Since I've been spending all my time trying to finish my school work and doing baby things, I haven't had nearly the time I would like to blog faithfully...especially on Thesis Tuesdays! But I had a special request for today's post- (simply) addressing the question of why there's dew on the ground in the morning.
The answer lies in saturation- a concept that every southerner knows about, even if you don't think you do:
When Mr. Poe and I make sweet tea, we mix in a heaping two cups of sugar. If we stir in the sugar while the tea is still warm, all of the sugar dissolves in the liquid. If we wait until after we put ice in the tea to stir in the sugar, we end up with undissolved sugar at the bottom of the pitcher because the tea is saturated- it's full of sugar and won't hold any more at this cooler temperature.
The same is true with the air. Also as every southerner knows, the air contains water (in gas form); the amount of water in the air at any given time is what we know as humidity.
As the temperature of the air drops at night, the air becomes saturated (just like the tea at a cooler temp), and can't hold the same amount of water as it could before. This extra water condenses, returning to it's liquid form on the ground as dew. The next day, when the temperature of the air increases again, the dew evaporates back into the air.
And THAT is why we have dew.
As a Christian, wife, mother of a little girl and two fur-babies, Mississippi State Bulldog, and PhD, I lead a very busy life. If you add on top of that my obsession for organization and loves for couponing, sewing, knitting, cooking, reading, gardening, and a host of other random hobbies, you've got the recipe for...well...variety. And variety is the spice of life!
Showing posts with label Fluid Mechanics. Show all posts
Showing posts with label Fluid Mechanics. Show all posts
Tuesday, September 11, 2012
Tuesday, January 31, 2012
MrsDrPoe: Examples of Lift and Drag
For this Thesis Tuesday here on the blog, I'd like to share an example of the concepts of lift and drag that we have been talking about for the past two weeks.
Example: The average pressure and shear stress acting on the surface of the 1m square flat plate are as indicated i nthe Figure. Determine the lift and drag generated if shear stress is and is not neglected.
Given: alpha = 7deg, pTa = -1.2 kN/m*m, pBa = 2.3 kN/m*m tauTa = 5.8e-2 kN/m*m, tauBa = 7.6e-2 kN/m*m
Find: D and L with and without shear
Solution:
Next week, get ready to study some internal flows!
Example: The average pressure and shear stress acting on the surface of the 1m square flat plate are as indicated i nthe Figure. Determine the lift and drag generated if shear stress is and is not neglected.
Given: alpha = 7deg, pTa = -1.2 kN/m*m, pBa = 2.3 kN/m*m tauTa = 5.8e-2 kN/m*m, tauBa = 7.6e-2 kN/m*m
Find: D and L with and without shear
Solution:
A = 1 m*m, Theta = 90deg - alpha = 83deg
D = int(p*cos(theta))dA + int(tauw*sin(theta))dA
Using average values to avoid integration:
DS = (pBa - pTa)*cos(theta)*A + (tauTa + tauBa)*sin(theta)*A = 559.544 N
Dns = (pBa - pTa)*cos(theta)*A = 426.543 N
L = -int(p*sin(theta))dA + int(tauw*cos(theta))dA
Again using average values to avoid integration:
LS = (pBa - pTa)*sin(theta)*A - (tauTa + tauBa)*cos(theta)*A = 3.458e3 N
Lns = (pBa - pTa)*sin(theta)*A = 3.474e3 N
Next week, get ready to study some internal flows!
Tuesday, January 24, 2012
MrsDrPoe: Lift
It's once again Thesis Tuesday here on the blog! Today we'll be discussing another external flow property, lift.
For asymmetrical objects moving through a fluid, there exists a resultant force from pressure and shear stress normal to the upstream velocity, termed lift. It can be calculated as:
however, as previously mentioned, these distributions are difficult to determine. Thus, the lift coefficient is often used:
As with the drag coefficient, CL = f(shape, Re, Ma, Fr, epsilon/l). The Froude number is only important if there is a free surface present. Surface roughness is often unimportant in therms of lift as well. The importance of Ma is low except for high-speed subsonic and supersonic flows. Reynolds number also yields no great impact; however, for high-Re flows, shear stress has little effect on the shape of an object is the key factor in the lift force placed on it.
Bodies designed to generate lift, like airfoils, typically do so by generating a pressure distribution that is different on the top (low) and bottom (high) surface of the body. For symmetric airfoils to generate lift, they must be moving through the fluid at some angle of attack. For asymmetric airfoils, there is some non-zero angle of attack for which no lift is generated. If the angle of attack is too large, however, the boundary layer along the airfoil separates and is unable to reattach to the body. This event is known as stall, and it is especially dangerous if it occurs for low-flying aircraft. For calculation of CL, the planform area (A = b*c) is used, where b is the length of the airfoil (into the page), and c is the chord length. Thus, the lift is the dynamic pressure times the planform area of the wing. The wing loading or average lift per unit area of the wing (L/A) is also useful for design, as is the aspect ratio (b*b)/A. From these characteristics, it can be determined that longer wings are more efficient, but more difficult to maneuver in flight.
Since, for most cases, the shear stress is not important for calculating lift, the potential flow solution can be employed to determine this force. For airfoils at angles of attack not equal to zero, the potential solution alone causes incorrect streamlines at the trailing edge of the airfoil. This can be corrected with the addition of circulation or clockwise swirl to the solution. While this may seem random, arbitrary, or inappropriate, this addition has well founded physical and mathematical grounds.
Next week, we'll look at calculating lift and drag. Happy studying!
For asymmetrical objects moving through a fluid, there exists a resultant force from pressure and shear stress normal to the upstream velocity, termed lift. It can be calculated as:
L = -int(p*sin(theta))dA + int(tauw*cos(theta))dA
however, as previously mentioned, these distributions are difficult to determine. Thus, the lift coefficient is often used:
CL = L/(0.5*rho*U*U*A)
As with the drag coefficient, CL = f(shape, Re, Ma, Fr, epsilon/l). The Froude number is only important if there is a free surface present. Surface roughness is often unimportant in therms of lift as well. The importance of Ma is low except for high-speed subsonic and supersonic flows. Reynolds number also yields no great impact; however, for high-Re flows, shear stress has little effect on the shape of an object is the key factor in the lift force placed on it.
Bodies designed to generate lift, like airfoils, typically do so by generating a pressure distribution that is different on the top (low) and bottom (high) surface of the body. For symmetric airfoils to generate lift, they must be moving through the fluid at some angle of attack. For asymmetric airfoils, there is some non-zero angle of attack for which no lift is generated. If the angle of attack is too large, however, the boundary layer along the airfoil separates and is unable to reattach to the body. This event is known as stall, and it is especially dangerous if it occurs for low-flying aircraft. For calculation of CL, the planform area (A = b*c) is used, where b is the length of the airfoil (into the page), and c is the chord length. Thus, the lift is the dynamic pressure times the planform area of the wing. The wing loading or average lift per unit area of the wing (L/A) is also useful for design, as is the aspect ratio (b*b)/A. From these characteristics, it can be determined that longer wings are more efficient, but more difficult to maneuver in flight.
Since, for most cases, the shear stress is not important for calculating lift, the potential flow solution can be employed to determine this force. For airfoils at angles of attack not equal to zero, the potential solution alone causes incorrect streamlines at the trailing edge of the airfoil. This can be corrected with the addition of circulation or clockwise swirl to the solution. While this may seem random, arbitrary, or inappropriate, this addition has well founded physical and mathematical grounds.
