# Momentum Conservation

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Momentum Conservation
Newton’s Second Law Force = Mass * Acceleration Alternate method: From Reynold’s theorem Fluid Flow Force = Momentum flux + Momentum Accumulation rate Macroscopic momentum conservation is discussed here. This is Newton’s second law, as applied to the fluid flow, in a particular control volume. By definition, Flux is rate of (OUT – IN) and the equation relating force and momentum (and its rate of change) are given here. If you consider a small area dA on the surface, and use the notation that going OUT is positive and coming IN is negative, then the mass flux is given by rho. Velocity. dA (you have to use vector dot product. So you get V-dot-n). The momentum flux is given by multiplying the mass flux by velocity vector. The surface integral gives the momentum flux for the control volumje. The mass in a small element of the control volume is given by rho * d(volume). Momentum is given by multiplying the mass by velocity. Take a volume integral to obtain the momentum of the control volume. Take a derivative w.r.t. time, to obtain the rate of change of momentum in the control volume. The final equation is the mathematical expression of Newton’s second law. Flux Accumulation Rate

Example Straight Pipe Steady State A2 V2 A1 V1 Assumption: No friction
For a fluid flowing in a pipe as shown in the figure, under steady state conditions, there is a change in pressure. The fluid is a liquid. So we assume it is not compressible. The equation of continuity (equation of conservation of mass) indicates that the volumetric flow rate is same in all the places (place 1 and place 2). If the area of cross section changes, then the velocity will change. Otherwise, the velocity is a constant. If we assume there is no friction, then the pressure will not change, unless the elevation and/or velocity changes. This is the case considered in this example. To find the force on the section (indicated by dotted lines), we can use the Eqn. of Conservation of Momentum. We first write the equation and the accumulation-rate term goes to zero (steady state). The force is only due to pressure (at least in the horizontal direction, which is of our main concern. In the vertical direction, the weight of the fluid should be considered. However, since the flow is only in horizontal direction, we consider it alone, for now). The change in fluid velocity gives a non-zero term for momentum flux. Substituting for these in the equation, we can get the force on the section. Assumption: No friction

Example 1 2 Flow in a straight pipe. Realistic case
Even in cases, where elevation and velocity does not change (e.g. flow in a straight pipe with constant cross section), there is a decrease in pressure, in real life. The pressure decreases because there is friction. There is friction between the fluid layers and also friction between the fluid and the solid wall. Some of the energy is converted to heat by the friction. Since the elevation is same and velocity is same (when you compare place 1 and place 2), the pressure is lower at place 2 because some energy is lost as friction between place 1 and place 2. The loss will depend on the flow characteristics (dia of pipe, average velocity, velocity profile…) and fluid characteristic (density, viscosity…). The pressure drop for a 50 m long pipe, with various diameters and same flow rates are given in the chart (for different fluids). For a 10” pipe, the velocity was 0.1 m/second. For smaller pipes, the velocity was increased to keep the flow rate constant. The pressure drop will increase when the velocity is increased (even if you hold the pipe diameter to be the same). It may also increase if you decrease the pipe diameter, because the ‘surface area of contact per unit fluid volume’ (i.e. ratio of perimeter to area) increases with decrease in diameter. So friction is more. Note that the trend is predictable qualitatively but the quantitative prediction is not necessarily ‘intuitive’. For phosphoric acid (85%), decreasing pipe dia by 2 increases pressure drop by 16 times (and all other fluids have pressure drop changed by more than 16). However, decreasing it by 2 times again, the pressure drop changes by 37 times (and this time, all other fluids have pressure drop changed by less than 37). Flow in a straight pipe. Realistic case

Example r1=1g/cm3 d1=8cm Steady State V1=5m/s d2=5cm Problem: 5.32
V1=5m/s Example d2=5cm Problem: 5.32 d2,V2 d1,V1 1 2 h=58 cm We know that there is friction in real life and we want to calculate the frictional force. If you have a manometer, you can measure the pressure difference between two locations. Here is an example. A fluid flows through a tube. The cross section of the tube changes between point 1 and point 2. Fluid is water (density = 1 g/cm3). D1 = 8 cm and D2 = 5 cm (all these refer to the inner diameters). The fluid velocity at location 1 is 5 m/s. The flow is steady. The manometer fluid is mercury (density 13.6 g/cm3). If the height difference in the manometer is h (58 cm), what is the frictional force? We write the Eqn of conservation of momentum and note that rate of accumulation term goes to zero. The force includes pressure force and frictional force (two pages back, we neglected friction. This time we do not). The flux term is also non zero because the velocity and area of cross section change. We can find the velocity at point 2 (using eqn of conservation of mass). We can find the pressure difference between P1 and P2 (using the manometer reading). Hence we can find the frictional force.

