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rusak2 [61]
3 years ago
11

A large part in a turbine-generator unit operates near room temperature and is made of ASTM A470-8 steel ( ). A surface crack ha

s been found that is roughly a semi-ellipse, with surface length 2c= 50 mm and a depth a= 15 mm. the stress normal to the plane of the crack is 250 MPa, and the member width and thickness are large compared to the crack size. What is the safety factor against brittle fracture? Should the power plant
Engineering
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer: safety factor= 1.26

safety factor does not exceed 2. Therefore, it is not safe to operate.

Explanation:

Express the stress infinity factor (k) that relate to the crack length (Q) applied stress for the centre crack of the plate.

K= Fsq √ πa/Q

Here,

the dimension less function quality is F

Crack length is a

Q= 1 + 1.48(a/c)^1.65

The surface length is c

c= 25mm

a= 15mm

Substitute c= 25mm and a=15mm into the equation

Q= 1 + 1.48(15mm/25mm)^1.65

Q= 1 + 1.48(0.6)^1.65

Q= 1 + 0.630

Q= 1.630

Find the stress intensity factor of the corresponding ratio of 0.4 and 0.12

Where

F= 1.12

S= 250mpa

Q= 1.63

a= 15mm

Substitute into the qequation

K= (1.12) (250mpa) √π(15mm× (1/1000mm))/1.63

K= (1.12) (250mpa)√47.1mm× (1/1000mm))/1.63

K= 2.80mpa √0.0471× (1m)/1.63

K= 280mpa × 0.170√m

K= 476mpa√m

Calculate the safety

Xk= Kk/K

Where fraction toughness is Kk

From the table of fraction toughness, corresponding tensile property for metal at room temperature, select this fraction toughness for ASTM A470-8 steel.

Substitute 47.6mpa√m for k

60mpa√m for Kk

Xk= 60mpa√m/47.6mpa√m

Xk= 1.26

In conclusion, from the above result, safety factor does not exceed 2. Therefore, it is not safe to operate.

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An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci
Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0
3 years ago
In digital communication technologies, what is an internal network also known as?
garri49 [273]
Intranet is a network that is internal and use internet technologies. It makes information of any company accessible to its employees and hence facilitates collaboration. Same methods can be used to get information, use resources, and update the data as that of the internet.
Hopefully this helped.
4 0
3 years ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
4 years ago
In your opinion...
ch4aika [34]

Answer:no

TTHANLS FOR FREE POINTS

Explanation:

8 0
3 years ago
An orchestra is having a recording done of 2 performances in the same concert hall. The first show is sold out. They struggled t
konstantin123 [22]

Answer:

yes, the recordings sound is same

Explanation:

given data

recording done = 2 performances

1st  show = sold out

2nd show =  lightly attended

to find out

recordings sound the same and why

solution

as per given in

  • 1st show is sold out it mean in this case concert hall is full so that recording sound should be high here
  • 2nd case only few people are attended and struggle for ticket  and orchestra

it mean it sound performance so in both case recording sound will be same

because we do not other all are sitting at front row or they sit as they want

4 0
3 years ago
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