1. Components of Thermodynamic Cycles
What can be the components of thermodynamic cycles?
Turbines, valves, compressors, pumps, heat exchangers (evaporators, condensers), mixers,

2. BUTTERFLY VALVE
GATE VALVE
SPHERE VALVE
PIPE CRACK
GLOBE VALVE

3. Throttling Devices (Valves)
Typical assumptions for throttling devices
No work
Potential energy changes are zero
Kinetic energy changes are usually small
Heat transfer is usually small
Two port device

4. Look at energy equation:
Apply assumptions from previous page:
We obtain: or
Does the fluid temperature:
increase,
decrease, or
remain constant
as it goes through an adiabatic valve?

5. Look at implications: If the fluid is liquid/vapor:
h const.
P const.
During throttling process:
The pressure drops,
The temperature drops,
Enthalpy is constant

6. Look at implications: if fluid is an ideal gas:
Cp is always a positive number, thus:

7. P1 > P2 P S T T v v s Pressão Entalpia Temperatura En. Interna
Entropia
diminui
constante
gás ideal h=h(T), portanto T fica constante
gás ideal u=u(T), portanto u fica constante
diminui, Ds = Cpln(T2/T1)-Rln(P2/P1)

8. 15MPa
611K (340oC)
s, kJ/(kgK)
T (K)
373K (100oC)

9. Cold embrittlement (dureza e rigidez mas baixa resistência a tensão)
T-1000 (Robert Patrick)

10. Cold embrittlement (dureza e rigidez mas baixa resistência a tensão)
Consequences of the Temperature Drop on Material Strenght
Cold embrittlement (dureza e rigidez mas baixa resistência a tensão)
Low temperature embrittlement does affect most materials more or less pronounced. It causes overloaded components to fracture spontaneously rather than accommodating the stress by plastic deformation. The picture shows a fractured fitting whose material was not suitable for low temperatures.

11. TEAMPLAY
Refrigerant 12 enters a valve as a saturated liquid at Mpa and leaves at MPa. What is the quality and the temperature of the refrigerant at the exit of the valve?
State (1)
Liquid saturated, x=0
Psat = MPa
Tsat = ?
Hliq = ?
State (2)
Liq+vap x=?
Psat = MPa
Tsat = ?
Hliq = ?
0.33
40oC
-15oC
75kJ/kg
22 & 180 kJ/kg

12. P-h Diagram of an Ideal Vapor-Compression Refrigeration Cycle
Throttling
Isentropic compression (1 to 2)
Constant pressure condensation (2 to 3)
Isenthalpic expansion (3 to 4)
Constant pressure evaporation (4 to 1)
P-h Diagram of an Ideal Vapor-Compression Refrigeration Cycle
The refrigerant enters the compressor as a saturated vapor and is cooled to the saturated liquid state in the condenser. It is then throttled to the evaporator
and vaporizes as it absorbs heat from the refrigerated space.

13. Heat exchangers are used in a variety of industries
Automotive - radiator
Refrigeration - evaporators/condensers
Power production - boilers/condensers
Power electronics - heat sinks
Chemical/petroleum industry- mixing processes

14. Something a little closer to home..

15. Heat Exchangers

16. Condenser/evaporator for heat pump

17. Heat Exchangers If we have steady flow, then:
Now, we must deal with multiple inlets and outlets:
If we have steady flow, then:

18. Heat Exchangers 0, (sometimes negligible) 0, (usually negligible)
0, (sometimes negligible)
0, (usually negligible)
0, (sometimes negligible)
0, (usually negligible)

19. Heat Exchangers And we are left with
The energy change of fluid A is equal to the negative of the energy change in fluid B.

