Flow and Drag Flashcards
What is drag
Forces acting opposite to the relative motion of the object moving through surrounding fluid
What is D Alambert’s paradox
Based on calculations an object in steady fluid flow will experience no drag. however an object in steady flow must experience drag because flow is viscid and there will be energy lost due to friction
Why did D Alambert calculate that objects in flow will experience no drag
because he assumed inviscid flow: no energy will be lost or gained ; total energy would remain constant ( no friction) and he said there will be no net pressure differences
Draw the pressure map experienced by an object in viscid flow
L12 slide 13
What is the boundary layer concept
- viscous effects of drag are confined within the boundary layer
- viscosity leads to a steep velocity gradient and high shear forces
- outside the layer viscosity is irrelevant and flow is inviscid
Describe what happens to the boundary layer of a moving sphere
- fluid hits stagnation point, no velocity high static pressure
- moves up the ball surface, gaining velocity because it is entering a constricted ‘area’ ( so static pressure decreases).
- favourable pressure gradient allows fluid to flow past ( while offsetting effects due to friction)
- when we move down the curve, velocity decreases due to frictional loses ( static pressure increases) and the increase in SA (continuity principle)
- dynamic pressure is converted back to static prssure but there is a loss of fluid momentum due to friction and the unfav. gradient
- there is an unfavourable pressure gradient. which causes flow to slow: leads to stagnation and flow separation and flow reversal
What is frictional drag ?
- occurs tangential to the surface of the object
- occurs as a direct result of viscosity
- arises due to shearing of the fluid in the boundary layer adjacent to the surface
- directly proportional to v
What is pressure drag?
- normal to the surface
- indirect result of the viscosity of the fluid reducing the momentum of the fluid
- leads to flow separation
- fluid collides with the plate area and exerts a force proportional to the density of the fluid and the surface area of the plate
- 1/2 * p * S* v^2
what is the maximum pressure acting on the object in a flowing medium
it is at the stagnation point where P_max = 1/2 * p * v^2
When we plot the pressure difference along an object length – how would we normalize this
y axis: take the dP/ P_max. (P_max= 1/2 * p * v^2) = the pressure coefficient
In a pressure coefficient vs length graph where do you expect to find
a. lens/eyes of the fish
b. the gills
a. lens will be found where C_p = 0 because there is no pressure difference for distortion
b. gills found at minimum Cp
Given that there is a fish swimming at 2m/s in salt water ( p =1027) what is the pressure gradient flowing through the gills if Cp = -0.4
Because we have to find the pressure gradient from mouth the gills so at mouth Cp =1 ( outermost point)
so we have 1-(-0.4) = 1.4 = dP / P_max
At high Reynold’s number what causes drag and what is the significance of drag
- drag is the net force exerted by the fluid on the object
- drag is due to the friction exerted by the fluid as it flows around the object AND the pressure drag due to the pressure differences around the object
How does the drag force change with increasing fluid speed
drag force increases ( since drag is proportional to fluid by v^2)
How do we standardize the drag force vs velocity graph
plot the coefficient of drag with respect to the Reynold’s number ( the greater the Reynold’s number the greater the drag since high Reynold’s number typically implies more velocity)
What is the coefficient of drag
- because drag is force = pressure* are = dynamic pressure * area so we take the drag force/ ( 1/2 * p *v ^2 * A)
- relative amount of drag experienced by an object in a flowing fluid relative to the amount of force exerted directly on some defined area of the object
What is the coefficient of drag for a perpendicular plate.
perpendicular plate will not have a flow behind the plate ( so no low pressure turbulent wake is made) will have Cd=1
For high drag body what is the area used for coefficient of drag calculations
cross sectional area
What is the reference area used for streamlined bodies
wetted area
What is the reference area used for objects with wings
planform area
What is the reference area used for Cd calc for blimp shaped objects
V^2/3
Draw how the coefficient of drag vs Re curve will look like
at low Re there is high viscosity and skin friction fominates
- at high Re pressure drag dominates
What is the drag crisis?
The minimum point in the coefficient of drag vs. Re curve.
It is when the boundary layer becomes turbulent upstream such that the boundary layer separates further back leading to a narrow wake and thus less pressure drag
Why is there flow separation of the boundary layer
drag present in the boundary layer reduces the momentum of the fluid and causes separation
How can we reduce the drag coefficient
Minimize pressure drag
- fill the space behind a bluff body to delay separation
- dimpling of the surface to induce turbulence in the boundary layer– increases the momentum exchanged between the high momentum free stream fluid and the no slip condition fluid, this adds momentum to the surface and delays boundary layer separation ( smaller wake = smaller pressure drag)
What happens to drag forces at low Re
drag depends solely on the viscous forces ( = uvS/l)
so the drag is proportional to the size and speed of the object ( greater size and speed= slower)
Why is a streamline slower than a sphere in low Re than in high Re
low Re: the streamline object has a greater surface area and so it experienes more viscous force than the sphere
high Re: because pressure forces are dominant so the streamline will have a narrow wake ( = less pressure force)
if heart volume scales with M, how will heart rate scale with animal mass
the blood volume should increase isometrically (M) while the rate at which heart is pumped should match the metabolic rate (M^2/3) so
blood flow = stroke volume * heart rate
M = M^2/3 * M^x
so M^x = M ^-1/3
Draw how the coefficient of drag vs Reynold’s number of a circular disc will look like
at high Re number the pressure drag will be dominant and the only contributing force. because the 1/2 pv^2*A will be the same so that is why it would be fixed at Cd =1 at high Re only ( at low Re it will have experienced frictional forces)
Compare the difference between the different drag components of a sphere and streamline body at high Re number
skin drag of sphere «_space;skin drag of streamline body
pressure drag of sphere»_space;»» pressure drag of streamline
so the sum of the pressure drag and skin drag causes
drag exp. by sphere > drag exp. by streamline
Given the Re and a Cd vs Re graph how would you find the drag experienced by the object
using Re read Cd off the graph and since Cd = Fd / 1/2pv^2*A we can find the area and backtrack to calculate Fd
What happens to the boundary layer if the fluid flow was inviscid
there would be no boundary layer as the no slip condition would not exist
