In this article “Drag coefficient of sphere” will be discuss and drag coefficient of sphere related detailed facts will be also explain. Drag coefficient of sphere is very important factor for a matter.

**Drag coefficient of sphere derive as, the ratio between the surface area of the sphere of the similar volume for the matter of the body comparative to the surface area for the matter of the body. Drag coefficient of sphere is a physical parameter which is deeply depends upon the shape and size of the matter.**

**What is the drag coefficient of a sphere?**

The drag coefficient of a sphere is depending upon the geometry of the matter and viscosity of the liquid substance through which the matter can flow.

**Drag coefficient of sphere derive as, any sphere shaped matter is move in a motion through a liquid substance is facing the physical parameter of drag. The total amount of force is applied in same direction where shear stress force and pressure acted on the plane of the matter.**

**How to calculate drag coefficient of a sphere?**

**Drag pressure coefficient for a sphere shaped matter can calculated using this formula,**

**Where,**

**c _{d} = Drag pressure coefficient for a sphere shaped matter**

**F _{d} = Drag force for a sphere shaped matter expressed in Newton**

**A = Plan form area for a for a sphere shaped matter expressed in square meter**

**ρ = Density of a sphere shaped matter express in kilogram per cubic meter**

**v = Viscosity a sphere shaped matter express in meter per second**

Using the eqn (1) putting the values of mass density, flow speed, drag force and area we can get the value of drag coefficient of a sphere.

For sphere matter the area can be calculated as, A = π r^{2} ……..eqn (2)

The eqn (2) is applicable for the surface area. Because surface area formula is, A = 4 πr^{2}

**Drag coefficient of sphere formula:**

**The formula of the coefficient of pressure drag is given below,**

**Where,**

**F _{d} = Drag force express in Newton**

**c _{d} = Drag coefficient**

**ρ = Density of a liquid substance express in kilogram per cubic meter**

** v = Flow velocity of a liquid substance express in meter per second**

**A =Reference area for a particular body shape substance expressed in square meter**

The pressure drag for sphere shape matter coefficient is dependent some several facts such as size and shape means geometry of the sphere body and viscosity of the liquid substance through which sphere shape matter can flow.

**Drag coefficient of a sphere vs. Reynolds number:**

**The Drag coefficient of a sphere is decreases with the Reynolds number and drag coefficient becomes almost a constant (C _{D} = 0.4) for a Reynolds number between 10^{3} and 2×10^{5}. As the Reynolds number increases (Re > 2×10^{5}), the boundary layer becomes thinner in the front of the sphere and begins its transition to turbulent.**

When a system is design which is flow in a motion through a liquid substance that time drag is best for measurement for the system or calculate drag by a simulation. Drag coefficient is frequently helpful revert the data back for a particular analytical model.

The only problem is arising during this process that is one model is not appropriate to derive all type of flow of motion in liquid substance in the region of transition and for both regimes.

Instead of particular practice is use to measure and simulated data to calculate models in each and every flow type regime and after that follow where the models are intersect for calculate the region of transition.

Now we are going to discuss the examples and the C_{f} expression is consider for both pressure drag and skin friction drag.

__Laminar Example:-__

__Laminar Example:-__

Drag coefficient of a sphere vs. Reynolds number for laminar flow can be written as,

Drag coefficient of a sphere vs. Reynolds number for laminar flow equation is perfectly goes with a wide range of Reynolds numbers in a closed system of geometries. Drag coefficient of a sphere vs. Reynolds number for laminar flow equation is not appropriate for low drag Reynolds number that could be less than 36 especially for incompressible or very near to incompressible flow.

In the incompressible or very near to incompressible flow we can observe drag coefficient is close to linear function of the velocity.

__Turbulent Example:-__

__Turbulent Example:-__

Drag coefficient of a sphere vs. Reynolds number for turbulent can be written as,

Drag coefficient of a sphere vs. Reynolds number for turbulent flow equation is perfectly goes with a one simple range of Reynolds numbers in a closed system of geometries.

**Drag force coefficient of sphere:**

The most studies case of the drag force coefficient is for sphere shaped matter of the body.

