Mechatronics Questions and Answers for Aptitude test on “Pneumatic Actuating Systems – Process Control Valves”.

1. What is the percentage change in amount of water flowing through a valve in one hour when originally the rate of flow from the valve was 15m^{3}/hr which was reduced to 5m^{3}/hr?

a) 50%

b) 60%

c) 20%

d) 66.67%

Answer: d

Clarification:

Given: Original flow rate: 15m^{3}/hr

Final flow rate: 5m^{3}/hr

Total volume of water flowing through the valve originally: 15m^{3}

Total volume of water flowing through the narrow valve: 5m^{3}

Percentage change in amount of water: [(Original-final)/Original]*100

Percentage change in amount of water: [(15-5)/15]*100 => 66.67%

The flow rate reduced by 66.67% of the original rate.

2. What is the rate of flow of water from a cylindrical valve if the cylindrical valve has a diameter of 10 cm and velocity of water is 5m/s?

a) 3.92 * 10^{-3}m^{3}/s

b) 3.92 * 10^{-2}m^{3}/s

c) 4.92 * 10^{-2}m^{3}/s

d) 4.92 * 10^{-3}m^{3}/s

Answer: b

Clarification:

Given: Diameter=10 cm

Radius = 5 cm or 0.05m

Area of cross section = π*(r)^{2}

Area of cross section = 3.14*(0.05)^{2}=7.85*10^{-3} m^{2}

Rate of flow = Area of cross section * Velocity of fluid

Rate of flow = 7.85*10^{-3}m^{2} * 5m/s

Rate of flow = 3.92 * 10^{-2}m^{3}/s.

3. What is the rate of flow of water in Litres per second from a cylindrical valve if the cylindrical valve has a diameter of 10 cm and velocity of water is 12m/s?

a) 100L/s

b) 127L/s

c) 157L/s

d) 94.2L/s

Answer: d

Clarification:

Given: Diameter=10 cm

Radius = 5 cm or 0.05m

Area of cross section = π*(r)^{2}

Area of cross section = 3.14*(0.05)^{2}=7.85*10^{-3} m^{2}

Rate of flow = Area of cross section * Velocity of fluid

Rate of flow = 7.85*10^{-3}m^{2} * 12m/s

Rate of flow = 0.0942m^{3}/s

Now, we know 1m^{3}/s=1000L/s

Therefore 0.0942m^{3}/s=94.2L/s.

4. What is the percentage change in amount of water flowing through a valve in one hour when originally the rate of flow from the valve was 10 m^{3}/hr which was reduced to 5m^{3}/hr?

a) 50%

b) 60%

c) 20%

d) 100%

Answer: a

Clarification:

Given: Original flow rate: 10m^{3}/hr

Final flow rate: 5m^{3}/hr

Total volume of water flowing through the valve originally: 10m^{3}

Total volume of water flowing through the narrow valve: 5m^{3}

Percentage change in amount of water: [(Original-final)/Original]*100

Percentage change in amount of water: [(10-5)/10]*100 => 50%

The flow rate reduced by 50% of the original rate.

5. In “two way ball valves” the flow can be switched and controlled by rotating the ball in 90 degrees.

a) True

b) False

Answer: a

Clarification: In “two way ball valves” the flow can be switched and controlled by rotating the ball in 90 degrees. These are the simplest type and commonly used control valves. It contains a ball that has a hole bored through it which used to turn ON and OFF the flow through the valve.

6. In “Butterfly valves” the flow can be switched and controlled by rotating the obstructer in 90 degrees.

a) True

b) False

Answer: a

Clarification: In “Butterfly valves” the flow can be switched and controlled by rotating the obstructer in 90 degrees. It is similar to the ball valves but fluid obstructer used in this is a flat disc rather than a ball. This disc is used to turn ON and OFF the flow through the valve.

7. Which type of valves contains tapered cylindrical obstructer?

a) Butterfly valves

b) Two way ball valves

c) Three way ball valves

d) Plug Valves

Answer: d

Clarification: Plug Valves contains tapered cylindrical obstructer. Two way and Three way ball valves contains a ball that has a hole bored through it which acts as an obstructer. Butterfly valves contains a flat disc which acts as an obstructer.

