written 6.7 years ago by | modified 2.5 years ago by |

**Subject:** Fluid Mechanics 2

**Topic:** Boundary layer theory

**Difficulty:** Low

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Explain displacement thickness, momentum thickness, energy thickness and shape factor.

written 6.7 years ago by | modified 2.5 years ago by |

**Subject:** Fluid Mechanics 2

**Topic:** Boundary layer theory

**Difficulty:** Low

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written 2.5 years ago by |

Displacement thickness is the distance measured perpendicular to the solid boundary surface by which the free stream lines are shifted or displaced on account of boundary layer formation.

Displacement thickness is defined as the distance measured perpendicular to the boundary of the solid surface by which the boundary of the solid surface should be displaced to compensate for the reduction in flow rate on account of boundary layer formation.

- Displacement thickness is defined as the thickness of flow (transverse distance measured perpendicular to the boundary of the solid surface) moving at the free stream velocity and having flow rate equal to the loss in flow rate on account of the boundary layer formation.

- Consider the flow of a fluid having free-stream velocity equal to U over a thin smooth plate.
- At a distance x from the leading edge consider a section 1-1.
- Distance BC is equal to the boundary layer thickness &
- At the section 1-1, consider an elemental strip.

y = Distance of elemental strip from the plate.

dy = Thickness of the elemental strip.

u = Velocity of fluid at the elemental strip,

b = Width of the elemental strip (= Plate width)

Then area of elemental strip,

dA = b x dy

Mass of fluid per second flowing through elemental strip

= Density x Velocity x Area of elemental the strip

= ρ x u x b x dy . . (1) ..........

- If there had been no plate, then the fluid would have been flowing with a constant velocity equal to free-stream velocity (U) at the section 1-1. Then the mass of fluid per second flowing through elemental strip would have been

= ρ x U x b x dy . (2) ...... ...

- As U is more than u, hence due to the presence of the plate and consequently due to the formation of the boundary layer, there will be a reduction in mass flowing per second through the elemental strip
- This reduction in mass/sec flowing through elemental strip = (2) - (1)

= ρ U b dy – ρ u b dy

= ρ b(U – u)dy

Total reduction in mass of fluid/sec flowing through BC due to plate

$$ \begin{gathered} =\int_{0}^{\delta} \rho b(U-u) d y \\ =\rho b \int_{0}^{\delta}(U-u) d y \end{gathered}$$

Displacement thickness is defined as the thickness of flow (transverse distance measured perpendicular to the boundary of the solid surface) moving at the free stream velocity and having flow rate equal to the loss in flow rate on account of the boundary layer formation.

Let the plate is displaced by a distance &* and velocity of flow for the distance 6* is equal to the free-stream velocity (i.e. U). Mass of the fluid/sec flowing through the distance 8*

= Density x Velocity x Area of elemental the strip

= ρ x U x (8* × b) ... ... (4)

Equating (3) and (4)

$$ \begin{gathered} \rho b \int_{0}^{\delta}(U-u) d y=\rho \times U \times\left(\delta^{*} \times b\right) \\ \int_{0}^{\delta}(U-u) d y=U \times \delta^{*} \\ \delta^{*}=\frac{1}{U} \int_{0}^{\delta}(U-u) d y \\ =\int_{0}^{\delta} \frac{(U-u) d y}{U} \\ \delta^{*}=\int_{0}^{\delta}\left(1-\frac{u}{U}\right) d y \end{gathered} $$

- Momentum thickness is defined as the distance measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in momentum of the flowing fluid on account of boundary layer formation.
- Momentum thickness is defined as the thickness of flow (transverse distance measured perpendicular to the boundary of the solid surface) moving at the free stream velocity and having momentum flux equal to the deficiency of momentum flux on account of boundary layer formation

- Consider the flow of a fluid having free-stream velocity equal to U over a thin smooth plate.
- At a distance x from the leading edge consider a section 1-1.
- Distance BC is equal to the boundary layer thickness &
- At the section 1-1, consider an elemental strip.

y = Distance of elemental strip from the plate.

dy = Thickness of the elemental strip.

u = Velocity of fluid at the elemental strip,

b = Width of the elemental strip (= Plate width)

