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algol [13]
4 years ago
5

Consider the series solution, Equation 5.42, for the plane wall with convection. Calculate midplane (x* = 0) and surface (x* = 1

) temperatures θ* for Fo = 0.1 and 1, using Bi = 0.1, 1, and 10. Consider only the first four eigenvalues. Based on these results, discuss the validity of the approximate solutions, Equations 5.43 and 5.44.

Engineering
1 answer:
VashaNatasha [74]4 years ago
8 0

Answer:

We conclude that the approximate series solution (with only one eigein value) provides systematically high results but by less than 1.5%, for the biot number range from 0.11 to 10. See attached image.

Explanation:

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A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is appli
Scrat [10]

Answer:

X_t=2.17391304*10^{-4}

X_r=2.89855072*10^{-4}

e_t=0.0026

e_r=0.0035

Explanation:

From the question we are told that:

Dimension 12*12

Thickness l_t=5mm=5*10^-3

Normal tensile force on top side F_t= 15kN

Normal tensile force on right side  F_r= 20kN

Elastic modulus, E=115Gpap=>115*10^9

Generally the equation for Normal Strain X is mathematically given by

 X=\frac{Force}{Area*E}

Therefore

For Top

 X_t=\frac{Force_t}{Area*E}

Where

 Area=L*B*T

 Area=12*10^{-2}*5*10^{-3}

 Area=6*10^{-4}  

 X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}

 X_t=2.17391304*10^{-4}

For Right sideX_r=\frac{Force_r}{Area*E}

Where

Area=L*B*T

 Area=12*10^{-2}*5*10^{-3}

 Area=6*10^{-4}  

 X_r=2.89855072*10^{-4}

 X_r=2.89855072*10^{-4}

Generally the equation for elongation is mathematically given by

 e=strain *12

For top

 e_t=2.17391304*10^{-4}*12

 e_t=0.0026

For Right

 e_r=2.89855072*10^{-4} *12

 e_r=0.0035

5 0
3 years ago
What does Clay say will happen if the system is rejected?
juin [17]

Answer:

the nation will suffer terrible consequences

Explanation:

I did that and got it right

6 0
3 years ago
Read 2 more answers
A sinusoidal voltage source produces the waveform, v t = 1 + cos 2πft. Design a system with v t as its input such that an LED wi
DerKrebs [107]

Answer:

See explaination

Explanation:

LM358 is the useful IC which works as buffer. It enables circuit to remove overloading effect on each other. Image is in attachment.

We can define a light-emitting diode (LED) as a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons

See attached file for detailed solution of the given problem.

3 0
3 years ago
(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and pr
Gnoma [55]

Answer:

See explanation below

Explanation:

Hypo-eutectoid steel has less than 0,8% of C in its composition.

It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.

Ferrite has a higher tensile strength than cementite but cementite is harder.

Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:

Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel

Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel

Hyper-eutectoid steel is harder than Hyper-eutectoid steel

Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.

When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because

1. It is harder

2. It has low cost

3. It is lighter

When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because

1. It is ductile

2. It has high tensile strength

3. It is durable

7 0
3 years ago
Read 2 more answers
9 Consider fully developed conditions in a circular tube with constant surface temperature Ts Tm. Determine whether a small- or
mojhsa [17]

Answer:

Hello your question is incomplete attached below is the complete question

answer:

Considering Laminar flow

Q ( heat )  will be independent of diameter

Considering Turbulent flow

The heat transfer will increase with decreasing "dia" for the turbulent

heat transfer = f(d^-0.8 )

Explanation:

attached below is the detailed solution

Considering Laminar flow

Q ( heat )  will be independent of diameter

Considering Turbulent flow

The heat transfer will increase with decreasing "dia" for the turbulent

heat transfer = f(d^-0.8 )

6 0
3 years ago
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