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Naya [18.7K]
3 years ago
14

6. What symptom will be exhibited on an engine equipped with a pneumatic governor system if the cooling fins

Engineering
1 answer:
anyanavicka [17]3 years ago
8 0
<h2>Symptom of Pneumatic Governor System</h2>

Explanation:

  • Engine speed will rise is right answer
  • Pneumatic Governor system that utilizations power from moving air delivered by pivoting flywheel blades to detect engine speed
  • Electronic Governor System Uses a limited point torque engine in place of the governor spring and -speed-sensing device utilized in a mechanical governor system
  • A governor is a system that is utilized to keep up the mean speed of a engine, inside specific cutoff points, under fluctuating burden conditions
  • It does this by managing and controlling the measure of fuel provided to the motor
  • Governors are likewise fitted in assistant diesel motors or generators, and alternators on the ship
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Find the mass if the force is 18 N and the acceleration is 2 m/s2
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Explanation:

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m = f/a

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A cylindrical tank is required to contain a gage pressure 520 kPa . The tank is to be made of A516 grade 60 steel with a maximum
enot [183]

Answer:

t= 4.5 mm

Explanation:

Given that

P = 520 KPa ( gauge)

Maximum allowable normal stress ,σ= 150

d= 2.6 m

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The normal stress for pressure vessel given as

\sigma=\dfrac{Pd}{2t}               ( hoop stress)

We always take maximum stress for safe design.

\sigma=\dfrac{Pd}{2t}

Now by putting the values

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So the minimum thickness, t, of the wall is 4.5 mm

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3 years ago
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3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10
Natali [406]

Answer:

14.52 minutes

<u>OR</u>

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

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3 years ago
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