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Analysis of heat exchangers: Use of the log mean temperature Difference

Analysis of heat exchangers: Use of the log mean temperature Difference. LMTD Method: Q= (m cp ∆T) h = (m cp ∆T) c Q= U A F∆T lm A=N װ DL ∆ T lm = ∆ T l -∆T2 / ln (∆ T l /∆T2). This law could be used to evaluate the following:

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Analysis of heat exchangers: Use of the log mean temperature Difference

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  1. Analysis of heat exchangers:Use of the log mean temperature Difference LMTD Method: Q= (m cp ∆T)h = (m cp ∆T)c Q= U A F∆Tlm A=N װDL ∆Tlm = ∆Tl -∆T2 / ln (∆Tl /∆T2)

  2. This law could be used to evaluate the following: 1- Area of heat exchanger 2-The length of heat exchanger 3- The diameter of heat exchanger 4- Number of tubes.

  3. For co-current: • ∆Tl = Th in – T c in • ∆T2 = Th out – T c out For current-current: • ∆Tl = Th in – T c out • ∆T2 = Th out – T c in

  4. Example (1): • A heat exchanger is required to cool 20 kg/s of water from 360 K to 340 K by means of 25 kg/s water entering at 300 K. If U= 2kW/m2 K, Calculate the area required in a counter - current and co- current using one shell pass and 2,4,6 tube passes. Cold water is used in tube and hot water in shell.

  5. Solution • Counter current: • Q= (m cp ∆T)h = (m cp ∆T)c • Q h= (m cp ∆T)h = 20*4.18* (360-340) = 1672 kW • Q c= (m cp ∆T)c = 25*4.18* (T2- 300) = 1672 kW • So T2= 316 K • ∆Tlm = ∆Tl -∆T2 / ln (∆Tl /∆T2) • ∆Tl = 44 • ∆T2 = 40, so ∆Tlm = 41.972

  6. Co-current: • Q= (m cp ∆T)h = (m cp ∆T)c • Q h= (m cp ∆T)h = 20*4.18* (360-340) = 1672 kW • Q c= (m cp ∆T)c = 25*4.18* (T2- 300) = 1672 kW • So T2= 316 K

  7. Tlm = ∆Tl -∆T2 / ln (∆Tl /∆T2) • ∆Tl = 60 • ∆T2 = 24, so ∆Tlm = 39.3 • To calculate F , R and P must be calculated: • P= (t2-t1/T1-t1) = (316-300/360-300 )= 0.267 • R= (T1-T2/t2-t1)= (360 – 340 / 316- 300 ) = 1.25 • So from chart, F = 0.97 For Counter and co current.

  8. Area for counter current: • Q= U A F∆Tlm • 1672 = 2 *A*0.97* 41.972 • A= 20.53 m2 • Area For co current: • Q= U A F∆Tlm • 1672 = 2 *A*0.97*39.3 • A= 21. 93m2

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