**When a sphere shaped body in solid state interact with the fluid that time drag force coefficient is produced on the sphere shaped solid matter. Drag force coefficient of sphere matter is not produced by any type of force field.**

__Frequent Asked Questions:-__

__Frequent Asked Questions:-__

__Question: – __

__Question: –__

**Write about Skin Friction drag Coefficient.**

** Solution: –**.

**Skin Friction drag coefficient is physical parameter which is dimensionless skin shear stress. It is mainly dimensionless because of the dynamic pressure is applied on the matter by the free stream.**

Skin Friction drag coefficient formula is,

Where,

C_{f} = Skin friction coefficient

T_{w}= Skin shear stress which is applied in the surface plane of the body

v_{∞} = Free stream speed for velocity of the body

ρ_{∞} = Free stream speed for density of the body

1/2ρ_{∞} v^{2}_{∞} ≡ q_{∞} = Free stream dynamic pressure for the body of the matter

Relation with the Reynolds number and the Skin Friction drag coefficient is indirectly proportional to each other. Means if Reynolds number is increases then skin Friction drag coefficient decreases and if the Reynolds number is decreases then skin Friction drag coefficient is increases.

__Laminar flow:-__

__Laminar flow:-__

Where,

Re_{x} = ρ vx/μ Represent the Reynolds number

x = Represent the distance particularly from point of the reference at which the layers of boundary is started.

__Transitional flow:-__

__Transitional flow:-__

Where,

y = Represent the distance from the wall

u = The speed for the fluid which is flow in a motion and given as y

K_{1} = **Karman Constant**

The value of Karman Constant is lower than the 0.41 and Karman Constant is the value for transitional boundary layer and turbulent boundary layer

K_{2}= Van Driest constant

K_{3}= Pressure parameter

p = Pressure

x = The coordinate along a surface where a boundary layers forms

__Turbulent flow:-__

__Turbulent flow:-__

__Question: –__

__Question: –__

**Riva drives her car daily Kolkata to Durgapur. When Riva drives her car that time the speed of the car is about 90 kilometres per hour and that time the drag coefficient is 0.35. The cross sectional area for the car is 6 square meter.**

**Now determine the amount of drag force**.

** Solution: – **Given data are,

Velocity of the car = 90 kilometres per hour

Drag coefficient of the car = 0.35

Cross sectional area of the car = 6 square meter

Density of the fluid of the car = 1.2 kilogram per cubic meter

We all are know that the velocity of the air is, 1.2 kilogram per cubic meter

Now, applied Drag force formula,

Where,

D = Pressure drag force

C_{d}= Pressure drag coefficient

ρ = Density

v = Velocity

A = Reference area

D = 0.35 * 1.2 * 8100 * 6 /2 * 3600

D = 2.8 Newton

Riva drives her car daily Kolkata to Durgapur. When Riva drives her car that time the speed of the car is about 90 kilometres per hour and that time the drag coefficient is 0.35. The cross sectional area for the car is 6 square meter.

The amount of drag force 2.8 Newton

__Question: –__

__Question: –__

**A plane is daily moves Mumbai to Katakana. When the plane moves that time the speed of the plane is about 750 kilometres per hour and that time the drag coefficient is 0.30. The cross sectional area for the car is 115 square meter.**

**Now determine the amount of drag force** **for the plane.**

** Solution: – **Given data are,

Velocity of the plane = 750 kilometres per hour

Drag coefficient of the plane = 0.30

Cross sectional area of the plane = 115 square meter

Density of the fluid of the plane = 1.2 kilogram per cubic meter

We all are know that the velocity of the air is, 1.2 kilogram per cubic meter

Now, applied Drag force formula,

D = C_{d} * ρ * A/2

Where,

D = Pressure drag force

C_{d} = Pressure drag coefficient

ρ = Density

v = Velocity

A = Reference area

D = 3234 Newton

A plane is daily moves Mumbai to Katakana. When the plane moves that time the speed of the plane is about 750 kilometres per hour and that time the drag coefficient is 0.30. The cross sectional area for the car is 115 square meter.

The amount of drag force for the plane is 3234 Newton

__Question: –__

__Question: –__

**What is the relationship between drag and Reynolds number?**

__Solution: – __**The relationship between drag and Reynolds number is directly proportional to each other. Means if the drag is increases then Reynolds number is also increases and if the drag is deceases then Reynolds number is also decreases.**

When Reynolds number is increases that time the viscous forces decreases relative to the internal forces so the point of separation moves in upward direction towards the equator.

Hi..I am Indrani Banerjee. I completed my bachelor’s degree in mechanical engineering. I am an enthusiastic person and I am a person who is positive about every aspect of life. I like to read Books and listen to music.

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