Example x y Example 1 (bent pipe), page 47 r1,V1 By Area=A1 Area=A2=A1
Q Bx Consider a general case of pipe bent at an angle theta. The are of the pipe is also a variable. If we take a control volume such that fluid goes in at point 1 and goes out at point 2, then Under steady state conditions, the rate of accumulation goes to zero. We can assume that there is no friction and that the pipe experiences a force because of change in pressure and momentum flux. The pipe is held in place by fixtures. That is why we have steady state conditions. If we find the force exerted by the fixtures on the pipe, then we can say that is the same force exerted by pipe on the fixture (due to fluid motion), just in opposite direction (Newton’s third law). Let us say that the force exerted by the fixture on the pipe is B. It has two components, Bx and By. For now, we assume that Bx is ‘up’ and By is ‘to the right’. If we find that the values are negative, then we of course realize that the forces are in the opposite directions. Force balance along x axis: Pressure, area (with cosine theta, because it is a vector dot product) and Bx. Force balance along y axis: Pressure, area (with sine theta), weight of fluid and By r2,V2 Weight of Fluid

Example out in out in Steady State d/dt =0
Continuing with the equation, Flux on the X axis: Going out, to the right, is positive. Coming in, to the right, is negative. Flux on the Y axis: Coming in is zero, so we don’t worry about it. Going out is positive, but since this goes out in the negative y direction, the term is negative (OR think of it as vector dot product and things will come out correctly). Since we know V1, we can find V2 using continuity equation. Steady State d/dt =0 From Eqn of Conservation of Mass

Example V1 By Area=A1 Area=A2=A1 Q Bx V2
Substituting in the Eqn of conservation of momentum we can get Bx and By in terms of quantities we know. If you assume fluid is incompressible (density does not change) and that the area is same (A1 = A2), then the equation becomes simpler. V2

Example Pipe with U turn P1,V1 F P2,V2 Use Gage Pressure!
What happens when there is a U-turn in the pipe? The ‘theta’ value is 180 degrees. We can use the above formula and find the effective force. Or we can simply look at the directions and say if the pressure force at 1 is positive, pressure force at 2 is also positive. Similarly, incoming flux is negative (direction is correct, since it is to the right) and outgoing flux is positive (direction is to the left, so we have to make it a negative term). The equations are written as above. Note: In the pressure term, always use the gage pressures. Do NOT use absolute pressures. That is because, the force on the pipe will depend on the pressure difference (between inside and outside). If the pipe direction does not change and if the location 1 and location 2 are at the same ambient (outside pressure), then using the absolute pressure will also give correct answer, because the errors cancel out. However, that is not the ‘right way’. In some cases, you may not be given the density of a fluid (e.g. gas). You have to use the pressure and temperature (and use ideal gas law or some such equation), to calculate the density. Use absolute pressure to calculate the density. Use Gage Pressure! In case of gas, use absolute pressure to calculate density

Example N “L” bend P1 E F Assume the force by the pipe on the fluid is in the positive direction P2 What happens when there is a 90 degree bend in the pipe? Let us consider two cases. Assume that the pipe is horizontal. The fluid flows from the west to the east and then after the bend, fluid flows towards south. What is the force on the pipe (note the magnitude as well as the direction)? The area is a constant and the pressures are given by P1 and P2. For case two, reverse the direction of fluid flow. (So, fluid flows from south to north and then takes a left turn and goes west). What is the direction and magnitude of force on the pipe? Does it make sense? In both cases, you don’t have to calculate the ‘downward’ force on the pipe, due to the weight of the fluid. Of course, the downward force exists, but the point is to bring your attention to the horizontal force in a ‘L’ bend (90 degree bend). As a related problem, think of a triangular shaped wooden object. If a ball is thrown at the center of the hypotenuse and it bounces at 90 degrees , which direction will the wooden object tend to move? (assume that there is no friction). What will happen if the ball is thrown from the destination in the reverse direction? What will the force be, if the flow is reversed (a) in a straight pipe? (b) in a L bend?

