Thermodynamic cycle
Thermodynamic cycle
Gas Power Cycles
Carnor cycle
**isothermal inversible**
The efficiency of the Carnor cycle
the heat transfer is during the 2 process **isothermal inversible**
so ,the amount of heat input and heat output for the cycle can be expressed as:
we see that the thermal efficiency of a Carnot cycle is:
Un Example of an Air-Standard Cycle
the ideal gas isentropic relation
andideal gas relation
Show the cycle on P-v and T-s Diagrams
For the P-v diagram, it's observed
that processes 2-3, 3-4, and 4-1 are isobaric
or isochoric
, represented by straight
lines. The 1-2 process is an **isentropic compression**
, where volume decreases and pressure increases, which is shown as a curve extending to the upper left direction.
For the T-s diagram, the 1-2 process is isentropic, reflected as
a **vertical line**
. From 2-3, heat is added at constant pressure, **causing**
the temperature and entropy **to rise**
. From 3-4, heat is removed at constant volume, **leading to a decrease**
in temperature and entropy. When heat is removed at constant pressure, both temperature and entropy decrease.
calculate the maximum temperature in the cycle
1-2
- $P_1 = 100kPa, T_1 = 300.15K$
- $P_2 = 1000kPa$,
for the ideal gas isentropic relation
2-3
- $q{in} = 2800kJ/kg = h{2-3},$
- $h_{2-3} = c_p(T_3-T_2)$
- $2800 \mathrm{~kJ} / \mathrm{kg}=(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(T3-579.2\right) \longrightarrow T{\max }=T_3=3360 \mathrm{~K}$
Determine the thermal efficiency
3-4
- form the
**ideal gas relation**
for a fixed mass
the total amount of
**heat rejected**
Note, the positive value of the heat radiated is taken here.
the thermal efficiency:
The ideal Otto Cycle
looking up Table A-17 &
**linear interpolation
& `power output**`
An ideal Otto cycle has a **compression ratio**
of 8. At the beginning of the compression process, air is at 100 kPa and 17°C, and 800 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Accounting for the variation of specific heats of air with temperature, determine
the maximum temperature and pressure that occur during the cycle
1-2
- $P_1 = 100kPa,T_1 = 290K$
- Because the heat capacity changes with temperature need to be considered, data is obtained by looking up Table A-17.
- heat is transferred during the constant volume heat addition process, so we look up u rather than h
by the compression radio: $V{r_2} = \frac {V{r1}} r = 84.51$
From the table A-17, By using the method of
**linear interpolation
** we get:**After obtaining**
the temperature, we can get the pressure using the ideal gas relation.
2-3
- $q_{in} = 800kJ/kg = c_v(T_3-T_2)$
- but the variation of specific can not be neglect, so we check the table A-17
- $u3 = u_2+q{in} = 1275.11kJ/kg$
- $T3 = 1575.1K, V{r 3}=6.108$ by the linear interpolation
the net work output
we need to find the internal energy
of the air at 4
3-4
4-1
由此得到净功:
the thermal efficiency
under the cold-aire-standard assumption:
the mean effective pressure
for the cycle.
