School of Management Indian Institute of Technology, Bombay
Managerial Accounting Prof. Dr. Vardaraj Bapat Department of School of Management Indian Institute of Technology, Bombay Lecture  26 Cost Volume Profit Analysis Dear participations in our early session,
More informationDIESEL EFFECT PROBLEM SOLVING DURING INJECTION MOULDING
RESEARCH PAPERS FACULTY OF MATERIALS SCIENCE AND TECHNOLOGY IN TRNAVA SLOVAK UNIVERSITY OF TECHNOLOGY IN BRATISLAVA 2014 Volume 22, Special Number DIESEL EFFECT PROBLEM SOLVING DURING INJECTION MOULDING
More informationNo Solution Equations Let s look at the following equation: 2 +3=2 +7
5.4 Solving Equations with Infinite or No Solutions So far we have looked at equations where there is exactly one solution. It is possible to have more than solution in other types of equations that are
More informationSection 7.2 Area. The Area of Rectangles and Triangles
Section 7. Area The Area of Rectangles and Triangles We encounter two dimensional objects all the time. We see objects that take on the shapes similar to squares, rectangle, trapezoids, triangles, and
More information