One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal
yst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher-temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15-nm-diameter silicon carbide nanowire onto a silicon carbide sururface. The surface is maintained at a temperature of Ts = 2400 K and the particular liquid catalyst that is used must be maintained in the range 2400 K ≤ Tc ≤ 3000 K to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by h = 105 W/m2.K and T[infinity] = 8000 KT. Assume properties of the nanowire are the same as for bulk silicon carbide.
1 answer:
Answer: maximum length of the nanowire is 510 nm
Explanation:
From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K
Thermal conductivity of silicon carbide k = 30 W/m.K
Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m
lets consider the equation for the value of m
m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )
m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04
now lets find the value of h/mk
h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353
lets consider the value θ/θb by using the equation
θ/θb = (T - T∞) / (T - T∞)
θ/θb = (3000 - 8000) / (2400 - 8000)
= 0.893
the temperature distribution at steady-state is expressed as;
θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]
so we substitute
0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]
L = 510 × 10⁻⁹m
L = 510 nm
therefore maximum length of the nanowire is 510 nm
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Answer:
1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.
2) The muzzle velocity of the ball is approximately 5.272 meters per second.
3) The maximum height of the ball is 1.417 meters.
Explanation:
1) Which of the following statements are true?
a) Mechanical energy is conserved because no dissipative forces perform work on the ball.
True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.
b) The forces of gravity and the spring have potential energies associated with them.
False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.
c) No conservative forces act in this problem after the ball is released from the spring gun.
False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.
2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:
(Eq. 1)
Where:
,
- Initial and final elastic potential energies of spring, measured in joules.
,
- Initial and final translational kinetic energies of the ball, measured in joules.
After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:
And the final velocity is cleared:
(Eq. 2)
Where:
,
- Initial and final velocities of the ball, measured in meters per second.
- Spring constant, measured in newtons per meter.
- Mass of the ball, measured in kilograms.
,
- Initial and final position of spring, measured in meters.
If we know that ,
,
,
and
, the muzzle velocity of the ball is:
The muzzle velocity of the ball is approximately 5.272 meters per second.
3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:
(Eq. 3)
Where:
,
- Initial and gravitational potential energies of the ball, measured in joules.
,
- Initial and final translational kinetic energies of the ball, measured in joules.
After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:
(Eq. 4)
,
- Initial and final heights of the ball, measured in meters.
,
- Initial and final velocities of the ball, measured in meters per second.
- Gravitational acceleration, measured in meters per square second.
If we get that ,
,
and
, the maximum height of the ball is:
The maximum height of the ball is 1.417 meters.