Example Pbm. 5.13 & 5.19 Conservation of Mass X Y V2 P2 r2 Vw r1 P1 V1
Stationary CV Movement of shock wave: Forces on the body If a shock wave moves through a fluid, we can calculate the force on the body. For example, if you abruptly close the water tap, a shock wave propagates through the pipe and sometimes you can hear a strong vibrating noise (This assumes that there is actually some water in the pipe). This is called ‘water hammer’. This is because a flowing water stream is suddenly stopped, the pressure builds up and the pressure wave propagates at the speed of sound in water. It can increase corrosion rate significantly and may also damage the pipe. In order to relieve the pressure, sometimes, special holding tanks are kept. Also many automatic valves are designed so that they will not completely close quickly even if the signal to them ask them to close abruptly. Using equation of conservation of mass, we can relate the velocities ‘before’ and ‘after’ shock wave. In case of water hammer, one of the velocities is zero.

Example Pressure difference? Pbm. 5.19 V1=-3 m/s V2=0 m/s X Y
Vw=Velocity of Sound in water V2 r2 Vw r1 P2 P1 V1 Stationary CV Take an example of water moving through a pipe at 3m/s. Suddenly the pipe is closed. So, water stops flowing (at the valve). A shock wave develops and moves through the pipe. Schematic is shown. At the left end, the valve (which is closed) is present. The right is the pipeline coming perhaps from a tank. The shock wave moves from left to right and the velocity of water to the left of the shock wave is zero. (i.e. water is continuing to flow at the same velocity, until it has ‘received the signal’ that the valve is closed. The speed of such a signal is more or less the speed of sound in the medium) The velocity to the left of the shock wave is zero. To the right, the fluid is continuing to come at the (previously steady state) velocity. Since it is to the left, the velocity is -3 m/s. This fluid has not ‘received the message’ that the valve is closed. Shock wave is traveling to the right at the speed of sound in water (approx 1.5 km per second). Take a control volume which is stationary. You can take the expression in the previous page and substitute the values. The answer you get will be approximately 41 atmospheres. This can damage pipeline easily. ~ 41 atm

Example V2=? Pbm. 5.1,5,3 V1=6 m/s 0.6 m 1.2 m x y
Consider a fluid flowing from left to right and a long cylinder is placed on the path. The cylinder blocks the flow and the velocity profile is disturbed. The velocity profile after a distance is known. Can we calculate the drag force on the cylinder? Can we experimentally measure it? We assume steady state conditions and say that the cylinder is very long (i.e. edge effects are negligible). Hence if we calculate the drag per unit length of cylinder, we can compare it with experimental data. If we know the shape of the velocity profile, we can estimate the magnitude of the velocity V2 (using equation of conservation of mass). We can measure it experimentally by hanging the cylinder using very thin and light wires, and measuring the ‘angle’ that the wires make. You know the weight of the cylinder and it is a matter of balancing the equations.

Momentum Conservation
Angular Momentum In a moving system Torque = Angular Momentum flux + Angular Momentum Accumulation rate The next conservation equation we see is for angular momentum. Replace ‘momentum’ by ‘angular momentum’ and ‘Force’ by ‘Torque’ and you get the corresponding equation. When a pipe is bent, the fluid flow in the pipe will exert a torque on the bend. The ‘holder’ has to be able to withstand the torque for the pipeline to remain stable. Flux Accumulation Rate

Example Example 4 in book Find the torque on the shaft
In a moving system Torque = Angular Momentum flux + Angular Momentum Accumulation rate Q One can calculate the torque on a turbine. For example, the torque on Pelton wheel is calculated (example in the book). We consider a single ‘cup’ or ‘vane’ and make some simplifying assumptions. The water flow is continuous and it hits the vane exactly in the middle. The stream gets divided into exactly two sections of equal weight and leave the vane exactly at an angle theta. We look at the steady state conditions. The vane is moving from left to right, because water is hitting the vane when the vane is at the bottom most location.

Example Approach-1. Find effective Force in X direction
Find the moment of Force Assume: No frictional loss, ignore gravity, steady state, atmospheric pressure everywhere Q We can solve this by at least two different approaches. One: find the force on the vane and calculate the torque. Second: Find the torque directly. First method: We ignore gravity. If you include gravity, then the results are exactly the same, because the force of gravity is acting downwards and it will not generate any torque. Anyway, consider the vane and imagine a control volume moving at the same speed as the vane (rw). Velocity of water coming in is called V-zero (wrt stationary coordinates). We can show that the V-out = V-in (eqn of conservation of mass, constant density, constant cross sectional area). We an also find the x direction momentum (in as well as out). Notice the signs carefully.

