جلسه یازدهم انتقال حرارت جابجایی- حل انتگرالی2

علی راد جمعه 8 خرداد‌ماه سال 1394 @ 22:48 چاپ

پست ناقص. ایشالا تا پایان هفته بعد کامل کنم.

جلسه هشتم کامل شد. فایل ارزشمندی رو اونجا قرار دادم..

برا جلسه ی دوم حل سرعتی و دمایی رو صفحه ی غیر افقی(عمودی یا زاویه با افق) رو قرار میدم. تو لیسانس هر چی خوندین برا ص تخت افقی بود با هر پروفیلی(یک هفتم سینوسی درجه2...) یه خورده دشوارتره از حل ص افقی..

Multiplying the continuity equation:

\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0

by u and adding the resulting equation to the momentum equation

u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})


\frac{\partial {{u}^{2}}}{\partial x}+\frac{\partial (uv)}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})

Integrating the above equation with respect to in the interval of (0, Y), where Y is greater than both δ and δt, one obtains:

\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=-\nu {{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}+g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy} \qquad \qquad(1)

By following a similar procedure, the integral energy equation can be obtained as follows

\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=-\alpha {{\left. \frac{\partial T}{\partial y} \right|}_{y=0}} \qquad \qquad(2)

In order to obtain the solution of a natural convection problem, the appropriate velocity and temperature profiles must be utilized together with the integral equations (1) and (2). The velocity and temperature profiles depend on the thicknesses of the momentum and thermal boundary layers, which in turn depend on the Prandtl number. Assuming the velocity profile is the third degree polynomial function of y in the boundary layer and using the boundary conditions to determine the unspecified constant, the velocity profile becomes (see Problem 6.8):

\frac{u}{U}=\frac{y}{\delta }{{\left( 1-\frac{y}{\delta } \right)}^{2}} \qquad \qquad(3)

where U is a characteristic velocity that is a function of x. Similarly, the temperature profile can be obtained by assuming a second degree polynomial function and the result is (see Problem 6.9):

\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}={{\left( 1-\frac{y}{{{\delta }_{t}}} \right)}^{2}} \qquad \qquad(4)

The following analysis will be based on the assumption that the momentum and thermal boundary layers have the same thickness, i.e. δt = δ. Substituting the velocity and temperature profiles into the integral form of the momentum equation (1) and energy equation (2) yields:

\frac{1}{105}\frac{d}{dx}({{U}^{2}}\delta )=-\nu \frac{U}{\delta }+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \qquad \qquad(5)
\frac{1}{30}\frac{d}{dx}(U\delta )=\frac{2\alpha }{\delta } \qquad \qquad(6)

At the leading edge of the vertical plate, the boundary layer thickness is zero and the characteristic velocity U is also zero:

U=\delta =0,\text{ }x=0 \qquad \qquad(7)

which are the initial conditions of eqs. (5) and (6). It is expected that as x increases, both U and δ should increase. Let us assume that they are functions of x such that

U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(8)

Substituting eq. (8) into eqs. (5) and (6), one obtains the following:

\frac{2m+n}{105}C_{1}^{2}{{C}_{2}}{{x}^{2m+n-1}}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu {{x}^{m-n}}+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}{{x}^{n}} \qquad \qquad(9)
\frac{m+n}{30}{{C}_{1}}{{C}_{2}}{{x}^{m+n-1}}=\frac{2\alpha }{{{C}_{2}}}{{x}^{-n}} \qquad \qquad(10)

The above two relations can be true for all x only if the indices of x for all terms in the same equation are the same, i.e. when the following equations hold:

  & 2m+n-1=m-n=n \\ 
 & m+n-1=-n \\ 

which are satisfied only if m = 1 / 2 and n = 1 / 4. This suggests that U\propto {{x}^{1/2}} and \delta \propto {{x}^{1/4}}; this is in agreement with the result of the scaling analysis. Substituting back the values of m and n into eqs. (9) and (10), one obtains

\frac{C_{1}^{2}{{C}_{2}}}{84}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu +\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}
\frac{{{C}_{1}}{{C}_{2}}}{40}=\frac{2\alpha }{{{C}_{2}}}

Solving for C1 and C2 from the above two equations yields:

{{C}_{1}}=4{{\left( \frac{5}{3} \right)}^{1/2}}\nu {{\left( \frac{20}{21}+\frac{\nu }{\alpha } \right)}^{-1/2}}{{\left[ \frac{g\beta ({{T}_{w}}-{{T}_{\infty }})}{{{\nu }^{2}}} \right]}^{1/2}}
{{C}_{2}}=4{{\left( \frac{15}{16} \right)}^{1/4}}{{\left( \frac{20}{21}+\frac{\nu }{\alpha } \right)}^{1/4}}{{\left[ \frac{g\beta ({{T}_{w}}-{{T}_{\infty }})}{{{\nu }^{2}}} \right]}^{-1/4}}{{\left( \frac{\nu }{\alpha } \right)}^{-1/2}}

The boundary layer thickness therefore becomes:

\frac{\delta }{x}=3.93{{\left( \frac{0.952+\Pr }{{{\Pr }^{2}}} \right)}^{1/4}}\text{Gr}_{x}^{-1/4} \qquad \qquad(11)

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