# UCL WIKI

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• Gas Power Cycles

# Carnot cycle

In 1824 Carnot suggested a particular cycle of operation for a CHPP which avoided all irreversibilities. It consisted of  four processes, two isothermal and two adiabatic. The process take place between a heat source at temperature Th and a heat sink at temperature Tc. The system is a mass of gas behind a piston. The cycle on a p-v diagram is shown below:

In state A the gas is temperature Th and the cylinder is fully insulated

The gas expands adiabatically and very slowly, i.e., quasi-statically (and therefore reversibly). As the gas expands its U decreases (dU=Q-W) and its temperature drops until it reaches Tc.

Isothermal Compression (B to C)

When the gas reaches temperatures Tc the heat reservoir at temperature Tc is brought into contact with the cylinder end. The gas is then compressed quasi statically from state B to C and at constant temperature with heat transfer from the gas to the reservoir through a negilible temperature difference. Thus process B to C is also reversible

At C the heat reservoir at temperature Tc is removed and the insulation put back. Then slow compression from C to D. At D the temperature reaches Th. Again process is reversible

Isothermal expansion (D to A)

At D the heat reservoir at temperature Th is brought into contact with the cylinder and as a result slow isothermal expansion occurs from state D to A, there by completing the cycle.
It is possible to show that the efficiency of this reversible CHPP depends only on temperature.
Thermodynamic temperature scale

Theorem
All reversible cyclic engines operating between the same two temperature level have the same (Maximum) efficiency.

Proof:
Assume opposite is true and show before that PMM2 is produced.
Because all reversible engines, whatever their internal processes or materials, operating between the same two thermal reservoirs have the same efficiency, then there must be some common factor which determines the efficiency. Clearly the only common factor is the temperature of the two reservoirs. Therefore the efficiency of a reversible heat engine depends upon the temperature of the thermal reservoirs that it exchanges energy with. (We shall show this more rigorously for a carnot cycle, with a perfect gas as working substance.)
We mentioned earlier that the zeorth law of TD provides the basis for temperature measurements, but that a temperature scale must be defined in terms of particular thermometer and device. Thus a temp scale that is independent of any particular substance is most desirable. Now the efficiency of a carnot cycle is independent of the working substance and depends only on temperature. This fact may be used to define the TD temperature scale. Now consider the following diagram:

Shows three heat reservoir and three reversible engines working on Carnot Cycle.

T1 is the highest temperature, T3 the lowest and T2 is the intermediate temperature.
Q1 is the same for engines A & C.
The two engines in conjunction (A&B) must have the same efficiency as (c) (they operate between the same temperature levels).
Therefore, 1W2 + 2W3 = 1W3 and  Q3 is the same for both sides.
Since efficiency of Carnot cycle is a function of temperature only then:

Applying these to the different engines we get:

But...
Thus...

Now, the LHS is a function of T1 and T3 (and not T2) and therefore the RHS of the equation must also be a function of T1 and T3 (and not T2) i.e. the function must be

Thus f(T2) will cancel out of the product
Therefore:
In general:
There are several relations that will satisfy this equation, eg:

Logarithmic

Linear   and so on...

Lord Kelvin proposed the Linear relationship ie.
(1)This relationship defines the Thermodynamic  (or Absolute) temperature scale.
With temperature so defined, the efficiency of a carnot cycle may be expressed as
(2)

Note:

(i ) It is impossible to attain negative temperatures on the absolute scale. If TL is negative > 1 i.e. a PMM2 which is impossible.
(ii) It is impossible (in the absence of a perfect insulator) for a finite system to attain zero temperature on the absolute scale. If T­L =0 then =1 i.e. Violates the second law. But we can get very close to T=0.

Units
Equation (1) only defines a ratio of absolute temperature but does not give information about the scale. In this case we need to define only one temperature point, since there is zero temperature.
Mark triple point of water 273.16K
Then temperature of melting ice is 273.16K and that of boiling water at 1 atm 373.15K, thus preserving the 100 units between ice and boiling point so that it corresponds in units to temperature difference in celcius scale.
T(oC) + 273.15 = T(K)
Example - maximum Efficiency
We said the maximum achievable efficiency of heat engine is that of the Carnot (or reversible) heat engine.
i.e.,

The maximum efficiency obtainted if we make TC as low as possible and TH as high as possible.
TC is practically limited by the temperature of the sea or ocean which is nearby the power plant ie 10-20oC or 283-303K. The upper limit is set by the metallurgical properties and at present it is limited to about 600oC for alloyed steel. So the maximum achievable efficiency of an engine is

Or 67%.

