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Ganezh [65]
3 years ago
5

Design a PLC ladder logic program to control the operation of a conveyor-storage system using the following sequence: - 1. Progr

am a count-up to keep track of the parts brought into the storage room by the CONVEYOR_IN and detected with a PROXIMITY_IN sensor.
2. Program a count-down to keep track of the parts taken out of the storage room by the CONVEYOR_OUT and detected with a PROXIMITY_OUT sensor.
3. Both conveyor motors are operated by one START/STOP seal-in pushbutton station.
4. A GREEN light must be on if the part count in the storage room is equal to zero (0).
5. A YELLOW light must be on if the part count in the storage room is between one (1) and ten (10) (inclusive).
6. A RED light must flash on and off in periods of 0.5 seconds if the part count in the storage room is greater than ten (10).
7. A RESET pushbutton is used to reset the count. - Program CONVEYOR_IN and CONVEYOR_OUT using bits 0 and 1 of the output module.
8. Program the GREEN, YELLOW, and RED lights using bits 2, 3, and 4 of the output module.
9. Program the START and STOP buttons as normally open in bits 0 and 1 of the input module.
10. Program PROXIMITY_IN and PROXIMITY_OUT using bits 2 and 3 of the input module.
11. Program the RESET pushbutton using bit 4 of the input module.

Engineering
1 answer:
Phantasy [73]3 years ago
8 0

Answer:

See explaination

Explanation:

Kindly check attachment for the step by step solution of the given problem.

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A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rota
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Answer:

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius.  The mass is computing by the rim density and its volume:

m=p.V

m=p*(A*R)

Where A is the cross-sectional area in meters:

m=((0.015m*0.200m)*0.6m)*(7800 kg/m^3)=28.08 kg

The angular speed in rad/s is:

ω = 800r/m . 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg)*(133.33 rad/s)^2*(0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π .R .b)

Where b is rim's with equal to 200mm :

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

Pt = 10MPa+397887Pa= 10000000Pa+397887Pa= 10397887Pa

Then for a thinner thick the stress is calculated by:

Pt*d =2σr*t

Take into account that the stress σr is over the radial direction. Then solving for o and by replacing the total pressure:

σr = Pt.d/(2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr =0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr =0.00498m= 4.98mm [/tex]

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