Next week, we'll look at calculating lift and drag. Happy studying!
Tuesday, January 17, 2012
MrsDrPoe: Drag
Good morning, and happy Thesis Tuesday! Today we'll be continuing our look at external flow with a brief discussion on drag.
Drag is a net force in the direction of the flow due to the pressure and shear forces on the surface of the object that is moving through a fluid. If the pressure distribution and wall shear stress are known, drag can be determined from:
however, there are very few cases for which this data can be found analytically. Typically, drag is determined from a given drag coefficient, CD:
As we will see, CD = f(shape, Re, Ma, Fr, epsilon/l), where epsilon is surface roughness.
Friction drag, Df, is the portion of drag due to the viscous shear stress on the object (the second term in the drag equation above). Typically, the surface of an object rotated in any fashion has parts parallel and perpendicular to the flow, but if we examine the case of a very thin flat plate, it can be seen that there is no friction drag force on the plate rotated perpendicularly to the flow. The friction drag on a flat plate can be determined by:
Pressure drag, Dp, is the portion of drag due to the pressure (or normal stresses) on an object (the first term in the drag equation above). The pressure drag can be found by:
Here, Cp is a pressure coefficient:
where po is a reference pressure, which does not influence the drag (the difference is important).
In most cases, the net effects of friction drag and pressure drag are considered instead of examining each type of drag individually as in the second equation above. As mentioned before, drag is influenced by many different aspects of the flow.
One of the most important contributors to the drag coefficient is the shape of the submerged body. Obviously, objects range in size from streamlined to blunt; more blunt objects (l/D -> 0, D >> l) result in larger drag coefficients. For extremely thin, streamlined bodies (l/D -> infinity, D << l) such as thin airfoils, the bodies are essentially tread as flat plates.
Reynolds number also affects the drag coefficient. At very low Re (<1), inertial effects are very small and CD = 2*C/Re, where C is a constant dependent upon size. For moderate Re, CD = Re^-0.5. For very high Re, CD increases for streamlined bodies and decreases for blunt bodies. For extremely blunt bodies, there CD depends little upon Reynolds number.
For sufficiently large object velocities (Ma > 0.5), compressibility effects become important and the drag coefficient becomes a function of Mach number (Ma = U/c). Sharp-pointed bodies develop their maximum drag coefficient around Ma = 1 (sonic flow), while that for blunt bodies increases with Ma far above Ma = 1.
For streamlined bodies, drag increases with increased surface roughness; however, for blunt bodies the opposite is true.
The Froude number (U/sqrt(g*l)) is the ratio of free-stream speed to a typical wave speed on the interface of two fluids, such as the surface of the ocean. Wave drag, Dw can be a complex function of the Froude number and body shape:
Often approximate drag calculations for a complex body can be obtained by examining the body as a collection of various parts. The drag on each simpler part can be calculated and added together to determine the overall drag on the body.
So that's drag! Until next week, happy studying!
Drag is a net force in the direction of the flow due to the pressure and shear forces on the surface of the object that is moving through a fluid. If the pressure distribution and wall shear stress are known, drag can be determined from:
D = int(p*cos(theta))dA + int(tauw*sin(theta))dA
however, there are very few cases for which this data can be found analytically. Typically, drag is determined from a given drag coefficient, CD:
CD = D/(0.5*rho*U*U*A)
As we will see, CD = f(shape, Re, Ma, Fr, epsilon/l), where epsilon is surface roughness.
Friction drag, Df, is the portion of drag due to the viscous shear stress on the object (the second term in the drag equation above). Typically, the surface of an object rotated in any fashion has parts parallel and perpendicular to the flow, but if we examine the case of a very thin flat plate, it can be seen that there is no friction drag force on the plate rotated perpendicularly to the flow. The friction drag on a flat plate can be determined by:
Df = 0.5*rho*U*U*b*l*CDf
Pressure drag, Dp, is the portion of drag due to the pressure (or normal stresses) on an object (the first term in the drag equation above). The pressure drag can be found by:
Dp = 0.5*rho*U*U*A*CDp
where: CDp = int(Cp*cos(theta))dA/A
Here, Cp is a pressure coefficient:
Cp = (p - po)/(0.5*rho*U*U)
where po is a reference pressure, which does not influence the drag (the difference is important).
In most cases, the net effects of friction drag and pressure drag are considered instead of examining each type of drag individually as in the second equation above. As mentioned before, drag is influenced by many different aspects of the flow.
One of the most important contributors to the drag coefficient is the shape of the submerged body. Obviously, objects range in size from streamlined to blunt; more blunt objects (l/D -> 0, D >> l) result in larger drag coefficients. For extremely thin, streamlined bodies (l/D -> infinity, D << l) such as thin airfoils, the bodies are essentially tread as flat plates.
Reynolds number also affects the drag coefficient. At very low Re (<1), inertial effects are very small and CD = 2*C/Re, where C is a constant dependent upon size. For moderate Re, CD = Re^-0.5. For very high Re, CD increases for streamlined bodies and decreases for blunt bodies. For extremely blunt bodies, there CD depends little upon Reynolds number.
For sufficiently large object velocities (Ma > 0.5), compressibility effects become important and the drag coefficient becomes a function of Mach number (Ma = U/c). Sharp-pointed bodies develop their maximum drag coefficient around Ma = 1 (sonic flow), while that for blunt bodies increases with Ma far above Ma = 1.
For streamlined bodies, drag increases with increased surface roughness; however, for blunt bodies the opposite is true.
The Froude number (U/sqrt(g*l)) is the ratio of free-stream speed to a typical wave speed on the interface of two fluids, such as the surface of the ocean. Wave drag, Dw can be a complex function of the Froude number and body shape:
CDw = Dw/(0.5*rho*U*U*l*l)
Often approximate drag calculations for a complex body can be obtained by examining the body as a collection of various parts. The drag on each simpler part can be calculated and added together to determine the overall drag on the body.
So that's drag! Until next week, happy studying!
Tuesday, January 10, 2012
MrsDrPoe: Momentum Integral Example
As promised last week, we'll look at an example problem using the momentum integral boundary layer equation.
Example: If the boundary layer velocity profile is approximated at u = U*y/delta in the boundary layer and u = U in the free stream, and the displacement thickness is measured at 6.7 cm, determine the boundary layer and momentum thicknesses.
Given: u(y) = U*y/delta, u(y) = U if y > delta, delta* = 6.7 cm
Find: boundary layer thickness, momentum thickness
Solution:
Choose some Y in the free stream:
Example: If the boundary layer velocity profile is approximated at u = U*y/delta in the boundary layer and u = U in the free stream, and the displacement thickness is measured at 6.7 cm, determine the boundary layer and momentum thicknesses.