Then area of elemental strip,

dA = b×dy

Mass of fluid per second flowing through elemental strip

= Density x Velocity x Area of elemental the strip

= ρ×u×b×dy

Momentum of this fluid/sec = Mass x Velocity

= ρ×u×b×dy

- Momentum of this fluid/sec in the absence of boundary layer

= ρubdy x U

- Loss of momentum/sec through elemental strip

= ρubdy x U – ρubdy x u

= ρbu(U – u)dy

- Total loss of momentum/sec through BC

$$ =\int_{0}^{\delta} \rho b u(U-u) d y $$

Let 0 distance by which plate is displaced where the fluid is flowing with a constant velocity U

Momentum/sec of fluid flowing through the distance $\theta$ with a velocity $U$ $$ \begin{gathered} =\text { Mass of fluid through } \theta \times \text { Velocity } \ =[\rho \times \text { area } \times \text { velocty }] \times \text { velocity } \ =[\rho \times(\theta \times b) \times U] \times U \ =\rho \theta b U^{2} \ldots \ldots \ldots(2) \end{gathered} $$

Equating (1) and (2)

$$ \begin{gathered} \rho \theta b U^{2}=\int_{0}^{\delta} \rho b u(U-u) d y \\ \rho \theta b U^{2}=\rho b \int_{0}^{\delta} u(U-u) d y \end{gathered} $$ $$ \begin{aligned} \theta U^{2}=\int_{0}^{\delta} u(U-u) d y & \theta=\frac{1}{U^{2}} \int_{0}^{\delta} u(U-u) d y \\ &=\int_{0}^{\delta} \frac{u(U-u) d y}{U^{2}} \\ \theta &=\int_{0}^{\delta} \frac{u}{U}\left[1-\frac{u}{U}\right] d y \end{aligned} $$

- Energy thickness is defined as the distance measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in kinetic energy of the flowing fluid on account of boundary layer formation.
- Energy thickness is defined as the thickness of flow (transverse distance measured perpendicular to the boundary of the solid surface) moving at the free stream velocity and having kinetic energy equal to the deficiency of kinetic energy on account of boundary layer formation

- Consider the flow of a fluid having free-stream velocity equal to U over a thin smooth plate.
- At a distance x from the leading edge consider a section 1-1.
- Distance BC is equal to the boundary layer thickness &
- At the section 1-1, consider an elemental strip.

y = Distance of elemental strip from the plate.

dy = Thickness of the elemental strip.

u = Velocity of fluid at the elemental strip,

b = Width of the elemental strip (= Plate width) Then area of elemental strip, $$ d A=b \times d y $$ Mass of fluid per second flowing through elemental strip $$ \begin{gathered} =\text { Density } \times \text { Velocity } \times \text { Area of elemental the strip } \\ =\rho \times u \times b \times d y \end{gathered} $$ Kinetic energy of this fluid per second $$ \begin{gathered} =\frac{1}{2} \times \text { mass } \times \text { Velocity }^{2} \\ =\frac{1}{2} \times \rho u b d y \times u^{2} \end{gathered} $$ Kinetic energy of this fluid per second in the absence of boundary layer $$ =\frac{1}{2} \times \rho u b d y \times U^{2} $$ Let $\delta^{* *}=$ distance by which plate is displaced where the fluid is flowing with a constant velocity U

Kinetic energy per second of fluid flowing through the distance $\delta^{* *}$ with a velocity $U$
$$
\begin{gathered}
=\frac{1}{2} \times \text { mass } \times \text { Velocity }^{2} \
=\frac{1}{2} \times[\rho \times \text { area } \times \text { velocty }] \times \text { Velocity }^{2} \
=\frac{1}{2} \times\left[\rho \times\left(\delta^{* *} \times b\right) \times U\right] \times U^{2} \
=\frac{1}{2} \rho b \delta^{* *} U^{3} \ldots \ldots \ldots(2)
\end{gathered}
$$
Equating (1) and (2)
$$
\begin{aligned}
\frac{1}{2} \rho b \delta^{* *} U^{3} &=\frac{1}{2} \rho b \int_{0}^{\delta} u\left(U^{2}-u^{2}\right) d y \\
\delta^{* *} &=\frac{1}{U^{3}} \int_{0}^{\delta} u\left(U^{2}-u^{2}\right) d y \\
\delta^{* *} &=\int_{0}^{\delta} \frac{u\left(U^{2}-u^{2}\right) d y}{U^{3}} \\
\delta^{* *} &=\int_{0}^{\delta} \frac{u}{U}\left(1-\frac{u^{2}}{U^{2}}\right) d y
\end{aligned}
$$

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