8. What is the velocity of water from a cylindrical valve if the cylindrical valve has a diameter of 10 cm and rate of flow of water in Litres per second is 157L/s?

a) 40m/s

b) 30m/s

c) 20m/s

d) 30m/s

Answer: c

Clarification:

Given: Diameter=10 cm

Radius = 5 cm or 0.05m

Area of cross section = π*(r)^{2}

Area of cross section = 3.14*(0.05)^{2}=7.85*10^{-3} m^{2}

Rate of flow of water = 157L/s

Rate of flow = Area of cross section * Velocity of fluid

157L/s = 7.85*10^{-3}m^{2} * Velocity of fluid

Now, we know 1m^{3}/s=1000L/s

Therefore, Velocity of fluid = 20m/s.

9. Which type of valves contains a flat disc obstructer?

a) Butterfly valves

b) Two way ball valves

c) Three way ball valves

d) Plug Valves

Answer: a

Clarification: Butterfly valves contain a flat disc obstructer. Plug Valves contains tapered cylindrical which acts as an obstructer. Two way and Three way ball valves contains a ball that has a hole bored through it which acts as an obstructer.

10. How many ports does a 3 way valve have?

a) 1

b) 2

c) 3

d) 4

Answer: c

Clarification: A 3 way valve has total 3 ports. Generally it is used when a single incoming input fluid has to be distributed into two directions. In this type of configuration, one port is used as input port and other two ports are used as exit ports.

11. What is the percentage change in amount of water flowing through a valve in one hour when originally the rate of flow from the valve was 10m^{3}/hr which was reduced to 8m^{3}/hr?

a) 50%

b) 60%

c) 20%

d) 800%

Answer: c

Clarification:

Given: Original flow rate: 10m^{3}/hr

Final flow rate: 8m^{3}/hr

Total volume of water flowing through the valve originally: 10m^{3}

Total volume of water flowing through the narrow valve: 8m^{3}

Percentage change in amount of water: [(Original-final)/Original]*100

Percentage change in amount of water: [(10-8)/10]*100 => 20%

The flow rate reduced by 20% of the original rate.

12. How many exit ports does a 2 way valve have?

a) 1

b) 2

c) 3

d) 4

Answer: a

Clarification: A 2 way valve has 1 exit port. It has total two ports, one entry port and one exit port. These ports are not fixed and can be interchanged. Both the port are separated by a different types obstructers.

13. Pneumatic actuator based valves use air to generate rotational force.

a) True

b) False

Answer: a

Clarification: Pneumatic actuator based valves use air to generate rotational force. This rotational force is used to turn the ball, disc or cylindrical type obstructer of the valves. The torque generated is directly proportional to the air pressure provided.

14. What is the rate of flow of water from a cylindrical valve if the cylindrical valve has a diameter of 10 cm and velocity of water is 10m/s?

a) 3.92 * 10^{-3} m^{3}/s

b) 3.92 * 10^{-2} m^{3}/s

c) 7.85 * 10^{-2} m^{3}/s

d) 7.85 * 10^{-3} m^{3}/s

Answer: c

Clarification:

Given: Diameter =10 cm

Radius = 5 cm or 0.05m

Area of cross section = π*(r) ^{2}

Area of cross section = 3.14*(0.05)^{2}=7.85*10^{-3} m^{2}

Rate of flow = Area of cross section * Velocity of fluid

Rate of flow = 7.85*10^{-3}m^{2} * 10m/s

Rate of flow = 7.85* 10^{-2} m^{3}/s.

15. What is the final amount of water flowing through a valve in one hour when originally the rate of flow from the valve was 10m^{3}/hr which was reduced by 80%?

a) 3m^{3}/hr

b) 4m^{3}/hr

c) 5m^{3}/hr

d) 2m^{3}/hr

Answer: d

Clarification:

Given: Original flow rate: 10m^{3}/hr

Reduction in flow = 80%

Total volume of water flowing through the valve originally: 10m^{3}

Percentage change in amount of water: [(Original-final)/Original]*100

Percentage change in amount of water: [(10-final)/10]*100 => 80/100

Final rate of flow of water = 2m^{3}/hr.