Given: u(y) = U*y/delta, u(y) = U if y > delta, delta* = 6.7 cm
Find: boundary layer thickness, momentum thickness
Solution:
Choose some Y in the free stream:
delta* = int(1 - (u(y)/Uinfinity))dy|0,Y
= int(1 - (u(y)/Uinfinity))dy|0,delta + int(1 - (u(y)/Uinfinity))dy|delta,Y
= int(1 - (U*y/delta)/Uinfinity)dy|0,delta + int(0)dy|delta,Y
= int(1 - (y/delta))dy|0,delta
= (delta - (delta*delta)/(2*delta))
= delta - delta/2 = delta/2
delta = 2*delta* = 13.4 cm
Theta = int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|0,Y
= int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|0,delta + int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|delta,Y
= int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|0,delta + int(0)dy|delta,Y
=int((y/delta)*(1 - y/delta))dy|0,delta
= (delta*delta)/(2*delta) - (delta*delta*delta)/(3*delta*delta)
= delta/6
Theta = delta/6 = 2.233 cm
Not too bad, huh? Until next week, happy studying!
Tuesday, January 3, 2012
MrsDrPoe: Momentum Integral Boundary Layer Equation
Happy Thesis Tuesday to all! Today we'll be looking at another case of external flow- the momentum integral boundary layer equation.
Defining the boundary layer height as the height from the plate where the velocity reached 99% of the free stream velocity is appealing, but often this is difficult to physically measure. Instead, we can measure the displacement thickness, delta*, which is how far the free stream streamlines are shifted (or displaced) due to the boundary layer. We can derive an expression for the displacement thickness by examining conservation of mass in a control volume that has a top control surface along a displaced streamline:
We will assume that the flow is steady and 1D with the flow crossing through two areas of the control surface with unit width (y(1) and H(1)) such that:
Assuming the flow is also incompressible, we can divide the rho out of both integrals:
We know that the free stream velocity is a constant and that y can be written as H + delta* and u can be written as Uinfinity + u - Uinfinity:
Further simplification yields:
As previously mentioned, this value is much easier to measure than delta.
Another boundary layer thickness definition commonly employed is the momentum thickness (Theta), which is related to the amount of momentum "lost" due to drag in the boundary layer. We can derive an expression for this thickness by applying the x-momentum equation to the control volume previously defined:
becomes:
Taking the x-momentum equation:
and applying the flow through an area of unit width with Fdrag = rho*Uinfinity*Uinfinity*Theta, then:
Employing the continuity equation:
We can divide both sides by -rho*Uinfinity*Uinfinity (assuming an incompressible fluid) to obtain:
or finally:
This momentum thickness is also more easily measured than delta. For most cases, Theta < delta* < delta. These integrals are valid for any y in the free stream. It can also be noted that the shear stress on a flat plate can be written in terms of the momentum integral:
Next week, we'll look at a quick example of these equations in action, but until then, happy studying!
Defining the boundary layer height as the height from the plate where the velocity reached 99% of the free stream velocity is appealing, but often this is difficult to physically measure. Instead, we can measure the displacement thickness, delta*, which is how far the free stream streamlines are shifted (or displaced) due to the boundary layer. We can derive an expression for the displacement thickness by examining conservation of mass in a control volume that has a top control surface along a displaced streamline:
d/dt(int(rho)dV) + int(rho*(V.n))dA = 0
We will assume that the flow is steady and 1D with the flow crossing through two areas of the control surface with unit width (y(1) and H(1)) such that:
int(rho*u)dy|0,y - int(rho*Uinfinity)dy|0,H = 0
Assuming the flow is also incompressible, we can divide the rho out of both integrals:
int(u)dy|0,y - int(Uinfinity)dy|0,H = 0
We know that the free stream velocity is a constant and that y can be written as H + delta* and u can be written as Uinfinity + u - Uinfinity:
Uinfinity*H = int(Uinfinity + u - Uinfinity)dy|0,y = int(Uinfinity)dy|0,H+delta* + int(u-Uinfinity)dy|0,y = Uinfinity*H + Uinfinity*delta* + int(u - Uinfinity)dy|0,y
Further simplification yields:
Uinfinity*delta* = int(Uinfinity - u)dy|0,y or delta* = int(1 -(u/Uinfinity))dy|0,y
As previously mentioned, this value is much easier to measure than delta.
Another boundary layer thickness definition commonly employed is the momentum thickness (Theta), which is related to the amount of momentum "lost" due to drag in the boundary layer. We can derive an expression for this thickness by applying the x-momentum equation to the control volume previously defined:
d/dt(int(rho*V)dV) + int(rho*V*(V.n))dA = Sum(F)
becomes:
int(rho*V*u)dA - int(rho*V*Uinfinity)dA = Fdrag
Taking the x-momentum equation:
int(rho*u*u)dA - int(rho*Uinfinity*Uinfinity)dA = -Fdrag
and applying the flow through an area of unit width with Fdrag = rho*Uinfinity*Uinfinity*Theta, then:
-rho*Uinfinity*Uinfinity = int(rho*u*u)dy|0,y - int(rho*Uinfinity*Uinfinity)dy|0,H = int(rho*u*u)dy|0,y - Uinfinity*int(rho*Uinfinity)dy|0,H
Employing the continuity equation:
-rho*Uinfinity*Uinfinity*Theta = int(rho*u*u)dy|0,y - Uinfinity*int(rho*u)dy|0,y
We can divide both sides by -rho*Uinfinity*Uinfinity (assuming an incompressible fluid) to obtain:
Theta = -int((u*u)/(Uinfinity*Uinfinity))dy|0,y + int(u/Uinfinity)dy|0,y = int((u/Uinfinity) - (u*u)/(Uinfinity*Uinfinity))dy|0,y
or finally:
Theta = int((u/Uinfinity)*(1 - (u/Uinfinity)))dy|0,y
This momentum thickness is also more easily measured than delta. For most cases, Theta < delta* < delta. These integrals are valid for any y in the free stream. It can also be noted that the shear stress on a flat plate can be written in terms of the momentum integral:
tauw = rho*U*U*d/dx(Theta)
Next week, we'll look at a quick example of these equations in action, but until then, happy studying!
Tuesday, December 27, 2011
MrsDrPoe: Blasius Boundary Layer Example
It's another Thesis Tuesday on the blog, and as promised, I'll be giving you an example of the Blasius Boundary Layer Solution.
Problem: A viscous fluid with a known density (rho = 1300 kg/m^3) flows past a flat plate such that the boundary layer thickness at a distance 1.3 m from the leading edge is 12 mm. Determine the boundary layer thickenss, shear stress and friction coefficient at a distance of 0.2 m from the leading edge if the free stream velocity is 1.5 m/s.
Given: xk = 1.3 m, deltak = 12 mm, rho = 1300 kg/m^3, U = 1.5 m/s
Find: delta, tauw, and Cf at 0.2 m
Solution:
Pretty simple, huh? Until next time...keep studying!
Problem: A viscous fluid with a known density (rho = 1300 kg/m^3) flows past a flat plate such that the boundary layer thickness at a distance 1.3 m from the leading edge is 12 mm. Determine the boundary layer thickenss, shear stress and friction coefficient at a distance of 0.2 m from the leading edge if the free stream velocity is 1.5 m/s.
Given: xk = 1.3 m, deltak = 12 mm, rho = 1300 kg/m^3, U = 1.5 m/s
Find: delta, tauw, and Cf at 0.2 m
Solution:
Rek = (5*xk/deltak)^2 = 2.934x10^5
nu = U*xk/Rek = 6.646x10^-6 m*m/s
mu = nu*rho = 8.64x10^-3 N*s/m*m
Rex = U*x/nu = 45,140
delta = 5*x/sqrt(Rex) =4.707 mm
tauw = 0.3332*U^(3/2)*sqrt(rho*mu/x) = 4.571 Pa
Cf = 0.664/sqrt(Rex) = 3.125x10^-3
Pretty simple, huh? Until next time...keep studying!