Now if we reduce TC  to the boiling point of liquid Helium ie.e.40K we get:

# Gas power cycles

Deal with systems that produce power in which the working fluid remains a gas throughout the cycle ( in other words, there is no change in phase).
Spark Ignition (gasoline) engines, Compression ignition (diesel) engines and conventional gas turbine engines (generally refer to as Internal Combustion engines or IC Engines) are some examples of engines that operate on gas cycles.

## Air standard cycles

Internal combustion engines: Combustion of fuel is non-cyclic process. Working fluid, air-fuel mixture undergoes permanent chemical change due to combustion Products are thrown out of the engine & Fresh charge is taken in.

Hence, the working fluid doesn't undergo a thermodynamic cycle.In order to analyze this complex gas power cycles, air standard cycles are conceived.
In air standard cycle a certain mass of air operates in a complete thermodynamic cycle where the heat is added and rejected using external reservoirs, and all the processes in the cycle are reversible.

Summary of assumptions made during such analysis:The working fluid, air behaves like an ideal gas (and specific heats are assumed to be constant)
Combustion process is replaced by heat addition and exhaust process by heat rejection
All the processes are reversible.
Internal combustion engines
There are two types of reciprocating engines:
Spark Ignition- Otto cycle
Compression Ignition-Diesel cycle

## IC Engines overview

Air and fuel mixture flows through inlet valve and exhaust leaves through exhaust valve
Converts reciprocating motion to rotary motion using piston and crank shaft
TDC; Top Dead Center: Position of the piston where it forms the smallest volume
BDC; Bottom Dead Center: Position of the piston where it forms the largest volume
Stroke: Distance between TDC and BDC
Bore: Diameter of the piston (internal diameter of the cylinder)
Clearance volume: minimum volume formed
Compression ratio: ratio of maximum volume to minimum volume VBDC/VTDC
Engine displacement = (# of cylinders) x (stroke length) x (bore area) (usually given in cc or liters)
MEP: mean effective pressure: A const. theoretical pressure that if acts on piston produces work same as that during an actual cycle
Wnet = MEP x Piston area x Stroke

= MEP x displacement volume
4 Stoke engine
Cycle consists of four distinct strokes (processes):
Intake
Compression stroke
Power stroke
Exhaust

## Otto cycle

Assumptions for Air standard cycle, as describe before:
Fixed amount of air (ideal gas) for working fluid
Combustion process replaced by constant volume heat addition with piston at TDC
Intake and exhaust not considered, cycle completed with constant volume heat removal with piston at BDC

All processes considered internally reversible

Air-Standard Otto cycle
Process 1- 2    Isentropic Compression
Process 2 - 3   Const. volume heat addition
Process 3 - 4   Isentropic expansion
Process 4 - 1  Constant volume heat removal

Otto Cycle- indicator diagram of otto cycle

Otto, P-V and T-S diagram

Compression ratio

From previous definition, compression ratio =
Since fixed mass:

1-2 Isentropic compression

Applying First law:

U2-U1 = Q - Win
Q = 0 (since, reversible adiabatic compression)
Win = U2-U1

Applying First law:

U3-U2 = +Qin - W
W = 0 (since, it is a constant volume process)
Qin = U3-U2

3-4 Isentropic Expansion

Applying First law:

U4-U3 = Q - Wout
Q = 0 (rev. adiabatic expansion)
Wout = U4-U3

4-1 Constant volume heat removal

Applying First law:
U1-U4 = - Qout + W
W = 0 (no piston work)
Qout = U4-U1

Otto cycle thermal efficiency

The thermal efficiency is given by:

The specific heats are assumed to be constant.

Here y=1.4 at ambient temperature

For higher efficiency, higher compression ratios are required, as shown below.

However, increase in pressure ratios, would increase the air-fuel temperature above the temperature at which the mixture can auto-ignite.

This would result in 'engine-knock', reducing the performance of the engine. In order to avoid such situations, additives are generally added which increases the auto-ignition temperature.

## 4 Stroke CI engine

Cycle consists of four distinct strokes (processes) as in the case of SI engines, except that the spark plug is replaced by a fuel injector
-        Intake
-        Compression stroke
-        Power stroke
-        Exhaust

Here the fuel is injected when the piston approaches TDC, ie when the air is at maximum temperature due to compression.
The combustion process starts now

The fuel is injected after the piston starts moving down The volume increases, on the other hand, the fuel evaporates to fill the volume. Thus keeping the pressure inside roughly the same.

Hence the combustion can be considered to occur at constant pressure.