20. SC Resp.: mw = 2.576 kg/s mw = ? kg/s hw = 100 kJ/kg mvap = 0.1 kg/s
hvap = 2776 kJ/kg
mc = ? kg/s
hc = 200 kJ/kg
Resp.: mw = kg/s

21. Nozzles and Diffusers V1, P1
Nozzle--a device which accelerates a fluid as the pressure is decreased.
V1, P1
V2, P2
Diffuser--a device which decelerates a fluid and increases the pressure.
V1, P1
V2, P2
These configuration is for sub-sonic flow.

22. For supersonic flow, the shape of the nozzle is reversed.
Nozzles

23. General shapes of nozzles and diffusers
Subsonic flow
nozzle
diffuser
Supersonic flow

24. conservation of energy
q = 0 (adiabatic)
w = 0 (these are not work producing devices; neither is work done on them)

25. Sample Problem Diffuser INLET T1 = 300C P1 = 100 kPa V1 = 250 m/s
An adiabatic diffuser is employed to reduce the velocity of a stream of air from 250 m/s to 35 m/s. The inlet pressure is 100 kPa and the inlet temperature is 300°C. Determine the required outlet area in cm2 if the mass flow rate is 7 kg/s and the final pressure is 167 kPa.
INLET
T1 = 300C
P1 = 100 kPa
V1 = 250 m/s
= 7 kg/s
Diffuser
OUTLET
P2 = 167 kPa
V2 = 35 m/s
A2=?

26. Conservation of Mass: Steady State Regime
solve for A2
But we don’t know v2!
Remember ideal gas equation of state?
and
We know T1 and P1, so v1 is simple. We know P2, but what about T2?
NEED ENERGY EQUATION!!!!

27. Energy Eqn
If we assumed constant specific heats, we could get T2 directly
The ideal gas law:
And the area:

28. Vapor Power Cycles
We’ll look specifically at the Rankine cycle, which is a vapor power cycle.
It is the primary electrical producing cycle in the world.
The cycle can use a variety of fuels.

32. We’ll simplify the power plant

33. compressor and turbine must handle two phase flows!
TH < TC TH 1 2
Carnot Vapor Cycle 4 TL 3 s
The Carnot cycle is not a suitable model for vapor power cycles because it cannot be approximated in practice.

34. Ideal power plant cycle is called the Rankine Cycle
The model cycle for vapor power cycles is the Rankine cycle which is composed of four internally reversible processes:
1-2 reversible adiabatic (isentropic) compression in the pump
2-3 constant pressure heat addition in the boiler.
3-4 reversible adiabatic (isentropic) expansion through turbine
4-1 constant pressure heat rejection in the condenser

35. Ideal power plant cycle is called the Rankine Cycle
The model cycle for vapor power cycles is the Rankine cycle which is composed of four internally reversible processes:
1-2 reversible adiabatic (isentropic) compression in the pump
2-3 constant pressure heat addition in the boiler.
3-4 reversible adiabatic (isentropic) expansion through turbine
4-1 constant pressure heat rejection in the condenser

36. What are the main parameters we want to describe the cycle?
Net Power
Net Specific Work
Efficiency or

37. Start our analysis with the pump
The pump is adiabatic, with no kinetic or potential energy changes. The work per unit mass is:

38. Boiler is the next component
Boilers do no work. In boilers, heat is added to the working fluid, so the heat transfer term is already positive. So,

39. Proceeding to the Turbine
Turbines are almost always adiabatic. In addition, we’ll usually ignore kinetic and potential energy changes:

40. Last component is the Condenser
Condensers do no work (they are heat exchangers), and if there is no KE and PE,

41. Efficiency

42. Start an analysis:
A Rankine cycle has an exhaust pressure from the turbine of 0.1 bars. Determine the quality of the steam leaving the turbine and the thermal efficiency of the cycle which has turbine inlet pressure of 150 bars and 600C.
P = 15 MPa
P = 0.01 MPa
Assumptions:
Pump and turbine are isentropic;
P2 = P3 = 150 bars = 15 MPa
T3 = 600C
P4 = P1 = 0.1 bars = 0.01 MPa
Kinetic and potential energy changes are negligible

