Question

Consider a large vertical plate with a uniform surface temperature of 130°C suspended in quiescent air at 25°C and atmospheric pressure. (a) Estimate the boundary layer thickness at a location 0.15 m measured from the lower edge. (b) What is the maximum velocity in the boundary layer at this location and at what position in the boundary layer does the maximum occur? (c) Using the similarity solution result, , determine the heat transfer coefficient 0.15 m from the lower edge. (d) At what location on the plate measured from the lower edge will the boundary layer become turbulent?

Answer 1

Answer:

a) 15.37 mm

b)

•   0.3659 m/s
•   3.074 mm

c) 5.7186 W/m². K

d) 0.60  m

Explanation:

Given that :

The surface temperature = 130°C = ( 130+ 273 ) K = 473 K

suspended in quiescent air at 25°C = ( 25 + 273 ) K = 298 K

$T_f = \frac{T_s + T \infty}{2 } \\ \\ T_f = {403+298}{2} \\ \\ T_f = \frac{701}{2} \\ \\ T_f = 350 \ K$

Atmospheric Pressure = 1 atm

The properties obtained from Table A - 4 include :

v = $20.92 *10^{-6} \ \ m^2 /s$

k = 0.03 W/m K

Pr = 0.700

η = 5

$Gr_x = 9.8[T_s - T \infty] \frac{x^3}{v^2}$

$Gr_x = 9.8*\frac{1}{350}[130-25] \frac{x^3}{20.92*10^{-6}^2}$

$Gr_x =6.718*10^9 \ x^3$

$\gamma = 5(0.15)((6.718*10^9)(0.15)^{3}/4)^{-1/4}$

$\gamma = 0.01537 \ m$

$\gamma = 15.37 \ mm$

Hence, the boundary layer thickness at a location 0.15 m measured from the lower edge is 15.37 mm

b) The maximum velocity  in the boundary layer  $\eta x_1$ with f'(n) = 0.275

$u = \frac{2v}{x} Gr_x^{1/2} f'(n)$

$u= \frac{2*20.92*10^{-6}}{0.75} [6.718*10^9(0.15)^3]^{1/2} * 0.275$

u = 0.3659 m/s

the maximum velocity in the boundary layer at this location is 0.3659 m/s

the position in the boundary layer where  the maximum occur is calculated as:

$y_{max} = \frac{1}{5}(15.37 \ mm)$

$y_{max} =$ 3.074 mm

c)  Using the similarity solution result, , determine the heat transfer coefficient 0.15 m from the lower edge.

we know that:

$Nu_x = \frac{h_x *x }{K} = (Gr_x/4)^{1/4} \ g (pr)$

$(Gr_x/4)^{1/4} \ g (pr)$ = $(6.718*10^9(0.15)^3/4)^{1/4} *0.586$

$(Gr_x/4)^{1/4} \ g (pr)$ = 28.593

Making $h_x$ the subject from the above formula:

$h_x = \frac{Nu_xK}{x}$

$h_x= \frac{28.593*0.03}{0.15}$

$h_x$ = 5.7186 W/m². K

d) to determine the location on the plate that the boundary layer we become turbulent ; we have the following:

$Ra _ {x,c} = Gr_x,c^{pr} \approx 10^9$

$x_c = [10^9/6.718*10^9(0.7)]^{1/3}$

$x_c =$ 0.60  m