Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The gas-turbine cycle of a combined gas–steam power plant has a pressure ratio of 8. Air enters the compressor at 290 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 15 MPa to 450°C in a heat exchanger. The combustion gases leave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500°C before it expands in a low-pressure turbine to 10 kPa. The mass flow rate of steam is 17 kg/s. Assume isentropic efficiencies of 100 percent for the pump, 85 percent for the compressor, and 90 percent for the gas and steam turbines.Determine the rate of total heat input.

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.

Answer: the thermal conductivity of the second material is 125.9 W/m.k

Explanation:

Given that;

The two rods could be approximated as a fins of infinite length.

TA = 75°C,    θA = (TA - T∞) = 75 - 25 = 50°C

TB = 70°C     θB = (TB - T∞) = 70 - 25 = 45°C

Tb = 100°C    θb = (Tb - T∞) = (100 - 25) = 75°C

T∞ = 25°C

KA = 200 W/m · K,   KB = ?

Now

The temperature distribution for the infinite fins are given by

θ/θb = e^(-mx)

θA/θb= e^-√(hp/A.kA) x 1  --------------1

θB/θb = e^-√(hp/A.kB) x 1---------------2

next we  take the natural logof both sides,  

ln(θA/θb) = -√(hp/A.kA) x 1 ------------3

In(θB/θb) = -√(hp/A.kB) x 1 ------------4

now we divide 3 by 4

[ ln(θA/θb) /in(θB/θb)] = √(KB/KA)

we substitute

 [ In(50/75) /In(45/75)] = √(KB/200)

In(0.6666) / In(0.6) = √KB / √200

-0.4055/-0.5108 = √KB / √200

0.7938 = √KB / 14.14

√KB = 11.22

KB = 125.9 W/m.k

So the thermal conductivity of the second material is 125.9 W/m.k

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.

In signal processing, a filter is a device or process that removes some unwanted components or features from a signal. Filtering is a class of signal processing, the defining feature of filters being the complete or partial suppression of some aspect of the signal. Most often, this means removing some frequencies or frequency bands. However, filters do not exclusively act in the frequency domain; especially in the field of image processing many other targets for filtering exist.

Explanation:

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 =   × d²

put the value of d here

area =   × 45²

swept area = 1590 m^2