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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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Aleksandr-060686 [28]

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

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Magnetic field for a straight wire is given as;

B = \frac{\mu I}{2 \pi r}

Where:

μ is permeability

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r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

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7 0
3 years ago
I don’t understand how to do this
anzhelika [568]

Explanation:

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5 0
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Studentka2010 [4]

Cool down the Slowest

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