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.

Answers: 2

Show answers

Answers

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

no simpler subs form complex ones. synthesise = make

Answer from: Quest

-first use of geothermal energy occurred more than 10,000 years ago in north america by american paleo-indians

-the first industrial use of geothermal energy began near pisa, italy in late 18th century

-in 1904, italian scientist piero ginori conti invented the first geothermal electric power plant in which steam was used to generate the power

-with the above experiment, the first geothermal plant in usa started in 1922 with a capacity of 250 kilowatts

-however, in 1946 first ground-source geothermal heat pump installed at commonwealth building in portland, oregon

-during the 1960’s, pacific gas and electric began operation of first large scale geothermal power plant in san francisco, producing 11 megawatts

-in 1973, when oil crisis began many countries began looking for renewable energy sources and by 1980’s geothermal heat pumps (ghp) started gaining popularity in order to reduce heating and cooling costs

-as effect of climate change started showing results, governments of various countries joined hands to fight against it, for which kyoto protocol was signed in japan in 1997, laid out emission targets for rich countries and required that they transfer funds and technology to developing countries, 184 countries have ratified it

hope this