Physical Properties of Fluids

Transcription

1 Physical Properties of Fluids Viscosity: Resistance to relative motion between adjacent layers of fluid. Dynamic Viscosity:generally represented as µ. A flat plate moved slowly with a velocity V parallel to itself past a stationary wall. Shear Stress τ: µdu/dy where u is the velocity profile and y is the axis along the line perpendicular to the two plates. Laminar and Turbulent flow At low flow rates when there is no mixing in between adjacent layers of fluid, the flow is said to be laminar. Turbulence produces a thorough mixing and the velocity at a point fluctuates very rapidly in a random manner. Fluids offer a resistance to change in shape. Internal shear stresses are set up when relative motion occurs between two layers of fluid that are next to each other. This property of resistance to shearing motion is called viscosity. Couette Flow: Consider laminar flow when a flat plate is moved with velocity V parallel to stationary wall. The fluid has zero velocity at the stationary wall and a velocity equal to V at the moving plate. There is a linear variation of velocity with distance y from the stationary wall. u = y/h*v. In order to maintain the motion it is necessary to apply a tangential force to the moving plate. The applied force is proportional to the area. By considering each element of the fluid it follows that tangential stress τ is proportional to du/dy and hence stress τ =µdu/dy where µ is the dynamic viscosity. Laminar and Turbulent Flows: At low flow rates there is no appreciable mixing between adjacent layers of fluid. Any flow visualization agent injected at a discrete point in the flow preserves its identity as the flow progresses downstream, except for a slight blurring owing to molecular diffusion. Turbulence produces a thorough mixing of the flow and any visualization agent is quickly distributed throughout the entire flow downstream of the point of injection.

2 Reynolds Number Reynolds number = ρud/µ where u is the mean flow velocity, d the diameter and ρthe density. µ is the dynamic viscosity. This is for a circular pipe for a constant area of cross-section. If p is the pressure drop between two axial positions, then the ratio of p to 1/2ρu 2 (the dynamic pressure) is a function of the Reynolds number, the Mach number and whether the flow is compressible. Here we are considering flows of very low velocity. Hence these flows are incompressible (compressibility effects are seen at around 0.3 Mach) and Mach numbers negligible. For Reynolds number below 2000 both small and large disturbances are damped out and in excess of 2000 a certain finite disturbance is needed to initiate transition from laminar to turbulent. For Re>3200 turbulence is spontaneous.

3 The coolants used The parameters to be considered for choosing the coolant include density, viscosity (which directly affects the Reynolds number) and the heat capacity and heat conductivity (for the transfer of heat to the coolant) for ethylene glycol and water Water Ethylene Glycol Density: kg/m 3 Viscosity:0.89* *10-3 Ns m -2 Heat capacity:4.19* *10 6 Wsm -3 K -1 Heat conductivity: Wm -1 K -1

4 Volume flow and velocity Assume the following parameters: Total dissipation per unit length: 5kW/m Temperature rise along the water channel: 10C/m Channel diameter: 8mm This calls for a volume flow of 3*10-5 m3/s and an average flow velocity of u = 4*dV/dt*d -2 /Π =0.593m/s. Re = udρ/µ = Since Re>2000 the flow is turbulent. A good coolant has high viscosity, low density and high heat capacity and conductivity. Water is thus the best coolant. The surface roughness also has a role in the transition from laminar to turbulent flow. To decrease the Reynolds number and hence achieve a laminar flow the given channel can be divided into 64 such subchannels of 1 mm diameter packed face to face and sealed around the circumference. In this case the Reynolds number is around 666. Using ethylene glycol the Re number can be reduced further. These cooling bars can be attached to the accelerating structure. However we shall try to estimate the vibrations assuming a turbulent flow.

