DESIGN AND COST ANALYSIS OF HEAT TRANSFER EQUIPMENTS

Transcription

1 DESIGN AND COST ANALYSIS OF HEAT TRANSFER EQUIPMENTS Md. Khairul Islam Lecturer Department of Applied Chemistry and Chemical Engineering. University of Rajshahi.

2 What is design? Design includes all the aspects involved in the development of either a new modified or expanded equipment or plant. Design of heat transfer equipment includes all the aspects involved in the development of either a new, modified or expanded heat transfer equipment. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 2

3 Heat Transfer Equipment Usual terminology: Exchanger: heat exchange between two process streams. Heater or Cooler: a process stream is heated/cooled by a utility stream. Vaporiser: a process stream is completely vaporised. Reboiler: vaporiser associated with a distillation column. Evaporator: used to concentrate a solution. Fired heater: heating is done by combustion. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 3

4 Tube and shell heat exchanger 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 4

5 Shell and Tube Heat Exchangers The most commonly used heat exchangers. Advantages: Large surface area in a small volume. Good mechanical layout. Uses well established fabrication methods. Can be constructed from a wide variety of materials. Easily cleaned and maintained. Well established design procedures. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 5

6 Approach to Heat-Exchanger Design Design of a heat exchanger is an iterative (trial & error) process. Here is a set of steps for the process: 1. Calculate the required heat transfer rate, Q, in Btu/hr from specified information about fluid flow rates and temperatures. 2. Make an initial estimate of the overall heat transfer coefficient, U, based on the fluids involved. 3. Calculate the log mean temperature difference, ΔTm, from the inlet and outlet temperatures of the two fluids. 4. Calculate the estimated heat transfer area required, using: A = Q/(U ΔTm). 5. Select a preliminary heat exchanger configuration. 6. Make a more detailed estimate of the overall heat transfer coefficient, U, based on the preliminary heat exchanger configuration. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 6

7 Approach to Heat-Exchanger Design (contd.) 7. Estimate the pressure drop across the heat exchanger. If it is too high, revise the heat exchanger configuration until the pressure drop is acceptable. 8. If the new estimate of U is significantly different than the previous estimate, repeat steps 4 through 7 as many times as necessary until the two estimates are the same to the desired degree of accuracy. 9. Input information needed. In order to start the heat exchanger design process, several items of information are needed as follows: The two fluids involved need to be identified. The heat capacity of each fluid is needed. The required initial and final temperatures for one of the fluids are needed. The design value of the initial temperature for the other fluid is needed. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 7

8 Considerations in selection of heat-transfer equipment Tube size: Length is standard, commonly 8, 12 or 16 ft. Diameter: most common 3/4 or 1 in OD. (Pipes are used to transport something, and tubes to construct something; hence, tubes are defined by the outside diameter and wall thickness (for construction stability), and pipes are measured by inside diameter to allow a calculation for transportation viz., speed, volumes etc) 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 8

9 Considerations in selection of heat-transfer equipment Tube pitch and clearance: Pitch is the shortest center-to-center distance between adjacent tubes. Commonly 1.25 to 1.5 time the tube diameter. Clearance is the distance between tubes. It should be larger than 25% of the tube diameter. Minimum 3/16 inch Triangular or square arrangement of tubes are quite common. P T = d o + C P T = tube pitch d o = tube outside diameter and C = clearance 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 9

10 Considerations in selection of heat-transfer equipment Baffles: Baffles are usually spaced between 20% and 100% of the ID of the shell. Shell: Up to 24 in nominal size, use standard pipes. In general, a shell thickness of 3/8 in. is used for shell diameters between 12 and 24 in. unless the fluids are extremely corrosive. Nominal Pipe Size (NPS) is a North American set of standard sizes for pipes. Nominal pipe size (NPS) for diameter based on inches. Nominal Pipe size NPS [inches] Nominal Diameter DN [mm] 1/ / / Outside Diameter OD [mm] 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 10

11 Considerations in selection of heat-transfer equipment Passes: Most usual pass is one. Split flow arrangement are used for pressure drop reduction, when the pressure drop is the controlling factor in the design. Fluid location: Corrosive fluids flow inside the tubes. Fluid with higher fouling tendency inside the tubes. High pressure fluid inside the tubes (if everything else the same). Hot fluid inside the tubes. Typical velocities: Liquids: 1-2 m/s in tubes, max 4 m/s to reduce fouling. 0.3 to 1 m/s in shell Vapors: m/s (vacuum), m/s (1 bar), 5-10 m/s (high P) 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 11

