Power Cycles and Entropy

Question 1

What is a cycle?

Answer 1

is a series of processes that eventually brings the system back to its original condition. Although heat can be extracted and work can be performed in a single process, a cycle is necessary to obtain work in a useful quantity and duration. When plotted, the area within the closed p-V and T-s curves represent both the net work and net heat of the cycle.

Question 2

What completely defines a cycle?

Answer 2

It is completely defined by the working substance, the high and low temperature reservoirs, the means of doing work on the system, and the means of removing energy from the system.

Question 3

Power Cycle

Answer 3

is a cycle that takes heat and uses it to do work on the surroundings.

Question 4

Thermal Efficiency (Power Cycle)

Answer 4

is defined as the ratio of useful net work output to the supplied input energy. The net work in the thermal efficiency equation is also equal to the net heat.

Question 5

Carnot Cycle

Answer 5

is an ideal power cycle that is impractical to implement. It’s theoretical work output sets the maximum attainable from any heat engine. the working fluid in a Carnot cycle is irrelevant.

Question 6

Thermal Efficiency (Carnot Cycle)

Answer 6

is the most efficient power cycle possible. The temperature is expressed in the absolute scale and the efficiency is equal to the ratio of the difference in reservoir temperatures to the temperature of the high reservoir.

Question 7

Rankin Cycle

Answer 7

is similar to the Carnot cycle except that the compression process occurs in the liquid region. The cycle consists of a feed pump (isentropic, work in), boiler and superheater (heat in), turbine (isentropic, work out), and condenser (heat out).

Question 8

Thermal Efficiency (Rankine Cycle)

Answer 8

The thermal efficiency of the Rankine cycle is given by the ratio of enthalpy differences in the numerator, which can be stated in terms of work or heat transfers. To the heat input into the system.

Question 9

Air Standard Cycle (Combustion Power Cycle)

Answer 9

An air standard cycle is a hypothetical closed system using a fixed amount of ideal air as the working fluid. (Combustion power cycles are analyzed as air standard cycles.)

Question 10

Air Standard Otto Cycle

Answer 10

an air standard Otto cycle consists of an isentropic compression, constant volume heat addition, isentropic expansion, constant volume heat rejection.

Question 11

Otto Cycle

Answer 11

The Otto cycle is a four stroke cycle because four separate piston movements are required to accomplish all of the processes: intake, compression, power, and exhaust strokes.

Question 12

Compression Ratio

Answer 12

is the ratio of the initial volume to final volume.

Question 13

Thermal Efficiency (Air Standard Otto Cycle)

Answer 13

The compression ratio and specific heat of the air is used to find the ideal thermal efficiency for the Otto cycle.

Question 14

Refrigeration Cycle

Answer 14

A refrigeration cycle are power cycles in reverse, with work being the input and cooling being the desired effect. The pump or compressor is the device that does work on the refrigerant and is necessary for the refrigerator to be able absorb heat from a low temperature area and reject it to a high temperature area. The pump work is also rejected to the high temperature area.

Question 15

Air Conditioner

Answer 15

If heat is being removed from air, to produce cold air. The device is known as a air conditioner.

Question 16

Chiller

Answer 16

If the heat is being removed from water, to produce cold water. The device is known as a chiller.

Question 17

Rate of Refrigeration

Answer 17

is the rate at which heat is removed. Measured in tons.

Question 18

Ton

Answer 18

A ton of refrigeration corresponds to 12,000 Btu/hr and is the heat flow required to melt a ton of ice in 24 hours.

Question 19

Cooling Effect (Refrigerator)

Answer 19

The main function of a refrigerator is to cool the low temperature area. The useful energy transfer is the heat removed from the low temperature area.

Question 20

Heating Effect (Heat Pump)

Answer 20

A heat pump is a device that also operates on a refrigeration cycle. There is no significant difference in the mechanism or construction of heat pumps and refrigerators. A heat pumps main function is to warm the high temperature area. The useful energy transfer is the heat rejected to the high temperature area.

Question 21

Coefficient of Performance (COP)

Answer 21

Is the ratio of useful energy transfer to the work input. The higher the coefficient of performance, the greater the effect for a given work input will be.

Question 22

Carnot Refrigeration Cycle

Answer 22

The Carnot Refrigeration Cycle is the Carnot Power Cycle in reverse. Because it is reversible, the Carnot Refrigeration Cycle has the highest COP for any given temperature limits of all the refrigeration cycles.

Question 23

COP (Carnot Refrigeration Cycle)

Answer 23

Is the ratio of useful temperature to the difference of the high and low temperature environments.

Question 24

Vapor Refrigeration Cycle (VRC)

Answer 24

This cycle begins with cold liquid refrigerant in a saturated mixture state. Passing through the evaporator, and absorbing heat from the environment becoming vaporized (saturated vapor). The vaporized refrigerant is then compressed, in a reciprocating compressor. Where it gains heat from the work of the compression, raising its temperature. It then passes through a condenser where the heat absorbed from the low temperature environment and the work of the compression is removed. Finally the pressure of the cooled vapor is reduced in a throttling process in the expansion valve, where some of the liquid refrigerant also flashes into a vapor.

Question 25

COP (Vapor Refrigeration Cycle)

Answer 25

Is the ratio of useful energy to work input expressed in differences of specific enthalpy.

Question 26

Two Stage Refrigeration Cycle

Answer 26

The condenser heat from the low temperature cycle evaporates a refrigerant in the evaporator of the high temperature cycle (through a heat exchanger). Among other advantages, a two stage cycle refrigerator can operate with a greater temperature difference between the hot and cold reservoir.

Question 27

COP (Two Stage Cycle)

Answer 27

The COP of a two stage cycle is equal to the useful energy transfer divided by the work input (compressor) at each stage. Can be expressed terms of enthalpy.

Question 28

Air Refrigerant Cycle

Answer 28

In an air refrigeration cycle, consists of the air conditioned space (heat absorbed), compressor (work in), heat exchanger (heat out), and turbine (work out).

Question 29

COP (Air Refrigeration Cycle)

Answer 29

Its COP is the ratio of useful energy to the difference of work input and output. Can be expressed in terms of enthalpy.

Question 30

Exergy (Availability)

Answer 30

Exergy is the amount of work obtainable when some matter is brought to a state of thermodynamic equilibrium with the local environment. If the processes that release energy are reversible (no friction and heat losses), all of the energy released will all be available to do work.

Question 31

Closed System Exergy

Answer 31

Is defined by the difference in internal energy, entropy and specific volume. The subscript L refers to the properties of the low temperature reservoir (often but not necessarily the local environment), which defines the limits of cooling and expansion.

Question 32

Maximum Work (Closed System Exergy)

Answer 32

Calculates the maximum useful work from the difference of the starting and ending availability functions. When the starting and ending conditions are not the local environment.

Question 33

Open System Exergy

Answer 33

For an open system, the steady state availability function, is given by the difference in enthalpies, entropy and incorporates terms for kinetic and potential energy (which cannot be extracted from a closed system). The L subscript refers to the properties of the low temperature reservoir (often the local environment), which defines the limits of cooling and expansion.

Question 34

Maximum Work (Open System Exergy)

Answer 34

When the staring and ending conditions are not the local environment. The maximum useful work is found from the difference of the steady state availability function of the starting and ending conditions.

Question 35

Irreversibility

Answer 35

To achieve maximum work output, both the process within the control volume and the energy transfers between the system and the environment must be reversible. The difference between the maximum and the actual work output, is known as the process irreversibility. It can also be found by multiplying the net entropy production, (considering both the substance and environment) by the temperature of the low temperature reservoir.