1 Lec 14: Heat exchangers. 2 Heat Exchangers and mixing devices Heat exchangers are devices which transfer heat between different fluids Mixing devices.

Slides:



Advertisements
Similar presentations
ME 200 L19: ME 200 L19:Conservation Laws: Cycles HW 7 Due Wednesday before 4 pm HW 8 Posted Start early Kim See’s Office ME Gatewood Wing Room
Advertisements

EGR 334 Thermodynamics Chapter 4: Section 6-8
ME 210 Advanced Thermodynamics
Entropy balance for Open Systems
The First Law of Thermodynamics
Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
First Law of Thermodynamics - Open Systems
Chapter 4 Mass and Energy Analysis of Control Volumes (Open Systems)
Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Lecture# 9 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
CHAPTER 5: Mass and Energy Analysis of Control Volumes
Lec 13: Machines (except heat exchangers)
Chapter 7 Entropy (Continue).
Components of Thermodynamic Cycles
Lec 19: Entropy changes, relative pressures and volumes, work
Lec 18: Isentropic processes, TdS relations, entropy changes
Exergy: A Measure of Work Potential Study Guide in PowerPoint
Mass and Energy Analysis of Control Volumes. 2 Conservation of Energy for Control volumes The conservation of mass and the conservation of energy principles.
Unit Four Quiz Solutions and Unit Five Goals
Quiz Twelve Solutions and Review for Final Examination
Lec 12: Closed system, open system
Chapter 5 Mass and Energy Analysis of Control Volumes Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 5th edition.
Lec 23: Brayton cycle regeneration, Rankine cycle
HEAT EXCHANGERS Day 2.
EGR 334 Thermodynamics Chapter 4: Section 9-10
Thermodynamics I Chapter 6 Entropy Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi Malaysia.
ME 200 L18: ME 200 L18:Conservation Laws: Heat Exchangers HW 7 Posted Due in One Week: Kim See’s Office ME Gatewood Wing Room
First Law of Thermodynamics
5. MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
4 CHAPTER The First Law of Thermodynamics: Control Volumes.
The First Law of Thermodynamics
CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
1 So far… We’ve developed a general energy balance We’ve developed a general material balance We’ve only actually looked at systems that are under steady.
CHAPTER 5: Mass and Energy Analysis of Control Volumes

Chapter 7 Energy and Energy Balance By : Mrs. Norazian Mohamed Noor
Chapter 4: Entropy: an Additional Balance Equation.
Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Basic Thermodynamics Chapter 2. We’ve been here before  The material in this chapter is a brief review of concepts covered in your Introductory Engineering.
The Second Law of Thermodynamics Entropy and Work Chapter 7c.
Mass and Energy Analysis of Control Volumes – Part 2 Chapter 5b.
WCB/McGraw-Hill © The McGraw-Hill Companies, Inc.,1998 Thermodynamics Çengel Boles Third Edition 4 CHAPTER The First Law of Thermodynamics: Control Volumes.
Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering Summary of Energy Topics Chapter 1: Thermodynamics / Energy Introduction Chapter 2: Systems.
Mass and Energy Analysis of Control Volumes Chapter 5a.
Chapter 3: Conservation of Energy. Important Notation 2.
Chapter 5 Part 2 Mass and Energy Analysis of Control Volumes Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 8th edition.
Chapter 12B: PROPERTY TABLES, REFRIGERATION CYCLES AND HX 1) Boiling of pure substances: water and steam tables 2) Refrigerant tables 3) Binary mixtures.
1 Chapter 5 Mass and Energy Analysis of Control Volumes.
First Law of Thermodynamics applied to Flow processes
Chapter 8 Exergy: A Measure of Work Potential Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 8th edition by Yunus.
Chapter 14 Gas-Vapor Mixtures and Air-Conditioning Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 8th edition.
Objectives Develop the conservation of mass principle.
Chapter: 06 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES.
Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
7–12 ISENTROPIC EFFICIENCIES OF STEADY-FLOW DEVICES
Chapter 8 Exergy: A Measure of Work Potential Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 5th edition by Yunus.
Chapter 6 Energy and Energy Balance
ES 211: Thermodynamics Tutorial 5 & 6
Chapter 5 The First Law of Thermodynamics for Opened Systems
Chapter 7 Entropy: A Measure of Disorder
Chapter 5 Mass and Energy Analysis of Control Volumes Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 6th edition.
Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Mass and Energy Analysis of Control Volumes (Open Systems)
Reading Materials: Chapter 9
Chapter Seven: Entropy
Energy Transfer by Heat and Work
Chapter Four: Energy transfer by Heat, Work and Mass
Chapter Seven: Entropy
4 CHAPTER The First Law of Thermodynamics: Control Volumes.
Internal Combustion Engineering Division
Presentation transcript:

1 Lec 14: Heat exchangers

2 Heat Exchangers and mixing devices Heat exchangers are devices which transfer heat between different fluids Mixing devices (also called open heat exchangers) combine two or more fluids to achieve a desired output, such as fluid temperature or quality

3 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

4 Heat exchangers can take a variety of shapes

5 Condenser/evaporator for heat pump

6 Cooling towers are a type of heat exchanger.

7 Something a little closer to home..

8

9 Heat Exchangers In a closed heat exchanger, the fluids do not mix. This is a shell-and-tube heat exchanger.

10 Heat Exchangers Your book has very simple examples of heat exchangers. One is counterflow where the fluids flow in opposite directions in the heat exchanger:

11 Heat Exchangers Another type is parallel flow, where the fluids flow in the same direction:

12 Heat Exchangers Yet another type is cross-flow, shown below. These are common in air conditioning and refrigeration systems.

