Ask question

Ask question

All categories

3 years ago

9

A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 1.0 kg

/s at a discharge velocity of 8 m/s. Is the claim reasonable? What if the electric power is 40 W?

1 answer:

solong [7]3 years ago

5 0

Answer:

No, the claim is not reasonable for 20 W electric power consumption.

It is reasonable for 40 W electric power consumption.

Explanation:

Power = (1/2)*mass flow rate*(square of velocity)

mass flow rate = 1 kg/s

velocity = 8 m/s

square of velocity = 64 m^2 / s^2

Power = (1/2)*(1)*(64)

Power = 32 W

For a fan that consumes 20 W power it is not possible to deliver more power than 20 W but this one is delivering 32 W hence it is a false claim.

For a fan that consumes 40 W it is indeed possible to deliver 32 W considering the efficiency. Hence this claim is reasonable.

You might be interested in

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

2 years ago

) A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging proc

bonufazy [111]

Answer:

it will be a scattered plot

Explanation:

5 0

2 years ago

Who's your favorite singer and WHT your favorite song

Anna007 [38]

Answer:

and my favorite song is popular loner

Explanation:

my favorite rapper is rod wave

6 0

3 years ago

Read 2 more answers

Select the statement that is false.

ra1l [238]

Answer:

D

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.

8 0

2 years ago

The advantages of solar cells include all of the following, except a.moderate net energy yield b.little or no direct emissions o

Xelga [282]

Answer:

C

Explanation:

One of the disadvantages of solar cells is that electricity storage systems are not readily available. Excess energy generated by the solar panels are wasted except they are stored by solar batteries for later use. There are various systems for storing electricity from solar cells apart from solar batteries which is the common storage system. An example of another electricity storage system for solar cell is using the water electrolyzer to store solar energy which can be used to later generate hydroelectricity.

Advantages of a solar cell includes Renewable energy, Economy-friendly and environmental-friendly energy and good durability

6 0

3 years ago