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3 years ago

What are the atomic binding force and energy? how do they relate to materials strength and thermal stability.

Engineering

1 answer:

Elanso [62]3 years ago

7 0

Answer:

As we know that every molecule is attached by a strong force .The force required to disassemble the atoms is know as atomic binding force or we can say that the force required to disassemble the electron from atoms is known as binding force.On the other hand the energy require to doing this is known as atomic binding energy.

If the binding force is high then it will become difficult to disassemble thermally as well as mechanically.So we can say that it have direct relationship with materials strength and thermal stability.

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The end of a cylindrical liquid cryogenic propellant tank in free space is to be protected from external (solar) radiation by pl

prohojiy [21]

If the temperature of the shield is 338 kelvin. Then the heat flux through the tank will be 25.3 Watt per square meter.

The increase in heat energy movement through a particular surface is known as heat flux, and the heat flux density is the absolute temperature per unit area.

Assume the view factor between the tank and the shield is unity; all surfaces are diffuse and gray, and the surroundings are at 0 K.

It is given that T= 100 K, ε₁ = ε₂ = 0. 10, $\varepsilon_t$ = 0.20, and GS= 1250 W/m².

Then we have

The temperature of the shield will be

$\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dot{q}_{ST} = 0$ ...1

and

$\rm q''_{12}=\dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}}$ ...2

Then from equations 1 and 2, we have

$\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}} = 0$

Then the value of $\rm T_s$ will be

$\rm T_s =\left [ \dfrac{\alpha _sGs+\left ( \dfrac{\sigma T_1^4}{\frac{1}{\varepsilon _1}+\frac{1}{\varepsilon _2} - 1} \right )}{\sigma \left ( \varepsilon _1 + \dfrac{1}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2}-1} \right )} \right ] ^{\dfrac{1}{4}}$

Put all the values, then we have

$\rm T_s = \left [ \dfrac{0.05 \times + \left ( \dfrac{\sigma (100)^4}{\frac{1}{0.1}+\frac{1}{0.05}-1} \right )}{\sigma \left ( 0.05 + \dfrac{1}{\frac{1}{0.1}+\frac{1}{0.05} - 1} \right )} \right ]^{\dfrac{1}{4}} \\\\\\T_s = 338 \ K$

Then the heat flux will be

$\rm q"_{ST}=\dfrac{\sigma (T_S^4 - T_t^4)}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2} - 1} \\\\\\q"_{ST}=\dfrac{5.67 \times 10^{-8}(388^4-100^4)}{\frac{1}{0.1}+\frac{1}{0.05}-1}\\\\\\q"_{ST} = 25.3 \ W/m^2$

More about the heat flux link is given below.

brainly.com/question/12913016

#SPJ1

8 0

3 years ago

The hull of a vessel develops a leak and takes on water at a rate of 57.5 gal/min. When the leak is discovered the lower deck is

leva [86]

Answer:

It will be around 146,27 min since the pump is turned on until the deck is clear of the water.

Explanation:

When the leak is discovered and the pump is turned on, the lower deck is already submerged and the leak is not fixed; then, in order to have the deck clear of water, the bilge pump has to remove the <em>accumulated water </em>( $V_{0}$ ) and the <em>water that is taking on</em> ( $r_{in}*t$ ) through the leak. We can represent this mathematically as follow:

$V_{0} +r_{in} *t-r_{out}*t=0$ <em>Equation 1</em>

Where:

$V_{0}$ : is the accumulated water when the leak was discovered

$r_{in}$ : is the takes on rate through the leak = 57.5 gal/min

$r_{out}$ : is the removing rate of the bilge pump = 73.8 gal/min

t= is the time since the pump is turned on until the deck is clear of water.

To calculate the accumulated water ( $V_{0}$ ), we will model the lower deck as a flat-bottomed container with a bottom surface area of 510 $ft^{2}$ and straight vertical sides. Knowing that the level submerged is 7.5 inches, and performing the corresponding unit conversions, we obtain:

$V_{0}$ = bottom surface area * lever submerged

$V_{0}= 510ft^{2}*7.5 in*\frac{1ft}{12in}=318.75 ft^{3}*7.48\frac{gal}{1ft^{3}}=2384.25 gal$ <em>Equation 2</em>

Solving equation 1 for time (t), and replacing the value obtained in equation 2, we get:

$t=\frac{V_{0}}{(r_{out}-r_{in})} =\frac{2384.25 gal}{(73.8-57.5)gal/min}=$ 146,27 min

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3 years ago

An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40

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Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

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g100num [7]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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