- the mean effective pressure:
where
**power output**
- Also, determine the power output from the cycle, in kW, for an engine speed of 4000 rpm (rev/min). Assume that this cycle is operated on an engine that has four cylinders with a total displacement volume of 1.6 L
the total aire mass taken by all four cylinders:
the net work:
there are two revolutions per thermodynamic cycle
The Ideal Diesel Cycle
the temperature and pressure of air at the end of each process
because the constant specific heats, we can use the ideal gaz isentropic relation and ideal gaz relation:
Stirling and Ericsson Cycle
**entropy change of an ideal gas**
thermal efficiency
- we know that
- so the entropy change of an ideal gas during an isothermal process:
- thermal efficiency
- for ecrisson cycle, we have$P1 = P_4,P_2 = P_3$ so we can get $\Delta s{3,4} = \Delta s_{1,2}$
- for Stirling cycle, we have $V1 = V4, V_2 = V_3$ so we can get$\Delta s{3,4} = \Delta s_{1,2}$
- at all, we can calcul the thermal efficiency
with the same result as the Carnot cycle
Brayton Cycle
**the back work radio**
the gas temperature at the exits of the compressor and the turbine
- the variation of specific heats with temperature is to be considered, so we need to look up the data in the table A-17
- Because heat addition and removal are isobaric processes, we use enthalpy instead pf internal
- we can easily know the pressure of $P1,P_2,P_3,P_4$,and we know $T_1$ so we can get $P{r1}$
- with the relation:
- we can get all the data we need
**the back work radio**
- we firstly find the work input to the
compressor
and the work output of theturbine
- Thus, the back work radio is defined as:
- That is , 40.3 percent of the turbine work output is used just to drive the compressror
Actual Gas-Turbine Cycle
**compressor efficiency**
and**turbine efficiency**
from actual work to actual temperature
back work ratio
For actual problems, we calculate the work using the standard model and deal with efficiency
- we dont need to memorize the calculation ordre, as long as we know that the actual input is greater than the theoretical input and the actual output is less than the theoretical output
- Thus:
The air temperature at the turbine exit
- The enthalpy of an isentropic process 3 and actual work can be used to calculate the enthalpy exit of a turbine, which can then be looked up in a table to obtain the temperature
- Then, from Table A–17,
Actual Gas-Turbine Cycle with Regeneration
**effectiveness of regenerator**
effectiveness of regeneration
It is important to understand the meaning of regeneration and calculate the effectiveness of regeneration
Regeneration improves efficiency by reducing the input heat.
Throughout this process, the net power remains unchanged.
This give the efficiency
A Gas Turbine with Reheating and Intercooling
**same pressure ratio**
&**same temperature and enthalpy**
same pressure ratio & same temperature and enthalpy
the work output is maximized when both stages of he compressor and the turbine have the same pressure ratio-the overall pressure radio
the temperature and enthalpy of the air at the exit of each compression stage will be the same:
At inlets: $\quad T_1=T_3, h_1=h_3$ and $T_6=T_8, \quad h_6=h_8$
At exits: $\quad T_2=T_4, h_2=h_4$ and $T_7=T_9, h_7=h_9$
same work input & output
as the result, the work input to each stage of the compressor and the work output from each stage of the turbine will be same:
Simultaneously, the added heat.
Thus:
and
an ideal regenerator with 100 percent effectiveness.
the net work don’ t change
The heat input and the thermal efficiency in this case are
The Ideal Jet-Propulsion Cycle
State 2: air exit the diffuser with a negligible velocity
