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azamat
3 years ago
10

There are 20 forging presses in the forge shop of a small company. The shop produces batches of forgings requiring a setup time

of 3.0 hours for each production batch. Average standard time for each part in a batch is 45 seconds, and there are an average of 600 parts in a batch. The plant workforce consists of two workers per press, two foremen, plus three clerical support staff. (a) Determine how many forged parts can be produced in 1 month, if there are 8 hours worked per day and an average of 21 days per month at one shift per day. (b) What is the labor productivity ratio of the forge shop, expressed as parts per worker-hour?
Engineering
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21  making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

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In the situation shown below, what would the Moon look like from Earth? Sun, Earth and Moon Four Moon Views A. View A B. View B
Blababa [14]

Answer:

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Explanation:

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3 years ago
A 860 kΩ resistor has 34 μA of current. What is the supply voltage for this electric circuit?
Juliette [100K]

Explanation:

first changing kilo ohm to ohm

860000 = 860 kΩ

and change 34 micro ampare to ampare

34 μA=3.4×10^-5

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8 0
3 years ago
Read 2 more answers
). 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
Where are the ar manufacturers not fitting the engine in the high end sport cars
fomenos

Answer:

it depends on the but i would recommend check in the front next to the turbo intake.

8 0
2 years ago
Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

6 0
3 years ago
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