43. Put together property data
Pump (1 to 2) -> isoentropic (const. volume)
Boiler [heat exchanger] (2 to 3) -> const. pressure
Turbine (3 to 4) -> isoentropic
Condenser [heat-exchanger] (4 to 1) -> const. pressure

44. P = 15 MPa
P = 0.01 MPa
Property Data

45. Let start with pump work
P = 15 MPa
P = 0.01 MPa
Let start with pump work

46. More calculations... Enthalpy at pump outlet:
Plugging in some numbers:

47. How Can I Get The Pump Outlet Temp?
If the Enthalpy at pump outlet is KJ/kg, then consider the compressed liquid a the same temperature of the saturated liquid which has h = KJ/kg
Interpolating from the saturated steam table one finds: 49oC

48. P = 15 MPa
P = 0.01 MPa
Calculate heat input

49. P = 15 MPa
P = 0.01 MPa
Turbine work
Isentropic:
Turbine work:

50. Overall thermal efficiency

51. Some general characteristics of the Rankine cycle
Low condensing pressure (below atmospheric pressure)
High vapor temperature entering the turbine (600 to 1000C)
Small backwork ratio (bwr)

52. Questions … Consider the ideal Rankine cycle from 1-2-3-4:
How would you increase its thermal efficiency ?
What determines the upper T limit?
What determines the lower T limit?
Improve the efficiency of the Rankine Cycle: decrease exhaust pressure of turbine
decreases condensing temperature
increases work output
increases heat input
decreases quality at turbine outlet

53. Lowering Turbine Exit Pressure
The average temperature during heat rejection can be decreased by lowering the turbine exit pressure.
Consequently, the condenser pressure of most vapor power plants is well below the atmospheric pressure.

54. Reducing Condenser Pressure
Notice that reducing the condenser pressure (which will lower the temperature of heat rejection and again increase the efficiency) will also reduce the quality of the steam exiting the turbine.
Turbines do not like to see water coming out the exhaust.
Lower qualities mean water droplets are forming before the steam leaves the turbine.
Water droplets lead to turbine blade erosion.
Efforts are made to keep the quality > 90%.

55. Raising Boiler Pressure
The average temperature during heat addition can be increased by raising the boiler pressure or by superheating the fluid to high temperatures.
There is a limit to the degree of superheating, however, since the fluid temperature is not allowed to exceed a metallurgically safe value.

56. Increase Superheat (Boiler Temperature)
Work output increases faster than heat input, so the cycle efficiency increases.
higher quality
Increasing Superheat
* increases heat input
* increases work output
* increases quality at turbine outlet
* may produce material problems
if temperature gets too high

57. Effect of Increasing Boiler Pressure on the Ideal Rankine cycle keeping constant the Boiler Outlet Temperature Tmax

58. TEAMPLAY
Consider a simple ideal Rankine cycle with fixed turbine inlet temperature and condenser pressure. What is the effect of increasing the boiler pressure on
Pump work input:
(a) increases, (b) decreases, (c) the same
Turbine work output:
Heat supplied:
Heat rejected:
Cycle efficiency:
Moisture content at turbine exit:

59. Simple schematic of Rankine reheat cycle

60. The Ideal Reheat Rankine Cycle

61. Reheat on T-s diagram:
Note that T5 < T3. Many systems reheat to the same temp (T5=T3).

62. Cogeneration
The production of more than one useful form of energy (such as process heat and electric power) from the same energy source is called cogeneration.
Cogeneration plants produce electric power while meeting the process heat requirements of certain industrial processes. This way, more of the energy transferred to the fluid in the boiler is utilized for a useful purpose.
The fraction of energy that is used for either process heat or power generation is called the utilization factor of the cogeneration plant.

63. An Ideal Cogeneration Plant

64. Schematic and T-s Diagram for Cogeneration

65. Combined Gas-Steam Power Plant