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Citrus2011 [14]
3 years ago
15

A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D

etermine the minimum required wire diameter assuming a factor of safety 5 and that the yield strength of the steel is 900MPa.
Engineering
1 answer:
zhuklara [117]3 years ago
7 0

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

d = \sqrt{4*P / (\pi*\sigma)}

d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm

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Water is the working fluid in a Rankine cycle with reheat. The turbine and the pump have isentropic efficencies of 80%. Superhea
zheka24 [161]

Answer:

A)  3783.952 kJ/kg

B)  34.5%

C)  2476.67 kJ/kg

Explanation:

A) Determine the rate of heat addition entering the first-stage turbine

Qin ( 1st stage ) =  ( h1 - h6 ) + ( h3 - h2 )

                         = ( 3321.4 - 164.07 ) + ( 3438.566 - 2811.944 )

                         = 3783.952 kJ/kg

<em>h values are gotten from super heated table </em>

B) Determine the thermal efficiency

n = 34.5%

attached below

C) Rate of heat transfer from working fluid passing through the condenser to the cooling water

Qout = h4 - h5

         = 2628.2 - 151.53

         = 2476.67 kJ/kg

8 0
3 years ago
1. Given: R= 25 , E = 100 V<br> Solve for I
Lyrx [107]

Answer:

4 amps

Explanation:

V = I*R

I / V/R

I = 100 / 25

I = 4 amps

5 0
3 years ago
A piston–cylinder assembly contains 5 kg of air, initially at 2.0 bar, 30 C. The air undergoes a process to a state where the pr
Ainat [17]

Answer: wor done is 145. 06kJ

Heat transfer is 135.53kJ

Explanation:

No of moles of air = mass/molar mass = 5000g/28gmol^-1 = 172.65mol

P1 = 2bar =2*101300 =202600pa

T1 = 30° +273k = 303k

P2 =p1 = 202600pa

V2 =? T2 =?

Using pV = nRT

R = 8.314 PA m^3 mol^-1 k^-1

V1 = (172.65*8.314*303)/202600

V1 = 2.146m^3

For second state, 1.5pv = const = P1V1

V2 = (202600*2.146)/(1.5*202600)

V2 = 1.43m^3

Volume change = 2.146 - 1.43 =0.715m^3

Word done = pressure* volume change

W = 202600*0.716 = 145061.6J

= 145.061kJ

Using V1/T1 = V2/T2

T2 = V2T1/V1

=(1.43*303)/2.146 = 201.9k

For internal energy U

U = nCv*(T2 - T1)

*CV is the heat capacity at const. vol approximately 0.718J mol^-1 k^-1

U = 172.65*0.718*(201.9-303)

U = -12532.6J = -12.532kJ

The -ve means the system lost internal energy.

Q = U+W = total heat energy of system

Q = - 12.532+145.061 = 132.52 kJ

7 0
4 years ago
The asymmetric roof truss is of the type used when a near normal angle of incidence of sunlight onto the south-facing surface AB
marysya [2.9K]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
4 years ago
The mean of 10 numbers is 9, then the sum (total) of these numbers will be​
qwelly [4]

Answer:

90

Explanation:

mean is basically taking the sum of all numbers and then dividing the sum with the number of all given numbers..

here, the mean is 9, total numbers are 10.. so the sum will be 9 multiplied by 10, that is 90.

5 0
3 years ago
Read 2 more answers
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