Thermodynamics, Refrigerator, and Air Conditioner

By Xinke Liu

Many people may have wondered if we could make a room cold by leaving a refrigerator door open? In fact, I tried it when I was a kid. But I never had the patience to see what happened. After learning the second law of thermodynamics, I thought this explains why a refrigerator cannot cool a room.

The second law of thermodynamics states:

Heat can flow spontaneously from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object.     

If a refrigerator cannot cool a room, how does air conditioner work? Does air conditioner work against the second law of thermodynamics?   

The second law of thermodynamics only offers partial explanation to the problem.

The Clausius statement of the second law of thermodynamics gives us a better idea, which states:

No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.

Both refrigerator and air conditioner are essentially ‘heat pumps!’ Work must be done to make heat flow from a body of lower temperature to a body of higher temperature.

Refrigerator and air conditioner are composed of evaporator, condenser, and compressor.

Evaporator transfers heat from hot air to fluid. Condenser transfers heat from fluid to outside air. Compressor does work on the fluid by compressing it and creating entropy.

If we leave a refrigerator door open, the room would warm up because the condenser is also inside the room. The condenser dumps heat out into the room all the while that it's running. And because the fridge is not 100% efficient, there will be more heat generated than there will be cooling.

How much heat can an air conditioner remove from a building’s air per second?

Let’s assume coefficient of performance is 2.0 and the air conditioner is rated to do work at 1000W.

QH= COP * W = 2.0 * 1000W = 2000 J.

An air conditioner with coefficient of performance is 2.0 and the air conditioner is rated to do work at 1000W can remove heat from a building’s air at a rate of 2000J, or at a rate of 2000W per second.

Scuba Diving Physics: Why people suffering from hearing loss cannot dive?

Written by Xinke Liu

Ever since Jacques-Yves Cousteau made the Great Blue Hole of Belize famous in 1971, this amazing wonder has lured thousands of scuba diving fans to explore it. The Great Blue Hole is said to be the largest blue hole in the world and has a perfect circular shape. This site is a great place to have a prehistoric journey to see remnants preserved from millions of years ago. You will have fascinating experience of swimming with SHARKS and freaky fish in the crystal clear water!!!! Myths say that the charming blue in the Great Blue Hole entices some divers to stay there FOREVER…. FOREVER…FOREVER…

Sadly, people suffering from ear diseases and hearing loss like me cannot dive. What’s keeping us away from this fascinating activity? One reason is that some of them may also have problem in balance that is controlled by the vestibular system in inner ear. But the major reason is, not surprisingly, pressure.

The Great Blue Hole is 412 feet deep. Most divers dive down to between 110 and 130 feet. How much pressure does the water exert on the eardrum when people are 130 feet deep into the sea?

The pressure P due to the weight of liquid is:

P = pgh.

We need to find out the density (p) for the water in the Great Blue Hole since density of seawater varies due to temperature, pressure, and salinity. Since the salinity of the seawater in the Great Blue hole is not as unusal as that of the seawater in Dead Sea, we would take the average value of density of surface surface seawater that is 1.025 *10³  kg·m³.

130 feet ≈ 39.6m.

g = 9.8 m/s².

P = (1.025 *10³  kg·m³) * (9.8 m/s²⁾ *(39.6m) ≈ 397782 N/m².

P ≈ 3.98*10⁵ Pa.

1 atm = 1.013 ×10⁵ Pa.

P/ 1 atm = (3.98*10⁵ Pa) / 1.013 ×10⁵ Pa ≈ 3.9 atm .

When people are 130 feet down in the Great Deep Hole, the water exerts 3.9 atm of pressure on the eardrum, almost 4 times of the average atm level on sea level. And this could be a disaster to people with hearing problem. And when diving, eardrum is in direct contact with the water.     

Some people may wonder why not divers just use earplugs to protect eardrum. However, in fact, earplugs are not recommended when diving. The graph below shows that fluids exert pressure on object from every direction. This means the seawater would push the earplug inside the ear.     

Physics of Flight

Physics of Flight

            For my physics news I decided to look at the physics of flight. There are four main forces involved in flight. They consist of an upward force, lift, a downward force, weight, a forward force, thrust, and a backward force, drag. In order for flight to occur, lift must be greater than or equal to the weight and thrust must be equal to or greater than drag.

            The wings of a plane are responsible for providing the lift. This concept can be explained by both Bernoulli’s principle and by Newton’s third law. As air currents pass by the wings of a plane, their shape, airfoil, establishes a differential pressure gradient. The distance the air has to travel is greater above the wing than below the wing. Since the air travels past the wing in the same amount of time, the air above the wing moves faster than the air below the wing. Bernoulli’s equation (P1 + .5pv1^2= P2 + .5pv2^2) indicates that this increase in speed causes a decrease in pressure above the wing and an increase in pressure below the wing. This unequal pressure causes dynamic lift, which pushes the plane upwards. The angle of the wing also plays a role in providing lift. The wing directs wind downwards as the plane flies. As established by Newton’s third law, the downward wind molecules push the wing in the opposite direction that they are traveling.

            The thrust of the plane comes from the engine or propeller. Like lift, it also utilizes Newton’s third law. Air is sucked in to the engine and pushed backward. As the air moves back at a rapid rate, the plane gets pushed forward. With the aide of an engine, the air traveling back causes a greater force than the wind on the body of the plane.