12 Considerations in selection of heat-transfer equipment Cleaning and Maintenance Use of Water in Heat Exchangers Use of Steam in Heat Exchangers Ineffective Surface Thermal Strains 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 12

13 Fouling Factor Fouling is the build up of sediments and debris on the surface area of a heat exchanger that inhibits heat transfer. Fouling will reduce heat transfer, impede fluid flow, and increase the pressure drop across the heat exchanger. The design engineer can include a resistance, called a dirt, scale, or fouling factor, when determining an overall coefficient of heat transfer. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 13

14 Types of Fouling There are several types of fouling, each forming depending on the type of fluid and conditions. Crystallization is one of the most common type of fouling. Certain salts commonly present in natural waters have a lower solubility in warm water than cold. Ex. Cerium (III) sulphate, lithium selenite etc [Common Solution: reducing the temperature of the heat transfer surface often softens the deposits] Sedimentation, the depositing of dirt, sand, rust, and other small matter is also common when fresh water is used. [Common Solution: velocity control] Biological Organic growth material occurs from chemical reactions, and can cause considerable damage when built up. [Common Solution: material selection] Chemical Reaction Coking appears where hydrocarbon deposits in a high temperature application. [Common Solution: reducing the temperature between the fluid and the heat transfer surface] Corrosion can destroy surface areas of the heat exchangers, creating costly damage. [Common Solution: material selection] Freezing Fouling results from overcooling at the heat transfer surface causing solidification of some of the fluid stream components. [Common Solution: reducing the temperature gradient between the fluid and the heat transfer surface] 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 14

15 Mechanisms of Heat Transfer: 1. Conduction: where Q is the amount of heat, Btu, transferred in timeθ, h k is the thermal conductivity, Btu/[h ft2 (of/ft)] A is the area of heat transfer normal to heat flow, ft 2 t is the temperature, of x is the thickness or length of the conduction path, ft. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 15

16 Mechanisms of Heat Transfer: 2. Convection: h is the heat transfer coefficient, Btu/[h ft 2 o F]. 3. Radiation: where σ is the Stefan-Boltzmann constant = Btu/(h ft 2 o R 4 ) ε is the emissivity of surface A is the exposed area for heat transfer, ft 2 T is absolute temperature, o R. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 16

17 Mechanisms of Heat Transfer: Rate of heat interchange between two bodys, Or Where, Fa is a correction factor based on the relative orientation of the two surfaces and the surface chosen for the evaluation of A, and Fe is a correction factor based on the emissivities and absorptivities of the surfaces. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 17

18 Mechanisms of Heat Transfer: q radiated from pipe= h r A 1 Δt f where h, is a fictitious heat-transfer coefficient based on the rate at which radiant heat leaves the surface of the pipe. q radiated from pipe= h r A 1 (T 1 -T 2 ) 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 18

19 Comparison For conduction for convection For radiation q= h r A 1 (T 1 -T 2 ) 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 19

20 Overall Heat Transfer Coefficient Definition of the overall heat transfer coefficient, U Or, U= q/aδt q =U A Δt U [=] Btu/(h ft 2 o F) 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 20

21 Overall Heat Transfer Coefficient Overall resistance, R=1/UA Resistances in series: Overall resistance = Sum of resistances In this case: U d : overall heat transfer coefficient h, h : outside/inside film heat transfer coefficient Af, Af outside/inside pipe diameter k: wall thermal conductivity h d, h d outside/inside fouling heat transfer coefficient 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 21

22 Temperature difference The log mean temperature difference (DTm or LMTD) can be calculated from the terminal temperatures of the two streams if the following assumptions are valid: All elements of a given fluid stream have the same thermal history in passing through the exchanger.* The exchanger operates at steady state. The specific heat is constant for each stream (or if either stream undergoes an isothermal phase transition). The overall heat-transfer coefficient is constant. Heat losses are negligible. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 22

