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guapka [62]
3 years ago
9

What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent leng

th of 150 ft
Engineering
1 answer:
skad [1K]3 years ago
6 0

Answer:

0.067wc

Explanation:

The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction

We substitute values

actual static pressure = 0.1

Total equivalent length = 150 ft

0.1 = (150ft/100) multiplied by Rate of friction

Friction rate at 100ft = 0.067

So we have that the required friction needed is 0.067wc

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A company intends to market a new product and it estimates that there is a 20% chance that it will be first in the market
Stells [14]

The EMV - Ending Market Value is given as:
$2,400,000.

<h3>How is the EMV Arrived At?</h3>

The EMV is given as:

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Hence the EMV = 2,000,000 x ( 1 + 20%)

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brainly.com/question/1350233

4 0
3 years ago
Sawing stock to reduce its thickness is known as __________ .
rjkz [21]

Answer:

resawing

Explanation:

8 0
4 years ago
Question # 3
tino4ka555 [31]

Answer:

I think it's False.

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4 0
3 years ago
Read 2 more answers
A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
Marrrta [24]

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

  diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0
3 years ago
You want a potof water to boil at 105 celcius. How heavy a
ankoles [38]

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

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to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = \frac{\pi }{4} d^2    ..................1

here d is diameter

put the value in equation 1

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area = 0.01767 m²

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Fnet = ( 120.8 - 101 ) × 0.01767

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we know

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mass  = \frac{350}{9.8}

mass = 35.7 kg

so 35.7 kg lid we put

3 0
3 years ago
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