ENGINEERING COLLEGE

The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms in any proportion without altering the crystal structure except by expanding it slightly: this is an example of a binary solid solution series. For different compositions, the lines in the powder diffraction patterns are in slightly different positions, because of the cell expansion, but the overall pattern remains basically the same. The spacing of the lattice planes varies linearly with composition, and this can be used in a rapid and non- destructive method of analysis. a. The (062) reflection from olivine is strong and well resolved from other lines. Calculate d062 for an olivine that displays its (062) reflection at a Bragg angle of 37.21° (i.e., a diffraction angle of 74.42°) when x-rays with a wavelength of 0.1790 nm are used. b. The d062 spacing as measured accurately for synthetic materials is 0.14774 nm for Mg2SiO4 and 0.15153 nm for Fe2SiO4. What would be the approximate composition, expressed in mol.% Mg2SiO4, of an olivine material for which do62 has the value obtained in part 2.1 above?

Answers

Answer 1

Answer:

Answer:

The answer is "0.147 nm and 99.63 mol %"

Explanation:

In point (a):

$\to nk1 = 062$

$\to \text{Bragg angle}$ $\theta =37.21^(\circ)$

$\to \text{diffraction angle}$ $2 \theta = 74.42^(\circ)$

$\to \lambda = 0.1790 nm$

find:

d(062)=?

formula:

$\to nx = 2d \sin \theta$

$\to d(062) = (1 * 0.1790^(\circ))/(2 * \sin 37.21^(\circ))\n$

$= (0.1790^(\circ))/(2 * 0.604738126)\n\n= (0.29599589)/(2)\n\n= 0.147 \n$

In point (b):

$\to Mg_2SiO_4\longleftrightarrow Fe_2SiO_4$

$d= 0.14774 \ \ \ \ \ olivine = 0.147 \ \ \ \ \ 0.15153$

formula:

$\to d=(a)/(√(n^2+k^2+i^2))\n$

that's why the composition value equal to 99.63 %

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What is the average distance in microns an electron can travel with a diffusion coefficient of 25 cm^2/s if the electron lifetime is 7.7 microseconds. Three significant digits and fixed point notation.

Answers

Answer: The average distance the electron can travel in microns is 1.387um/s

Explanation: The average distance the electron can travel is the distance an exited electron can travel before it joins together. It is also called the diffusion length of that electron.

It is gotten, using the formula below

Ld = √DLt

Ld = diffusion length

D = Diffusion coefficient

Lt = life time

Where

D = 25cm2/s

Lt = 7.7

CONVERT cm2/s to um2/s

1cm2/s = 100000000um2/s

Therefore D is

25cm2/s = 2500000000um2/s = 2.5e9um2/s

Ld = √(2.5e9 × 7.7) = 138744.37um/s

Ld = 1.387e5um/s

This is the average distance the excited electron can travel before it recombine

Describe how a feeler gauge can be used to assist in the adjustment of a spark plug electrode gap

Answers

Answer:

Explanation:

Adjusting the distance between the two electrodes is called gapping your spark plugs. You need a feeler gauge to gap your spark plugs properly If you're re-gapping a used plug, make sure that it's clean (gently scrub it with a wire brush)

hope this you

The hot water needs of an office are met by heating tab water by a heat pump from 16 C to 50 C at an average rate of 0.2 kg/min. If the COP of this heat pump is 2.8, the required power input is: (a) 1.33 kW (d) 10.2 kW (b) 0.17 kW (c) 0.041 kW

Answers

Answer:

option B

Explanation:

given,

heating tap water from 16° C to 50° C

at the average rate of 0.2 kg/min

the COP of this heat pump is 2.8

power output = ?

$COP = (Q_H)/(W_(in))\nW_(in) = (Q_H)/(COP)\nW_(in) = ((0.2)/(60)* 4.18* (50-16))/(2.8)\nW_(in) = 0.169$

the required power input is 0.169 kW or 0.17 kW

hence, the correct answer is option B

An inventor claims that he wants to build a dam to produce hydroelectric power. He correctly realizes that civilization uses a lot more electricity during the day than at night, so he thinks he has stumbled upon a great untapped energy supply. His plan is to install pumps at the bottom of the dam so that he can pump some of the water that flows out from the generators back up into the reservoir using the excess electricity generated at night. He reasons that if he did that, the water would just flow right back down through the generators the next day producing power for free. What is wrong with his plan?

Answers

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.

Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressure of 300 kPa is required to move the piston. Initially, the air is at 100 kPa and 27°C and occupies a volume of 0.4 m^3. A) Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K. Assume air has constant specific heats evaluated at 300 K.

Answers

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

for a rankine cycle with one stage of reheat between turbines, there are how many relevant pressures?

Answers

The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.

For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:

Boiler pressure (P1): This is the pressure at which the water is heated in the boiler before entering the first turbine.
High-pressure turbine outlet pressure (P2): This is the pressure at the outlet of the first turbine before the steam is sent to the reheater.
Reheat pressure (P3): This is the pressure at which the steam is reheated before entering the second turbine.
Low-pressure turbine outlet pressure (P4): This is the pressure at the outlet of the second turbine, which is also the condenser pressure.

To know more about Rankine cycle visit:

brainly.com/question/30985136

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