23 Mean area of heat transfer 1. Conduction of heat through a solid of constant cross section (example, a large flat plate). A m =A arith.avg = (A 1 +A 2 )/2 2. Conduction of heat through a solid when the cross-sectional area of heat transfer is proportional to the radius (example, a long hollow cylinder). 3. Conduction of heat through a solid when the cross-sectional area of heat transfer is proportional to the square of the radius (example, a hollow sphere). A m =A geom.mean = (A 1. A 2 ) 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 23

24 Film Coefficients for Fluids in Pipes and Tubes (No Change in Phase) 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 24

25 Film Coefficients for Fluids Flowing Outside Pipes and Tubes (No change in Phase) where a, = 0.33 if tubes in tube bank are staggered and 0.26 if tubes are in line F, = safety factor to account for bypassing effects G, = shell-side mass velocity across tubes, based on minimum free area between baffles at shell axis? lb/(hxft 2 > Subscript f refers to properties at average film temperature 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 25

26 Tube-side Pressure Drop f i is the friction factor for isothermal flow at the mean temperature n p is the number of tube passes g c is the conversion factor in Newton s law of motion, (32.17) X (3600) 2 ft.lb m /h 2 lb f φ i is a correction factor for non-isothermal flow φ i =1.1( μ i / μ w ) 0.25 for Re < 2100 φ i =1.02( μ i / μ w ) 0.14 for Re > 2100 B i is a correction factor for friction due to contraction, expansion and reversal of flow direction = = 1 + (Fe + Fe + F r )/(2f i G 2 L/g c p i2 D i φ i ). G is the mass velocity inside the tube. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 26

27 Shell-side Pressure Drop Where f = special friction factor for shell-side flow N r = number of rows of tubes across which shell fluid flows B 0 = correction factor to account for friction due to reversal in direction of flow, recrossing of tubes, and variation in cross section; when the flow is across unbaffled tubes, B 0 can be taken as 1.0 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 27

28 HEAT-TRANSFER EQUIPMENT COSTS The major factors that can influence the cost for heat transfer equipment are indicated in the following list: 1. Heat-transfer area 2. Tube diameter and gauge 3. Tube length 4. Pressure 5. Materials of construction for tubes and shell 6. Degree and type of baffling 7. Supports, auxiliaries, and installation 8. Special features, such as floating heads; removable bundles; multipass, finnedsurfaces; and U bends 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 28

29 Problems and Solutions Calculate a preliminary estimate of the heat exchanger area needed to cool 55,000 lb/hr of a light oil (specific heat = 0.74 Btu/lb. F) from 190 F to 140 F using cooling water that is available at 50 F. The cooling water can be allowed to heat to 90 F. An initial estimate of the Overall Heat Transfer Coefficient is 120 Btu/hr.ft². F. Also estimate the required mass flow rate of cooling water. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 29

30 Problems and Solutions Solution: required heat transfer rate for the above indicated light oil: Q = m.cp. (T 2 T 1 ) Imperial Units: Q = (55,000 lb/hr) (0.74 Btu/lb. F) ( ) F = 2,035,000 Btu/hr. Next calculate the log mean temperature difference: T 1 = Inlet tube side fluid temperature (light oil hot side = 190 F) t 2 = Outlet shell side fluid temperature (water cold side = 90 F) T 2 = Outlet tube side fluid temperature (light oil cold side = 140 F) t 1 = Inlet shell side fluid temperature (water cold side = 50 F) ΔTm=(190-90)-(140-50)/[ ln (190-90)/(140-50)] 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 30

31 Problems and Solutions The preliminary area estimate of the heat exchanger can now be calculated as A = Q / (U x ΔTm) = A = [2, 035, 000 Btu/hr]/[ (120Btu/h.ft². F).(94.9 F)]= ft² The required mass flow rate of water can be calculated from Q = m.cp.δtm. Rearranging: m = Q /Cp. ΔTm m = [(2,035,000 Btu/hr)]/[(0.74Btu/lb. F)(94.9 F)]= 28,978 lb/hr 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 31

