# Integral solution of boundary layer equation

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As noted before, the similarity solution provides an exact analytical solution for laminar boundary layer conservation equations; however, there are limitations in terms of geometry and boundary conditions, as well as laminar flow restrictions. The integral method gives approximately closed form solutions, which have much less limitation in terms of their geometry and boundary conditions. It can also be applied to both laminar and turbulent flow situations. The integral method easily provides accurate answers (not exact) for complex problems.

Figure 1: Momentum and heat transfer over a wedge with an unheated starting length.

Using the integral method, one usually integrates the conservative differential boundary layer equation over the boundary layer thickness by assuming a profile for velocity, temperature, and concentration, as needed. The better the approximate shape for the profile, such as velocity and temperature, the better the prediction of drag force and heat transfer (friction coefficient or heat transfer coefficient). The integral methodology has been applied to a variety of configurations to solve transport phenomena problems (Schlichting and Gersteu, 2000). To illustrate the integral methodology, it will be applied to flow and heat transfer over a wedge with non-uniform temperature and blowing at the wall. Consider two dimensional laminar steady flow with constant properties over a wedge, as shown in Figure 4.16, with an unheated starting length, x0.

The governing boundary layer equations for mass, momentum and energy for constant property, steady state, and laminar flow as well as boundary conditions for convective heat transfer over a wedge are presented below: Continuity equation $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ (4.151) Momentum equation

$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{\partial ^{2}u}{\partial y^{2}}+U\frac{dU}{dx}$ (4.152) Energy equation $u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\alpha \frac{\partial ^{2}T}{\partial y^{2}}$ (4.153) Boundary conditions \begin{align} & u\left( 0,x \right)=0 \\ & v\left( 0,x \right)=v_{w} \\ & u\left( \delta ,x \right)=U\left( x \right) \\ & T\left( \infty ,x \right)=T_{\infty } \\ & T\left( 0,x \right)=T_{\infty }\quad \text{for}\ xx_{0} \\ \end{align} (4.154) It should be noted that U is known from potential flow theory:

$-\frac{1}{\rho }\frac{\partial p}{\partial x}=U\frac{dU}{dx}$

If U is constant, then $\frac{dU}{dx}=0$ (4.155) As for the case of flow over a flat plate, If x0 = 0, U = constant and vw = 0, the problem will be similar to the case presented using the similarity solution before. In integral methods, it is customary to assume a profile for u and obtain v from the continuity eq. (4.151). Let us integrate eq. (4.151) with respect to y from y = 0 to y = δ. The velocity and temperature field outside δ is uniform.

$\int_{0}^{\delta }{\frac{\partial u}{\partial x}dy+}\int_{0}^{\delta }{\frac{\partial v}{\partial y}dy=0}$ (4.156) The second term can be easily integrated.

$\int_{0}^{\delta }{\frac{\partial v}{\partial y}dy=\left. v \right|_{y=\delta }-}\left. v \right|_{y=0}=v_{\delta }-v_{w}$ (4.157) Combining (4.156) and (4.157) we get:

$\int_{0}^{\delta }{\frac{\partial u}{\partial x}dy=v_{w}-v_{\delta }}$ (4.158) Applying Leibnitz’s formula to the left hand side of (4.158) yields:

$\frac{\partial }{\partial x}\int_{0}^{\delta }{udy}-u\left( x,\delta \right)\frac{d\delta }{dx}=v_{w}-v_{\delta }$ (4.159) which can be rearranged to obtain:

$v_{\delta }=v_{w}+U\frac{d\delta }{dx}-\frac{\partial }{\partial x}\left( \int_{0}^{\delta }{udy} \right)$ (4.160) The momentum equation (4.152) can be rearranged to the following form:

$\frac{\partial u^{2}}{\partial x}+\frac{\partial vu}{\partial y}-u\left( \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y} \right)=U\frac{dU}{dx}+\nu \frac{\partial ^{2}u}{\partial y^{2}}$ (4.161) where term in parenthesis on the left-hand side is zero because of the continuity equation. The integration of eq. (4.161) from y = 0 to y = δ gives:

$\int_{0}^{\delta }{\frac{\partial u^{2}}{\partial x}dy}+\int_{0}^{\delta }{\frac{\partial vu}{\partial y}dy}=\int_{0}^{\delta }{U\frac{dU}{dx}dy}+\nu \int_{0}^{\delta }{\frac{\partial ^{2}u}{\partial y^{2}}dy}$ (4.162) Upon further integration and simplification, the above equation reduces to

$\int_{0}^{\delta }{\frac{\partial u^{2}}{\partial x}dy}+\left. vu \right|_{\delta }-\left. vu \right|_{0}=\int_{0}^{\delta }{U\frac{dU}{dx}dy}+\nu \left. \frac{\partial u}{\partial y} \right|_{\delta }-\nu \left. \frac{\partial u}{\partial y} \right|_{0}$ (4.163) Using eq. (4.160) for vδ, the no slip boundary condition at the wall u(0,x) = 0, and assuming no velocity gradient at the outer edge of the boundary layer at y = δ we get:

$\int_{0}^{\delta }{\frac{\partial u^{2}}{\partial x}}dy+Uv_{w}+U^{2}\frac{d\delta }{dx}-U\frac{\partial }{\partial x}\left( \int_{0}^{\delta }{udy} \right)=-\frac{\tau _{w}}{\rho }+\int_{0}^{\delta }{U\frac{dU}{dx}dy}$ (4.164) where $\tau _{w}=\left. \mu \frac{\partial u}{\partial y} \right|_{0}$ (4.165) is the shear stress at the wall. Applying Leibnitz’s rule and rearranging will provide the final form. $\frac{\partial }{\partial x}\left[ \int_{0}^{\delta }{u\left( u-U \right)dy} \right]+\left( \int_{0}^{\delta }{udy} \right)\frac{dU}{dx}-\int_{0}^{\delta }{U\frac{dU}{dx}dy=-\frac{\tau _{w}+\rho Uv_{w}}{\rho }}$ (4.166) The only dependent unknown variable in the above equation is u since v is eliminated using continuity. U, τw, and vw should be known quantities. Equation (4.166) can be further rearranged. $\frac{\partial }{\partial x}\left[ U^{2}\int_{0}^{\delta }{\frac{u}{U}\left( 1-\frac{u}{U} \right)dy} \right]+\left[ \int_{0}^{\delta }{\left( 1-\frac{u}{U} \right)dy} \right]U\frac{dU}{dx}=\frac{\tau _{w}+\rho Uv_{w}}{\rho }$ (4.167) It is customary to assume a third order polynomial equation for the velocity profile in order to obtain a reasonable result, u = c1 + c2y + c3y2 + c4y3 (4.168) where c1, c2, c3, and c4 are constants and can be obtained from boundary conditions for velocity and shear stress at the wall and outer edge. Once the constants are obtained, they are substituted into the momentum integral equation (4.167) and are used to solve for the momentum boundary layer thickness, δ. Similarly, the energy equation (4.153) can be rearranged into the following form to make the integration process easier: $\frac{\partial uT}{\partial x}+\frac{\partial vT}{\partial y}-T\left( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \right)=\alpha \frac{\partial ^{2}T}{\partial y^{2}}$ (4.169) Let’s integrate the above equation from y = 0 to y = δT, knowing that the term in parenthesis on the left hand side is zero because of the continuity equation. $\int_{0}^{\delta _{T}}{\frac{\partial uT}{\partial x}dy+\int_{0}^{\delta _{T}}{\frac{\partial vT}{\partial y}dy}}=\alpha \int_{0}^{\delta _{T}}{\frac{\partial ^{2}T}{\partial y^{2}}}dy$ (4.170) Using integration and the continuity equation to obtain vδ, eq. (4.160), and assuming there is no temperature gradient at the outer edge of the thermal boundary layer, we get the following equation:

$\int_{0}^{\delta _{T}}{\frac{\partial }{\partial x}\left( uT \right)dy-T_{\infty }\frac{\partial }{\partial x}\int_{0}^{\delta _{T}}{udy+T_{\infty }v_{w}}}-v_{w}T_{w}+T_{\infty }u_{\delta _{T}}\frac{d\delta _{T}}{dx}=\left. \frac{-k}{\rho c_{p}}\frac{\partial T}{\partial y} \right|_{y=0}$ (4.171) Using Leibnitz’s rule and rearranging gives:

