1. Chapter 4.1: Design and Rating of Double Pipe Heat Exchangers

2. Parallel-Flow Regime: 1) Tube Side: Energy balance Parallel-Flow Regime: 1) Tube Side: Energy balance

3. Since, Multiplying the two sides by Let, Since, Multiplying the two sides by Let,

4. 2) Shell Side: Energy balance Since, 2) Shell Side: Energy balance Since,

5. or Multiplying the two sides by Let, or Multiplying the two sides by Let,

6. From eqn.(1), we have then Substituting the previous equation into eqn.(2), one obtains By differentiating eqn.(1), we reach to The substitution of eqn.(3) into eqn.(4) results in From eqn.(1), we have then Substituting the previous equation into eqn.(2), one obtains By differentiating eqn.(1), we reach to The substitution of eqn.(3) into eqn.(4) results in

7. Similarly, differentiating eqn.(2) yields Substituting eqn.(3) into eqn.(6), gives Solving eqns.(5) and (7) using the differential operator as follows: where,. The auxiliary eqn. is Therefore, the solution is given by Similarly, differentiating eqn.(2) yields Substituting eqn.(3) into eqn.(6), gives Solving eqns.(5) and (7) using the differential operator as follows: where,. The auxiliary eqn. is Therefore, the solution is given by

8. Similarly, the solution of eqn.(7) results in The boundary conditions: 1) At x * = 0,  c = 0, 2) At x * = 0,  h = 1 3) At x * = 0, d  c /dx*= NTU [from eqn.(1)] 4) At x * = 0, d  h /dx*= - NTU.C r [from eqn.(3)] Applying eqns.(9) and (8), conditions (3) and (1) give Applying eqns.(11) and (10), conditions (4) and (2) yield Similarly, the solution of eqn.(7) results in The boundary conditions: 1) At x * = 0,  c = 0, 2) At x * = 0,  h = 1 3) At x * = 0, d  c /dx*= NTU [from eqn.(1)] 4) At x * = 0, d  h /dx*= - NTU.C r [from eqn.(3)] Applying eqns.(9) and (8), conditions (3) and (1) give Applying eqns.(11) and (10), conditions (4) and (2) yield

9. and Heat Exchanger Effectiveness: and Heat Exchanger Effectiveness:

10. Counter-Flow Regime: 1) Tube Side: Energy balance In this case, the cold fluid has the same flow direction as that of parallel-flow, therefore the differential equation for the tempera- ture distribution in the cold fluid takes the same form as eqn.(1): Counter-Flow Regime: 1) Tube Side: Energy balance In this case, the cold fluid has the same flow direction as that of parallel-flow, therefore the differential equation for the tempera- ture distribution in the cold fluid takes the same form as eqn.(1):

11. 2) Shell Side: Energy balance In this case, the hot fluid has an opposite flow direction as that of parallel-flow, therefore the differential equation for the temperature distribution in the hot fluid takes the same form as eqn.(2) but with changing the sign of C r as follows: The substitution of eqn.(15) into eqn. (16) gives Similarly, 2) Shell Side: Energy balance In this case, the hot fluid has an opposite flow direction as that of parallel-flow, therefore the differential equation for the temperature distribution in the hot fluid takes the same form as eqn.(2) but with changing the sign of C r as follows: The substitution of eqn.(15) into eqn. (16) gives Similarly,

12. The solution of eqns.(18) and (19) using the differential operator results in The boundary conditions: 1) At x * = 0,  c = 0, 2) At x * = 1,  h = 1 3) At x * = 0, d  c /dx*= NTU.  h,o [from eqn.(15)] 4) At x * = 0, d  h /dx*= C r d  c /dx*= NTU.C r  h,o [from eqn.(17)] The solution of eqns.(20–23) under these boundary conditions are given by The solution of eqns.(18) and (19) using the differential operator results in The boundary conditions: 1) At x * = 0,  c = 0, 2) At x * = 1,  h = 1 3) At x * = 0, d  c /dx*= NTU.  h,o [from eqn.(15)] 4) At x * = 0, d  h /dx*= C r d  c /dx*= NTU.C r  h,o [from eqn.(17)] The solution of eqns.(20–23) under these boundary conditions are given by

