Question

P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20C. Compute the total friction drag if the stream is parallel to (a) the long side and (b) the short side.

Answer 1

Answer:

a

The total friction drag for the long side of the plate is 107 N

b

The total friction drag for the long side of the plate is 151.4 N

Explanation:

The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate

The block representation of the this problem is shown on the first uploaded image  

Where the U is the initial velocity = 6 m/s

    So the equation we will be working with is

               $F = \frac{1}{2} \rho C_fAU^2$

    Where $\rho$ is the density of SAE 10W = $870\ kg/m^3$ This is obtained from the table of density at 20° C

                $C_f$ is the friction drag coefficient

   This coefficient is dependent on the Reynolds number if the Reynolds number is less than $5*10^5$ then the flow is of laminar type and

          $C_f$  = $\frac{1.328}{\sqrt{Re} }$

But if the Reynolds number is greater than $5*10^5$ the flow would be of Turbulent type and

         $C_f = \frac{0.074}{Re_E^{0.2}}$

Where Re is the Reynolds number

   To obtain the  Reynolds number  

                                      $Re = \frac{\rho UL}{\mu}$

          where L is the length of the long side = 110 cm = 1.1 m

 and $\mu$ is the Dynamic viscosity of SAE 10W oil $= 1.04*10^{-1} kg /m.s$

  This is gotten from the table of Dynamic viscosity of oil

  So        

                    $Re = \frac{870 *6*1.1}{1.04*10^{-1}}$

                          $= 55211.54$

Since            55211.54 < $5.0*10^5$

Hence

                    $C_f = \frac{1.328}{\sqrt{55211.54} }$

                          $= 0.00565$

                 $A$ is the area of the plate  = $\frac{ (110cm)(55cm)}{10000}$ =$0.55m^2$

Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid

                $F = \frac{1}{2} \rho C_f(2A)U^2$

                $F =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2$

                 $F = 107N$

Considering the short side

            To obtain the Reynolds number

                      $Re = \frac{\rho U b}{\mu}$

Here b is the short side

                        $Re =\frac{870*6*0,55}{1.04*10^{-1}}$

                              $=27606$

Since the value obtained is not greater than $5*10^5$ then the flow is laminar

   And

              $C_f = \frac{1.328}{\sqrt{Re} }$

                    $= \frac{1.328}{\sqrt{27606} }$

                   $= 0.00799$

The next thing to do is to obtain the total friction drag

             $F = \frac{1}{2} \rho C_f(2A)U^2$

      Substituting values

           $F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2$

                $= 151.4 N$