$\frac{\partial }{\partial x}\left[ \int_{0}^{\delta _{T}}{u\left( T-T_{\infty } \right)dy} \right]+\left( \int_{0}^{\delta _{T}}{udy} \right)\frac{dT_{\infty }}{dx}=\frac{{q}''_{w}}{\rho c_{p}}+v_{w}\left( T_{w}-T_{\infty } \right)$ (4.172) Once again, the integral form of the energy equation is in terms of the unknown temperature, assuming the velocity profile is known. Similar to the momentum integral equation, a temperature profile should be assumed and substituted into the integral energy equation (4.172) to obtain δT. To illustrate the procedure, we use the above approximation to solve the classic problem of flow and heat transfer over a flat plate when U = U∞ = constant, with no blowing or suction at the wall, and with constant wall and flow stream temperature. The momentum and energy integral equations (4.167) and (4.172) reduce to the following forms using the above assumptions: $\frac{\partial }{\partial x}\left[ U^{2}\int_{0}^{\delta }{\frac{u}{U}\left( 1-\frac{u}{U} \right)dy} \right]=\frac{\tau _{w}}{\rho }$ (4.173) $\frac{\partial }{\partial x}\left[ \int_{0}^{\delta _{T}}{u\left( T-T_{\infty } \right)dy} \right]=\frac{{q}''_{w}}{\rho c_{p}}$ (4.174) Let’s assume a polynomial velocity profile is a third degree polynomial function with the following boundary conditions. $u\left( 0 \right)=0$ (4.175) $u\left( \delta \right)=U$ (4.176) $\left. \frac{\partial u}{\partial y} \right|_{y=\delta }=0$ (4.177)

$\left. \frac{\partial ^{2}u}{\partial y^{2}} \right|_{y=\delta }=0$ (4.178) It is assumed that shear stress at the boundary layer edge is zero, which is a good approximation for this configuration. Equation (4.178) is obtained by using eq. (4.177) and applying the x-direction momentum equation at the boundary layer edge. Upon applying eqs. (4.175) – (4.178) in eq. (4.168), one obtains four equations containing four unknowns (c1, c2, c3, and c4). The final velocity profile is

$\frac{u}{U}=\frac{3}{2}\left( \frac{y}{\delta } \right)-\frac{1}{2}\left( \frac{y}{\delta } \right)^{3},\text{ }y\le \delta$ (4.179) Shear stress at the wall, τw, is calculated using eq. (4.179)

$\tau _{w}=\mu \left. \frac{\partial u}{\partial y} \right|_{y=0}=\frac{3\mu U}{2\delta }$ (4.180) Substituting eqs. (4.179) and (4.180) into eq. (4.173), and performing the integration we get

$\frac{d}{dx}\left( \frac{39U^{2}\delta }{280} \right)=\frac{3\nu U}{2\delta }$ (4.181) Integrating the above equation and assuming δ = 0 at x = 0, we get

$\delta =\left( \frac{280\nu x}{13U} \right)^{1/2}$ (4.182) or

$\frac{\delta }{x}=\frac{4.64}{\operatorname{Re}_{x}^{1/2}}$ (4.183) The friction coefficient is found as before

$c_{f}=\frac{\tau _{\omega }}{\rho \frac{U_{\infty }^{2}}{2}}=\frac{\mu \left. \frac{\partial u}{\partial y} \right|_{y=0}}{\rho \frac{U_{\infty }^{2}}{2}}$ (4.184) or using eqs. (4.180) and (4.182)

$\frac{c_{f}}{2}=\frac{0.323}{\operatorname{Re}_{x}^{1/2}}$ (4.185) The predictions of the momentum boundary layer thickness and friction coefficient, cf, by the integral method are 7% and 3% lower than the exact solution obtained using the similarity method, respectively. Using the same general third order polynomial equation for a temperature profile with the following boundary conditions:

$T\left( 0 \right)=T_{w}=\text{constant}$ (4.186)

$T\left( \delta _{T} \right)=T_{\infty }=\text{constant}$ (4.187)

