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2 years ago

What are the constraints of the problem for thermoforming

Engineering

1 answer:

Inga [223]2 years ago

6 0

Answer:

here are four options so you can choose the ones of your choice.

Explanation:

1. warpage

2 Dimensional Inconsistencies

3 Part Thickness Inconsistencies

4 Lack of Detail in Part Geometry and Aesthetics

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Plot the function for . Notice that the function has two vertical asymptotes. Plot the function by dividing the domain of x into

elena-s [515]

This is a very very difficult one for me, let me get back to you with the proper answer.

8 0

3 years ago

(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1

olya-2409 [2.1K]

Answer:

The answer is " $4.35 \times 10^{-3}\ mm$ and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

$y = yo + \frac{k}{\sqrt{x}}$

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.

Replacing the equation for each condition:

$y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k$

People can get yo = 275 MPa with both equations and k= 15.5 Mpa $mm^{\frac{1}{2}}$ .

To substitute the answer,

$310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm$

In point b:

The equation is $\sigma y = \sigma 0 + k y d^{\frac{1}{2}}$

equation is:

$75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\$

by putting the above value in the formula we get the $\sigma y$ value that is= 157.5 MPa

5 0

3 years ago

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

3 years ago

Describe at least one way you take advantage or could take advantage of each of the different forms of energy

Scorpion4ik [409]

Answer:

Kinetic energy can be used to develop electric energy which can be used as electricity.

Explanation:

The kinetic energy can be harnessed; much like some hydro power technologies harness water movement. A way to convert this kinetic energy into electric energy is through piezoelectric. By applying a mechanical stress to a piezoelectric crystal or material an electric current will be created and can be harvested.

Kinetic energy is also generated by the human body when it is in motion. Studies have also been done using kinetic energy and then converting it to other types of energy, which is then used to power everything from flashlights to radios and more.

4 0

3 years ago

Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?

Kisachek [45]

Answer:

the current consumed is 3.3 A

Explanation:

Given;

resistance, R = 30 ohms

inductance, L = 200 mH

Voltage supply, V = 230 V

frequency of the coil, f = 50 Hz

impedance, Z = 69.6 Ohms

The current consumed is calculated as;

$I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A$

Therefore, the current consumed is 3.3 A

4 0

3 years ago