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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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Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp
telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

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