32 Problems and Solutions A manufacturer is to design a heat exchanger in which the specification takes some steam at 4 bar g (58 psi g) to heat secondary water from 10 C (50 F) to 60 C (140 F). It is assumed that 1 litre of water has a mass of 1 kg. Mass flow rate = 1.5 l/s (24 gpm) x 1 kg/l = 1.5 kg/s (3.30 lb/s). The manufacturer uses a Heat Transfer Coefficient 'U' for the heat exchanger of 2500 W/m2 C (440 Btu/hr.ft². F). Take the Specific Heat of water (cp) as 4.19 kj/kg C = 4190 J/ kg C (1.0 Btu/lb. F). Determine: The design heat load. The corresponding steam flow rate. The minimum heating area required. When the minimum heat load occurs when the inlet water temperature rises to 30 C (86 F), determine: The minimum heat load. The corresponding steam pressure in the heat exchanger. The corresponding steam flow rate. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 32

33 Problems and Solutions A) Find the design heat load using the heat transfer flow rate equation: Q = m.cp. ΔT m Where: Q = Heat transfer flow rate (kj/s) (kw) (Btu/s) (HP) m = mass of steam (kg/s) (lb/s) cp = Specific heat capacity of the secondary fluid (kj/kg C) (Btu/lb. F) ΔT m = Temperature rise of the secondary fluid (K or C) - Q = 1.5 kg/s x 4.19 kj/kg C (60 10) C = kj/s = kw Or Q = 3.30 lb/s x 1.0 Btu/lb. F x (140 F - 50 F) = 297 Btu/s = 420 HP 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 33

34 Problems and Solutions (B) Find the corresponding steam flow rate: At 4 bar g, (39 F) the saturation temperature (Ts) is 152 C (305 F), and he = kj/kg = 2,108,100 J/kg = 906 Btu/lb (from steam tables). Calculate the required steam flow at the design condition using equation below: Metric Units: Steam flowrate (m) =[ kw x 3600 (kg/h)]/ Steam flowrate (m) = [ x 3600 (kg/h) ]/ Steam flowrate (m) = kg/h he Imperial Units: Steam flowrate (m) =[297 x 3600 Steam flowrate (m) = 1180 lb/h (lb/h)]/906 Btu/lb 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 34

35 Problems and Solutions ( C) Use the LMTD (ΔTm) to calculate the minimum amount of heating area to satisfy the design rating: ΔTm = [ (T2 T1)]In (Ts T1)/(Ts T2) ΔTm = Logarithmic Mean Temperature Difference (LMTD) T s = Steam temperature = (152 C) (305 F) T 1= Secondary fluid in temperature = (10 C) (50 F) T 2= Secondary fluid out temperature = (60 C) (140 F) In = The mathematical function known as 'natural logarithm' ΔT m = 115 C= 206 F By re-arranging the general heat transfer equation (Q = U x A x ΔT m ): A = Q U.ΔTm Where: A = Heating area (m ² ) Q = Mean heat transfer rate (W) U = Heat transfer coefficient (W/m ² C) ΔT M = Mean Temperature Difference. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 35

36 Problems and Solutions Metric Units: A = [314,250 W]/[2500 W/m². C x C] A = 1.09 m² Imperial Units: A = [1,072,800 Btu/h]/[440 Btu/hr.ft². Fx 239 F] A = 10.2 ft² (D) Find the minimum heat load, when the inlet water temperature is 30 C: Q = m.cp. ΔT Metric Units: Q = (1.5 kg/s).(4.19 kj/kg. o C) (60 o C - 30 o C) = kj/s = kw Imperial Units: Q = 3.30 lb/s x 1.0 Btu/lb. F x (140 F - 86 F) =178.2 Btu/s = 252 HP 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 36

37 Problems and Solutions (E) Find the corresponding heat exchanger steam pressure and enthalpy at minimum load: 1) A steam temperature of C (239.3 F) corresponds a steam pressure of 0.7 bar g. 2) The Specific Enthalpy of evaporation at 0.7 bar g (he) = kj/kg (see steam tables). (F) Find the steam flow rate at minimum load: From (D) the minimum heat load is kw = 252 HP From (E) the he is kj/kg = 952 Btu/lb Steam flowrate (m) =[kw x 3600kg/h]/he Steam flowrate (m) = [188.5 kw x 3600kg/h]/2 215 kj/kg Steam flowrate (m) = kg/h at minimum load Imperial Units: Steam flowrate (m) = 952 Btu/lb 178 Btu/s x 3600 lb/h Steam flowrate (m) = 673 lb/h - at minimum load. 4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 37