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Veseljchak [2.6K]
2 years ago
11

A wheel spins at a constant angular speed of 24rad/s.How many revolutions will the dosk go through in 5minutes?​

Engineering
1 answer:
Margaret [11]2 years ago
5 0

<u>Answer:</u>

The wheel will go through 1146 revolutions in 5 minutes.

<u>Explanation:</u>

We can the formula:

\boxed{\omega = \frac{2 \pi}{T}}

where

ω ⇒ angular speed (24 rad/s),

T ⇒ time period (? s),

and solve for T to find the time it takes for the wheel to complete one revolution.

⇒  24 = \frac{2 \pi}{T}

⇒  T = \bf \frac{2 \pi}{24}  s

This means it takes \bf \frac{2 \pi}{24} seconds for the wheel to complete one revolution.

Now, using the unitary method,

In \frac{2 \pi}{24} seconds          ⇒    1 revolution completed

In 1 second              ⇒    1 ÷ \frac{2 \pi}{24}   =  \frac{24}{2 \pi}  revolutions completed

In (5 × 60 = ) 300s  ⇒     \frac{24}{2 \pi}   ×  300   = 1145.9

                                                            ≅ 1146 revolutions completed

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Answer:

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Explanation:

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7 0
3 years ago
Consider a fan located in a 3 ft by 3 ft square duct. Velocities at various points at the outlet are measured, and the average f
natulia [17]

Answer:

minimum electric power consumption of the fan motor is 0.1437 Btu/s

Explanation:

given data

area = 3 ft by 3 ft

air density = 0.075 lbm/ft³

to find out

minimum electric power consumption of the fan motor

solution

we know that energy balance equation that is express as

E in - E out  = \frac{dE \ system}{dt}    ......................1

and at steady state  \frac{dE \ system}{dt} = 0

so we can say from equation 1

E in = E out

so

minimum power required is

E in = W = m \frac{V^2}{2} = \rho A V \frac{V^2}{2}  

put here value

E in =  \rho A V \frac{V^2}{2}  

E in =  0.075 *3*3* 22 \frac{22^2}{2}  

E in = 0.1437 Btu/s

minimum electric power consumption of the fan motor is 0.1437 Btu/s

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3 years ago
. En la facultad de Ingeniería Industrial se realizó una encuesta a 200 personas para saber que lenguaje de programación preferí
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3 years ago
Explain why advances in technology can actually create more problems for engineers.
andrew11 [14]

Explanation:

Engineering is science in practical terms. It is the application of scientific findings in problem solving and creating a better world.

How does technological advancements create more problems for engineers?

  • Loss of job to automation: the world is driving at automating work processes through the use of  specially designed and crafted machinery. Work is now properly being done using machines with little to no human input in the whole process. This is a huge let off for engineers. Engineers have to compete with machines which are their own inventions for jobs now.
  • Fast paced work environment: machines can handle work more efficiently and faster than the people making them. There is an increasing race between engineers and their own inventions today for better product delivery. Unless a machine is faulty, they are more productive and efficient than man. This can cause engineers to want to catch up with their own inventions leading to a work life of stress.
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Learn more:

New technology brainly.com/question/5768621

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3 0
3 years ago
An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f
Gelneren [198K]

Answer:

(a) Q=332 kvar and C=5.66 uF

(b) pf=0.90 lagging

Explanation:

Given Data:

P=600kW

V=12.47kV

f=60Hz

pf_{old} =0.75

pf_{new} =0.95

(a) Find the required kVAR rating of a capacitor

\alpha _{old}=cos^{-1}(0.75) =41.41°

\alpha _{new}=cos^{-1}(0.95) =18.19°

The required compensation reactive power can be found by

Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))

Q=600(tan(41.41) - tan(18.19))

Q=332 kvar

The corresponding capacitor value can be found by

C=Q/2\pi fV^{2}

C=332/2*\pi *60*12.47^{2}

C=5.66 uF

(b) calculate the resultant supply power factor

First convert the hp into kW

P_{mech} =250*746=186.5 kW

Find the electrical power (real power) of the motor

P_{elec} =P_{mech}/n

where n is the efficiency of the motor

P_{elec} =186.5/0.80=233.125 kW

The current in the motor is

I_{m} =(P/\*V*pf)

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

I_{m} =(233.125/12.47*0.85)

I_{m}=18.694+11.586j A

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

I_{load} =(600/12.47*0.75)

I_{load} =48.115-42.433j A

Now the supply current is the current flowing in the load plus the current flowing in the motor

I_{supply} =I_{m} + I_{load}

I_{supply}= (18.694+11.586)+(48.115-42.433)

I_{supply} =66.809-30.847j A

or in polar form

I_{supply} =73.58°

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

pf=cos(24.78)=0.90 lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0
4 years ago
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