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1 year ago

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There are lots of examples of ideal gases in the universe, and they exist in many different conditions. In this problem we will

examine what the temperature of these various phenomena are. Give an expression for the temperature of an ideal gas in terms of pressure P, particle density per unit volume rho and fundamental constants.

1 answer:

elena-14-01-66 [18.8K]1 year ago

5 0

Answer:

P = ρRT/M

Explanation:

Ideal gas equation is given as follows generally:

PV = nRT (1)

P = pressure in the containing vessel

V = volume of the containing vessel

n = number of moles

R = gas constant

T = temperature in K

n = m/M

m = mass of the gas contained in the vessel in g

M = molar mass in g/mol

ρ = m/V

Density of the gas = ρ

Substituting for n in (1)

PV = mRT/M. (2)

Dividing equation (2) through by V

P = m/V ×RT/M

P = ρRT/M

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Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

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1 year ago

A 26 cm object is 18 cm in front of a plane mirror. A ray of light strikes the object and is reflected off the mirror at a 42-de

matrenka [14]

Answer:

42 degrees, virtual image, same size as the object (26 cm)

Explanation:

The law of reflection states that:

- When a ray of light is incident on a flat surface (such as the plane mirror), the angle of reflection is equal to the angle of incidence

So, since in this case the angle of incidence is 42 degrees, the angle of reflection is also 42 degrees.

Moreover, the image formed by a plane mirror is always:

- Virtual (on the same side as the object)

- The same size as the object

So in this case, since the object's size is 26 cm, the image's size is also 26 cm.

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2 years ago

A particle moving in the x direction is being acted upon by a net force F(x)=Cx2, for some constant C. The particle moves from x

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Answer:

Change in kinetic energy is ( 26CL³)/3

Explanation:

Given :

Net force applied, F(x) = Cx² ....(1)

Displacement of the particle from xi = L to xf = 3L.

The work-energy theorem states that change in kinetic energy of the particle is equal to the net amount of work is done to displace the particle.

That is,

ΔK = W = ∫F·dx

Substitute equation (1) in the above equation.

ΔK = ∫Cx²dx

The limit of integration from xi = L to xf = 3L, so

$\Delta K=\frac{C}{3}(x_{f} ^{3} - x_{i} ^{3})$

Substitute the values of xi and xf in the above equation.

$\Delta K=\frac{C}{3}((3L) ^{3} - L ^{3})$

$\Delta K=\frac{C}{3}\times26L^{3}$

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1 year ago

Read 2 more answers

1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

azamat

Answer:

$\dot{W} = 339.84 W$

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

$\dot{W} =\dot{m}\frac{V^{2}}{2}$

here $\dot{m}$ is mass flow rate and given as

$\dot{m} = \rho*Q$

$\dot{W} =\rho*Q\frac{V^{2}}{2}$

Putting all value to get minimum power

$\dot{W} =1.18*9*\frac{8^{2}}{2}$

$\dot{W} = 339.84 W$

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1 year ago