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2 years ago

How can a heat pump warm a house by causing a refrigerant to evaporate and condense

1 answer:

4 0

The hot discharge gas from the refrigerant compressor is normally cooled and condensed at high pressure. This is then passed through an 'Expansion' valve which decreases the pressure to a low level causing expansion of the refrigerant liquid.
<span>The liquid partially vapourises causing a 'Joule's/Thompson' refrigeration effect' which decreases temperature of the refrigerant which then passes to an evaporator coil in the air circulation system of the building. </span>
<span>In the evaporator coil, the heat exchange between the cold refrigerant and the warm air of the building, vaporises and heats the refrigerant which returns to the compressor. </span>
<span>The cycle is repeated until the air temperature reaches the thermostat set-point and switches off the system. </span>

<span>As a Heat pump, the hot refrigerant gas is not evaporating and condensing. </span>
<span>From the compressor discharge, the hot gas is by-passing the cooler/condenser unit and the expansion valve and passes directly to the 'evaporator' coils but now, as the heating medium for the air circulation system where it's cooled by the heat exchange between the hot gas and the cooler air in the building and returns to the compressor in a continuous cycle. </span>

<span>A Thermostat in the system starts and stops the compressor motor according to the heat or cool temperature settings.</span>

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

2 years ago

The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.

ollegr [7]

(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

f = 3v/4L

where;

v is speed of sound
L is length of the pipe

f = (3 x 340) / (4 x 0.44)

f = 579.55 Hz

<h3>level of the water</h3>

wave equation for first harmonic of a closed pipe is given as

f = v/(4L)

251.1 = 340/(4L)

4L = 340/251.1

4L = 1.35

L = 1.35/4

L = 0.34 m

level of water = 0.44 m - 0.34 m = 0.1 m

Thus, the level of the water inside the vertical pipe is 0.1 m.

Learn more about harmonics of closed pipes here: brainly.com/question/27248821

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3 0

11 months ago

Using one or more of your senses to gather information is

valentinak56 [21]

It is called observation because you observe with your nose by semlling things your eyes by looking at things your ears by hearing things your tongue by tasting things and your hands by feeling things

3 0

2 years ago

Read 2 more answers

While i was jogging, i wad able to run 5 miles in thirty minutes. from this information, we can determine my?

Llana [10]

You can only determine the speed since the only info we know is how much you ran in how long of a time.

7 0

3 years ago

Read 2 more answers

If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 wil

Lina20 [59]

Answer:

The answer to this question can be defined as follows:

Explanation:

Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.

Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.

4 0

2 years ago