Tuesday, December 20, 2011
MrsDrPoe: The Blasius Boundary Layer Solution
It's Thesis Tuesday on the blog, and today we'll be looking at the Blasius Boundary Layer solution.
Often when discussing boundary layer flow over a flat plate, the Blasius boundary layer solution is examined. This solution is based on the fact that, even though the boundary layer profile changes along the plate, its general shape remains the same. To capture this relationship, Blasius defined a similarity variable, eta = y*sqrt(Uinfinity/(2*nu*x)). We can define a stream function for the flow based on the volumetric flow rate between streamlines: psi = sqrt(2*nu*Uinfinity*x)*f(eta). Using our stream function/velocity relations and the product rule, we can see that:
Since we now have our velocity components in terms of x and y, we can plug these expressions into the boundary layer momentum equation to obtain (with MUCH manipulation):
The boundary conditions for this differential equation are f = f' = 0 at eta = 0 and f' -> 1 as eta -> infinity. Check out the figure here.
From the computed solution, it can be seen that u = 0.99*Uinfinity when eta = 5.0. This gives us: delta = 5*sqrt(nu*x/Uinfinity) = 5*x/sqrt(Rex), where Rex is the local Reynolds number at any x-location along the plate (Rex = Uinfinity*x/nu). Furthermore, since we now know the velocity profile in the boundary layer, we can calculate the shear stress at any x-location along the plate by tauw = 0.332*Uinfinity^(3/2)*sqrt(rho*mu/x). We can also determine the friction coefficient at a particular x-location Cf = 0.664/sqrt(Rex).
Next week we'll look at an example problem involving these equations. Happy studying!
Often when discussing boundary layer flow over a flat plate, the Blasius boundary layer solution is examined. This solution is based on the fact that, even though the boundary layer profile changes along the plate, its general shape remains the same. To capture this relationship, Blasius defined a similarity variable, eta = y*sqrt(Uinfinity/(2*nu*x)). We can define a stream function for the flow based on the volumetric flow rate between streamlines: psi = sqrt(2*nu*Uinfinity*x)*f(eta). Using our stream function/velocity relations and the product rule, we can see that:
u = d/dy(psi) = Uinfinity*f'(eta)
v = -d/dx(psi) = sqrt((nu*Uinfinity)/(4*x))*(eta*f'(eta) - f(eta)).
f''' + f*f'' = 0.
The boundary conditions for this differential equation are f = f' = 0 at eta = 0 and f' -> 1 as eta -> infinity. Check out the figure here.
From the computed solution, it can be seen that u = 0.99*Uinfinity when eta = 5.0. This gives us: delta = 5*sqrt(nu*x/Uinfinity) = 5*x/sqrt(Rex), where Rex is the local Reynolds number at any x-location along the plate (Rex = Uinfinity*x/nu). Furthermore, since we now know the velocity profile in the boundary layer, we can calculate the shear stress at any x-location along the plate by tauw = 0.332*Uinfinity^(3/2)*sqrt(rho*mu/x). We can also determine the friction coefficient at a particular x-location Cf = 0.664/sqrt(Rex).
Next week we'll look at an example problem involving these equations. Happy studying!
Tuesday, December 13, 2011
MrsDrPoe: Boundary Layer Equations
It's once again Thesis Tuesday on the blog, and today we'll look some more at external flow.
One of the most important considerations of external flow is this region around the body in the flow that is affected by viscous stresses. This region, called a boundary layer, possesses a large velocity gradient as the velocity transitions from 0 at the wall (from the no-slip condition) to 0.99*Uinfinity at the top of the boundary layer. Since tau = mu*d/dy(u), we know that it is because of this velocity gradient that viscous stresses are important! Typically, the height (distance from the surface to the top of the boundary layer) is denoted delta.
Fluid flow in the boundary layer is governed by a special reduced form of the governing equations. We can find these equations using simple scaling analysis. We will begin with the continuity equation for flow over a flat plate:
Reducing for incompressible, 2D flow:
We can say that du scales with Uinfinity (Uinfinity is a representative x-velocity), dx scales with L, dv scales with vs, and dy scales with delta. For continuity to be satisfied, these two scaled terms must be of the same magnitude. Thus, setting them equal, we find that vs = U*delta/L.
Next we will look at the Navier-Stokes y-momentum equation (so we've made assumptions of Newtonian Fluid, steady, constant properties, laminar, incompressible) for 2D flow:
Applying our scaling arguments as before with the exception of p scaling with rho*U*U: rho*U*vs/L, rho*vs*vs/delta, rho*U*U/delta, v*vs/(L*L), v*vs/(delta*delta). If we substitute in the value for vs that we found from the continuity equation:
rho*U*(1/L)*(U*delta/L), rho*(U*delta/L)*(1/delta)*(U*delta/L), rho*U*U/delta, v*(1/(L*L))*(U*delta/L), v*(1/delta)*(U*delta/L)
We can make these terms dimensionless by multiplying both sides by delta/(U*U*rho): (delta*deta)/(L*L), (delta*delta)/(L*L), 1, (delta*delta)/(L*L), 1/Re, 1/Re. If we then take the limit as delta/L (the boundary layer is VERY thin) -> 0 and Re -> infinity, we see that the only term that does not go to zero is the pressure term, d/dy(p). From these arguments, we can see that d/dy(p) = 0, or the pressure does not vary significantly in the direction normal to the wall.
Finally, we will apply scaling arguments to the Navier-Stokes x-momentum equation (again already reduced for 2D flow): rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*(d/dx(d/dx(u)) + d/dy(d/dy(u))). The terms become: rho*U*U/L, rho*vs*U/delta, rho*U*U/L, v*U/L, v*U/delta. Substituting our expression for vs: rho*U*U/L, rho*(U*delta/L)*U/delta, rho*U*U/L, v*U/L, v*U/delta and non-dimensionalizing by multiplying by L/(U*U*rho): 1, 1, 1, 1/Re, (L*L/delta*delta), 1/Re. If we again take the limit as delta/L -> 0 and Re -> infinity, we can see that the fourth term goes to zero; however, it is unclear what the last term becomes. We can examine three possibilities: a) it's less that 1, b) it's greater than 1, or c) it's equal to 1.
For case (a), scaling allows us to eliminate this term (since it is insignificant compared to the terms that reduced to 1) leaving: rho*(u*d/dx(v) + v*d/dy(v)) = -d/dx(p). This equation is true for potential flow (irrotational, no viscous stresses), but since we definitely have these stresses present in the boundary layer, this option cannot be true.
For case (b), the fifth term is the most significant term in the momentum equation so it would become: 0 = v*d/dy(d/dy(u)). While this equation accounts for viscous stresses, it is purely diffusive and incorrect (we can see that d/dy(u) changes along the plate in the x-direction and is not constant).
So case (c) must be correct, which means our boundary layer x-momentum equation can be written: rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*d/dy(d/dy(u)).
A few miscellaneous notes:
Our full set of boundary layer equations consists of:
The boundary conditions for these equations are: u(x,0) = 0 (no-slip), v(x,0) = 0 (no-slip), u(x,delta) = Uinfinity, and u(x0,y) = uin(y) (starting velocity profile).
