Pressure drag acting on a sphere

In summary: If you want to use a vector integral, then you would have to multiply the area element dS by the unit normal vector.
  • #1
Zar139
2
0

Homework Statement


I am trying to show that the pressure drag acting on a sphere is 2πμaU by integrating around the surface of the sphere where U is the speed of the fluid the sphere is in, a is its radius, and μ is the viscosity.

Homework Equations


The pressure at a given position r can be written as:
P-P0 = -1.5μUacos(Θ)/r2

The Attempt at a Solution


I have tried integrating the above equation for Θ from 0 to 2π over the surface of the sphere but my answer is 0 which is clearly not the correct answer. I feel like this should be a fairly easy integration but I do not know how else to go about it. Any suggestions would be greatly appreciated.

Thank you!
 
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  • #2
Zar139 said:

Homework Statement


I am trying to show that the pressure drag acting on a sphere is 2πμaU by integrating around the surface of the sphere where U is the speed of the fluid the sphere is in, a is its radius, and μ is the viscosity.

Homework Equations


The pressure at a given position r can be written as:
P-P0 = -1.5μUacos(Θ)/r2

The Attempt at a Solution


I have tried integrating the above equation for Θ from 0 to 2π over the surface of the sphere but my answer is 0 which is clearly not the correct answer. I feel like this should be a fairly easy integration but I do not know how else to go about it. Any suggestions would be greatly appreciated.

Thank you!
The pressure acts normal to the surface of the sphere at all locations. So, you have to include this directionality in your determination of the drag force.
 
  • #3
Does this mean that I should multiply the area element dS by the unit normal vector?
 
  • #4
Zar139 said:
Does this mean that I should multiply the area element dS by the unit normal vector?
You definitely have to integrate the forces vectoriallly. How you do this depends on how you want to approach it.
 

1. What is pressure drag and how does it affect a sphere?

Pressure drag is a type of drag force that occurs on an object moving through a fluid. It is caused by the difference in pressure between the front and back of the object. In the case of a sphere, pressure drag is the force that acts in the opposite direction of the sphere's motion, slowing it down.

2. How is the pressure drag on a sphere calculated?

The pressure drag on a sphere can be calculated using the formula: Dp = 0.5 * ρ * V^2 * Cd * A, where ρ is the density of the fluid, V is the velocity of the sphere, Cd is the coefficient of drag (dependent on the shape of the sphere), and A is the cross-sectional area of the sphere.

3. What factors affect the pressure drag on a sphere?

The pressure drag on a sphere is influenced by several factors, including the velocity of the sphere, the density of the fluid, the shape of the sphere, and the roughness of its surface. Additionally, the presence of any obstacles or disturbances in the fluid flow can also affect the amount of pressure drag experienced by the sphere.

4. How does the size of a sphere affect the pressure drag?

The size of a sphere can have a significant impact on the pressure drag experienced. In general, larger spheres will experience higher pressure drag due to their larger cross-sectional area. This means that a larger force will be acting against the sphere's motion, resulting in a greater slowing effect.

5. Can pressure drag be reduced on a sphere?

Yes, pressure drag can be reduced on a sphere through various methods. One way is by streamlining the shape of the sphere, such as using a teardrop shape instead of a round shape. Additionally, adding a smooth or streamlined coating to the surface of the sphere can also help reduce pressure drag. Finally, reducing the speed or velocity at which the sphere is moving through the fluid can also decrease the amount of pressure drag experienced.

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