Process 1-2 isentropic compression of an ideal gas in a diffuser
- $T_2 = 480R$
we can look up the pressure of state 2, but as the unite is not standard, we use the isentropic relations:
State 3: The pressure ratio
Process 2-3 isentropic compression of an ideal gaz
State 5 : the turbine work to be equal to the compressor work
State 6: the aire velocity at the nozzle exit
Efficiency
Vapor and Combined Power Cycles
The Simple Ideal Rankine Cycle
Four Stage
State 1-2
State 2-3
State 3-4
Efficiency
An Actual Steam Power Cycle
isentropic efficiency
The Ideal Reheat Rankine Cycle
Cycle
- State 1:
- $\begin{aligned}& h1=h{f @ 10 \mathrm{psia}}& U1=U{f @ 10 \mathrm{psia}}\end{aligned}$
- State 2:
- $w_{\text {pump,in }}=U_1\left(P_2-P_1\right)$
- $h2=h_1+w{\text {pump,in }}$
- State 3:
- $\left.\begin{array}{l}P_3=600 \mathrm{psia} \T_3=600^{\circ} \mathrm{F}\end{array}\right} \begin{aligned}& h_3=1289.9 \mathrm{Btu} / \mathrm{lbm} \& s_3=1.5325 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\end{aligned}$
- State 4:
- $x4=\frac{s_4-s_f}{s{f g}}=\frac{1.5325-0.54379}{1.00219}=0.9865$
- $h4=h_f+x_4 h{f g}$
- State 5:
- $\left.\begin{array}{l}P_5=200 \mathrm{psia} \T_5=600^{\circ} \mathrm{F}\end{array}\right} \begin{aligned}& h_5=1322.3 \mathrm{Btu} / \mathrm{lbm} \& s_5=1.6771 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\end{aligned}$
- State 6:
- $x6=\frac{s_6-s_f}{s{f g}}=\frac{1.6771-0.28362}{1.50391}=0.9266$
- $h6=h_f+x_6 h{f g}$
Efficiency
Ideal Regenerative Rankine Cycle
Cycle
- state 1:
- $P_1 = 10kPa$ , State 1 is the saturated liquid, we can look up the table
- $h_1 = 191.81kJ/kg$, $v_1 = 0.001010m^3/kg$
- state 2:
- $h_2 = h_1+v_1(P_2-P_1) = 193.01kJ/kg$
- state 3:
- $P_2 = P_3 = 1.2MPa$, State 3 is the saturated liquid, we can look up the table
- $h_3 = 798.33kJ/kg, v_3 = 0.001138M^3/kg$
- state 4:
- $h_4 = h_3+v_3(P_4-P_3) = 814.0344kJ/kg$
- state 5:
- $P_5 = 15MPa, T_5 = 600\degree C$, state 5 is the process overheated,
- $h_5 = 3539.0kJ/kg, v_5 = 0.018185m^3/kg$
- $s_5 = 6.6796
kJ/kg.K$ - $h_5=3583.1 \mathrm{~kJ} / \mathrm{kg}$
- state 6
- from state 5 to state 6, the process is isentropic
- $s_6 = s_5 = 6.6796
kJ/kg.K$ - we have $P_6 = 1.2MPa$, by looking up table $T_6 = 218.45\degree C$
- $h_6=2860.2 \mathrm{~kJ} / \mathrm{kg}$
- state 7
- from state 6 to state 7, the process is isentropic
- $s_7 = s_5 = 6.6796
kJ/kg.K$ - The pressure of state 7: $P_7 = 10kPa$
- we can look up the table: $sf = 0.6492, s{fg} = 7.4996$
- Thus, the quality $x = \frac{s7-s_f}{s{fg}} = 0.8041$
- We can get the enthalpy: $h_7 = 191.81+0.8041\cdot 2392.1 = 2115.29$
The fraction of steam extracted from the turbine
Efficency
Refrigration Cycles
The Ideal Vapor-Compression Refrigeration Cycle
throttling
COP
A Two-Stage Cascade Refrigeration Cycle
steady-flow energy balance
COP
A Two-Stage Refrigeration Cycle with a Flash Chamber
Quality of state 6
Enthalpy at state 9
Work in
$Q_L$
COP
The Simple Ideal Gas Refrigeration Cycle
- with the assumption of ideal gaz, we can calculate all proprieties with one propriety.
- $T1=460 \mathrm{R} \longrightarrow h_1=109.90 \mathrm{Btu} / \mathrm{lbm} \text { and } P{r 1}=0.7913$
- $P{r 2}=\frac{P_2}{P_1} P{r 1}=(4)(0.7913)=3.165 \longrightarrow \begin{aligned}& h_2=163.5 \mathrm{Btu} / \mathrm{lbm} \& T_2=683 \mathrm{R}\left(\text { or } 223^{\circ} \mathrm{F}\right)\end{aligned}$
- $T3=540 \mathrm{R} \longrightarrow h_3=129.06 \mathrm{Btu} / \mathrm{lbm} \text { and } P{r 3}=1.3860$
- $P{r 4}=\frac{P_4}{P_3} P{r 3}=(0.25)(1.386)=0.3456 \longrightarrow \begin{aligned}& h_4=86.7 \mathrm{Btu} / \mathrm{lbm} \& T_4=363 \mathrm{R}\left(\text { or }-97^{\circ} \mathrm{F}\right)\end{aligned}$