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3 years ago

How are the particles moving and and arranged in a gas?

Engineering

1 answer:

alexandr402 [8]3 years ago

7 0

Answer:

The particles in gas do not have any particular arrangement and there are very, very weak forces between them. So, the particles in a gas can easily move around and fill the shape of the container they are in, meaning they have no fixed shape.

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A school is playing $0.XY per kWh for electric power. To reduce its power bill, the school installs a wind turbine with a rated

Semmy [17]

Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved

answer : Electric power generated = 216 * 10^6 kJ

money saved = $0.XY * 60000 kwh

Explanation:

<u>Calculating the amount of electric power generated by wind turbine</u>

power generated = ( 30 * 2000 ) kWh = 60000 kWh

Electric energy generated = 60000 kWh * 3600 kJ = 216 * 10^6 kJ

<u>Calculate money saved by school per year </u>

$0.XY * 60000 kwh

5 0

3 years ago

A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as

Ulleksa [173]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ = 254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa

4 0

3 years ago

An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°

Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

$D = 7.5 \times 10^{-11} m^2/s$

for 0.225% carbon concentration following formula is used

$\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})$

where, erf stand for error function

$\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248$

$0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})$

$erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248$

$erf(\frac{x}{2\sqrt{DT}}) = 0.751$

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

$\frac{x}{2\sqrt{DT}} = 0.815$

$x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)$

$x = 2.39\times 10^{-3} m$

x = 0.002395 mm

8 0

3 years ago

It is appropriate to use the following yield or failure criterion for ductile materials (a) Maximum shear stress or Tresca crite

Nataly [62]

Answer:

(b)Distortion energy theory.

Explanation:

The best suitable theory for ductile material:

(1)Maximum shear stress theory (Guest and Tresca theory)

It theory state that applied maximum shear stress should be less or equal to its maximum shear strength.

(2)Maximum distortion energy theory(Von Mises henkey's theory)

It states that maximum shear train energy per unit volume at any point is equal to strain energy per unit volume under the state of uni axial stress condition.

But from these two Best theories ,suitable theory is distortion energy theory ,because it gives best suitable result for ductile material.

6 0

3 years ago

In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had

kipiarov [429]

Answer:

Moisture/ water content w = 26%

Void ratio , e = 0.73

Explanation:

Initial mass of saturated soil w1 = mass of soil - weight of container

= 113.27 g - 49.31 g = 63.96 g

Final mass of soil after oven w2 = mass of soil - weight of container

= 100.06 g - 49.31 g = 50.75

Moisture /water content, w = $\frac{w1-w2}{w2}$ = $\frac{63.96-50.75}{50.75}$ = 0.26 = 26%

Void ratio = water content X specific gravity of solid

= 0.26 X 2.80 =0.728

5 0

3 years ago