What innovation drove the industrial revolution in the 1800’s?

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What innovation drove the industrial revolution in the 1800’s?. THE STEAM ENGINE. A steam engine is an example of a heat engine. In goes heat. Q in. And the difference is work. Out goes less heat. Q out. W out. Heat engines have been harnessed to do all kinds of work for us. - PowerPoint PPT Presentation

Transcript of What innovation drove the industrial revolution in the 1800’s?

Page 3: What innovation drove the industrial revolution in the 1800’s?

A steam engine is an example of a heat engine.

In goes heat

Qin

QoutOut goes less heat

Wout

And the difference is work

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Heat engines have been harnessed to do all kinds of work for us.

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They come in all kinds of designs but basically do the same thing.

RX-7 Rotary Engine

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Ever since they were invented, it was desired to improve their efficiency.

To use less coal to do the same job saves $$.

Enter Sadi Carnot, a French Engineer early 1800’s

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Most engineers at the time were looking for ways to tweak designs for better efficiency.

Carnot wanted to find the maximum efficiency nature would allow. In the process, he founded the field of

THERMODYNAMICS

As an engineer he was lousy and never built anything of note,but as a scientist his contributions have long outlasted any designs because developed the fundamental understanding.

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Thermodynamics:

The study of heat and its transformation into mechanical energy.

Most of modern thermodynamics can be understood with a little common sense and an understanding of conservation of energy.

heat motion

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1st off, heat flows from high temperature objects to low temperature objects, until they are at the same temperature.

They are now at the same temperature or THERMAL EQUILIBRIUM

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The 0th law of thermodynamics.

A

B

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If both A and C are in thermal equilibrium with B, then A and C are in thermal equilibrium with each other.

This is the common sense bit….

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The 1st law of thermodynamics: (in laymen's terms)

When energy is added to a system, it has to be somewhere!

Q

Now remember a hot cup of coffee is not going to suddenly shoot up a hill. Even though the atoms have a lot of kinetic energy. This is because the atoms are bouncing around randomly

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The 1st law of thermodynamics: (in laymen's terms)

When energy is added to a system, it has to be somewhere!

Q

In this case heat was added and the internal energy of the system increased. The atoms move faster. But the cup of coffee doesn’t take off. Why?

U

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Since this is not A ChE class,

U ONLY if there is a change in TEMPERTURE

Change in internal energy means change in temperature. T =0 U =0

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Hint: W__ __ __ = E

What causes the transfer of energy???

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An expanding piston of gas can do work on a wheel or something.In doing so energy has left the gas.

dW = F d How does the gas

exert a force on the piston?

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d

W = F d

F = P A

A W = P (A d)d

V

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W = PVLet’s not worry about signs just yet.

But was work done ON the gas, or did the gas DO work in this case?Did the gas gain or lose energy?

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A gas is compressed at a constant pressure (I guess it is was cooled)How much work is done on the gas?

W = PV

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The work done is equal to the AREA UNDER THE CURVE in a P vs V graph!!!!

W = PV

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W = PV

If no heat is added to the gas while the piston expands is the pressure of the gas constant?

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How much work did the gas do on the piston here?

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The 1st law of thermodynamics can be summed up two ways.

The change in energy of a system is equal to how much it gains minus how much it loses! AKA Conservation of Energy

OR……

U = Q + W

The equation sheet has Q – W. BUT let’s just use common sense and take the perspective of the gas. Remember Fs = -kx

PV

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U = Q + W

A system gains 50 J of heat and 20 J of work is done on it by compressing it.

Find U, does the temperature increase or decrease?

A system gains 50 J of heat and does 20 J of work while expanding.

A system loses 50 J of heat and does 20 J of work while expanding.

A system loses 50 J of heat and 20 J of work is done on it by compressing it.

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Some terms for processes:

Isothermal: ___________ doesn’t change

Isobaric:_____________ doesn’t change

Isochoric:____________ doesn’t change

Adiabatic: ___________ does not flow in or out.Q= 0

Temperature

Pressure

Volume

Heat

Leave some space to add notes

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In an adiabatic process, let’s say a piston of air is compressed. What is zero?

U = Q + W0 +

Is work positive or negative?

What does this tell us?

