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2 years ago

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Thirty-six grams of air in a piston–cylinder assembly undergo a Stirling cycle with a compression ratio of 7.5. At the beginning

of the isothermal compression, the pressure and volume are 1 bar and 0.03 m3, respectively. The temperature during the isothermal expansion is 1200 K. Assuming the ideal gas model and ignoring kinetic and potential energy effects, determine the net work, in kJ.

1 answer:

MatroZZZ [7]2 years ago

8 0

Answer:

W = 6044.709 J

Explanation:

Given

Mass of air: m = 36.00 g

P₁/P₂ = 7.5

P₁ = 1 bar = 1 bar = 10⁵ Pa

V₁ = 0.03 m³

T = 1200 K

R = 8.314472 J/(mol*K)

W = ?

We have the get n (number of moles) as follows:

P*V = n*R*T ⇒ n = P₁*V₁ / (R*T₁)

⇒ n = (10⁵ Pa)*(0.03 m³) / (8.314472 J/(mol*K)*1200 K)

⇒ n = 0.3 mol

Then, we apply the equation

W = n*R*T₁*Ln (P₁/P₂)

⇒ W = (0.3 mol)*(8.314472 J/(mol*K))*(1200 K)*Ln (7.5)

⇒ W = 6044.709 J

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Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b

grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975K W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5 W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5 dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2 dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5 = 3.48K(0.1)^-0.5 W/m^1.5

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5 = 3.48K(1)^-0.5 W/m^1.5

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975K W/m^1.8

3 0

2 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

2 years ago

You should be extra careful during the hours of sunrise, sunset, and nighttime because _____. A. of increased law enforcement ac

torisob [31]

Answer:

B, its the only valid answer

5 0

2 years ago

A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$

Explanation:

Given data:

P_1 power = 20 dBm = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor $= 10 log \frac{P_1}{P_f}$

solving for final power

$20 = 10 log\frac{P_1}{P_f}$

$2 = log \frac{P_1}{P_f}$

$100 = \frac{0.1}{P_f}$

$P_f = 0.001 watt = 0 dBm$

Directivity $D = 10 \frac{P_f}{P_b}$

$35 = 10 \frac{0.001}{P_b}$

$P_b = 3.162 \times 10^{-7} wattt$

output Power $= P_1 -P_f - P_b$

$= 0.1 - 0.001 - 3.162 \times 10^{-7}$

$P_O = 0.989 watt = 19.9 dBm$

6 0

2 years ago

Liquid oxygen is stored in a thin-walled, spherical container 0.75 m in diameter, which is enclosed within a second thin-walled,

nadezda [96]

Answer:

Explanation:

the solution is well stated

3 0

2 years ago