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3 years ago

Evaluate to three significant figures and using appropriate prefix: (354 mg)(45 km)/(0.0356 kN)

Engineering

1 answer:

frutty [35]3 years ago

5 0

Answer:

0.447 s²

Explanation:

First, convert to SI units.

(354 mg) (45 km) / (0.0356 kN)

(0.354 g) (45000 m) / (35.6 N)

One Newton is kg m/s²:

(0.354 g) (45000 m) / (35.6 kg m/s²)

(0.000354 kg) (45000 m) / (35.6 kg m/s²)

Simplify:

0.447 s²

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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

Which of the following identifies the beginning phase of the engineering design process?

lora16 [44]

Answer:

<❤)structural analysi(❤>

Explanation:

(♨️)BRAINLEIST PLEASE(♨️)

5 0

3 years ago

In a river reach, the rate of inflow at any time is 350 cfs and the rate of outflow is 285 cfs. After 90 min, the inflow and out

Semenov [28]

Answer:

change in storage = -310,500 ft^3

intital storage= 3.67 acre ft

Explanation:

Given data:

Rate of inflow = 350 cfs

Rate of outflow = 285 cfs

After 90 min, rate of inflow = 250 cfs

Rate of outflow = 200 cfs

final storage = 10.8 acre-ft

calculating the average inflow and outflow

average inflow $= \frac{(350+250)}{2} = 300 cfs$

average outlow $= \frac{(285+200)}{2} = 242.5 cfs$

total amount of water drain during the period of one hour

= (average outflow - average inflow) *60*90

= (242.5 - 300)*60*90 = -310,500

change in storage is calculate as

= -310,500 ft^3

in cubic meter

= -310500/35.315 = 8792.30 cm^3

in acre-ft

= -310,500/43560 = 7.13 acre ft

initial storage = 10.8 - 7.13 = 3.67 acre ft

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Keith_Richards [23]

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A cyclone is operated in a closed circuit with a ball mill. The cyclone is feed from a rod mill with a slurry that has a density

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