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Elena-2011 [213]
4 years ago
7

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

kinetic and potential energy effects are negligible. Assuming the ideal gas model for the air, determine the maximum theoretical pressure at the exit, in lbf/in.2 p2,max = lbf/in.2
Physics
1 answer:
vova2212 [387]4 years ago
5 0

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}

Here

P = Pressure

T = Temperature

\gamma =The ratio of specific heats. For air normally is 1.4.

Our values are given as,

P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K

Therefore replacing we have,

(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}

(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

Solving for P_2,

P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}

P_2 = 44.15Lbf/in^2

Therefore the maximum theoretical pressure at the exit is 44.15Lbf/in^2

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A cubic sample of a new kind of artificial tissue is subject to an increase in pressure of 160 kPa which results in a reduction
grin007 [14]

Answer:

0.82 MPa

Explanation:

the change in pressure 'σ'=160kPa

K= σ/∈Ф_v => σ/3∈Ф_L

K= 160/(3 x 0.065)

K=820 kPA=0.82 MPa

Thus,the bulk modulus of the tissue 'K' is 0.82 MPa

3 0
3 years ago
Air resistance is an example of what type of friction?
Sav [38]
Well, Air resistance is a special type of friction (you cannot classify it in other categories). That force of air-resistance is often observed to oppose the motion of the object,( like every other frictional forces)

Hope this helps!
4 0
4 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0
4 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
4 years ago
In models of magnetic and electric fields, why are field vectors depicted by arrows?
taurus [48]

The arrows in models of magnetic and electric fields show  both their magnitude and direction.

In Physics, a vector refers to a quantity that has both magnitude and direction. Hence, a vector always points in a given direction. The direction in which the arrow points is the direction of the vector in space.

In models of magnetic and electric fields, field vectors depicted by arrows  because they represent both their magnitude and direction. The length of the arrow shows magnitude.

Learn more: brainly.com/question/102477

7 0
2 years ago
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