Physics, 03.03.2020 02:17 skrillex88

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.

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Answer from: groundcontrol

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

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

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

= 55211.54

Since 55211.54 < 5.0*10^5

Hence

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

= 0.00565

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

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

Answer from: Quest

trig 15.3 cos

Answer from: Quest

the answer is 3.1 his answer is wrong mines is right

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