+

It gets hotter, remember the fire syringe??It happened too quickly for the heat to escape…

Adiabatic, Q=0

PV

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1.) Fuel and air in

2.) Compression

3.) Ignition and expansion

4.) Exhaust

Diesel engines don’t have spark plugs, how does the fuel ignite?

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A sample of gas contained in a piston is very well insulated. It is allowed to expand adiabatically.

U = Q + W

No heat can come in or out.0 J

Gas does work on surroundings.(-)

Must lose Internal Energy. (-, Cools)

Try breathing slowly on your hand and the blowing through pursed lips

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How does doing work on a piston reduce the temperature of an expanding gas? Consider the elastic collision below

Gas atom Piston

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Compressing a gas and doing work on it increases its temperature..

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Fire syringe demo

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Warm humid air

Air expands and cools forming rain.

Dry air forms desert.

Mojave Desert

Pacific Ocean

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In an isothermal process, 50 J of heat is added to a gas, what is zero

U = Q + W0 +

Is work positive or negative?

What does this tell us?

?

Heat is added so it must do work (negative work to lose energy)

Isothermal, U=0

PV

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In an isochoric process, 50 J of heat is added to a gas, what is zero

U = Q + W0+

Is work positive or negative?

What does this tell us?

Heat is added, it can’t do work, so the energy just goes to increasing temperature

Isochoric, V=0

PV+

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In an isobaric process, what is zero

U = Q + W

What does this tell us?

Just crunch the numbers. You know P

Isobaric, P=0

PV

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What type of process is this

Isochoric, V=0

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What type of process is this

Isobaric, P=0

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In an isothermal change we know that what 2 things about a gas are constant?

U = Q + W

Temperature & Internal Energy

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If the temperature of a gas is constant, then we know based on the ideal gas law….

P V = n R T

Constant

Ideal Gas Constant

Constant assuming that ….

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P V = constant

In an isothermal change

The value of a the constant depends on the starting conditions of the gas:Temperature, moles, pressure etc….

Let’s pick a constant and see what a P vs V looks like for it.

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P V = constantI like easy numbers, Let’s make the starting conditions for everything 1.Here we assume temperature is constant.What direction is heat flow?

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A gas is allowed to expand isothermally and adiabatically. In which case does it do more work on its surroundings?

Isothermal

Adiabatic

Heat is absorbed andmore work is done.

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A gas is put through a process that starts and ends at point A.Determine the following for the completed cycle.a.) The change in internal energy.b.) The net work on the gas.c.) The heat flow

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Note that for a cyclical process Q = W watch signs

Or in the case of a heat engine the net work out is equal to the net heat in.

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A gas can be brought from point A to B two ways.AD – gas is compressed near a heat sink to remove heatDB- gas is heated but volume is held constant.

Or the gas can be heated isothermally, allowed to expand to cool doing work on it surroundings to keep the temperature down

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Isobaric

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How much work is done on the gas going from ADB?

& Heat flow?

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The 2nd law of thermodynamics.

It has to do with things that tend to occur spontaneously,that is naturally withoutsome guided help.

Things naturally go from “ordered” states to disordered states unless some WORK is put in.ENTROPY

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The 2nd law of thermodynamics.

Heat flows naturally from a hot object to a cold object.Heat will not flow spontaneously from a cold object to a hot one.

And it is impossible to have a heat engine with 100% efficiency.

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How hard is it to convert work to heat with 100% efficiency?

Easy rub your hands together.

Now do the reverse, get the heat to move your hands back and forth?

The randomness of heat makes it hard to corral.

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A heat engine attempts to take the randomness of heat energy and squeezes useful work out of it.

Heat source

Heat sink

Heat engine

QH

QC

WAll we can hope to do is skim some off the top before the heat does what it does naturally.

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Heat source

Heat sink

Heat engine

QH

QC

W

A car engine is an example

Heat from burning fuel

Work to get the car moving

Heat out exhaust pipe and radiator

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QH

W

QC

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Their must be a heat sink create a pressure difference and keep the fluid moving.

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Just like there needs to be a height difference to keep the water moving. And note we only can capture some of the water’s energy here as well.

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If the top were higher or the bottom were lower we could capture more energy from the water.

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Recall that for a cyclical process,

Net Work Out = Net Heat In

W= QH - QC

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The efficiency of an engine is how much work we get out vs. how much heat we must feed it.

e =W

QH

W= QH - QC

e =QH

QH - QC

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An engine pulls 450 J of input heat and releases 250 J of waste heat. What is its efficiency and how much work did it do?

