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Derive Von Karman Momentum Integral equation

Subject: Fluid Mechanics 2

Topic: Boundary layer theory

Difficulty: Medium

Consider a control volume $A B D C$ in a boundary layer region. $\because$ Rate of mass flow through $A B\left(\dot{m}_{A B}\right.$ (entry)) $=\int_{0}^{\delta} \rho u d y$ (for unit width) $$\dot{m}_{C D}\left(\text { exit) }=\dot{m}_{A B}+\frac{\partial \dot{m}_{A B}}{\partial x} d x\right.$$ since, mass is conserve, then $$\dot{m}_{A B} \text { (entry) }+\dot{m}_{A C} \text { (entry) }=\dot{m}_{C D} \text { (exit) }$$ From above equation, we get $\Rightarrow \quad \dot{m}_{A C}=\dot{m}_{C D}-\dot{m}_{A B}=\left(\dot{m}_{A B}+\frac{\partial \dot{m}_{A B}}{\partial x} d x\right)-\left(\dot{m}_{A B}\right)$ $\Rightarrow \quad \dot{m}_{A C}=\frac{\partial}{\partial x}\left[\int_{0}^{\delta} \rho u d y\right] d x$ From Newton's second law Change in momentum $=$ Force $(\dot{m} v)_{C D}-(\dot{m} v)_{A B}-(\dot{m} v)_{A C}=-\tau_{0} d x$ $\Rightarrow \quad(\dot{m} v)_{A B}+(\dot{m} v)_{A C}-(\dot{m} v)_{C D}=\tau_{0} d x$ $$\begin{array}{l} \Rightarrow \quad \dot{m}_{A C}=\dot{m}_{C D}-\dot{m}_{A B}=\left(\dot{m}_{A B}+\frac{\partial \dot{m}_{A B}}{\partial x} d x\right)-\left(\dot{m}_{A B}\right) \\ \Rightarrow \quad \dot{m}_{A C}=\frac{\partial}{\partial x}\left[\int_{0}^{\delta} \rho u d y\right] d x \\ \text { From Newton's second law } \end{array}$$ Figure , Putting the value of $(\dot{m} v)_{C D}$ from Equation $to$ the given Equation, we get $(\dot{m} v)_{A B}+(\dot{m} v)_{A C}-\left[(\dot{m} v)_{A B}+\frac{\partial}{\partial x}(\dot{m} v)_{A B} d x\right]=\tau_{0} d x$ $\because \quad \dot{m}_{A C}=\frac{\partial}{\partial x}\left[\int_{0}^{\delta} \rho u d y\right] d x$ and, velocity at boundary surface $A C$ is $0.99 U_{\infty} \approx U_{\infty}$, then Equation reduces to $\Rightarrow \quad \frac{\partial}{\partial x}\left[\int_{0}^{\delta} \rho u U_{\infty} d y\right] \cdot d x-\frac{\partial}{\partial x}\left[\int_{0}^{\delta} \rho u^{2} d y\right] \cdot d x=\tau_{0} d x$ $\Rightarrow \quad \frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left(\rho u U_{\infty}-\rho u^{2}\right) d y\right]=\tau_{0}$ $\Rightarrow \quad \frac{\partial}{\partial x}\left[\int_{0}^{\delta} \frac{\rho u U_{\infty}}{\rho U_{\infty}^{2}}\left(1-\frac{u}{U_{\infty}}\right) d y\right]=\frac{\tau_{0}}{\rho U_{\infty}^{2}}$ $\Rightarrow \quad \frac{\partial}{\partial x}\left[\int_{0}^{\delta} \frac{u}{U_{\infty}}\left(1-\frac{u}{U_{\infty}}\right) d y\right]=\frac{\tau_{0}}{\rho U_{\infty}^{2}}$ $\Rightarrow \quad \frac{\partial \theta}{\partial x}=\frac{\tau_{0}}{\rho U_{\infty}^{2}} \quad\left\{\because \theta=\int_{0} \frac{u}{U_{\infty}}\left(1-\frac{u}{U_{\infty}}\right) d y\right\}$ or $\frac{d \theta}{d x}=\frac{\tau_{0}}{\rho U_{\infty}^{2}}$ Hence, Verified.