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3 years ago

8

Which one is your choice and Why?

1 answer:

marishachu [46]3 years ago

3 0

Poopy poggers

Explanation Weeb#7777

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An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What

Jet001 [13]

Answer:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

Explanation:

Given

$Volume = 9m^3$

Represent the height as h, the length as l and the width as w.

From the question:

$Length = 2 * Width$

$l = 2w$

Volume of a box is calculated as:

$V = l*w*h$

This gives:

$V = 2w *w*h$

$V = 2w^2h$

Substitute 9 for V

$9 = 2w^2h$

Make h the subject:

$h = \frac{9}{2w^2}$

The surface area is calculated as:

$A = 2(lw + lh + hw)$

Recall that: $l = 2w$

$A = 2(2w*w + 2w*h + hw)$

$A = 2(2w^2 + 2wh + hw)$

$A = 2(2w^2 + 3wh)$

$A = 4w^2 + 6wh$

Recall that: $h = \frac{9}{2w^2}$

So:

$A = 4w^2 + 6w * \frac{9}{2w^2}$

$A = 4w^2 + 6* \frac{9}{2w}$

$A = 4w^2 + \frac{6* 9}{2w}$

$A = 4w^2 + \frac{3* 9}{w}$

$A = 4w^2 + \frac{27}{w}$

To minimize the surface area, we have to differentiate with respect to w

$A' = 8w - 27w^{-2}$

Set A' to 0

$0 = 8w - 27w^{-2}$

Add $27w^{-2}$ to both sides

$27w^{-2} = 8w$

Multiply both sides by $w^2$

$27w^{-2}*w^2 = 8w*w^2$

$27 = 8w^3$

Make $w^3$ the subject

$w^3 = \frac{27}{8}$

Solve for w

$w = \sqrt[3]{\frac{27}{8}}$

$w = \frac{3}{2}$

Recall that : $h = \frac{9}{2w^2}$ and $l = 2w$

$h = \frac{9}{2 * \frac{3}{2}^2}$

$h = \frac{9}{2 * \frac{9}{4}}$

$h = \frac{9}{\frac{9}{2}}$

$h = 9/\frac{9}{2}$

$h = 9*\frac{2}{9}$

$h= 2$

$l = 2w$

$l = 2 * \frac{3}{2}$

$l = 3$

Hence, the dimension that minimizes the surface area is:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

6 0

3 years ago

At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0

3 years ago

I'm really bad at measurements so I don't understand this.

cricket20 [7]

Answer:

the answer should be 16

6 0

4 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

Why does my man bun not have its own erodynamics

Aloiza [94]

Answer:

umm okay for starters I have no clue lol.

7 0

3 years ago

Read 2 more answers