Incidentally, for our scaling: delta/L scales with 1/sqrt(Re). Also, just outside the boundary layer, Bernoulli's equation applies.
One of the most important considerations of external flow is this region around the body in the flow that is affected by viscous stresses. This region, called a boundary layer, possesses a large velocity gradient as the velocity transitions from 0 at the wall (from the no-slip condition) to 0.99*Uinfinity at the top of the boundary layer. Since tau = mu*d/dy(u), we know that it is because of this velocity gradient that viscous stresses are important! Typically, the height (distance from the surface to the top of the boundary layer) is denoted delta.
Fluid flow in the boundary layer is governed by a special reduced form of the governing equations. We can find these equations using simple scaling analysis. We will begin with the continuity equation for flow over a flat plate:
d/dt(rho) + d/dx(rho*u) + d/dy(rho*v) + d/dz(rho*w) = 0
Reducing for incompressible, 2D flow:
d/dx(u) + d/dy(v) = 0
We can say that du scales with Uinfinity (Uinfinity is a representative x-velocity), dx scales with L, dv scales with vs, and dy scales with delta. For continuity to be satisfied, these two scaled terms must be of the same magnitude. Thus, setting them equal, we find that vs = U*delta/L.
Next we will look at the Navier-Stokes y-momentum equation (so we've made assumptions of Newtonian Fluid, steady, constant properties, laminar, incompressible) for 2D flow:
rho*(u*d/dx(v) + v*d/dy(v)) = -d/dy(p) + v*(d/dx(d/dx(v)) + d/dy(d/dy(v)))
Applying our scaling arguments as before with the exception of p scaling with rho*U*U: rho*U*vs/L, rho*vs*vs/delta, rho*U*U/delta, v*vs/(L*L), v*vs/(delta*delta). If we substitute in the value for vs that we found from the continuity equation:
rho*U*(1/L)*(U*delta/L), rho*(U*delta/L)*(1/delta)*(U*delta/L), rho*U*U/delta, v*(1/(L*L))*(U*delta/L), v*(1/delta)*(U*delta/L)
We can make these terms dimensionless by multiplying both sides by delta/(U*U*rho): (delta*deta)/(L*L), (delta*delta)/(L*L), 1, (delta*delta)/(L*L), 1/Re, 1/Re. If we then take the limit as delta/L (the boundary layer is VERY thin) -> 0 and Re -> infinity, we see that the only term that does not go to zero is the pressure term, d/dy(p). From these arguments, we can see that d/dy(p) = 0, or the pressure does not vary significantly in the direction normal to the wall.
Finally, we will apply scaling arguments to the Navier-Stokes x-momentum equation (again already reduced for 2D flow): rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*(d/dx(d/dx(u)) + d/dy(d/dy(u))). The terms become: rho*U*U/L, rho*vs*U/delta, rho*U*U/L, v*U/L, v*U/delta. Substituting our expression for vs: rho*U*U/L, rho*(U*delta/L)*U/delta, rho*U*U/L, v*U/L, v*U/delta and non-dimensionalizing by multiplying by L/(U*U*rho): 1, 1, 1, 1/Re, (L*L/delta*delta), 1/Re. If we again take the limit as delta/L -> 0 and Re -> infinity, we can see that the fourth term goes to zero; however, it is unclear what the last term becomes. We can examine three possibilities: a) it's less that 1, b) it's greater than 1, or c) it's equal to 1.
For case (a), scaling allows us to eliminate this term (since it is insignificant compared to the terms that reduced to 1) leaving: rho*(u*d/dx(v) + v*d/dy(v)) = -d/dx(p). This equation is true for potential flow (irrotational, no viscous stresses), but since we definitely have these stresses present in the boundary layer, this option cannot be true.
For case (b), the fifth term is the most significant term in the momentum equation so it would become: 0 = v*d/dy(d/dy(u)). While this equation accounts for viscous stresses, it is purely diffusive and incorrect (we can see that d/dy(u) changes along the plate in the x-direction and is not constant).
So case (c) must be correct, which means our boundary layer x-momentum equation can be written: rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*d/dy(d/dy(u)).
A few miscellaneous notes:
Our full set of boundary layer equations consists of:
d/dx(u) + d/dx(v) = 0
rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*d/dy(d/dy(u))
0 = d/dy(p)
The boundary conditions for these equations are: u(x,0) = 0 (no-slip), v(x,0) = 0 (no-slip), u(x,delta) = Uinfinity, and u(x0,y) = uin(y) (starting velocity profile).
Incidentally, for our scaling: delta/L scales with 1/sqrt(Re). Also, just outside the boundary layer, Bernoulli's equation applies.
Tuesday, December 6, 2011
MrsDrPoe: Characteristics of Flow Past an Object
Happy Thesis Tuesday to you all! Today we'll be starting a series on external flow by examining some of this flow's general characteristics.
Flows in which an object is completely submerged in a fluid are termed external flows; however, flows such as those around buildings are also considered external flows, even though buildings aren't completely submerged. We can consider cases where the object or body is stationary and fluid is flowing around it, where the object is moving through a stationary fluid, or some combination of the two. For all of these situations, if we fix our coordinate system with the body, we can analyze these scenarios as if fluid were flowing over a stationary body. We will consider the velocity upstream of the body (Uinfinity) to be constant with respect to time and space. The bodies in the flow can be classified using one of two systems:
1a) 2D object extending infinitely in a third direction
1b) axi-symmetric bodies formed by rotating a cross-section about an axis of symmetry
1c) 3D bodies
2a) streamlined bodies
2b) blunt bodies
As you can imagine, flow past an object is influenced by both the fluid properties and the size and shape of the object. These characteristics are typically grouped in dimensionless parameters; those used most often for external flows are Reynolds number, Mach number, and the Froude number.
We can examine some general differences in flows for a range of Reynolds number values by taking a closer look at flows over a flat plate. Here we consider a plate of length, l, in the same fluid with the same viscosity and density, but we will continue to increase the upstream velocity, Uinfinity, so that Re = 0.1, 10, and 10^7. As the Reynolds number increases, the region around the plate where the viscous forces/stresses are important shrinks considerably, causing outer streamlines to be deflected from the plate less and less. The wake region behind the plate also shrinks as the Reynolds number increases. It should be noted that flows with Re < 1 are dominated by viscous effects while flows with Re > 1 are dominated by inertia.
If we look at flow over a cylinder, we can make some more generalizations about how flow over blunt bodies is affected by changes in Reynolds number. For low Reynolds number flows (0.1), again we see large deflection of the streamlines far away from the body. These streamlines appear to be symmetric about the center of the cylinder (the stagnation streamline) as we saw when we investigated the velocity potential function. As the Reynolds number increases (50, 10^5), we see this area affected by the viscous stresses again shrinking. We can also see that the flow separates from the surface of the clyinder, creating a recirculation bubble or wake behind the cylinder.
Next week we'll continue looking at external flows, focusing on the region affected by the viscous forces. Until then, happy studying!
