User:GUPTA SHUBHAM/sandbox

INTRODUCTION
Whenever heat transfer phenomena takes place it is always accompanied by entropy generation which is useful in determining how much useful work is being destroyed Thermodynamic design take into consideration of the entropy generation. For example a good heat exchanger design ,it should have least generation of entropy or destruction of available work.

In this case we consider forced convection in laminar boundary layer flow over a thin plate and how its geometry is to be altered to minimize the destruction of available work .The heat flux $$\ddot{q}$$ is assumed to be uniform on both side and free stream is parallel to the plate as shown in fig (1).

Fig 1.Laminar boundary layer flow on a flat plate with uniform heat flux on both sides.

METHODOLOGY
Consider a thin plate and the heat flux$$\ddot{q}$$ is assumed on both side and the free stream is parallel to the plate  as shown in the fig (1).To calculate the entropy generation is similar to the entropy generation in surrounding flow along the plate

The total rate of entropy generation is given by the equation 1.1


 * $$S_{gen}=\frac{1}{T_\infty^2}.\int_{A}^{} \ddot{q}.(T-T_\infty).dA +\frac{F_D.U_\infty}{T_\infty} $$ (1.1)

In this expression A is the surface area .T is the surface temperature and $${F_D}$$ is the drag force experienced by the body. The entropy generation has two terms because it is the integral of the volumetric rate of equation over the space occupied by the fluid .The first term account for the irreversibility of heat transfer and second represent the irreversibility of fluid flow.

In equation (1.1) the integral is replaced by $$\ddot{q}.(\overline{T}-{T_\infty}).(2LW)$$ in which $$(\overline{T}-{T_\infty})$$ is the wall temperature difference averaged from X=0 to X=L. The drag force over the entire length of the plate is calculated as $$F_D=2LW{\tau_{0-L}}$$.

The average wall shear stress is over the entire length is given by $${\tau_{0-L}}	=.664{\rho}.{U_\infty}^2.{Re_L}^{-1/2}$$.

Substituting these value in equation (1.1) .The equation obtained is


 * $$\frac{S_{gen}}{W}=\frac{.736.\ddot{q}^2}{{{T_\infty}^2}.{Pr}^{1/3}.{Re_L}^{1/2}}+1.328 (\frac{\mu}{T_\infty}).{U_\infty}^2.{Re_L}^{1/2}$$(1.2)

The $$\dot{q}$$represents the total rate of heat transfer between the plate and the fluid ,per unit width,$$\dot{q}$$ =2L$$\ddot{q}$$ .

The length L appears as $${Re_L}={U_\infty} L/{\nu} $$in both part of the equation (1.2). In the equation the irreversibility due to heat transfer decreases as the plate is made longer where as the fluid flow irreversibility increases. This behavior means that Sgenis minimum when L has a value that is neither too small nor too large. That value is obtained by solving

$$\frac{\partial S_{gen}}{\partial{Re_L}}=0 $$(1.3)

CONCLUSION
After solving this equation ,we derive at the conclusion that if the plate is to transfer heat at a given rate $$\dot{q}$$to a stream with specified velocity$${U_\infty}$$ ,the length for minimum irreversibility is given by

$$ {L_{opt}}=.554 \frac{\dot{q}^2}{{k}.{T_\infty}.{\rho}{{U_\infty}^3}.{Pr^{1/3}}}$$ (1.4)

The minimum entropy generation is $${S_{gen,min}}=1.98 \frac{{q}.{U_\infty}}{(\frac{k}{\mu})^{1/2}.{T_\infty^{3/2}}.Pr^{1/6}} $$  (1.5)