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kozerog [31]
3 years ago
15

Para uma dada velocidade do vento, o gerador de turbina mostrado na figura produz 22 kW de potência elétrica. Como um sistema co

nversor de energia a turbina tem um rendimento de 30%. Determine (a) a energia cinética por unidade de tempo do ar que penetra a seção circular de 7,80 m de diâmetro e (b) a velocidade do vento.
Engineering
1 answer:
____ [38]3 years ago
4 0

Answer:

b

Explanation:

You might be interested in
A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he
sergejj [24]

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

8 0
3 years ago
The simple majority decision rule may generate results that are
Arisa [49]

Answer:

Efficient when the marginal benefits of project = marginal costs of project.

Explanation:

Majority Decision Rule:

Majority decision rule is based on the notion of equality. An alternative is selected which has majority of votes. The simple majority decision rule may generate efficient results if the marginal benefits of a project are equal or greater than the marginal costs of the project.

5 0
3 years ago
A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum
adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress \sigma = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder  there exist both the circumferential stress and the longitudinal stress.

Circumferential stress \sigma = \dfrac{pd}{2t}

Making thickness t the subject; we have

t = \dfrac{pd}{2* \sigma}

t = \dfrac{560000*3}{2*150000000}

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

\sigma = \dfrac{pd}{4t}

t= \dfrac{pd}{4*\sigma }

t = \dfrac{560000*3}{4*150000000}

t = 0.0028  mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value  with the maximum thickness = 5.6 mm

8 0
4 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C.
Maksim231197 [3]

Answer:

(d) all of the above

Explanation:

before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial

so option (D) will be correct because we need lubricate in all the given parts  

8 0
3 years ago
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