Example Approach-2. Using conservation of angular momentum
Stationary CV If you use the second approach, the calculation is made easier by choosing a stationary control volume. Velocity of water coming in is V-zero. Velocity of water going out is not V-zero. So, what happens to the eqn of conservation of mass?

Example Consider a jet hitting a moving plate After 1 second
First, we find the velocity of water coming out. From the above picture, it looks like there is accumulation of water. But then we said ‘steady state’. What we are missing is that there is water to the right of the vane (water is coming in continuously and before the vane entered the CV, water must have been there). It is exiting the CV as the plate moves. And it is exiting the at the velocity of V-plate. To obtain the velocity of water exiting at the bottom, in a stationary CV, we split the task into following parts. Vnoz water has entered into the CV Plate has moved by Vplate In a control volume which moves with the plate, Vnoz-Vplate water has entered the CV (and exited at the bottom)

Example First we find the velocity of water coming in and going out, wrt a moving CV. (CV moves at V-plate). We can obtain the x and y components of the velocity. Then change the velocities back to stationary coordinates.

Example Consider plate moving @ half Vnoz, alpha = 180 degrees
To emphasize this, consider an example, where the plate is moving at half the V-nozzle. Here the angle alpha is 180 degrees. No water will exit at the bottom (here we should say to the left), as long as the vane is in CV. To the right of the vane, there is water exiting at V-plate. However, the cross sectional area is twice that of A-nozzle. So, V-plate = half of V-nozzle and area = twice A-nozzle add up correctly. Intuitively also, you can look at the above image and convince yourself that the V-out will be zero (in stationary coords). So, the equation that we derived predicts the exit velocity correctly.

Example Pbm 5.24 Thickness of slit =t, vol flow rate =Q, dia of pipe=d, density given Ignore gravity effects 3ft 6ft If you consider a pipe, which is bent (as shown in the picture) and has water flowing out from an arm, you can calculate the torque. This can perhaps be considered as an approximation for flow in a pipe with holes (used in agriculture). Flux Accumulation Rate

Example 1 3ft Pbm 5.31 P1, P2, density, dia, vol flow rate given 2
Pipelines frequently have bends in them and we can calculate the torque exerted on a particular location (where the pipe is ‘held’, perhaps using a block). For the pbm 5.31, British units are given and I haven’t used them in the calculation. To find the torque at the point, under steady state conditions, we can use the Eqn of cons of angular momentum. Since the line of force at point 1 (and the flux) go through the point of interest, the torque due to pressure force is zero and the flux component is also zero. For the outlet, the flux component is positive. (to the right is positive). We move on to eqn of conservation of energy Calculate velocity at 1 (=2)

Energy Conservation Friction Loss (Viscous) Mechanical Work Heat
done by the system Work done by pressure force Friction Loss (Viscous) From Reynolds’ theorem, you can obtain the above equation. The energy accumulation is further divided into various parts, so that we can calculate the components and express them in terms of various kinds of energy terms used in ‘every day’ language.

Energy Conservation No Frictional losses Incompressible Steady
No heat, work No internal energy change In the beginning, we consider simplified equation, where there is no loss due to friction and no heat is added and no work is done. Along a stream line, the Bernoulli’s equation says the sum of kinetic, pressure and ‘head’ (potential) energies is a constant.

Example Flow from a tank Toricelli’s Law Sections 2 and 3 Dia = d1 1
h1 2 Dia = d2 Pressure = atm at the top and at the outlet h3 3 You can use this to calculate the velocity of fluid draining from a large tank. If the tank is large, velocity at location 1 can be assumed to be zero (negligible). Velocity at 2 can be calculated by Toricelli’s law and from there, the velocity at 3. One can calculate the area of the jet at various locations, using eqn of continuity. Velocity at 1 ~ 0 Toricelli’s Law Sections 2 and 3

Example How long does it take to empty the tank?
What if you had a pipe all the way upto level 3? Dia = d1 1 h1 2 Dia = d2 h3 3 And you can calculate how long it will take to empty the tank. If you add a pipe to the outlet at the bottom, will it drain faster or slower? (assuming no viscosity). If you put a pipe all the way upto level 3, you will actually increase the flow rate (assuming friction is negligible). section 2 != atm section 3 = atm

