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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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determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004
Afina-wow [57]

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 ,  To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

6 0
3 years ago
Explain the difference in the heat transfer modes of conduction and convection.
LiRa [457]

Explanation:

Conduction is the heat transfer through a stationary matter by the physical contact.  

For example, the heat transferred between electric burner of stove and bottom of pan is transferred by the process of conduction.

Convection is heat transfer by macroscopic movement of the fluid. The particles of the fluid carry the current within themselves.  

For example, the water is the pot is warmed overall by heat transferred by the process of convection.

4 0
3 years ago
A First Stage in a turbine receives steam at 10 MPa, 800 C with an exit pressure of 800 KPa. Assume the stage is adiabatic and r
9966 [12]

Answer

given,

P₁  = 10 MPa                 T₁ = 800

P₂ = 800 KPa               T₂ = ?

Using formula

\dfrac{T_2}{T_1} = (\dfrac{P_2}{P_1})^{\dfrac{n-1}{n}}

For steam   n = 1.33                    

\dfrac{T_2}{800+273} = (\dfrac{800\times 10^{3}}{10\times 10^{6}})^{\dfrac{1.33-1}{1.33}}

T₂ = 573.368 K                    

T₂ = 573.368 - 273 = 300.368 °C

W = \dfrac{P_1V_1-P_2V_2}{n-1}

          =\dfrac{mR(T_1-T_2)}{n-1}

          =\dfrac{1\times 0.287\times (800 - 300.368)}{1.33-1}

      W = 434.53 kJ/kg

7 0
3 years ago
A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory
Molodets [167]

Answer:

Explanation:

Mean temperature is given by

T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}

Tmean = (Ti + T∞)/2

T_mean = 107.5^{0}

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

\mu = 221.6 × 10⁻⁷N.s/m²

\kappa = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α  = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

              = 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/\mu

Substituting for Pv

Re = 4m/(πD\mu)

     = (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

     = 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0
3 years ago
Oil system cleaning products should not use solvents problem
pantera1 [17]
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7 0
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