Presentation on theme: "ENERGY CONVERSION MME 9617A Eric Savory"— Presentation transcript:
1ENERGY CONVERSION MME 9617A Eric Savory Lecture 8 – Basics of heat exchangersDepartment of Mechanical and Material EngineeringUniversity of Western Ontario
2Heat ExchangersThe most common types of energy conversion systems (e.g. internal combustion engines, gas/steam turbines, boilers) consist of three parts:1. a combustion process generating heat and kinetic energy (K.E.)2. a device for converting K.E. to mechanical(useful) energy3. heat exchangers to recuperate the heateither for heating purposes or to increase efficiency.
3The different applications of heat exchangers require different designs (geometries):
4Heat Exchangers are classified according to their function and geometry: Recuperative: two fluids separated by a solid wall (this is the most common type)Evaporative: enthalpy of evaporation of one fluid is used to heat or cool the other fluid (condensers/evaporators and boilers)Regenerative: use a third material which stores/releases heatGeometry: 1. Double Tube2. Shell and Tube3. Cross-flow Heat Exchangers4. Compact Heat Exchangers
5Underlying calculation approach The heat transfer rate for most heat exchangers can be calculated using the LMTD-method (Log Mean Temperature Difference), if the inlet (T1) and outlet (T2) temperatures are known:U = Overall heat transfer coefficient [ W/m2-oC ]A = Effective heat transfer surface area [ m2 ]F = Geometry correction factor= Log mean temperature difference
6Otherwise, the Effectiveness () – Number of Transfer Units (NTU) method may be used:
7General Formulation for Heat Exchanger Analysis (LMTD-method) Most heat exchangers are characterized relative to a double-pipe heat exchanger (H = Hot, C = Cold):T1T2
8This will be done by considering the first law (for counter flow): We now want to derive the expression for LMTD for a counter-flow double-pipe heat exchanger.This will be done by considering the first law (for counter flow):First globally:Then locally: Apply the first law between points 1 and 2 (for counter-flow)Heat lost by hot side = Heat gained by cold side
9For counter-flow:By using the notation 1 and 2, as shown on the graphs, this definition is valid for both Counter-current and Co-flow (parallel) double-pipe heat exchangers.
10ξ-NTU (Effectiveness – Number of Transfer Units) Method If the inlet or outlet temperatures are not given, the LMTD-method becomes cumbersome to use. It is thus advisable to use the Effectiveness-NTU method. The method can be formulated fromthe following definitions:Effectiveness:Minimum thermal capacity max. temp. difference
11In general:Actual heat transfer is given byTheoretical maximum heat transfer by:Hence, we obtain the effectiveness as:
12For a counter-flow heat exchanger: LetandWhich, on using the definition for LMTD, leads to an expression for the effectiveness as:
13If, insteadthenWe end up with the same effectiveness:
15Similar expressions are used for other types of geometry. For example, for a parallel double-pipe heat exchanger, the effectiveness is:Next we shall look at some applications of these concepts.
16Typical thermal design problems Given the entrance temperature of the two streams, given one exit temperature;Find heat transfer area, A.Problem #2Given entrance temperature of the two streams, given the heat transfer area, A;Find the exit temperatures of the two streams.
17Objective: Calculation procedure and advantages / disadvantages of: Double pipeShell and tubeCross flow heat exchangers1. Double Pipe Heat Exchangers:
20Advantages:- low pressure loss- small applications (simple, cheap to build)- counter flow: high effectiveness; parallel flow:quick (short) fetches.Disadvantage:requires large surface area (footprint on floor) iflarge heat transfer rates are needed.