Flows in which an object is completely submerged in a fluid are termed external flows; however, flows such as those around buildings are also considered external flows, even though buildings aren't completely submerged. We can consider cases where the object or body is stationary and fluid is flowing around it, where the object is moving through a stationary fluid, or some combination of the two. For all of these situations, if we fix our coordinate system with the body, we can analyze these scenarios as if fluid were flowing over a stationary body. We will consider the velocity upstream of the body (Uinfinity) to be constant with respect to time and space. The bodies in the flow can be classified using one of two systems:
1a) 2D object extending infinitely in a third direction
1b) axi-symmetric bodies formed by rotating a cross-section about an axis of symmetry
1c) 3D bodies
2a) streamlined bodies
2b) blunt bodies
As you can imagine, flow past an object is influenced by both the fluid properties and the size and shape of the object. These characteristics are typically grouped in dimensionless parameters; those used most often for external flows are Reynolds number, Mach number, and the Froude number.
We can examine some general differences in flows for a range of Reynolds number values by taking a closer look at flows over a flat plate. Here we consider a plate of length, l, in the same fluid with the same viscosity and density, but we will continue to increase the upstream velocity, Uinfinity, so that Re = 0.1, 10, and 10^7. As the Reynolds number increases, the region around the plate where the viscous forces/stresses are important shrinks considerably, causing outer streamlines to be deflected from the plate less and less. The wake region behind the plate also shrinks as the Reynolds number increases. It should be noted that flows with Re < 1 are dominated by viscous effects while flows with Re > 1 are dominated by inertia.
If we look at flow over a cylinder, we can make some more generalizations about how flow over blunt bodies is affected by changes in Reynolds number. For low Reynolds number flows (0.1), again we see large deflection of the streamlines far away from the body. These streamlines appear to be symmetric about the center of the cylinder (the stagnation streamline) as we saw when we investigated the velocity potential function. As the Reynolds number increases (50, 10^5), we see this area affected by the viscous stresses again shrinking. We can also see that the flow separates from the surface of the clyinder, creating a recirculation bubble or wake behind the cylinder.
Next week we'll continue looking at external flows, focusing on the region affected by the viscous forces. Until then, happy studying!
Tuesday, November 22, 2011
MrsDrPoe: Dimensional Analysis, Part II
Happy Thesis Tuesday! Today we'll resume our discussion on dimensional analysis by further investigating the Buckingham Pi Theorem.
This theorem is applied to a problem through several simple steps:
Step 1: List all the variables that are involved in the problem. This step is crucial- without properly identifying all the necessary variables, we will not be able to obtain the correct dimensionless groups. Each of the properties should be independent. For example, if density and specific weight are important, we could use rho and gamma, rho and g, or gamma and g, but to use all three would be redundant (since gamma = rho*g).
Step 2: Express each of the variables in terms of basic dimensions. We will be looking at these quantities in terms of length (L), time (T), mass (M), and temperature (theta), since these are the basic dimensions for SI and BG unit systems.
Step 3: Determine the required number of pi terms. This number is equal to k-r where k is the number of variables in the problem (from Step 1), and r is the number of reference dimensions required to describe these variables (from Step 2). Note: if k-r = 1, skip to Step 5.
Step 4: Select a number of repeating variables, where the number required is equal to the number of reference dimensions. Here we should choose r number of variables from our list of independent variables; there are our "repeating variables." It is best to choose ones with the simplest representation in basic dimensions (for instance: choose a distance, L, instead of a velocity, L/T).
Step 5: Form a pi term by multiplying one of the non-repeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless. For example: Pii = ui*u1^(ai)*u2^(bi)*u3^(ci), where ui is a non-repeating variable, u1, u2, and u3 are repeating variables, and a, b, and c are exponents. Note: it is acceptable for a, b, or c to equal zero as long as at least one is non-zero.
Step 6: Repeat Step 5 for each of the remaining non-repeating variables.
Step 7: Check all resulting pi terms to make sure they are dimensionless. It's easy to make mistakes through this process; check your work by writing each of the variables in terms of their basic dimensions (or even units) to see if all the basic dimensions cancel out.
Step 8: Express the final form as a relationship among the pi terms, and think about what it means. For example: Pi1 = f(Pi2, Pi3, ..., Pik-r), where Pi1 is the pi term containing the dependent variable.
While at this point you may not believe what I said about this process being simple, next week, I'll prove it to you with an example problem. We'll also look at some common pi terms/dimensionless number in fluid mechanics. Happy studying!
This theorem is applied to a problem through several simple steps:
Step 1: List all the variables that are involved in the problem. This step is crucial- without properly identifying all the necessary variables, we will not be able to obtain the correct dimensionless groups. Each of the properties should be independent. For example, if density and specific weight are important, we could use rho and gamma, rho and g, or gamma and g, but to use all three would be redundant (since gamma = rho*g).
Step 2: Express each of the variables in terms of basic dimensions. We will be looking at these quantities in terms of length (L), time (T), mass (M), and temperature (theta), since these are the basic dimensions for SI and BG unit systems.
Step 3: Determine the required number of pi terms. This number is equal to k-r where k is the number of variables in the problem (from Step 1), and r is the number of reference dimensions required to describe these variables (from Step 2). Note: if k-r = 1, skip to Step 5.
Step 4: Select a number of repeating variables, where the number required is equal to the number of reference dimensions. Here we should choose r number of variables from our list of independent variables; there are our "repeating variables." It is best to choose ones with the simplest representation in basic dimensions (for instance: choose a distance, L, instead of a velocity, L/T).
Step 5: Form a pi term by multiplying one of the non-repeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless. For example: Pii = ui*u1^(ai)*u2^(bi)*u3^(ci), where ui is a non-repeating variable, u1, u2, and u3 are repeating variables, and a, b, and c are exponents. Note: it is acceptable for a, b, or c to equal zero as long as at least one is non-zero.
Step 6: Repeat Step 5 for each of the remaining non-repeating variables.
Step 7: Check all resulting pi terms to make sure they are dimensionless. It's easy to make mistakes through this process; check your work by writing each of the variables in terms of their basic dimensions (or even units) to see if all the basic dimensions cancel out.
Step 8: Express the final form as a relationship among the pi terms, and think about what it means. For example: Pi1 = f(Pi2, Pi3, ..., Pik-r), where Pi1 is the pi term containing the dependent variable.
While at this point you may not believe what I said about this process being simple, next week, I'll prove it to you with an example problem. We'll also look at some common pi terms/dimensionless number in fluid mechanics. Happy studying!
Tuesday, November 15, 2011
MrsDrPoe: Dimensional Analysis, Part I
Happy Thesis Tuesday on the blog! Today we'll look at a brief introduction to the interesting topic of dimensional analysis.
As with any type of experimentation, fluid flow experiments should be performed in a meaningful and systematic manner. If we consider, for instance, the pressure change across a section of pipe in Poiselle flow, we can see that this change in pressure is affected by fluid properties (density, rho; and viscosity, mu), pipe diameter, and fluid velocity. By running countless numbers of measurement experiments, each time varying a single parameter, we could determine how the pressure change is affected by each of these items; however, these relationships would only be valid for this pipe with this fluid and this experimental setup.
In order to improve the quality, applicability, and repeatability of our experimental results, we use dimensionless products or dimensionless groups. These terms are combinations of variables necessary for a problem and often employ basic dimensions (mass, M; length, L; and time, T) or units (pounds, feet, and seconds, etc.).
The method we use to find these dimensionless terms is called the Buckingham Pi Theorem - "If an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the variables."
Next week, we'll look at an application of this theorem, which is much simpler than it may sound!