13 With heat exchangers, we have to deal with multiple inlets and outlets

14 TEAMPLAY At steady flow, what is the relationship between

15 What assumptions to make? Ask yourself: See any devices producing/using shaft work? What about potential energy effects? What about kinetic energy changes? Can we neglect heat transfer?

16 Apply conservation of mass on both streams... If we have steady flow, then: AndFluid A Fluid B

17 Conservation of energy can be a little more complicated... I’ve drawn the control volume around the whole heat exchanger. Implications: No heat transfer from the control volume. Fluid A Fluid B

18 Conservation of energy looks pretty complicated: We know from conservation of mass:

19 Conservation of energy equation for the heat exchanger Apply what we know about the mass flow relationships:

20 Heat Exchangers Generally, there is no heat transfer from or to the heat exchanger, except for that leaving or entering through the inlets and exits. So, And, because the device does no work, Also, potential and sometimes kinetic energy changes are negligible.

21 Heat Exchangers - apply assumptions

22 Heat Exchangers After throwing away a bunch of terms, we’re left with: The energy change of fluid A is equal to the negative of the energy change in fluid B.

23 TEAMPLAY How would the energy equation differ if we drew the boundary of the control volume around each of the fluids?

24 Heat Exchangers Now if we want the energy lost or gained by either fluid we must let that fluid be the control volume, indicated by the red.

25 Heat Exchangers The energy equation for one side: Or dividing through by the mass flow:

26 Example Problem Refrigerant 134a with a mass flow rate of 5 kg/min enters a heat exchanger at 1.2 MPa and 50C and leaves at 1.2 MPa and 44C. Air enters the other side of the heat exchanger at 34 C and 1 atmosphere and leaves at 42 C and 1 atmosphere. Calculate: a) the heat transfer from the refrigerant in (kJ/min) b) the mass flow rate of the air (kg/min)

27 Draw diagram R134a INLET T 1 =50C P 1 = 1.2 MPa R134a OUTLET T 2 =44C P 2 = 1.2 MPa AIR INLET T 3 =34C P 3 = 101 kPa AIR OUTLET T 3 =42C P 3 = 101 kPa

28 State assumptions Steady state, steady flow No work Air is ideal gas Kinetic energy change is zero Potential energy change is zero

29 Start analysis with R134a Apply assumptions We can get h 1 and h 2 from tables. The refrigerant mass flow is given.

30 From R134a tables h 1 = kJ/kg h 2 = kJ/kg Plugging back into energy equation:

31 On to part (b) of the problem. We want to get the mass flow of the air... Start by writing the energy equation for the air side: Simplify 000

32 Sample Problem, Con’t If air is an ideal gas, then we can rewrite the enthalpy difference as: Rearrange to solve for mass flow: How do we get the heat transfer rate to/from the air?

33 Almost there!!!!! We can write: so Get specific heat from table: Plug in numbers from here:

34 TEAMPLAY Work problem 5-100E

35 Open heat exchangers (mixers) In an open heat exchanger, the fluids mix.

36 The mixer may look more like a tank

37 Apply conservation equations for steady state, steady flow Mass Energy

38 Look at some typical assumptions Steady state, steady flow No work No heat transfer - not always true Kinetic energy change is zero - usually Potential energy change is zero - usually

39 Look at the impact on the energy equation Physically, what does this equation tell you?

40 Sample Problem Water is heated in an insulated tank by mixing it with steam. The water enters at a rate of 200 lbm/min at 65°F and 50 psia. The steam enters at 600°F and 50 psia. The mixture leaves the tank at 200°F and 48 psia. How much steam is needed?

41 Draw system with knowns: T 1 = 65°F P 1 = 50 psia Water Steam T 2 = 600°F P 2 = 50 psia T 3 = 100°F P 3 = 50 psia Outlet

42 Assumptions Steady state, steady flow No work No heat transfer Kinetic energy change is zero Potential energy change is zero

43 Applying assumptions gives us: Mass Energy We have two unknowns (m 2 and m 3 ) and two equations.

44 Get some properties and complete the solution h 1 = 33.1 btu/lb m h 2 = btu/lb m h 1 = 33.1 btu/lb m

45 TEAMPLAY Work problem 5-96