Example What if you had a pipe all the way upto level 3? Dia = d1 1 h1
2 Dia = d2 h3 3 That is according to calculation. However, you will get turbulence and vortex formation and the flow is likely to be unsteady in reality. More flow with the pipe Turbulence, friction Unsteady flow Vortex formation

Example Height is known Moving reference; Aircraft 60 km/h
Find P and r (eg from tables) 150 km/h Flight as Reference 1 2 3

Flow through a siphon vs constriction
We can use the Bernoulli’s equation to determine the flow through a siphon; and more importantly the pressure at various points. If the pressure at a particular point becomes equal to the vapor pressure, then the liquid will evaporate and the equation is not valid anymore. Likewise, we can calculate the pressure at a constriction and the equation is valid as long as the pressure it predicts is greater than the vapor pressure. 1 2

Example Pbm. 6.4 Steady flow through pipe , with friction
Friction loss head = 10 psi Area, vol flow rate given Find temp increase Assume no heat transfer One can account for frictional losses if one has an equation to predict the energy converted to heat. OR if you can measure all the relevant parameters (pressure, velocity and height), one can calculate the frictional loss. In a very simple case, for flow through a horizontal pipe of constant area, under steady state condition, the velocity and elevation are the same. Hence any drop in pressure is coming only due to friction. We can use this to calculate the increase in temperature (assuming all the heat is retained by the fluid and there is no loss of heat or addition of heat to/from the pipe). We find that this increase in temperature is negligible.

Example D2 Pbm. 6.10 Fluid entering from bottom, t P2=atm
exiting at radial direction Steady, no friction t P2=atm h2 D1 Find Q, F on the top plate P1=10 psig When you use Bernoulli’s equation (or for that matter any equation), always remember the physical significance. You cannot use it arbitrarily and expect to get results. Consider an example (pbm 6.10). The fluid is entering the tube at the bottom and exiting at the periphery of the top plate. The exit velocity is uniform. Steady state, no friction. Can you use Bernoulli’s equation to calculate the pressure on the plate and hence the force on the plate? We will first solve it by eqn of conservation of momentum. We can get the velocity (eqn of conservation of mass). WE assume steady state conditions and so the d/dt term goes to zero.

Example F y D1 P1=10 psig Force on the fluid = force on the plate + weight of plate (OR if the plate has negligible weight, it become simpler). Incoming momentum is known. Outgoing momentum is zero (in the vertical direction). So we can get the force. How about using Bernoulli’s equation? The pressure on the plate is going to depend on the location. Very near the edge (where velocity is V2), the pressure will be close to atmospheric. As you move towards the center, pressure will decrease, because velocity is higher (and the elevation is the same as location 2). So all we need to do is to obtain the velocity profile and from that get the pressure profile. Integrate it (area integral) and we are done. The problem is , at the exact center, the velocity will be infinity (according to our assumption of ideal radial flow ), pressure will be minus-infinity and we don’t get useful result. We have to resort to eqn of conservation of momentum, to find the force. If the velocity distribution just below the top plate is known, then P can be found using Bernoulli’s eqn

Modifications to Eqn Unsteady state, for points 1 and 2 along a stream line We have seen one modification to the eqn already (for frictional loss). Another one is for unsteady state flow (as long as we stick to stream line). The equation is given above. For now, take this for granted. We will derive this a bit later, where I believe it fits in well.

Draining of a tank We can obtain the time it takes to drain a tank
(i) Assume no friction in the drain pipe (ii) Assume you know the relationship between friction and velocity L H R D 1 2 Le us take that the bottom location is 2 and the top fluid surface is 1 Incompressible fluid Problem in BSL. First we will work on the problem that assumes that there is no friction in the drain tube. This we will sub divide into two sections, one with quasi steady state assumption and other with out that (i.e. unsteady state problem). Next we will work on the quasi steady state problem but we will incorporate an expression for frictional losses. First we use the eqn of continuity for incompressible fluids

Draining of a tank: Quasi steady state
Quasi steady state assumption Velocity at fluid surface at 1 is very small i.e. R >> D No friction : L is negligible P1 = P2 = Patm For the case when the tank is very large compared to the tube dia, we can obtain the Toricelli’s equation. We also take L to be very small, so we are justified in ignoring frictional losses

Draining of a tank: Quasi steady state
Original level of liquid is at H = H0 Integrating above equation from t=0, H=H0 to t=tfinal, H =0, we can find the efflux time By combining the Bernoulli’s equation (approximated to Toricelli’s) and Mass conservation, we get the first order ODE which relates the height of the liquid level with time. Integrating between the appropriate limits, one can find the time it takes to drain the tank.

