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ahrayia [7]
3 years ago
7

Technician A says that the most efficient method of EVAP system leak detection is introducing smoke under low pressure from a ma

chine specifically designed for this purpose. Technician B says that nitrogen gas pressurization uses nitrogen gas under a very low pressure (lower than 1 psi) in the fuel system; you listen for escaping air with amplified headphones. Who is correct?
Engineering
1 answer:
Inessa05 [86]3 years ago
3 0

Answer:

does anyone know this answer???

Explanation:

You might be interested in
How do you make a 3d print
yulyashka [42]

Answer:you need a 3d printer

Explanation:

5 0
3 years ago
). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially
Delicious77 [7]

Answer:

\frac{e'_z}{e_z} = 0.87142

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            \frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ]  }{-\frac{s_z}{E}}  \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142... Answer

5 0
3 years ago
Describe the algorithm you use for looking up a person’s telephone number in the phone book. The input is person’s name; the out
Stella [2.4K]

Answer:

The Algorithm for finding a number from a phone book with the person's name as the input and the phone number as output is as follows:

1. Try to remember the name, i.e last name first and first name last, Also make sure you get the spelling right.

2. Using the first letter of the last name, locate the appropriate alphabetical section in which the name should appear.

3. Using the second letter of the last name, find the subsection of first and second letters combined, in the appropriate order, in which the name should appear. (If the last name consists of only two letters, find the appropriate first name.)

4. Using the third letter, find the possible names in a subsection of the first three letters in the correct order. Continue this step with x+1 letters of the name until you have a subsection of names exactly matching the last name of the person whose number you are trying to locate. (x is the number of letters used in the previous step, consistently.) If there is only one of the last name, (check for duplicates) identify the number, and return phone number information.

5. Begin the second step using the first letter of the first name, but limit the section to only those exactly matching the last name. Continue to step 4, again focusing on the first name only within the set of exactly matching last names.

6. When both first and last name match the name you are locating, check for duplicates. IF there are no duplicates, return phone number information.

Explanation:

People's names are generally arranged in phone books in alphabetical order by the last name of the person. The first name of the person is listed after the last name so that people of the same last name can be differentiated.

7 0
3 years ago
Read 2 more answers
A step-up transformer has 20 primary turns and 400 secondary turns. If the primary current is 30 A, what is the secondary curren
-BARSIC- [3]
150
A
Explanation:
V
s
V
p
=
N
s
N
p

(
1
)
N
refers to the number of turns
V
is voltage
s
and
p
refer to the secondary and primary coil.
From the conservation of energy we get:
V
p
I
p
=
V
s
I
s

(
2
)
From
(
1
)
:
V
s
V
p
=
900
00
3
00
=
300
∴
V
s
=
300
V
p
Substituting for
V
s
into
(
2
)
⇒
V
p
I
p
=
300
V
p
×
0.5
∴
I
p
=
150
A
Seems a big current.
3 0
3 years ago
What is the fastest motorcycle in the world ?
givi [52]

Answer:

Kawasaki Ninja H2R – top speed: 222 mph. This one is another beast in the form of a bike. ...

MTT Turbine Superbike Y2K – top speed: 227 mph. This bike is one of the most powerful production motorcycles. ...

Suzuki Hayabusa – top speed: 248 mph. 1340cc

8 0
3 years ago
Read 2 more answers
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