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Mariana [72]
3 years ago
7

What are the benefits of using the engineering design process

Engineering
1 answer:
Shalnov [3]3 years ago
7 0

Answer:

Some of the benefits are tangible for they are visible in the design and production process, while the other benefits are intangible which may not be visible directly but result in improvement in the quality of product, better control over designing and production process, reduction of stress on the designers etc.

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A thin 20-cm*20-cm flat plate is pulled at 1m/s horizontally through a 4-mm thick oil layer sandwiched between two stationary pl
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A circular section of material is tested. The original specimen is 200 mm long and has a diameter of 13 mm. When loaded to its p
n200080 [17]

Answer:

modulus of elasticity = 100.45 Gpa,

proportional limit = 150.68 N/mm^2.

Explanation:

We are given the following parameters or data in the question as;

=> "The original specimen = 200 mm long and has a diameter of 13 mm."

=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."

=> " The total axial load is 20 kN"

Step one: Calculate the area

Area = π/ 4 × c^2.

Area = π/ 4 × 13^2 = 132.73 mm^2.

Step two: determine the stress induced.

stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.

Step three: determine the strain rate:

The strain rate = change in length/original length = 0.3/ 200 = 0.0015.

Step four: determine the modulus of elasticity.

modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.

Step five: determine the proportional limit.

proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.

7 0
3 years ago
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Define Plastic vs elastic deformation.
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Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

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3 years ago
1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diame
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3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

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2 years ago
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