As with any type of experimentation, fluid flow experiments should be performed in a meaningful and systematic manner. If we consider, for instance, the pressure change across a section of pipe in Poiselle flow, we can see that this change in pressure is affected by fluid properties (density, rho; and viscosity, mu), pipe diameter, and fluid velocity. By running countless numbers of measurement experiments, each time varying a single parameter, we could determine how the pressure change is affected by each of these items; however, these relationships would only be valid for this pipe with this fluid and this experimental setup.
WHEW!
In order to improve the quality, applicability, and repeatability of our experimental results, we use dimensionless products or dimensionless groups. These terms are combinations of variables necessary for a problem and often employ basic dimensions (mass, M; length, L; and time, T) or units (pounds, feet, and seconds, etc.).
The method we use to find these dimensionless terms is called the Buckingham Pi Theorem - "If an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the variables."
Next week, we'll look at an application of this theorem, which is much simpler than it may sound!
Tuesday, November 8, 2011
MrsDrPoe: The First Law of Thermodynamics, Part 6
Hello and welcome to Thesis Tuesday on the blog! As promised, today we will be finishing our look at the first law of thermodynamics by applying the differential form of the first law to an example problem.
Problem: A liquid is flowing downward along an inclined plane surface, as shown in the figure:
The free liquid surface (y = h) is maintained at temperature Th, and the solid surface (y = 0) is maintained at To. Determine an expression for the temperature distribution in the film, recognizing that the viscous heating effects can be ignored.
Given: non-isothermal film flow, T(0) = T0 and T(h) = Th
Find: An expression for the temperature distribution in the fluid film
Assumptions: steady, laminar, incompressible, Newtonian fluid, ignore viscous heating effects, constant thermal conductivity, assume T = f(y), constant properties
Solution:
Starting with the general form of the differential energy equation for an incompressible Newtonian fluid:
Steady flow eliminates the first term on the left side; the last term on the right side is eliminated because we are ignoring viscous heating effects.
Our assumption of laminar flow means that v = w = 0; the first term on the left side is also zero since T is in not a function of x:
The flux terms can be related to the temperature gradient using Fourier's law:
Furthermore, since the thermal conductivity (k) is constant, our energy equation becomes:
We can divide both sides by k; the first and third terms cancel since T is not a function of x or z:
If we separate and integrate this equation twice, we end up with the expression:
We can solve for c1 and c2 by applying the given boundary conditions at 0 and h:
So we now know the expression that shows us the temperature distribution in the fluid film:
Not so bad, huh? Next week we'll begin anew with a fresh subject, but until then, happy studying!
Problem: A liquid is flowing downward along an inclined plane surface, as shown in the figure:
The free liquid surface (y = h) is maintained at temperature Th, and the solid surface (y = 0) is maintained at To. Determine an expression for the temperature distribution in the film, recognizing that the viscous heating effects can be ignored.
Given: non-isothermal film flow, T(0) = T0 and T(h) = Th
Find: An expression for the temperature distribution in the fluid film
Assumptions: steady, laminar, incompressible, Newtonian fluid, ignore viscous heating effects, constant thermal conductivity, assume T = f(y), constant properties
Solution:
Starting with the general form of the differential energy equation for an incompressible Newtonian fluid:
rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz)) + mu*Phiv
Steady flow eliminates the first term on the left side; the last term on the right side is eliminated because we are ignoring viscous heating effects.
rho*cp*(u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz))
Our assumption of laminar flow means that v = w = 0; the first term on the left side is also zero since T is in not a function of x:
0 = (d/dx(qx) + d/dy(qy) + d/dz(qz))
The flux terms can be related to the temperature gradient using Fourier's law:
qx = k*d/dx(T), qy = k*d/dy(T), and qz = k*d/dz(T)
Furthermore, since the thermal conductivity (k) is constant, our energy equation becomes:
0 = k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))
We can divide both sides by k; the first and third terms cancel since T is not a function of x or z:
0 = d/dy(d/dy(T))
If we separate and integrate this equation twice, we end up with the expression:
T(y) = c1*y + c2
We can solve for c1 and c2 by applying the given boundary conditions at 0 and h:
c2 = T0 and c1 = (Th - T0)/h
So we now know the expression that shows us the temperature distribution in the fluid film:
T(y) = (Th - T0)/h*y + T0
Not so bad, huh? Next week we'll begin anew with a fresh subject, but until then, happy studying!
Tuesday, November 1, 2011
MrsDrPoe: The First Law of Thermodynamics, Part 5
Happy Thesis Tuesday to you all! Today we'll be continuing our discussion of the First Law of Thermodynamics by examining its final form- the differential form. For an incompressible, Newtonian fluid in Cartesian coordinates, this is:
Typically, Phiv is neglected. It is only important for flows with large velocity gradients or very large viscosities. Furthermore, it should be noted that if the fluid has a constant thermal conductivity, the first term on the right-hand side of the equation becomes:
rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz)) + mu*Phiv
where
qx = d/dx(k*T), etc.
Phiv = 2*((d/dx(u))*(d/dx(u)) + (d/dy(v))*(d/dy(v)) + (d/dz(w))*(d/dz(w))) + (d/dx(v) + d/dy(u))*(d/dx(v) + d/dy(u)) + (d/dy(w) + d/dz(v))*(d/dy(w) + d/dz(v)) + (d/dx(w) + d/dz(u))*(d/dx(w) + d/dz(u)) - (2/3)*(d/dx(u) + d/dy(v) + d/dz(w))*(d/dx(u) + d/dy(v) + d/dz(w))
Typically, Phiv is neglected. It is only important for flows with large velocity gradients or very large viscosities. Furthermore, it should be noted that if the fluid has a constant thermal conductivity, the first term on the right-hand side of the equation becomes:
(d/dx(qx) + d/dy(qy) + d/dz(qz)) = k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))
So the final equation that we will be dealing with is:
rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))
In these equations, T is the temperature of the fluid and q is the heat flux. Next week we'll conclude our look at the first law with an example employing this differential equation.
Tuesday, October 25, 2011
MrsDrPoe: The First Law of Thermodynamics, Part IV
Happy Thesis Tuesday! Today we'll continue our discussion of the head form of the First Law of Thermodynamics with an example problem:
Water is moved from one large reservoir to another at a higher elevation. The loss of available energy associated with 2.5 ft*ft*ft/s being pumped from sections (1) to (2) is 30.5 |V|*|V|/g, where |V| is the average velocity of water in the 8 in inside diameter piping involved. Determine the amount of shaft power required to pump the water.
Given: Q = 2.5 ft*ft*ft/s, losses = 30.5 *|V|*|V|/g, zdelta = 50 ft, D = 8 in
Find: shaft power required
Assumptions: steady, laminar, 1D, incompressible, constant average velocity, and constant properties
Solution:
For this case, we do not need to consider the continuity equation if we set our problem up correctly. So we will start with the head form of the energy equation:
We will chose "in" as a point at the surface of the lower tank and "out" as a point at the surface of the upper tank, which tells us:
In putting these values into the equation, we have:
Since our equation for the losses is given as a function of the average velocity, we know that these losses are primarily due to the friction in the pipe. We can determine this velocity since we know both the flow rate and the pipe diameter. NOTE: Q = |V|*A is only valid for use with uniform velocity or average velocity. We know that since there are friction losses, mu is not zero, our velocity is not uniform, so in order to use this equation, we must use average velocity, which we are given.
Solving for hP:
Now we can determine the necessary power from the pump head:
And that's the head form of the equation. Next week, we'll look at the differential form of the First Law!