## Diesel Cycle

Assumptions for Air standard cycle, as describe before:
-        Fixed amount of air (ideal gas) for working fluid
-        Combustion process replaced by constant pressure heat addition
-        Intake and exhaust not considered, cycle completed with constant volume heat removal with piston at BDC
-        All processes considered internally reversible

Air-Standard Otto cycle
Process 1- 2    Isentropic Compression
Process 2 - 3   Const. pressure heat addition
Process 3 - 4   Isentropic expansion
Process 4 - 1  Constant volume heat removal

Diesel T-S and P-V diagram

Three Volume Ratios
From previous definition:

Thermal Efficiency of Diesel Cycle

Given:

(1)

Process 1-2: Isentropic compression
(2)
(3)
Process 3-4: Isentropic expansion
(4)
Thermal efficiency

From 2, 3 and 4 all temperatures can be expressed in terms of T3.

Otto and Diesel Cycle Comparison
For given rc higher thermal efficiency is obtained via higher compression ratio rv  and for  a given rv higher thermal efficiency is achieved by lowering
the cut-off ratio rc

However a smaller rc yields less net work per cycle, so to achieve the same power at lower rc  values higher engine speeds are required.

Otto and Diesel cycle comparison

Therefore, the efficiency of the diesel cycle is less than that of the otto cycle for the same compression ration. However, the advantages of Diesel over petrol engines is that we can operate at higher compression ratios without auto ignition and fuel is less expensive.

## Gas Turbine Power Plants

Gas turbine power plants are lighter and compact when compared to power plants running on vapour cycles. The power to weight ratios are generally high for high throughout Gas turbine power plants and hence are favoured for the aviation and also for power generation.
A simple GT power plant is shown in the image below.Air is first compressed
The compressed air enters the combustion chamber where fuel is injected and burned, essentially at constant pressure
The combustion products expand in turbine to the ambient pressure and thrown out to the surroundings.

## Air Standard Brayton Cycle

Brayton cycle is the air standard for GT power plant.

Air is first compressed reversibly and adiabatically

Heat is added to it reversibly at constant pressure

Air expands reversibly, adiabatically in the turbine The heat is removed from the system reversibly at constant pressure to bring it to original state

Brayton cycle therefore consists of two isobars and two reversible adiabatics (isentropics):

Air is first compressed reversibly and adiabatically

Heat is added to it reversibly at constant pressure

Air expands reversibly, adiabatically in the turbine The heat is removed from the system reversibly at constant pressure to bring it to original state
P-V, T-S diagram of ideal Brayton Cycle

1 - 2    Isentropic compression
2 - 3    Constant pressure heat addition
3 - 4    Isentropic expansion
4 - 1    Constant pressure heat removal

Thermal efficiency:
The thermal efficiency of the ideal Brayton cycle is

Since processes 1-2 & 3-4 are isentropic between the same pressures :-

Where rv is the pressure ratio

Hence, substituting in the efficiency expression

This is the efficiency for ideal Joule/Brayton Cycle.

Work Ratio

It may easily be shown from the expression,
Work ratio =

And a  similar approach to that above, that work ratio =
What we deduce from the above equations above improvements that we might make?
h is increased by :-
increasing T3
decreasing T4 or
increasing the pressure ratio

We also know that a high work ratio is desirable in order to minimize the effect of irreversibilities in real gas turbines. This depends on the temperature limits and the pressure ratio for constant gamma.

Consider the T-S diagram below for the ideal cycle & the dotted cycles.

T3 is usually fixed by metallurgical limits on turbine blading & T1 is the natural sink temperature for an ideal cooler.  The two dotted cycles show the limits of operation. Consider left hand dotted cycle.  Here the pressure ratio is large & the cycle efficiency approaches the Carnot Efficiency ie T2 has been raised.  Unfortunately the net work output is approaching zero. The other dotted cycle has a reduced T2 &  again net work output is approaching zero. It can be shown that for an ideal cycle with fixed T1 and T3, the value of T2 for maximum work output is:

Irreversibilities and isentropic efficiencies

We  shall only consider the effect of irreversibilities upon compression and expansion processes. Irreversibilities in heaters and coolers who up as pressure drops and are not considered here.
The two T-S diagrams, show the effect on compression and expansion processes in general from state 1 to state 2. These are analogous to the similar diagrams for the Rankine cycle except that they are processes of a perfect gas. Then for the steady flow compression process:-

For the steady flow compression process:

For the steady flow expansion process:

Note that Celsius temperatures may also be used in these expressions.

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