$\left. \frac{\partial T}{\partial y} \right|_{y=\delta _{T}}=0$ (4.188)

$\left. \frac{\partial ^{2}T}{\partial y^{2}} \right|_{y=0}=0$ (4.189) the temperature profile can be obtained as:

$\frac{T-T_{\infty }}{T_{w}-T_{\infty }}=1-\frac{3}{2}\frac{y}{\delta _{T}}+\frac{1}{2}\left( \frac{y}{\delta _{T}} \right)^{3},\text{ }y\le \delta _{T}$ (4.190) and the heat flux at the wall is

${q}''_{w}=-k\left. \frac{\partial T}{\partial y} \right|_{y=0}=\frac{3}{2}\frac{k}{\delta _{T}}\left( T_{w}-T_{\infty } \right)$ (4.191) Substitution of eqs. (4.189), (4.190) and (4.191) into eq. (4.174), and approximate integration for δT / δ < 1

yields:


\begin{align} & \frac{\partial }{\partial x}\int_{0}^{\delta _{T}}{u\left( T-T_{\infty } \right)}\text{ }dy=\frac{\partial }{\partial x}\left[ U_{\infty }\left( T_{w}-T_{\infty } \right)\delta _{T}\int_{0}^{1}{\frac{u}{U_{\infty }}\left( \frac{T-T_{\infty }}{T_{w}-T_{\infty }} \right)d\left( \frac{y}{\delta _{T}} \right)} \right] \\ & \text{ }=\frac{\partial }{\partial x}\left[ \frac{3}{20}\frac{\delta _{T}^{2}}{\delta }\left( 1-\frac{\delta _{T}^{2}}{14\delta ^{2}} \right)U\left( T_{w}-T_{\infty } \right) \right] \\ & \text{ }=\frac{{q}''_{w}}{\rho c_{p}}=\frac{3}{2}\frac{\alpha }{\delta _{T}}\left( T_{w}-T_{\infty } \right) \\ \end{align} (4.192) Upon further simplification we get

$\frac{d}{dx}\left[ \frac{\delta _{T}^{2}}{\delta }\left( 1-\frac{\delta _{T}^{2}}{14\delta ^{2}} \right) \right]=\frac{10\nu }{\Pr U}\frac{1}{\delta _{T}}$ (4.193) The only unknown in the above equation is δT since δ is known. Assuming $\varsigma =\delta _{T}/\delta$ , eq. (4.193) becomes

$\frac{d}{dx}\left[ \varsigma ^{2}\delta \left( 1-\frac{\varsigma ^{2}}{14} \right) \right]=\frac{10}{\Pr }\frac{\nu }{U_{\infty }}\frac{1}{\varsigma \delta }$ (4.194) where term $\varsigma ^{2}/14$ can be neglected since it is much less than 1. The solution of eq. (4.194) for ζ = 0 at x = x0 yields

$\varsigma =\frac{\Pr ^{-1/3}}{1.026}\left[ 1-\left( \frac{x_{0}}{x} \right)^{3/4} \right]^{1/3}$ (4.195) The local heat transfer coefficient can now be calculated since δT is known.

$h_{x}=\frac{{q}''_{w}}{T_{w}-T_{\infty }}=\frac{-k\left. \frac{\partial T}{\partial y} \right|_{y=0}}{T_{w}-T_{\infty }}=\frac{3}{2}\frac{k}{\delta _{T}}$ (4.196) or

$h_{x}=\frac{3}{2}\frac{k}{\varsigma \delta _{T}}$ (4.197) Using ζ from eq. (4.195) and δ from eq. (4.69) to calculate the local Nusselt number, $\text{Nu}_{x}=\frac{h_{x}x}{k}$

$\text{Nu}_{x}=\frac{0.332\Pr ^{1/3}\operatorname{Re}_{x}^{1/2}}{\left[ 1-\left( \frac{x_{0}}{x} \right)^{3/4} \right]^{1/3}}$ (4.198) The above equation without unheated starting length (x0 = 0) reduces to the exact solution obtained by the similarity solution.

$\text{Nu}_{x}=0.332\Pr ^{1/3}\operatorname{Re}_{x}^{1/2}$ (4.199)