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Cyclical processes occur in all heat engines. In a combustion engine it is a batch process.

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In the case of a power plant it is a continuous process in a loop.

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The cycle for the heat engine that would have the Highest Efficiency nature would allow was devised by Sadi Carnot. (remember him?)

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1.) Isothermal Expansion

QH

2.) Adiabatic Expansion

3.) Isothermal Compression

QC

4.) Adiabatic Compression

The Carnot Cycle

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The Highest possible efficiency is given by the Carnot Efficiency

ec

=TH - TC

TH

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A natural gas power plant will generate high pressure superheated steam at about 500oC to drive a turbine and then can cool the condensate back to 25oC using outside air temperature. What is the ideal (Carnot) efficiency of the power plant?

How could the efficiency be improved?

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ec

=TH - TC

TH

According to Carnot’s theory, what is the only way to get a heat engine to have an efficiency of 100%? (ec = 1)

Can we ever do this?

Would it make financial sense to use liquid nitrogen as a heat sink?

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Heat source

Heat sink

Heat engine

QH

QC

W

Heat source

Heat engine

QH

W

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Remember the 2nd law of thermodynamics.

Heat flows naturally from a hot object to a cold object.Heat will not flow spontaneously from a cold object to a hot one.

Is it possible to pump heat from inside a cold container to the warmer outside surroundings?

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Heat engine

QH

QC

W

A heat pump (like a refrigerator or air conditioner) works like a heat engine in reverse.

Work is added and heat is pumped from a cold source to a warmer source

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1.) Cool low pressure Freon gas enters the compressor

2.) Hot high pressure Freon gas exits the compressor ~110oF

3.) The gas is cooled to outside temperature and liquefies under the high pressure

4.) The cool liquid flows through a restriction. The other side is low pressure and the liquid expands to form a mixture of COLD Gas and liquid ~40oF

5.) The COLD liquid boils absorbing heat from inside the house.

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the “A” coil

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Translational KE

Vibrational KE

Rotational KE

PE between units

+-

PE within bonds

+-

Recall that the internal energy (U) of a substance in “stored” in a variety of ways.

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However, the change in internal of a substance (U),can be measured very easily by measuring a change in its….

Temperature

U = Q + WPV

If I add some heat (Q) to a substance when do I know that all of it will go to increasing the internal energy of a substance.

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If its volume is constant… V = 0

U = Q + WPV

All of the heat added, increases the internal energy of the gas.

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Q = U

Q = m cV T

U = m cV T

For a gas

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The heat capacity of a gas at constant pressure (Cp) is more because it will do work and therefore lose energy from its internal energy

Q

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For a solid CP and CV are about the same because when you heat a solid…

Q

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Now Heat Capacity can be expressed in different ways

cv

Cv

Jkg oC

Jmol oC

U = m cV T

U = n CV T

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2.1 moles of a gas is brought from 2.3x105 Pa to 4.8x105 Pa at a constant volume of 12 x 10-3 m3. If the gas has a CV of 15 J/mol K, determine the temperature of each state and its change in internal energy.

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P

V

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c b

d

In going from state a to state b along acb, the system does 60 J of work and absorbs 90 J of heat. What is the change in internal energy between points a & b

Uab = +30

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P

V

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c b

d

a.) If the system goes from state b to state a along the curved path it absorbs 20 J of heat in the process, what is the work in the same process?

Uab = +30

Uba = -30 J = Q + W+20

W=-50

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P

V

a

c b

d

b.) If the system absorbs 50 J of heat while moving along the path along the path of adb, what is Wadb?

Uab = +30

Uab = +30 = Q + W+50

Wad = -20 J

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P

V

a

c b

d

If Pc = 10 Pa, What is Wacb

Uab = +30

Wad = -20 J

Wad = -20 J = Pa* V

Wbc= PC* V Wbc= 200 J

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P

V

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c b

d

c.) If Ua = 0 and Ud = 10, what is Qdb?

Uab = +30

Ubd = +20 = Q + W

Qdb = + 20

Wad = -20 J

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P

V

a

c b

d

In going from state a to state b along acb, the system does 60 J of work and absorbs 90 J of heat.

d.) For the cyclical process adbda, determine the signs of …. U _____, W ______, and Q_______.0 -+