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Neko [114]
1 year ago
6

for a parallel flow double pipe heat exchanger insulated from its surroundings, the mean outlet temperature of the initially col

d fluid can exceed the mean outlet temperature of the initially hot fluid.
Physics
1 answer:
Xelga [282]1 year ago
7 0

Heat flows from the hot fluid to the cold fluid, and the amount of heat transferred is directly dependent on the temperature difference between the two sides.

What is heat exchange?

In a heat exchanger, heat flows from the hot fluid to the cold fluid, and the amount of heat transferred is directly dependent on the temperature difference between the two sides.

In a parallel flow arrangement, the cold inlet and hot inlet are interacting with each other. At that entry point, the temperature difference is wide, and heat transfers quickly.

As the streams pass through the parallel-flow exchanger, they start to approach each other’s temperature. The heat transfer rate drops in line with the reduction in temperature difference.

Now we have the answer to your question. It is impossible for the exit temperature of the cold liquid to exceed the exit temperature of the hot liquid. This is because once the same temperature is reached, all heat transfer stops because there is no temperature difference.

This limitation of parallel flow heat exchangers is the reason most heat exchangers are arranged in counterflow. The counter-flow design results in a higher average temperature difference and greater heat transfer.

Therefore, heat flows from the hot fluid to the cold fluid, and the amount of heat transferred is directly dependent on the temperature difference between the two sides.

To know more about heat exchanger, visit:

brainly.com/question/17029788

#SPJ4

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Technically, we have no way of knowing that without seeing Figure 16-2.
So the question should be reported for incomplete content.  But I'm
going to take a wild stab at it anyway.

There's so much discussion of 'cylinder' and 'strokes' in the question,
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combustion gasoline engine. 

If I'm right, then the temperature of the material within the cylinder is
greatest right after the spark ignites it.  At that instant, the material burns, 
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This is obviously happening because of the great, sudden increase in
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5 0
3 years ago
An object is 45.0 kg here on Earth. What is its mass on Jupiter if the acceleration
puteri [66]

Answer:

B. 45.0 kg

Explanation:

On earth the object has ;

Mass = 45 kg

acceleration = 9.81 m/s²

On Jupiter, acceleration is  24.79 m/s²

The mass of this object on Jupiter will be 45 kg. It will not change. Mass of an object will only change when you remove part of the object from it or add to it another part. The mass is the same on Earth and on Jupiter. However, due to increased acceleration on Jupiter , the weight will change/ increase because;

Weight = mass * acceleration

<u>On Earth </u>

Weight of the object will be : 45 *  9.81 = 441.45 kg

<u>On Jupiter</u>

Weight of the object will be : 45*24.79 =1115.55 kg

8 0
3 years ago
Which statement is true about the type of light used by plants to make sugar during photosynthesis? Group of answer choices
denis-greek [22]

Answer:

4. It is infrared.

Explanation:

the photosystems on the chlorophyll absorb light at 600-700nm

3 0
3 years ago
A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel
g100num [7]

Answer:

option A

Explanation:

Given,

wavelength of light,\lambda = 560\ nm

refractive index of gasoline, n₁ = 1.40

Refractive index of water, n₂ = 1.33

thickness of the film, t = ?

Condition of constructive interference is given by

2 n t = (m+\dfrac{1}{2})\lambda

For minimum thickness of the film m = 0

From the question we can clearly observe that phase change from gasoline to air

so, n = 1.4

2 \times 1.4 \times t = \dfrac{560}{2}

t = 100\ nm

Hence, the correct answer is option A

7 0
4 years ago
A cubic box of volume 5.1 x 10 -2m3 is
I am Lyosha [343]

Answer:

21544 N

Explanation:

Let the atmospheric pressure be 101325 Pa

The side length of the cubic box:

V = d^3 = 0.051m^3

d = \sqrt[3]{V} = \sqrt[3]{0.051} = 0.371 m

The area of the cubic box:

A = d^2 = 0.371^2 = 0.1375 m^2

20 C = 20 + 273 = 293 K

180 C = 180 + 273 = 453 K

As the volume of the air inside the closed cube is not changed, assume the ideal gas law we have

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Where P1 = 101325 Pa and T1 = 293K are the original atmospheric pressure and temperature. P2 and T2 = 453 are the new pressure and temperature after the cube gets heat up

P_2 = P_1\frac{T_2}{T_1} = 101325\frac{453}{293} = 156656 Pa

The net force on each side of the box it its pressure times side area

F = P_2A = 156656 * 0.1375 = 21544 N

8 0
3 years ago
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