Draining of a tank: Unsteady state
BSL eg.7.7.1 At any point of time, the kinetic + potential energy of the fluid in tank is converted into kinetic energy of the outgoing fluid We still neglect friction Now, you may not be sure that neglecting V1 in the Bernoulli’s equation was such a good idea. How about solving the ‘real’ equation? We will go about it here, but we will not go into solving the final equation. The details are given in the ‘worked out example’ of BSL 7.7.1 First determine the potential energy and kinetic energy of fluid in the tank. The fluid is moving at a velocity V1 at any time. The kinetic energy and potential energy can be found out easily. Note that the kinetic energy of a thin disk of fluid is the same regardless of its location and hence you don’t need to integrate. However, the potential energy of the disk changes and you have to integrate to find the total potential energy. Potential Energy of a disk at height z and thickness dz

Draining of a tank: Unsteady state
Also, using continuity equation This is conservation of energy and mass. This is still not to be considered as microscopic balance, because we are not looking into the flow patterns inside the tank. The differential equations come because of unsteady state nature of the problem, not because of microscopic study. The equation is non-linear, because the derivative of ‘H’ wrt ‘t’, occurs in second power. (dH/dt- whole square). It is second order, because you will get ‘d-square-H/d-t-square). This can be solved with two initial conditions. At time t=0, the height is known (H0) and the instantaneous velocity (dH/dt) is also known. Please refer to BSL for solution Substituting, you get a 2nd order non linear ODE with two initial conditions. Please refer to BSL for solution

Draining of a tank (accounting for friction)
What if the flow in the tube is laminar and you want to account for friction? Bernoulli’s eqn is not used (friction present) Continuity L H R D 1 2 Hagen-Poiseuille’s eqn BLS pbm 7B.9. You should not use Bernoulli’s equation (and you don’t have to). Using the Hagen Poisseulle’s equation (relating flow rate vs pressure difference) and equation of continuity, you get the first order ODE with initial condition as t=0, H=H0.

Draining of a tank (accounting for friction)
Substituting and re arranging, Integrating with limits The book uses slightly different notation (it uses R for radius of tank and D for dia of tube) and hence the answer is in a different notation. However it is a reasonably straight forward problem. Note: The answer is given in terms of diameter of tube, so that it is easier to compare with the answer given in the book

Example A1,A2, initial height h1 known A1 >> A2 1 L
Consider section 3 and 2 h1 If a system as shown in the figure is disturbed by suddenly opening the pipe, what is the velocity of the liquid at the outlet as a function of time? We can use the modified Bernoulli’s equation. Remember, in all these things, we are assuming that the velocity profile is ‘flat’ and also that when the fluid is behaving as ‘one unit’ up to the tube entrance, it is behaving differently in the tube (i.e. streamlines start and end from 3 and 2 respectively). This example is from Fox and McDonald 3 2 Pseudo Steady state ==> Toricelli’s law

Example Rearranging and solving, we get
The differential equation is first order with initial condition t=0, V=0 and is easy to solve. The hyperbolic tangent approaches unity, as time increases and hence the solution approaches the Toricelli’s equation. This serves as a ‘cross check’ for our solution As t increases, the solution approaches the Toricelli’s equation

Appendix:Example Oscillating fluid in a U-tube Let h=h1-h2 1 h1 2 h2
This is likely a JEE problem, where again the modified Bernoulli’s eqn can be used. (There are perhaps other methods to solve it also and they are also fine). L3

Appendix:Example Blood Flow in vessels Minimization of ‘work’
Murray’s Law: Laminar Flow, negligible friction loss (other than that due to viscous loss in laminar flow) , steady The blood flow in vessels is studied and an empirical law proposes that the capillaries are designed (by the body) such that the work necessary to pump blood is minimum. The model is based on laminar flow, negligible friction loss (other than due to viscous loss in laminar flow) and steady flow. In reality, flow is turbulent, pulsating. Still the law comes out to be reasonable in terms of predicting the radii of branches. This problem is discussed in Stanley Middleman Turbulent, pulsating flow Assume

Appendix:Example If the ratio of ‘smaller’ to larger capillary is constant And Metabolic requirement =m= power/volume Work for maintaining blood vessel Total work There is some work done in ‘maintaining’ the blood vessels and that is captured by the second term beta-r-square. Total work is the sum of ‘maintenance’ work and pumping work. Optimum radius