23Advantages:- ideal for large scale applications- commonly used in petrochemical industry wheredangerous substances are present (protectiveshell)- compact design or double tube heat exchanger.Disadvantages:- very bulky (heavy construction), baffles are usedto increase mixing- subject to water hammer and corrosion (behindbaffles)- high pressure loses (recirculation behind baffles)
24Heat transfer calculations: Using counter flow, double pipe heat exchanger definition for the temperatures
25Heat exchanger correction factor plot for one shell pass and an even number of tube passes =+
26Heat exchanger correction factor plot for two shell passes and twice an even number of tube passes
27For n-shell passes with an even number of tubes: Again, for boiling or evaporation R 0so that = 1 – e-NTU
28Cross flow and compact heat exchangers Overview: Cross-flow and compact heat exchangers are used where space is limited. These aim to maximize the heat transfer surface area.Cross-flow Heat Exchangers:Commonly used in gas (air) heating applications. The heat transfer is influenced by whether the fluids are unmixed (i.e. confined in a channel) or mixed (i.e. not confined, hence free to contact several different heat transfer surfaces).e.g.: both fluids unmixed: air-conditioning devicese.g.: both fluids mixed: boilers
29Advantage: large surface area-good for transferring heat to gases Disadvantages: heavy, high pressure lossesCross-Flowsmay be mixedor unmixedIn a cross-flow heat exchanger the direction of fluids are perpendicular to each other. The required surface area, Across for this heat exchanger is usually calculated by using tables. It is between the required surface area for counter-flow (Acounter) and parallel-flow (Aparallel) i.e. Acounter< Across <Aparallel
31Cross-flow heat exchangers have the same analysis equations as before: with F as the correction factor (see graphs). The -NTU method may also be used
32Heat exchanger correction factor plot for single pass, cross-flow with one fluid mixed
33Heat exchanger correction factor plot for single pass, cross-flow with both fluids unmixed
34Compact heat exchangers: These are cross-flow heat exchangers characterized by very large heat transfer area per unit volume. In fact, the contact area is so large that much of the flow behaves as duct or channel flow.For this reason, the heat-transfer is dominated by wall effects and the characteristics cannot be evaluated as for the other types.For these heat exchangers, the heat transfer rate is directly related to pressure loss.
35Advantages:very smallideal for transferring heat to / from fluids withvery low conductivity or where theheat transfer must be done in very smallspaces (e.g. electronic component cooling,cryogenic cooling, domestic furnaces).Disadvantages:high manufacturing costsvery heavyextremely high pressure losses.
37To solve problems involving design and selection (sizing) of compact heat exchangers it is first required to find the effective pressure (static) loss. This loss can be shown, based on fundamental heat transfer principles, to be directly related to the heat transfer rate based on Colburn’s analogy:f – friction factor, St – Stanton number,Pr – Prandtl number and jH = Colburn factor
38These calculations can be quite involved and so most design or sizing applications use data in tables and graphs.All material properties are calculated at the bulk average temperature, i.e. at (T1+T2)/2, if T1 = inlet, T2 = exit
39Reynolds number at the smallest diameter: DH = hydraulic diameter at smallest cross-section= 4 Ac / PAc = smallest cross-sectional areaP = perimeter (circumference) of tubeµ = dynamic viscosity = thermal diffusivityG = maximum mass flow rate flux= mass flow rate
40 = ratio of open area to total frontal area (A) h = heat transfer coefficientCp = specific heat capacityp = pressure loss through heat exchangerVm = (V2 + V1) / 2
41Overall heat transfer coefficient UA is computed from: (h A)h = hot fluid(h A)c = cold fluidA = effective heat transfer areaThen the heat transfer Q is:
42Heat transfer and friction factor for a finned flat tube heat exchanger
43Heat transfer and friction factor for a finned circulator-tube heat exchanger (details on next slide)
50Example 1 – Finned flat tube heat exchanger Air at 1 atm and 300 K enters a finned flat tube heat exchanger (as in graph in an earlier slide) with a velocity of 15 m/s. Calculate the heat transfer coefficient (h).Note at this temperature the air properties (found from tables) are: = kg/m3 = x 10-5 kg/msCp = kJ/KgoCPr = 0.708
51Example 2 – Shell and tube heat exchanger Hot oil at 100oC is used to heat air in a shell and tube heat exchanger. The oil makes 6 tube passes and the air makes 1 shell pass. 2.0 kg/s of air (specific heat of 1009 J/kgoC) is to be heated from 20 to 80oC. The specific heat of the oil is 2100 J/kgoC and its flow rate is 3.0 kg/s. Calculate the area required for the heat exchanger for U = 200 W/m2oC.
52Example 3 – Finned-tube (both fluids unmixed) cross-flow heat exchanger A finned-tube exchanger is used to heat 2.36 m3/s of air (specific heat of 1006 J/kgoC) at 1 atm from to 29.44oC. Hot water enters the tubes at 82.22oC and the air flows across the tubes, producing an average overall heat transfer coefficient of 227 W/m2oC. The total surface area of the exchanger is 9.29m2. Calculate the heat transfer rate (kW) and the exit water temperature.Note: We don’t know whether the air or the water is the minimum thermal capacity fluid. So try with the air as the minimum fluid first and see if the -NTU equations give a possible solution. If not then we have to use water as the minimum and iterate to a solution.