5 Turbulent Flow Using Navier Stokes Momentum equations and boundary conditions at r (radius) = 0, an expression for the shear stress (varying with the radius) can be derived. We also assume that the flow is fully developed. τ/ τ w = r/a = (1-y/a) where a is the radius of the pipe y is the distance measured away from the pipe wall and τ w =-a/2*dp/dx. It can be shown that, just like laminar flow, the pressure at a fixed cross-section is a constant. It has also been demonstrated that it is very difficult to calculate the velocity profile of a turbulent flow. Hence we make use of experimental data and empirical results. The Power-law relations An examination of measured velocity profiles has shown that the distribution of velocity in fully developed turbulent flow can be represented by an equation of the form u/u max = (y/a) 1/n where u max is the velocity along the axis of the pipe and hence the mean velocity U can be derived by πa 2 U = 2 π 0 a urdr = 2 π 0 a Umax ((a-r)/a) 1/n r dr

6 From the definition of friction factor τ w = f*1/2ρu 2 where f is the friction factor. Friction velocity is defined is defined by u τ = (τ w / ρ) 1/2 From dimensional considerations u/ u τ = φ(y u τ ρ/µ) Comparing with the power law u/ u τ = K1(y u τ ρ/µ) 1/n The following table shows the variation of Re, n and K1 Re n K1 < * * * From the above equations it can be shown that f = K4*(Re) -2/(n+1) where K4 = 2/K3; K3=(K2 n /2) 2/(n+1) ; and K2 = K1*2n 2 / ((n+1)*(2n+1)) Hence f = 0.079*(Re) -1/4 in this case as Re < 10 5

7 Trying to estimate the transverse vibrations. The pressure drop along a circular pipe p between any two points is directly proportional to the axial distance between the two points. To write this nondimensionally we have p / 1/2ρu 2 = λ(x 2 -x 1 )/d. Where 1/2ρu 2 is the dynamic pressure and d the diameter of the pipe. x 2 and x 1 are the two points. λ is the friction factor. In literature a variety of terms are used to describe f and λ. However the more commonly used term is f and λ is equal to 4*f. So the pressure drop is equal to p =1/2ρu 2 * 4fl/d where l is the length of the pipe Or p =2ρu 2 * fl/d where f has been defined above. f = 0.079*(Re) -1/4 Suppose we look at a frame moving with velocity u (the average velocity of the fluid otherwise), the only velocity terms in the turbulent flow are the random perturbations along the three axes. Thus the turbulent kinetic energy is 1/2ρu 2 where u is the instantaneous local velocity. Thus the average pressure drop p is converted entirely to the irrecoverable turbulent kinetic energy. So 2 p / ρ = u x 2 + u y 2 + u z 2

8 Thus u 2 = 3* u y 2 = 4fl/d* u 2 assuming isotropy Thus ρu rms y = ρu(4fl/3d) 1/2 We make an assumption to estimate P rms y. We assume that the kinetic energy is concentrated in local eddies or vortices popularly known as coherent structures. We try to estimate the size of the coherent structures. The length of the coherent cell is equal to half the pipe diameter. So the vertical momentum gained by the coherence cell of volume Ad/2 is P rms y = Ad ρ/2 u rms y = Ad ρ/2 u(4fl/3d) 1/2 There are 2l/d cells in the structure of length l and their momenta is added up P rms ytot = Al ρu(4f/6) 1/2 If the accelerating structure including the support is expressed by an equivalent rigid mass M then the velocity induced by turbulence is U rms ytot=um/m* (4f/6) 1/2 and ω = 2πu/d. The assumption of a coherence length implies the existence of a dominant frequency. The coherence cells of length d/2 pass through a fixed location at velocity u with assumed parameters ω/ 2π=74Hz Thus the RMS vibration amplitude is given by S rms y = 1/ 2π (2f/3)1/2 md/m

9 Conclusions and Further discussions 1) The vibration amplitude depends on f 1/2 and hence Re -1/8. 2) It also depends on the ratio of the mass per unit length of water column to that of the structure. 3) The diameter of the cooling pipes. This could be reduced as mentioned earlier. 4) If the number of channels used for cooling are increased, the term within the square root should be multiplied by the number used. 5) If the mass m of the water column per unit length is 50g/m and the structure is of copper with mass (M) per unit length of 11kg/m, then S rms y = 0.94µm. 6) The basis of employing the given coherence length dimensions must be studied to get a clear perspective of estimating the pressure and rms velocities.