13. Heat Exchanger Effectiveness: Case of Evaporation or Condensation (C r = 0): Heat Exchanger Effectiveness: Case of Evaporation or Condensation (C r = 0):

14. Case of Equal Capacity Rates (C r = 1): The differential equations (18) and (19) become Their solutions are The boundary conditions: 1) At x * = 0,  c = 0, 2) At x * = 1,  h = 1 3) At x * = 0, d  c /dx*= NTU.  h,o [from eqn.(15)] 4) At x * = 0, d  h /dx*= d  c /dx*= NTU.  h,o [from eqn.(17)] Applying the previous boundary conditions, the dimensionless temperature distributions in case of equal capacity rates are Case of Equal Capacity Rates (C r = 1): The differential equations (18) and (19) become Their solutions are The boundary conditions: 1) At x * = 0,  c = 0, 2) At x * = 1,  h = 1 3) At x * = 0, d  c /dx*= NTU.  h,o [from eqn.(15)] 4) At x * = 0, d  h /dx*= d  c /dx*= NTU.  h,o [from eqn.(17)] Applying the previous boundary conditions, the dimensionless temperature distributions in case of equal capacity rates are

15. Heat Exchanger Effectiveness: Heat Exchanger Effectiveness:

16. Solved Example: A counter-flow heat exchanger heaving a heat transfer area 12.5 m 2 is to cool oil of specific heat 2000 J/kg.K with water of specific heat 4170 J/kg.K. The oil enters at a temperature 100 o C and a mass flow rate 2 kg/s, while the water enters at 20 o C and a mass flow rate 0.48 kg/s. The overall heat transfer coefficient is 400 W/m 2.K. Calculate the exit temperatures of water and oil as well as the total heat transfer rate. Determine also the temperature distribution of both hot and cold fluids and draw them in dimensionless form. Data: Counter-flow heat exchanger, A = 12.5 m 2, U = 400 W/m 2.K, Hot fluid: oil, c p,h = 2000 J/kg.K, T h,i = 100 o C, m h = 2 kg/s, Cold fluid: water c p,c = 4170 J/kg.K, T c,I = 20 o C, m c = 0.48 kg/s. Find: T c,o, T h,o,, T c = T c (x * ), T h = T h (x * ). Solved Example: A counter-flow heat exchanger heaving a heat transfer area 12.5 m 2 is to cool oil of specific heat 2000 J/kg.K with water of specific heat 4170 J/kg.K. The oil enters at a temperature 100 o C and a mass flow rate 2 kg/s, while the water enters at 20 o C and a mass flow rate 0.48 kg/s. The overall heat transfer coefficient is 400 W/m 2.K. Calculate the exit temperatures of water and oil as well as the total heat transfer rate. Determine also the temperature distribution of both hot and cold fluids and draw them in dimensionless form. Data: Counter-flow heat exchanger, A = 12.5 m 2, U = 400 W/m 2.K, Hot fluid: oil, c p,h = 2000 J/kg.K, T h,i = 100 o C, m h = 2 kg/s, Cold fluid: water c p,c = 4170 J/kg.K, T c,I = 20 o C, m c = 0.48 kg/s. Find: T c,o, T h,o,, T c = T c (x * ), T h = T h (x * ).

17. Solution: For counter-flow regime Solution: For counter-flow regime

18. Or Since Temperature distribution: For counter-flow regime Or Since Temperature distribution: For counter-flow regime

19. 10.80.60.40.20X*X* 0.8330.7380.6160.4590.2580 cc 10.9530.8920.8130.7130.583 hh 86.6479.0469.2856.7240.6420T c ( o C) 10096.2491.3685.0477.0466.64T h ( o C)

20. Temperature distribution In a counter-flow heat exchanger Temperature distribution In a counter-flow heat exchanger