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Naya [18.7K]
2 years ago
6

External crack of length of 3.0 mm was detected on the surface of the shaft of wind turbine made from 4340 steel. The diameter o

f the shaft of wind turbine is equal to 32 mm. Assume the shaft is subjected to a maximum load of 50,000 N during operation and the radius of the curvature of the crack is 3 x 10-2 mm. Determine the Fracture toughness KC of the shaft Express your answer in to four significant figures. Do not include the units.
Engineering
1 answer:
sveticcg [70]2 years ago
4 0

Answer:

The correct answer is "K_c=6.0369 \ MPa\sqrt{m}".

Explanation:

Given:

Maximum load,

P = 50,000 N

Crack length,

a = 3mm

or,

  = 3×10⁻³ m

Diameter,

d = 32 mm

As we know,

⇒  Maximum stress, \sigma=\frac{P}{A}

                                      =\frac{50000}{(\frac{\pi}{4}\times 32^2)}

                                      =62.20 \ N/mm^2

Now,

⇒  Fracture tougness, K_c=Y \sigma\sqrt{\pi a}

On substituting the values, we get

                                           =1\times 62.20\times \sqrt{3.14\times 3\times 10^{-3}}

                                           =6.0369 \ MPa\sqrt{m}

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A cylindrical bar of metal having a diameter of 20.9 mm and a length of 205 mm is deformed elastically in tension with a force o
natita [175]

Answer:

a. 1.91 b.  -8.13 mm

Explanation:

Modulus =stress/strain; calculating stress =F/A, hence determine the strain

Poisson's ratio =(change in diameter/diameter)/strain

8 0
3 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
2 years ago
3. Which of these instruments is used to measure wind speed? A. anemometer C. wind sock B. thermometer D. wind vane It is an ins
siniylev [52]

Answer:

wind vane if it can be used to show wind speed and the other is a

Explanation:

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5 0
2 years ago
Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A
uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

7 0
3 years ago
Question text
lisabon 2012 [21]

Answer:

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3 0
3 years ago
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