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heat exchanger
This document discusses heat exchangers and calculating the log mean temperature difference (LMTD). It describes different types of heat exchangers including double-pipe, shell and tube with various shell and tube passes. It explains how to calculate the LMTD for countercurrent and co-current flow and introduces correction factors for multi-pass exchangers. An example problem demonstrates calculating the LMTD, heat transfer rate, and area required for a given problem.
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heat exchanger
- 1. DKK1493 HEAT TRANSFER CHAPTER4 HEAT EXCHANGER
- 2. Heat exchangertype The Heat Exchanger Analysis: Log Mean Temperature Difference (LMTD) CONTENTS
- 3. It is expectedthat students will be able: to classify different types of heat exchangers to calculate the log mean temperature difference (LMTD) and determine correction factor for multi-passes heat exchanger. TOPIC OUTCOMES
- 4. TYPE OF HEATEXCHANGER
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- 6. Shell and TubeHeat Exchanger 1 shell pass I tube pass (1-1 exchanger) 1 shell pass 2 tube passes (1-2 exchanger)
- 7. Shell and tubeexchanger Shell inlet Shell outlet Tube inlet Tube outlet Shell inlet Tube inlet Tube outlet Shell outlet
- 8. Cross-Flow Heat exchangers Onefluid mixed, one fluid unmixed Both fluids unmixed Gas flow Gas flow
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- 12. Energy Balances andHeat Transfer Rate Energy balance on hot fluid: Energy balance on cool fluid: Heat transfer rate: ohhihp TTcmq cicocp TTcmq m TUAq
- 13. This Tlm holdsfor a double-pipe heat exchanger and 1-1 exchanger with 1 shell pass and 1 tube pass in parallel or counter-flow. When the hot and cold fluids in a heat exchanger are in true counter-current flow or in co-current (parallel) flow, LOG MEAN TEMPERATURE DIFFERENCE (LMTD) 2 1 21 1 2 12 lnln T T TT T T TT Tlm lm TUAq
- 14. Example An oil whichhas a cpm = 2.30 kJ/kg.K is being cooled in a heat exchanger from 371.9 K to 349.7 K and flows inside the tube at a rate of 3630 kg/h. A flow of water enters at 288.6 K for cooling out and flows outside the tube at 319.1 K. Calculate the heat-transfer area if the U = 340 W/m2.K for each counter-current and co- current flow. Discuss your result. (Ans: 2.66 m2; 2.87 m2)
- 15. For multiple-passheat exchanger, the correction factor FT used to calculate the true mean temperature drop. (under assumption of counter-flow conditions) Two dimensionless ratios are used as follows: For counter-flow: Correct mean temperature: Heat transfer rate: LOG MEAN TEMPERATURE DIFFERENCE (LMTD) CORRECTION FACTORS ciho cohi cihocohi lm TT TT TTTT T ln )( Tlmm FTT cico hohi TT TT Z cihi cico TT TT Y moomii TAUTAUq
- 16. Correction factor FTfor shell and tube heat exchanger with one shell pass and any multiple of two tube passes (2, 4, 6, etc, tube passes) Correction factor FT for shell and tube heat exchanger with two shell pass and any multiple of four tube passes (4, 8, 12, etc, tube passes)
- 18. EXAMPLE 4.9-1: TemperatureCorrection Factor for a Heat Exchanger A 1-2 heat exchanger containing one shell pass and two tubes passes heats 2.52 kg/s of water from 21.1 to 54.4oC by using hot water under pressure entering at 115.6 and leaving at 48.9oC. The outside surface area of the tubes in the exchanger is Ao = 9.30 m2 a) Calculate the mean temperature difference ΔTm in the exchanger and the overall heat-transfer coefficient Uo b) For the same temperature but using a 2-4 exchanger, what would be the ΔTm ? 1 T 2 T hi TC15.61 o C9.48 o ho T C1.21 o ci T co TC4.54 o hot water waterkg/s2.52
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- 21. (b) ExchangerHeat42 W348200 )1.214.54)(4187(52.2 cicopm TTcmq .KW/m1196 )3.31(30.9 3482002 mo o TA q U 94.04b,-4.9FigtheFrom T F K39.8C39.8 )94.0(3.42 o Tlmm FTT
- 22. PROBLEM 4.9-2: CoolingOil by Water in an Exchanger Oil flowing at the rate of 5.04 kg/s (cpm = 2.09 kJ/kg.K) is cooled in a 1-2 heat exchanger from 366.5 K to 344.3 K by 2.02 kg/s of water entering at 283.2 K. The overall heat transfer coefficient Uo is 340 w/m2.K. Calculate the area required. (Hint: A heat balance must first be made to determine the outlet water temperature)