ENGINEERING COLLEGE

A wind turbine system has the following specifications: Diameter:45 m Rated power 700 kW at the wind speed of 12 m/s Turbine speed 1500 rpm Determine the swept area of the wind turbine. a)- 1640 m^2 B)- 1690 m^2 c)- 1590 m^2 d)- 1540 m^2

Answers

Answer 1

Answer:

Answer:

1590 m^2

Explanation:

Given data in this question

Diameter = 45 m

power = 700 kW

wind speed = 12 m/s

turbine speed = 1500 rpm

To find out

swept area of the wind turbine

Solution

we know wind turbine is rotate circular form

and diameter is given so by the area of circular swept we will calculate it

we know area = $\pi /4$ × d²

put the value of d here

area = $\pi /4$ × 45²

swept area = 1590 m^2

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Air is compressed in a piston-cylinder device. List three examples of irreversibilities that could occur

Answers

Answer:

While air is compressed in a piston cylinder there are following types of irreversibilities

1.Due to finite temperature difference heat transfer take place between cylinder and surrounding.

2.Due friction force between cylinder and piston .

3.Compression process is so fast due to this ,it leads in the irreversibility of system.

Consider the binary eutectic copper-silver phase diagram in Fig. P8.22. Make phase analyses of an 88 wt % Ag−12 wt % Cu alloy at the temperatures (a) 1000°C, (b) 800°C, (c) 780°C + ΔT, and (d) 780°C − ΔT. In the phase analyses, include: (i) The phases present (ii) The chemical compositions of the phases (iii) The amounts of each phase (iv) Sketch the microstructure

Answers

Answer:

answer is 0 ok answer is 0

Explanation:

Musk is working on developing reusable rockets in order to further us as a space-faring civilization. Given his life dedication to energy production, conservation, and efficiency, can you think of any other reasons that Musk may want to expand our ability to exist on other planets and in space?

Answers

Answer:

exploration

Explanation:

Other than the scientific reasons listed in the question, one of the main reasons why people all over the world are pursuing this endeavor is exploration. As human beings, we love to imagine new worlds and life-forms that we have never seen before. This fuels our need for exploration. Scientists throughout generations have dedicated their entire lives to learning and creating newer and better technology in order for humans to take that next step in exploring and learning the secrets of the universe.

Why or why not the following materials will make good candidates for the construction of a) Turbine blade for a jet turbine and
b) Thermal barrier coating. Cement, aluminum, engineering ceramic, super alloy, steel and glass

Answers

Answer:

Answer explained below

Explanation:

3.] a] A turbine blade is the individual component which makes up the turbine section of a gas turbine. The blades are responsible for extracting energy from the high temperature, high pressure gas produced by the combustor.

The turbine blades are often the limiting component of gas turbines. To survive in this difficult environment, turbine blades often use exotic materials like superalloys and many different methods of cooling, such as internal air channels, boundary layer cooling, and thermal barrier coatings. The blade fatigue failure is one of the major source of outages in any steam turbines and gas turbines which is due to high dynamic stresses caused by blade vibration and resonance within the operating range of machinery.

To protect blades from these high dynamic stresses, friction dampers are used.

b] Thermal barrier coatings (TBC) are highly advanced materials systems usually applied to metallic surfaces, such as on gas turbine or aero-engine parts, operating at elevated temperatures, as a form ofexhaust heat management.

These 100μm to 2mm coatings serve to insulate components from large and prolonged heat loads by utilizing thermally insulating materials which can sustain an appreciable temperature difference between the load-bearing alloys and the coating surface.

In doing so, these coatings can allow for higher operating temperatures while limiting the thermal exposure of structural components, extending part life by reducing oxidation and thermal fatigue.

In conjunction with active film cooling, TBCs permit working fluid temperatures higher than the melting point of the metal airfoil in some turbine applications.

Due to increasing demand for higher engine operation (efficiency increases at higher temperatures), better durability/lifetime, and thinner coatings to reduce parasitic weight for rotating/moving components, there is great motivation to develop new and advanced TBCs.

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace

Answers

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

$(T_(0)-T_(x) )/(T_(1)-T_(x) ) = C_(1) e^{(-0.4888^(2)*Fo )}$

= 0.4167 = $1.0396e^(-0.4888*Fo)$

therefore Fo = 3.8264

Now determine how long the plate should be left in the furnace

Fo = $((k)/(pc_(p) ) ) ( (t)/((L/2)^2) )$

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

or a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 164 mm2 without plastic deformation

Answers

Answer:

The maximum load (in N) that may be applied to a specimen with this cross sectional area is $F=43788 N$

Explanation:

We know that the stress at which plastic deformation begins is 267 MPa.

We are going to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:

$P=(F)/(A)$ (equation 1)

We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.

Before apply the equation, we need to convert the units of area in $m^(2)$ . So,

$A[m^(2)]=A[mm^(2) ]((1 m)/(1000 mm)) ^(2)\nA[m^(2)]=164 mm^(2)((1 m)/(1000 mm))^(2) \nA[m^(2)]=0.000164 m^(2)$

And then, from the equation 1,

$F=PA\nF=(267 MPa)(0.000164 m^(2))\nF=(267x10^(6) Pa)(0.000164 m^(2))\nF=(267x10^(6) N/m^(2))(0.000164 m^(2))\nF=43788 N$

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