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lina2011 [118]
4 years ago
15

Why or why not the following materials will make good candidates for the construction of

Engineering
1 answer:
zvonat [6]4 years ago
3 0

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.

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Estimate the energy (head) loss a short length of a pipe conveying 300 litres of water per second and suddenly enlarging from a
Ludmilka [50]

Known :

Q = 300 L/s = 0.3 m³/s

D1 = 350 mm = 0.35 m

D2 = 700 mm = 0.7 m

g = 9.81 m/s²

Solution :

A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²

A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²

hL = (kL / 2g) • (U1² - U2²)

hL = (kL / 2g) • Q² (1/A1² - 1/A2²)

hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))

hL = 0.467 m

5 0
3 years ago
A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a
statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0
3 years ago
When using fall arrest, free fall must be kept at or below how many feet
SashulF [63]
<h3>Answer:</h3>

two feet or less

<h3>Explanation:</h3>

8 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 875 K to a cold reservoir at 300 K. Calculate the rate at which th
blsea [12.9K]

Answer:

The correct answer is 0.004382 kW/K, the second law is a=satisfied and established because it is a positive value.

Explanation:

Solution

From the question given we recall that,

The transferred heat rate is = 2kW

A reservoir cold at = 300K

The next step is to find the rate  at which the entropy of the two reservoirs changes is kW/K

Given that:

Δs =  Q/T This is the entropy formula,

Thus

Δs₁ = 2/ 300 = 0.006667 kW/K

Δs₂ = 2 / 875 =0.002285

Therefore,

Δs = 0.006667 - 0.002285

= 0.004382 kW/K

Yes, the second law is satisfied, because it is seen as positive.

8 0
4 years ago
Read the following statement.
alekssr [168]
The correct answer is A
5 0
4 years ago
Read 2 more answers
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