IDEAL CYCLE - SIMPLIFIED
Internally Reversible Processes (not exactly REAL, but solvable)
HEAT ENGINES = Power Cycles
IDEAL, SIMPLIFIED CYCLE:
Ts Diagrams are used to study heat engines. Recall:
Increase efficiency by raising TH and/or by lowering TL.
Practical Limits:
ASSUMPTIONS:
COLD AIR STD Assumption:
specific heat is constant (evaluated at 25oC)
Compression Ratio: 
[Eqn 7-3]
Mean Effective Pressure, MEP WNET = (MEP) DV
Increasing MEP gives more work for a given engine displacement.
Processes (Fig. 7-13):
Two of these processes are isometric (constant v)
Two of these processes are isentropic (constant s)
On a Ts diagram (Fig. 7-15)
First Law Analysis:
qIN = u3-u2 = CV(T3-T2)
qOUT = -(u1-u4) = CV(T4-T1)
[Eqn
7-7]
and since v2=v3 and v4=v1
(always use absolute
temperatures)
so the compression ratio, 
[Eqn
7-9]
and 
[Eqn 7-8]
To increase hth,OTTO, just raise the compression ratio,r
which also give more wNET per cycle,
BUT, to raise r, you must raise P3, which also increases T3. This can cause autoignition (engine knock)
Leaded Fuels allowed higher pressure operation, without engine knock, but with pollution
Unleaded Fuel required lower compression ratios. After that, performance improvements came from lower weight, drag, friction, etc.
On a Pv diagram:
However, since process 2-3 is constant
pressure (not constant v):
BE CAREFUL, the compression ratio,
The Cutoff Ratio relates the volume in the cylinder before and after the combustion process.
Power is produced by expanding combustion gas in
a turbine
Both the heat addition (2-3) and the heat rejection (4-1) processes occur at constant pressure
Since both these processes are steady-flow devices:
qIN = Dh = CP (T3 - T2)
qOUT = Dh = CP (T4 - T1)
For the Brayton cycle (with its two constant
pressure processes), use the pressure ratio:
and calculate its thermal efficiency:
For gas turbines, the "back work" ratio:
rBW is high because (working with air) the compressor work requirement is high.
Gas Turbines are used:
How do we make processes occur at constant pressure or constant temperature ? (as a saturated mixture)
Carnot cycle under the saturation curve doesn’t work very well.
Processes:
Notice:
ANALYSIS of the CYCLE:
Pump: wPump,IN = h2 - h1 = v (P2 - P1)
(use: h1 = hf @P1 and v = vf @P1)
Boiler: qIN = h3 - h2
Turbine: wTurbine,OUT = h3 - h4
Condenser: qOUT = h4 - h1
for the cycle: wNET = qIN - qOUT = wTurbine,OUT - wPump,IN
Increasing the efficiency (to get more wNET for
the same qIN)
i.e., increase the area inside the process polygon on the Ts diagram
1) Lower the pressure in the condenser
limited by leakage (low, vacuum pressure), more liquid in turbine, available cooling medium
2) Higher superheated temperature in the boiler
limited by material restrictions at high operating temps.
3) Higher pressure in the boiler
limited by pressure vessel code (safety) limits, and gives lower quality, more liquid in the turbine
Notice:
ANALYSIS of the CYCLE:
Pump:
wPump,IN = h2 - h1 = v (P2 - P1)
Boiler: qIN = (h3 - h2) + (h5 - h4)
Turbine:
wTurbine,OUT = (h3 - h4) + (h5 - h6)
Condenser: qOUT = h6 - h1
for the cycle: wNET = qIN - qOUT = wTurbine,OUT - wPump,IN
How do we make processes occur at constant pressure or constant temperature ? (as a saturated mixture)
The Carnot cycle under the saturation curve is unsuitable for refrigeration cycles.
Processes:
Typical Assumptions:
Notice:
ANALYSIS of the CYCLE:
Compressor: wIN = h2 - h1
use: h1 = hg @P1
Condenser: qOUT = h2 - h3
use: h3 = hf @P3
Expansion Valve: h3 = h4
use: h4 to find x4
Evaporator: qIN = h1 - h4
