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4 years ago

What different tests did the team perform to come up with a workable design?

1 answer:

5 0

They ran different shapes and materials through a wind tunnel to see which shape and material would decrease energy output so that it takes in equal COthan it puts out.

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Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr

katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

$\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}$

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

$\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m$

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

$n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times$

c) By using the equation of the ideal gases you obtain:

$PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules$

5 0

3 years ago

Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

3 years ago

Explain how feedback control is used to<br> adjust robotic movements.

LuckyWell [14K]

Answer:

Feedback control of arm movements using Neuro-Muscular Electrical Stimulation (NMES) combined with a lockable, passive exoskeleton for gravity compensation

6 0

2 years ago

What is the governing ratio for thin walled cylinders?

Ann [662]

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words: