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Hunter-Best [27]
2 years ago
10

(1) Estimate the specific volume in cm3 /g for carbon dioxide at 310 K and (a) 8 bar (b) 75 bar by the virial equation and compa

re to the experimental values of 70.58 and 3.90, respectively.
(2) Estimate the specific volume in cm3/g for carbon dioxide at 310 K and (a) 8 bar (b) 75 bar by the compressibility factor charts and compare to the experimental values2 of 70.58 and 3.90, respectively.
Engineering
1 answer:
ludmilkaskok [199]2 years ago
3 0

Answer:

70.66 cm^3

The specific volume for P = 70.66  is within 1% of the experimental value while the viral equation will be inaccurate when the second viral coefficient is used )

Explanation:

Viral equation : Z = 1 + Bp + Cp^2 + Dp^3 + -----

Viral equation can also be rewritten as :

Z = 1 + B ( P/RT )

B ( function of time )

Temperature = 310 K

P1 = 8 bar

P2 = 75 bar

<u>Determine the specific volume in cm^3 </u>

V = 70.66 cm^3

<u>b) comparing the specific volumes to the experimental values </u>

70.58 and 3.90

The specific volume for P = 70.66  is within 1% of the experimental value while the viral equation will be inaccurate when the second viral coefficient is used )

attached below is the detailed solution

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The current through a 10-mH inductor is 10e−t∕2 A. Find the voltage and the power at t = 8 s.
NNADVOKAT [17]

Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L\frac{di(t)}{dt}             -----------(i)

Where;

L = inductance of the inductor

i(t) = current through the inductor at a given time

t = time for the flow of current

From the question:

i(t) = 10e^{-t/2}A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = (10*10^{-3})\frac{d(10e^{-t/2})}{dt}

Solve the differential

v(t) = (10*10^{-3})\frac{-1*10}{2} (e^{-t/2})

v(t) = -0.05 e^{-t/2}

At t = 8s

v(t) = v(8) = -0.05 e^{-8/2}

v(t) = v(8) = -0.05 e^{-4}

v(t) = -0.05 x 0.223

v(t) = -0.01116V

(b) To get the power, we use the following relation:

p(t) = i(t) x v(t)

Power at t = 8

p(8) = i(8) x v(8)

i(8) = i(t = 8) = 10e^{-8/2}

i(8) = 10e^{-4}

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

7 0
3 years ago
Read 2 more answers
Javier’s class visited a power plant near his city, and they learned how it produced electricity. What does this form of power d
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The electricity is generated by the burning of fuel in the power plant that increases the kinetic energy of water vapor which rotates the turbine and shaft of the generator.

<h3>What is a power plant?</h3>

A power plant is a place where heavy machinery objects are installed to produce electricity.

The power is generated in the power plant with the help of the burning of fuel that increases the temperature of the boiler. There are so many tubes in which water is present. Water absorbs heat and increases its kinetic energy. And this kinetic energy help to rotate the turbine and the shaft of the generator is attached to the turbine as result electricity is generated.

More about the power plant link is given below.

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2 years ago
Engineered lumber should not be used for
Dimas [21]

Answer:

Composite panel garage doors

Explanation:

8 0
2 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
Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_to_celsius, and modifying th
aleksandr82 [10.1K]

Answer:

# kelvin_to_celsius function is defined

# it has value_kelvin as argument

def kelvin_to_celsius(value_kelvin):

   # value_celsius is initialized to 0.0

   value_celsius = 0.0

   

   # value_celsius is calculated by

   # subtracting 273.15 from value_kelvin

   value_celsius = value_kelvin - 273.15

   # value_celsius is returned

   return value_celsius

   

# celsius_to_kelvin function is defined

# it has value_celsius as argument

def celsius_to_kelvin(value_celsius):

   # value_kelvin is initialized to 0.0

   value_kelvin = 0.0

   

   # value_kelvin is calculated by

   # adding 273.15 to value_celsius

   value_kelvin = value_celsius + 273.15

   # value_kelvin is returned

   return value_kelvin

   

value_c = 0.0

value_k = 0.0

value_c = 10.0

# value_c = 10.0 is used to test the function celsius_to_kelvin

# the result is displayed

print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')

value_k = 283.15

# value_k = 283.15 is used to test the function kelvin_to_celsius

# the result is displayed

print(value_k, 'is', kelvin_to_celsius(value_k), 'C')

Explanation:

Image of celsius_to_kelvin function used as guideline is attached

Image of program output is attached.

4 0
3 years ago
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