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∆T lm A=N װ DL ∆ T lm = ∆T l -∆T2 / ln (∆T l /∆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: ∆ T l = T h in – T c in ∆ T2 = T h out – T c out For current-current: ∆ T l = T h in – T c out ∆ T2 = T h 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/m 2 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 ∆T lm = ∆T l -∆T2 / ln (∆T l /∆T2) ∆T l = 44 ∆T2 = 40, so ∆T lm = 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. T lm = ∆T l -∆T2 / ln (∆T l /∆T2) ∆T l = 60 ∆T2 = 24, so ∆T lm = 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∆T lm 1672 = 2 *A*0.97* 41.972 A= 20.53 m2 Area For co current: Q= U A F∆T lm 1672 = 2 *A*0.97* 39.3 A= 21. 93m2