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Lelu [443]
2 years ago
10

two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl

ether's lower explosive level? (LEL=1.9%, SG=.71)
Engineering
1 answer:
lesya692 [45]2 years ago
7 0

The air flow necessary to remain at the lower explosive level is 4515. 04cfm

<h3>How to solve for the rate of air flow</h3>

First we have to find the rate of emission. This is solved as

2pints/1.5 x 1min

= 2/1.5x60

We have the following details

SG = 0.71

LEL = 1.9%

B = 10% = 0.1 a constant

The molecular weight is given as 74.12

Then we would have Q as

403*100*0.2222 / 74.12 * 0.71 * 0.1

= Q = 4515. 04

Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm

Read more on the rate of air flow on brainly.com/question/13289839

#SPJ1

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3 years ago
A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T th
fiasKO [112]

Answer:

Explanation:

Using equation of pure torsion

\frac{T}{I_{polar} }=\frac{t}{r}

where

T is the applied Torque

I_{polar} is polar moment of inertia of the shaft

t is the shear stress at a distance r from the center

r is distance from center

For a shaft with

D_{0} = Outer Diameter

D_{i} = Inner Diameter

I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}

Applying values in the above equation we get

I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\I_{polar}= 1.74 x 10^{-7} m^{4}

Thus from the equation of torsion we get

T=\frac{I_{polar} t}{r}

Applying values we get

T=\frac{1.74X10^{-7}X100X10^{6}  }{.021}

T =829.97Nm

7 0
3 years ago
The criminal and traffic code requires that a driver must have a valid driver's license in his/her
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Answer:

true

Explanation:

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5 0
3 years ago
Question Set 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.2.2 In
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Answer:

# Program is written in python

# 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.

# initializing string

Stringtocheck = "mississippi"

# using count() to get count of s

counter = Stringtocheck.count('s')

# printing result

print ("Count of s is : " + str(counter))

# 2.2 In the string 'mississippi', replace all occurrences of the substring 'iss' with 'ox

# Here, we'll make use of replace() method

# Prints the string by replacing iss by ox

print(Stringtocheck.replace("iss", "ox"))

#2.3 Find the index of the first occurrence of 'p' in 'mississippi'

# declare substring

substring = 'p'

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index = Stringtocheck.find(substring)

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8 0
3 years ago
A metal specimen with an original diameter of 0.50 in. and a gauge length of 2.75 in. is tested in tension until a fracture occu
tatiyna

Answer:

Percent Elongation = 52.72%

Percent Reduction in Area = 64%

Explanation:

First we find percent elongation:

Percent Elongation = {Final Gage Length - Initial Gauge Length/Initial Guage Length} x 100%

Percent Elongation = {(4.20 in - 2.75 in)/2.75 in} x 100%

<u>Percent Elongation = 52.72%</u>

Now, for the percent reduction in area:

Percent Reduction in Area = {Final Cross Sectional Area - Initial Cross Sectional Area|/Initial Cross Sectional Area Length} x 100%

Percent Reduction in Area = {π(0.3 in)² - π(0.5 in)²/π(0.5 in)²} x 100%

<u>Percent Reduction in Area = - 64%</u>

here, negative sign shows a decrease in area.

5 0
3 years ago
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