It is defined as the distance measured perpendicular to the boundary of solid body, due to which the boundary should be displayed so as to compensate for the reduction in kinetic energy of flowing fluid on account of boundary layer formation.

It is denoted by $\delta^{\text{**}}$

Consider the flow over the plate, having sections 1-1 at distance x from leading edge. Mass of fluid flowing per second through elemental strip of thickness (dy) at a distance y from the plate.

$mass = (i)=\rho u\text{bdy}$

Kinetic energy of this fluid:

$=\dfrac{1}{2}\times m\times \text{velocity}^2$

$=\dfrac{1}{2}\times (\rho ubdy)\times u^2$

Kinetic energy in absence of boundary layer:

$\dfrac 12\times (\rho ubdy)\times V^2$

Therefore loss of kinetic energy through elemental strip

$\dfrac{1}{2}\times (\rho ubdy)\times V^2-\dfrac {1}{2} \times (\rho ubdy)\times u^2$

$\dfrac{1}{2}\times (\rho ubdy)\times [V^2-u^2]$

Therefore total loss of kinetic energy of fluid passing through 'BC'

$=\int_0^\delta \dfrac{1}{2}\times (\rho ub)\times [V^2-u^2] dy$

$=\dfrac{1}{2}\rho b\int_0^\delta u [V^2-u^2] dy.................(1)$

[If fluid is considered incompressible]

Let $\delta^{**}$ = distance by which the plate is displayed to compensate for the reduction of K.E

$=\dfrac {1}{2}\times (\text{mass})\times \text{velocity}^2$

$=\dfrac{1}{2}(\rho\times b\times \delta^{**}\times V)V^2$

$=\dfrac {1}{2}\rho b\delta^{**}V^3.................(2)$

Equating equation (1) and (2) we get

$\dfrac {1}{2}\rho b\delta^{**}V^3=\dfrac 12 \rho b\int_0^\delta u [V^2-u^2] dy$

Cancelling $\dfrac 12\rho b$, we get

$\delta^{**}=\dfrac 1{V^3}\int_0^\delta u [V^2-u^2] dy$

$\delta^{**}=\int_0^\delta \dfrac u{V} \left[1-\dfrac {u^2}{V^2}\right] dy$

Numericals

Q1) Derive Momentum thickness and energy thickness for the given velocity profile.

$\dfrac uV=2\left(\dfrac y\delta\right)-\left(\dfrac y\delta\right)^2$

Solution:- Given:

Velocity Distribution, $\dfrac uV=2\left(\dfrac y\delta\right)-\left(\dfrac y\delta\right)^2$

(i) Displacement thickness $\delta^{**}$ is given by

Momentum thickness $\theta$ is given by,

$\theta=\int_0^\delta \dfrac u{V} \left[1-\dfrac {u}{V}\right] dy$

$\theta=\int_0^\delta \left[2\left(\dfrac y\delta\right)-\left(\dfrac y\delta\right)^2 \right]\left[1-2\left(\dfrac y\delta\right)-\left(\dfrac y\delta\right)^2\right] dy$

$\theta=\int_0^\delta \left[\dfrac {2y}\delta-\dfrac {y^2}{\delta^2}-\dfrac {4y^2}{\delta^2}+\dfrac {2y^3}{\delta^3}+\dfrac {2y^3}{\delta^3}-\dfrac {y^4}{\delta^4}\right] dy$

$\theta=\int_0^\delta \left[\dfrac {2y}\delta-\dfrac {5y^2}{\delta^2}+\dfrac {4y^3}{\delta^3}-\dfrac {y^4}{\delta^4}\right] dy$

$\theta=\left[\dfrac {2y^2}{2\delta}-\dfrac {5y^3}{3\delta^2}+\dfrac {4y^4}{4\delta^3}-\dfrac {y^5}{5\delta^4}\right]_0^\delta$

$\theta=\left[\dfrac {2\delta^2}{2\delta}-\dfrac {5\delta^3}{3\delta^2}+\dfrac {4\delta^4}{4\delta^3}-\dfrac {\delta^5}{5\delta^4}\right]$

$\theta=\left[\delta-\dfrac {5\delta}{3}+\delta-\dfrac {\delta}{5}\right]$

$\theta=\left[\dfrac{15\delta-25\delta+15\delta-3\delta}{15}\right]$

$\theta=\left[\dfrac{2\delta}{15}\right]$ - Momentum thickness

(ii) Energy thickness $\delta^{**}$ is given by

$\delta^{**}=\int_0^\delta \dfrac u{V} \left[1-\dfrac {u^2}{V^2}\right] dy$

$\delta^{**}=\int_0^\delta \left[2\left(\dfrac y\delta\right)-\left(\dfrac y\delta\right)^2\right] \left[1-\left(2\left(\dfrac y\delta\right)-\left(\dfrac y\delta\right)^2\right)^2\right] dy$

$\delta^{**}=\int_0^\delta \left[\dfrac {2y}\delta-\dfrac {y^2}{\delta^2}\right] \left[1-\dfrac {4y^2}{\delta^2}+\dfrac {4y^3}{\delta^3}-\dfrac {y^4}{\delta^4}\right] dy$

$\delta^{**}=\int_0^\delta \left[\dfrac {2y}\delta-\dfrac {y^2}{\delta^2}-\dfrac {8y^3}{\delta^3}+\dfrac {4y^4}{\delta^4}-\dfrac {2y^5}{\delta^5}+\dfrac {y^6}{\delta^6}+\dfrac {8y^4}{\delta^4}-\dfrac {4y^5}{\delta^5}\right] dy$

$\delta^{**}=\int_0^\delta \left[\dfrac {2y}\delta-\dfrac {y^2}{\delta^2}-\dfrac {8y^3}{\delta^3}+\dfrac {12y^4}{\delta^4}-\dfrac {6y^5}{\delta^5}+\dfrac {y^6}{\delta^6}\right] dy$

$\delta^{**}= \left[\dfrac {2y^2}{2\delta}-\dfrac {y^3}{3\delta^2}-\dfrac {8y^4}{4\delta^3}+\dfrac {12y^5}{5\delta^4}-\dfrac {6y^6}{6\delta^5}+\dfrac {y^7}{7\delta^6}\right]_0^\delta$

$\delta^{**}= \left[\dfrac {2\delta^2}{2\delta}-\dfrac {\delta^3}{3\delta^2}-\dfrac {8\delta^4}{4\delta^3}+\dfrac {12\delta^5}{5\delta^4}-\dfrac {6\delta^6}{6\delta^5}+\dfrac {\delta^7}{7\delta^6}\right]$

$\delta^{**}= \left[\delta-\dfrac {\delta}{3}-2\delta+\dfrac {12\delta}{5}-\delta+\dfrac {\delta}{7}\right]$

$\delta^{**}= \left[-\dfrac {\delta}{3}-2\delta+\dfrac {12\delta}{5}+\dfrac {\delta}{7}\right]$

$\delta^{**}= \left[\dfrac {-210\delta-35\delta+252\delta+15\delta}{105}\right]$

$\delta^{**}= \left[\dfrac {22\delta}{105}\right]$ - energy thickness