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just olya [345]
3 years ago
7

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12

bar, 400 K. Heat transfer occurs at a rate of 14 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Engineering
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

W = - 184.8 kW

Explanation:

Given data:

P_1 = 1.05 bar

T_1 = 300K

\dot V_1 = 84 m^3/min

P_2 = 12 bar

T_2 = 400 K

We know that work is done as

W = - [ Q + \dor m[h_2 - h_1]]

forP_1 = 1.05 bar,  T_1 = 300K

density of air is 1.22 kg/m^3 and h_1 = 300 kJ/kg

for P_2 = 12 bar, T_2 = 400 K

h_2 = 400 kj/kg

\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

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Explanation:

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5 0
3 years ago
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

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The pic shown can help to understand the question.

5 0
4 years ago
Which of the following is a possible unit of ultimate tensile strength?
levacccp [35]

Answer:

Newton per square meter (N/m2)

Explanation:

Required

Unit of ultimate tensile strength

Ultimate tensile strength (U) is calculated using:

U = \frac{Ultimate\ Force}{Area}

The units of force is N (Newton) and the unit of Area is m^2

So, we have:

U = \frac{N}{m^2}

or

U = N/m^2

<em>Hence: (c) is correct</em>

4 0
3 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
3 years ago
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