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3 years ago

Identify three material considerations an engineer would need to consider when working on a design process.

Engineering

1 answer:

Anika [276]3 years ago

4 0

Answer:

Three material considerations are;

1. Identify and appraise the attainment of the goal of the with the design specification

2. Ascertain the required load the product being designed will experience and the suitability of the design specification to that load

3. Review the producibility of the design to ensure that it can be produced with the available technology

Explanation:

1. The appraisal of the design includes the consideration of the factors of the design and the building of reliability and efficiency into the design from the beginning

2. Ascertain if the product will require toughness, elasticity, and if will be subject to sudden or repeated loading conditions

3. Ensure that the design can be readily produced with the accessible manufacturing equipment during the conceptualization stage of the design.

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Alla [95]

Answer:

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3 years ago

A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above

creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation $N_{o} = N_{i} * \frac{E_{f}-E_{i} }{KT}$

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, $N_{d} = 1.65*10^{16} - 10^{16} \\ = 6.5 * 10^{15} per cm cube$

5 0

2 years ago

Read 2 more answers

Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

3 years ago

A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum

antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

Density = 61 veh/km.

Speed = 59 km/hr.

Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

$q_{max}=\frac{V_f \times K_i}{4}$

<u>Where:</u>

$V_f$ is the free flow speed.

$K_i$ is the Jam density.

In order to calculate the free flow speed, we would use this formula:

$V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr$

Substituting the parameters into the model, we have:

$q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}$

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

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2 years ago

The maximum stress that a bar will withstand before failing is called • Rapture Strength • Yield Strength • Tensile Strength • B

konstantin123 [22]

Answer: Rupture strength

Explanation: Rupture strength is the strength of a material that is bearable till the point before the breakage by the tensile strength applied on it. This term is mentioned when there is a sort of deformation in the material due to tension.So, rupture will occur before whenever there are chances of failing and the material is still able to bear stresses before failing.

7 0

3 years ago