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ELEN [110]
3 years ago
9

7 One lb of Refrigerant 134a contained within a piston–cylinder assembly undergoes a process from a state where the temperature

is 60F and the refrigerant is saturated liquid to a state where the pressure is 140 lbf/in 2 and quality is 50%. Determine the change in specific entropy of the refrigerant, in Btu/lbR. Can this process be accomplished adiabatically?
Physics
1 answer:
julsineya [31]3 years ago
7 0

Answer:

\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb ^ {\circ} R}

Explanation:

A) In order to solve the table it is necessary to consult tables A11-E and A10E for refrigerant R134-a

In this way we obtain that:

S_1 = S_f = 0.0648 \frac {BTU} {Lb. ^ {\circ} R}

In this way,

S_2 = S_f + x_2 (S_g-S_f) = 0.0902 + 0.5 (0.2161-0.0902)

S_2 = 0.15315 \frac {BTU} {Lb ^ {\circ} R}

In this way the entropy change is,

\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb. ^ {\°} R}

B) Whenever entropy yields a positive result, the process can be carried out adiabatically.

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Hang two sheets of paper vertically from adjacent corners. The sheets should be parallel and close to each other with a small ga
Mrrafil [7]

Answer:

a. The sheets move toward each other and the gap narrows.

Explanation:

This exercise is related to fluid mechanics, when blowing between the two sheets, we can apply Bernoulli's equation, where the index 2 is the space between the two sheets

       P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y²

if the two leaves are at the same height

                      y₁ = y₂

 whereby

         P₁ + ½ ρ g v₁² = P₂ + ½ ρ v₂²

for the air velocity between the leaves let us use the continuity equation

        A₁ v₁ = A₂ v₂

the area between the leaves is less than the external area, so the air speed must increase. If we use this in Bernoulli's equation, increasing the speed 2 (between the leaves) to maintain equality the pressure must decrease.

If the pressure decreases, the blades should move closer

When resisting the answers, the correct one is  a

4 0
3 years ago
Two objects are dropped from different heights at the same instant. If the first object takes 10.7 seconds to hit the ground and
Lubov Fominskaja [6]

Answer:

The answer to your question is : 521.8 m

Explanation:

Data:

Different heights

Time first object (tfo) = 10.7 s

Time second object (tso)= 14.8 s

Initial speed of both objects(vo) = 0 m/s

a = 9.81 m/s²

Formula:

h = vot + 1/2 (a)(t)² but vo = 0    so, h = 1/2 (a)(t)²

Then, height fo      h = 1/2 (9.81)(10.7)² = 561.6 m

          height so     h = 1/2(9,81)(14.8)² = 1074.4 m

Difference in their heights =  1074.4 m - 561.6 m = 521.8 m

5 0
3 years ago
How does gas exert pressure? Explain using the change in momentum.
san4es73 [151]

Answer:

The molecules are continually colliding with each other and with the walls of the container. When a molecule collides with the wall, they exert small force on the wall The pressure exerted by the gas is due to the sum of all these collision forces. The more particles that hit the walls, the higher the pressure.

4 0
2 years ago
Read 2 more answers
Just think about this.
diamong [38]
This ain’t the place, bud. If you have a QUESTION, then you can post it here.
8 0
2 years ago
Read 2 more answers
Suppose you want to operate an ideal refrigerator with a cold temperature of − 15.5 °C , and you would like it to have a coeffic
tensa zangetsu [6.8K]

Answer:

15.65 °C

Explanation:

cold temperature (Tc) = -15.5 degree C = 273.15 - 15.5 = 257.65 kelvin

minimum coefficient of performance (η) = 8.25

find the maximum hot reservoir temperature of such a generator (Th)

η = \frac{Tc}{Th-Tc}

Th = Tc x (\frac{1}{η} + 1)

Th = 257.65 x (\frac{1}{8.25} + 1)

Th = 288.8 K

Th = 288.8 - 273.15 = 15.65 °C

8 0
3 years ago
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