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Dimas [21]
3 years ago
14

What impact does modulus elasticity have on the structural behavior of a mechanical design?

Engineering
1 answer:
devlian [24]3 years ago
8 0

Answer with Explanation:

The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:

1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.

2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.

3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity  the resistance it offer's to shear forces and the torques is more.

While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.

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The products of combustion from burner are routed to an industrial application through a thin-walled metallic duct of diameter D
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Answer:

Start by calculating the heat lost falling from Tm,i to Tm,o

As a first approximation use an average Cp of (Tm,i + Tm,o)/2 and an average density at (Tm,i + Tm,o)/2

Cp air at (Tm,i + Tm,o)/2 ----(1)

Density of air at (Tm,i + Tm,o)/2 : PV=nRT, V=T1/T2, air density at 293K = 1.204kg/m^3

Air density at (Tm,i + Tm,o)/2 1.204×293/(Tm,i + Tm,o)/2

Energy lost per kg for (Tm,i - Tm,o)K drop: (Tm,i - Tm,o)K × (1) =

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Energy lost per kg drop/t seconds = 24.3 kW

Volume occupied by 1kg air at at mean tempera ture = 1/Air density at mean temperature

Length of pipe ( Di m diameter) needed to hold Volume occupied by 1 kg of air at mean temperature :

Cross section = π/4 Di^2, Volume occupied by 1 Kg of air at mean temp. ÷ π/4Di^2

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Q/A=k (delta T)/ thickness,

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Explanation:

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3 years ago
A 10-m-long countercurrent-flow heat exchanger is being used to heat a liquid food from 20 to 808C. The heating medium is oil, w
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Answer:

mlf=0.5038kg/s

Explanation:

a. Please kindly check attachment for the step by step solution

b. B) in concurrent flow heat exchanger same exit temperatures for both fluids cannot be obtained. Since tho<tco

Case ii) only liquid food is considered tci=20 &tco=80

Cp=3.9kj/kgk

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Determine the following for a south facing surface at 30� slope in
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Answer:

z=60.32°, i=0.32°, Beam Radiation = 1097.2 W/m²,  Id = 94.2 W/m², Ir=14.1W/m², total radiation = 1205.4 W/m², Local time=1:21PM

Explanation:

A. Zenith Angle:

As we know that,

Zenith angle=z=90⁰-α=L(latitude)=29.68⁰

Another way to do it is to find α first,

At solar time hour angle is 0⁰. So, solar altitude becomes equal to latitude which could be written as

sinα=cosL

α=sin⁻¹(cosL)=sin⁻¹(cos29.68⁰)=60.32°

B. Angle of incidence:

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First we need to calculate extra terrestrial radiations

Iext.=1353[1+0.034cos(360n/365)]

where n=264

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Now,

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where β=30°

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E. Reflected Radiations:

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= (0.2)(1097.1)(sin60.32+0.092)sin²(30/2)

= 14.1W/m²

F. Total Radiation:

total radiation = beam radiation + diffuse radiation + reflected raddiation

= 1205.4 W/m²

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LST= ST-ET-(lₓ-l(local))4min/₀

     = 12:00-7.9min-(75°-82.27°)4min/₀

     =12:21PM

Local time

LDT=LST+=12:21+1:00=1:21PM

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