PHYSICS COLLEGE

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer 1

Answer:

Answer:

$\dot W_(in) = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_(u)\cdot T$

$P\cdot V = (m)/(M)\cdot R_(u)\cdot T$

$\rho = (P\cdot M)/(R_(u)\cdot T)$

$\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))$

$\rho = 1.2\,(kg)/(m^(3))$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )$

$\dot m = 0.18\,(kg)/(s)$

The work input is:

$\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)$

$\dot W_(in) = 49.386\,kW$

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A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18m to the plate before it is caught. How long does the ball stay in the air?

A.0.41 sec
B.41 sec
C.4.1 sec
D.4 sec

Answers

A horizontal baseball pitch is launched at 44 m/s. The ball will stay for 4.1 sec (approx) in the air. Hence, option C is correct.

What is Velocity?

The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.

Its SI unit is represented as m/s, and it is a vector quantity, it means that it has both magnitude and direction.

According to the question, the given values are :

Initial Velocity, u = 44 m/s,

Distance travelled, s = 18 m and,

Final velocity, v = 0.

Use equation of motion :

v = u + at

0 = 44 + (-9.8)t

t = 44 / 9.8

t = 4.3 (approx)

Hence, the time for which the ball stay in the air is 4.1 sec (approx).

To get more information about velocity :

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Answer:

a 0.41

plug number into equation

On a map each inch represents 25 miles what is the length of a highway if it is 6 inches long on a map

Answers

Answer:

150 hope this helps

Explanation:

Answer:

150

Explanation:

Why does an astronaut in a spacecraft orbiting Earthexperience
a feeling of weightlessness?

Answers

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

Learn more about energy of photons here:

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(a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 ? 106 V/m). The plates are separated by2.98 mm and a potential difference of 5575 V is applied. (b) How close together can the plates be with this applied voltage without exceeding the breakdown strength?

Answers

Answer:

(a) 1.87×10⁶ V/m

(b) 1.12 mm closer

Explanation:

(a)

Electric Field = Electric potential/distance.

E = V/d ................... Equation 1

Where E = Electric Field, V = Electric potential, d = distance.

Given: V = 5575 V, d = 2.98 mm = 0.00298 m.

Substitute into equation 1

E = 5575/0.00298

E = 1.87×10⁶ V/m

(b)

without exceeding the breakdown strength,

make d the subject of equation 1

d = V/E.............. Equation 2

Given: E = 3×10⁶ V/m, V = 5575 V

Substitute into equation 2

d = 5575/3000000

d = 1.86 mm.

the plate will be = 2.98-1.86 = 1.12 mm closer

According to American Heart Association, your target heart rate is__________. Group of answer choices

A .all of the above

B. 85% - 95% of your max heart rate

C. 20%-30% of your max heart rate

D. 50%-85% of your max heart rate

Answers

Answer:

Explanation:

A normal heart rate is from 85-95% the other heart rate is not normal because your heart is beating more than normal

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