Water is moved from one large reservoir to another at a higher elevation. The loss of available energy associated with 2.5 ft*ft*ft/s being pumped from sections (1) to (2) is 30.5 |V|*|V|/g, where |V| is the average velocity of water in the 8 in inside diameter piping involved. Determine the amount of shaft power required to pump the water.
Given: Q = 2.5 ft*ft*ft/s, losses = 30.5 *|V|*|V|/g, zdelta = 50 ft, D = 8 in
Find: shaft power required
Assumptions: steady, laminar, 1D, incompressible, constant average velocity, and constant properties
Solution:
For this case, we do not need to consider the continuity equation if we set our problem up correctly. So we will start with the head form of the energy equation:
(pout/gamma) + 0.5*|V|*|V|/g + zout = (pin/gamma) + 0.5*|V|*|V|/g + zin + hP - hL
We will chose "in" as a point at the surface of the lower tank and "out" as a point at the surface of the upper tank, which tells us:
pin = 0, pout = 0 since both surfaces are exposed to the atm
vin = 0, vout = 0 since we can assume that the water on the surfaces is still
zin = 0, zout = zdelta setting z = 0 at the surface of the bottom tank
In putting these values into the equation, we have:
zout = hP - hL where hL = losses
Since our equation for the losses is given as a function of the average velocity, we know that these losses are primarily due to the friction in the pipe. We can determine this velocity since we know both the flow rate and the pipe diameter. NOTE: Q = |V|*A is only valid for use with uniform velocity or average velocity. We know that since there are friction losses, mu is not zero, our velocity is not uniform, so in order to use this equation, we must use average velocity, which we are given.
|V| = 4*Q/(pi*D*D) = 7.162 ft/s
hL = 30.5*|V|*|V|/g = 48.625 ft
Solving for hP:
hP = zout + hL = 98.625 ft
(since this answer is positive, work is introduced INTO the flow)
Now we can determine the necessary power from the pump head:
Wreq = hP*gamma*Q = 20,850 W = 27.958 hp
And that's the head form of the equation. Next week, we'll look at the differential form of the First Law!
Tuesday, October 11, 2011
MrsDrPoe: The First Law of Thermodynamics, Part II
Happy Thesis Tuesday to you! Last week, we started talking about the integral form of the first law of thermodynamics. Today we'll look at a simple example of how this equation is applied.
Problem: Suppose water is flowing through a 0.4 m*m pipe (1) with a velocity of 25 m/s and a static pressure of 940 kPa. The pipe splits into two branches, one that is 0.18 m*m (3) and one that is 0.28 m*m (2). At section (3), the static pressure is measured to be 570 kPa, and the velocity is 30 m/s; at section (2), the static pressure is 1140 kPa, and the velocity is unknown. Determine the amount of available power lost in this horizontal y-connection.
Given: we know the areas and pressures at each location, the velocity at two locations, and the fluid properties.
Find: amount of available power lost
Assumptions: steady, incompressible, uniform properties and pressure across the pipe cross-section, horizontal pipe, no pumps/turbines, constant average velocity
Solution:
After making our assumptions and drawing our control volume, we must apply the continuity equation:
For steady, incompressible flow with three surfaces where flow is crossing the CS:
Since (1) is an inlet, (2) and (3) are outlets, and the velocities are constant averages, and dividing out density:
Now that we have determined the velocity at (2), we can apply the first law:
For steady, incompressible flow with no pumps or turbines, and three surfaces where flow crosses the CS:
Since (1) is an inlet, (2) and (3) are exits, and we have uniform properties and constant average velocities:
Consider the terms:
Since the y is horizontal, there is no difference in z1, z2 or z3:
The terms in the parentheses above is the same as those in expression (I), which we know is equal to zero; therefore, (II) is equal to zero.
Our losses for this problem are defined by:
So we end up with:
Problem: Suppose water is flowing through a 0.4 m*m pipe (1) with a velocity of 25 m/s and a static pressure of 940 kPa. The pipe splits into two branches, one that is 0.18 m*m (3) and one that is 0.28 m*m (2). At section (3), the static pressure is measured to be 570 kPa, and the velocity is 30 m/s; at section (2), the static pressure is 1140 kPa, and the velocity is unknown. Determine the amount of available power lost in this horizontal y-connection.
Given: we know the areas and pressures at each location, the velocity at two locations, and the fluid properties.
Find: amount of available power lost
Assumptions: steady, incompressible, uniform properties and pressure across the pipe cross-section, horizontal pipe, no pumps/turbines, constant average velocity
Solution:
After making our assumptions and drawing our control volume, we must apply the continuity equation:
d/dt(int(rho)dV) + int(rho*(V.n))dA = 0
For steady, incompressible flow with three surfaces where flow is crossing the CS:
rho*int(V1.n1)dA1 + rho*int(V2.n2)dA2 + rho*int(V3.n3)dA3 = 0
Since (1) is an inlet, (2) and (3) are outlets, and the velocities are constant averages, and dividing out density:
-|V1|*A1 + |V2|*A2 + |V3|*A3 = 0 (I)
|V2| = (|V1|*A1 - |V3|*A3)/A2 = 16.429 m/s
Now that we have determined the velocity at (2), we can apply the first law:
d/dt(int(e*rho)dV) + int((u + (p/rho) + 0.5*|V|*|V| + g*z)*rho*(V.n))dA = Qdot,net_in + Wdot,shaft_in
For steady, incompressible flow with no pumps or turbines, and three surfaces where flow crosses the CS:
int((u1 + (p1/rho) + 0.5*|V1|*|V1| + g*z1)*rho*(V1.n1))dA1 + int((u2 + (p2/rho) + 0.5*|V2|*|V2| + g*z2)*rho*(V2.n2))dA2 + int((u3 + (p3/rho) + 0.5*|V3|*|V3| + g*z3)*rho*(V3.n3))dA3= Qdot,net_in
Since (1) is an inlet, (2) and (3) are exits, and we have uniform properties and constant average velocities:
-rho*(u1 + (p1/rho) + 0.5*|V1|*|V1| + g*z1)*|V1|*A1 + rho*(u2 + (p2/rho) + 0.5*|V2|*|V2| + g*z2)*|V2|*A2 + rho*(u3 + (p3/rho) + 0.5*|V3|*|V3| + g*z3)*|V3|*A3 = Qdot,net_in
Consider the terms:
-rho*g*z1*|V1|*A1 + rho*g*z2*|V2|*A2 + rho*g*z3*|V3|*A3
Since the y is horizontal, there is no difference in z1, z2 or z3:
rho*g*z*(|V2|*A2 + |V3|*A3 - |V1|*A1) (II)
The terms in the parentheses above is the same as those in expression (I), which we know is equal to zero; therefore, (II) is equal to zero.
Our losses for this problem are defined by:
-Qdot,net_in - rho*u1*|V1|*A1 + rho*u2*|V2|*A2 + rho*u3*|V3|*A3
So we end up with:
losses = rho*((p1/rho) + 0.5*|V1|*|V1|)*|V1|*A1 -rho*((p2/rho) + 0.5*|V2|*|V2|)*|V2|*A2 -rho*((p3/rho) + 0.5*|V3|*|V3|)*|V3|*A3
losses = 1.152 x 10^6 W
And there ya go. Next week we'll look at the head form of the first law, but until then, happy studying!
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