            Both Drag and weight limit the ability of the plane to fly. As wind hits the plane, it pushes the plane in the opposite direction that it wants to travel. In order to limit drag, planes have been designed to lower the surface area of the body. By doing this, the path of the air is less hindered, therefore causing less drag. The weight force is simply the force of gravity and the mass of the plane.

            Now that I have explained the physics behind flight I will look specifically at the Boeing 747 jet airplane to mathematically show these principles.

Force Weight= mg

Force Thrust= ma

Force Lift= C* (.5pv^2) *wing A

Force Drag= C* (.5pv^2)*Total A

FW=(333,390kg)(9.8m/s^2)=3267222N

FT=(333,390kg)(a) FT= 4(223000N)

(4*223,000N)/333,390kg = 2.7m/s^2

Acceleration seemed slow to me but this is due to the fact that it is a commercial plane. They accelerate very slowly over a long period of time.

FW=FL 3267222N= C* (.5*1.29*(265.3m/s)^2*510.95m^2 C=.14 constant

FD=FT 892000N= .022* (.5*1.29*(265.3m/s)^2 *A   A=893m^2

How much tension is there on a guitar string?

By David Haimes

The lowest note possible on a guitar is the open low E string. Using a guitar-tuning app
on my phone, I measured the frequencies to be about 80 Hz.

Given
Frequency: 80 Hz
Length: About 25 inches from fixed point to fixed point
Mass: 10 grams (very rough estimate/guess)

The Force due to tension equation is

FT = ( m / L ) * v2

We need to find v.

The equation for v is: v = ƒλ
And λ = 2L = 2 (25 inches * (.0254 m / 1 inch) ) = 1.27 m

Now we can solve for v : v = ƒλ = 80 Hz (1.27 m) = 101.6 m/s

Now we can solve for FT= ( .010 kg / (25 inch (.0254m/1 inch) ) * (101.6 m/s)2 = 65 N

So the approximate tension on the string is 65 N

The website lists the recommended tension for the string on my guitar to be 27 lbs which
is about 120 N. Assuming all the calculations are correct, the one big assumption/
approximation I made was the mass of the string. Working backwards, a tension of 120 N
would require a mass of…

m = FT / (v2*L) = 120 N / (101.6 m/s * 1.27m) = .0092 kg

I plan to bring the string to lab on Thursday to see how close the mass of the string is to
this! However, the string is larger than length used in the equation, due to excess beyond
the fixed points.

Paintball Physics

By Michelle Lu

Kids often bruise when they are hit by a paintball if they’re not wearing a vest.
Let’s see how much force a paintball would be exerting on you if you were hit by a
paintball.

I chose to assume there is no air resistance. Target is at a distance of 50 ft and you
are shooting at an angle of 15° at 300 ft/s. The ball stops in 0.001s once it hits the
target.

m = 3.2 g = 0.0032 kg
vo = 300 ft/s = 91.4 m/s
v0x = 91.4cos15 = 88.3 m/s
voy = 91.4 sin15 = 23.67 m/s
Θ = 15°
Δx = 50 ft = 15.24 m

Δx = voxt + 1/2at2
15.24 = 88.3 t + 0
t = 0.17 s

Δy = vot + 1/2at2
Δy = (23.67 m/s)(0.17s) + ½(-9.8 m/s2) (0.17s)2
Δy = 3.88 m

vfy2 = voy2 + 2aΔy
vfy2 = (23.67 m/s)2 + 2(-9.8 m/s2)(3.88 m)
vfy = 22.0 m/s

vf = (vfx2 + vfy2)1/2
vf = (88.32 + 22.02)1/2
vf = 91 m/s

F = Δp/t
F =m(vf – vi)/t
F = [(0.0032 kg)(0 m/s – 91 m/s)]/0.001t
F = 291 N

Analyzing traffic patterns using fluid motion

By Lauren Charette

By using the average speed of cars on a highway and the number of lanes as the radius we can
get an idea about the ration of the number of cars on the road in a given area and understand the
amount of traffic.

Given the equation:

Q=AV

If the cars are moving at a speed of 27 m/s And the road is four lanes wide a lane being 3ms so
we’ll give it an r value of 12m and we’ll look at a 10m length so for this part the area is 120m2

Q=27*120

Q=3240
If the velocity is consistent but the number of lanes is reduced to 3 what happens to the speed?

Q=90*27

Q=2430

However because bottlenecking causes cars to slow down if the speed is reduced five m/s what is
the rate of flow?

90*22=Q

Q=1980

Physics of fighter jets

By Paul Donohue

I’ve always been interested in fighter jets, particularly of how they are able to take off and land on an aircraft carrier. They are able to accelerate so rapidly with the aid of a steam powered catapult. See below:

The release of a great deal of pressure in the form of steam drives the piston of the catapult forward, bringing the plane with it. In finding just how large the magnitude of work done by the catapult might be, I had to do some research. I found that the standard length of a catapult track is 72.8 m and the thrust of a F-18 super hornet’s engines is around 3.9x105 N collectively. I estimated that the fighter would have 6.03x107 J of Kinetic Energy at takeoff meaning the work done by the catapult is:
=6.03 x 107 - 3.90 x105*72.8=3.2*107J.