All categories

3 years ago

Which is a common refrigerant for domestic air conditioners?

Engineering

2 answers:

mihalych1998 [28]3 years ago

6 0

Answer:

The most common refrigerants used for air conditioning over the years include

Explanation:

The most common refrigerants used for air conditioning over the years include.

Reptile [31]3 years ago

4 0

Answer:

Explanation:

The most common HFC used in air conditioners is R-410A. This refrigerant is better than R-22 in terms of “Ozone Depletion” potential and energy efficiency, but it still causes global warming. A few more HFCs that are commonly used are: R-32 in Air Conditioners and R-134A in refrigerators.

You might be interested in

Why do giant stars become planetary nebulas while supergiant stars become supernovas when their nuclear fusion slows and is over

sashaice [31]

The reason why giant stars become planetary nebulas is Supergiant stars do not have enough mass to generate the gravity necessary to cause a planetary nebula.

<h3>Why do giant stars become planetary nebulae?</h3>

A planetary nebula is known to be formed or created by a dying star. A red giant is known to be unstable and thus emit pulses of gas that is said to form a sphere around the dying star and thus they are said to be ionized by the ultraviolet radiation that the star is known to releases.

Learn more about giant stars from

brainly.com/question/27111741

#SPJ1

3 0

2 years ago

Two pressure gauges measure a pressure drop of 16.3 psi (lb/in.2) at the entrance and exit of an old buried pipeline. The origin

Sholpan [36]

Answer: 866.25ft

Explanation:

Pressure drop= 16.3 psi

1 feet of water column= 0.4335psi

x= 16.3psi

Therefore, head loss of water= 16.3/0.4335=37.6ft

In inches, head loss of water= 37.6 × 12=451.2 inches

Q= 1.64ft^2/s

A= pi/4 D^2

Where D= 6 inches, 0.5ft

A= 3.142/4 × (0.5)^2

A= 0.1962ft^2

V= Q/A= 1.64/ 0.1962

V= 8.35ft/s, 100.28 in/s

REGNOLD'S NO

Re= SvD/ μ=

Where, D= 6 inches, V= 100.28 inches/s, S=62.4lb/ft^3= 0.03611lb/inches^3

At 68F, Dynamic viscosity μ= 1.0016× 10^-3NS/m^2

1psi= 6895N/m^3

μ= 1.0016× 10^-3NS/m^2/6895N/m^3

μ= 1.452× 10^-7lbs/ins^2

Re= 0.03611× 100. 28×6/1.42×10^-7

Re= 1.49×10^8

e/D----- Relative roughness

e= 0.0005ft

e/D= 0.001

Therefore, for Re= 1.49×10^8 and e/D= 0.001

F=0.02

Head loss is given by Dancy-Weisbach formula

hL= fLV^2/2gD

Wherr g= 9.81m/s^2= 386.12inches/s^2

451.2= 0.02×L×(100.28^2)/2 × 386.12inches/s^2× 6

L= 10395inches

L= 10395/12= 866.25ft

Therefore, the length of pipe is 866.25ft.

7 0

4 years ago

Using data in Appendix A ,calculate the number of atoms in 1 tonne of iron

maksim [4K]

<h2>Answer:</h2>

$1.0783*10^{28}(atoms).$

<h2>Explanation:</h2>

<em>Since I don't have access to "Appendix A", I'll solve the problem using data from the periodic table.</em>

<em />

<h3>1. Determine the molar mass of iron.</h3>

<em>According to the periodic table, the molar mass of iron is:</em>

55.845g/mole.

<h3>2. Convert 1 tonne to grams.</h3>

$1(tonne)*1000=1000kg\\1000kg*1000=1000000g=10^6g$

<h3>3. Apply rule of 3.</h3>

$55.845g$ ----------- $1 mole$

$10^6g$ ----------- $x$

$x=\frac{10^6*1}{55.845}=17906.7061(moles)$

<h3>4. Determine the amount of atoms.</h3>

<em>Considering that there are, approximately, </em> $6.022*10^{23}$ atoms in a mole of any element, apply another rule of 3.

1 mole --------------------- $6.022*10^{23}(atoms)$

$17906.7061(moles)$ --------------------- x

$x=\frac{17906.7061*6.022*10^{23}}{1}=1.0783*10^{28}$ .

4 0

2 years ago

I need solution for this question please<br>Select the right answer

Ostrovityanka [42]

Explanation:

the 7th one answer is beacause mercury is bad at sharing electrons

the 8th one's answer is Rhodium

3 0

4 years ago

Read 2 more answers

Quinn’s relatives relayed a story about putting on a headset and seeing a digital world that they could walk around in and explo

Kryger [21]

Answer:

I know it is C)Virtual reality

Explanation:

Look at the clues

story about putting on a headset ( virtual reality head set!)

seeing a digital world (A virtual reality world)

they could walk around in (Fake walking you are basically jogging in place)

explore in order to see what ancient Benin looked like (Looking at a real place only digitally)

as if they were really there ( they think they are actually there)

The only reason I know all of this is because I have done virtual reality multiple times and I LOVED it SUPER fun ( I was doing archery) :) Hope this helps!

6 0

3 years ago

Read 2 more answers