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Komok [63]
3 years ago
14

Various factors to be considered in deciding the factor of safety?

Engineering
1 answer:
Eduardwww [97]3 years ago
4 0

Answer with Explanation:

There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:

1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

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The major resisting force in gravity dam? ​
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The self weight of dam.

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A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
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Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

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Work = (8000 - 6080)/ -0.3

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A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
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Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

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specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

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W = 6927.415 kW

power developed by the turbine = 6927.415 kW

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