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# Second Low of Thermodynamics PowerPoint PPT Presentation

Second Low of Thermodynamics. Example 1. Heat always flows from high temperature to low temperature. So, a cup of hot coffee does not get hotter in a cooler room. Yet, doing so does not violate the first low as long as the energy lost by air is the same as the energy gained by the coffee.

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Second Low of Thermodynamics

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## Second Low of Thermodynamics

### Example 1

• Heat always flows from high temperature to low temperature.

• So, a cup of hot coffee does not get hotter in a cooler room.

• Yet, doing so does not violate the first low as long as the energy lost by air is the same as the energy gained by the coffee.

### Example 2

• The amount of EE is equal to the amount of energy transferred to the room.

### It is clear from the previous examples that..

• Processes proceed in certain direction and not in the reverse direction.

• The first law places no restriction on the direction of a process.

• Therefore we need another law (the second law of thermodynamics) to determine the direction of a process.

### Thermal Energy Reservoir

• It is defined as a body to which and from which heat can be transferred without a change in its temperature.

• If it supplies heat then it is called a source.

• If it absorbs heat then it is called a sink.

• Some obvious examples are solar energy, oil furnace, atmosphere, lakes, and oceans

• Another example is two-phase systems,

• and even the air in a room if the heat added or absorbed is small compared to the air thermal capacity (e.g. TV heat in a room).

### Heat Engines

• We all know that doing work on the water will generate heat.

• However transferring heat to the liquid will not generate work.

• Yet, doing so does not violate the first low as long as the heat added to the water is the same as the work gained by the shaft.

• Previous example leads to the concept of Heat Engine!.

• We have seen that work always converts directly and completely to heat, but converting heat to work requires the use of some special devices.

• These devices are called Heat Engines and

• can be characterized by the following:

### Characteristics of Heat Engines..

High-temperature Reservoir at TH

• They receive heat from high-temperature source.

• They convert part of this heat to work.

• They reject the remaining waste heat to a low-temperature sink.

• They operate on (a thermodynamic) cycle.

QH

HE

W

QL

Low-temperature Reservoir at TL

### Difference between Thermodynamic and Mechanical cycles

• A heat engine is a device that operates in a thermodynamic cycle and does a certain amount of net positive work through the transfer of heat from a high-temperature body to a low-temperature body.

• A thermodynamic cycle involves a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid.

• Internal combustion engines operate on a mechanical cycle (the piston returns to its starting position at the end of each revolution) but not on a thermodynamic cycle.

• However, they are still called heat engines

### Thermal efficiency

Thermal Efficiency

< 100 %

### Thermal efficiency

QH= magnitude of heat transfer between the cycle

device and the H-T medium at temperature TH

QL= magnitude of heat transfer between the cycle

device and the L-T medium at temperature TL

Thermal Efficiency

< 100 %

### thermal efficiency can not reach 100%

Even the Most Efficient Heat Engines Reject Most Heat as Waste Heat

Automobile Engine 20%

Diesel Engine 30%

Gas Turbine 30%

Steam Power Plant 40%

### Can we save Qout?

• Heat the gas (QH=100 kJ)

• Load is raised=> W=15 kJ

• How can you go back to get more weights (i.e. complete the cycle)?

• By rejecting 85 kJ

• Can you reject it to the Hot reservoir? NO

• What do you need?

• I need cold reservoir to reject 85 kJ

A heat- engine cycle cannot be completed without rejecting some heat to a low temperature sink.

Example 5-1: Net Power Production of a Heat Engine

Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine.

<Answers: 30 MW, 0.375>

### The Second Law of Thermodynamics: Kelvin-Plank Statement (The first)

• The Kelvin-Plank statement:

It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.

• It can also be expressed as:

No heat engine can have a thermal efficiency of 100%, or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace.

• Note that the impossibility of having a 100% efficient heat engine is not due to friction or other dissipative effects.

• It is a limitation that applies to both idealized and the actual heat engines.

• Example 1 at the beginning of the notes leads to the concept of Refrigerator and Heat Pump..

• Heat can not be transferred from low temperature body to high temperature one except with special devices.

• These devices are called Refrigerators and Heat Pumps

• Heat pumps and refrigerators differ in their intended use. They work the same.

• They are characterized by the following:

### Refrigerators

High-temperature Reservoir at TH

QH

W

QL = QH - W

Ref

QL

Low-temperature Reservoir at TL

Objective

### An example of a Refrigerator and a Heat pump ..

Coefficient of Performance of a Refrigerator

The efficiency of a refrigerator is expressed in term of the coefficient of performance (COPR).

### Heat Pumps

Objective

High-temperature Reservoir at TH

QH

QH = W + QL

Read to parts of

pp 259 and 260

W

HP

QL

Low-temperature Reservoir at TL

Heat Pump

Coefficient of Performance of a Heat Pump

The efficiency of a heat pump is expressed in term of the coefficient of performance (COPHP).

Relationship between Coefficient of Performance of a Refrigerator (COPR) and a Heat Pump (COPHP).

The second Law of Thermodynamics: Clausius Statement

The Clausius statement is expressed as follows:

It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.

Both statements are negative statements!

Read pp 262

### Equivalence of the Two Statements

High-temperature Reservoir at TH

Net QOUT = QL

QH

QH + QL

HE

Ref

HE + Ref

W = QH

QL

Net QIN = QL

Low-temperature Reservoir at TL

Example (5-5): Heating a House by a Heat Pump

A heat pump is used to meet the heating requirements of a house and maintain it at 20oC. On a day when the outdoor air temperature drops to -2oC, the house is estimated to lose heat at rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air.

Sol:

### Perpetual Motion Machines

• Any device that violates the first or second law is called a perpetual motion machine

• If it violates the first law, it is a perpetual motion machine of the first type (PMM1)

• If it violates the second law, it is a perpetual motion machine of the second type (PMM2)

• Perpetual Motion Machines are not possible

• The second law of thermodynamics state that no heat engine can have an efficiency of 100%.

• Then one may ask, what is the highest efficiency that a heat engine can possibly have.

• Before we answer this question, we need to define an idealized process first, which is called the reversible process.