Effectiveness and number of transfer units for Parallel flow
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1. The document summarizes the effectiveness and number of transfer units (NTU) method for analyzing heat exchangers operating with parallel flow. It defines heat exchanger effectiveness, NTU, and provides equations relating these parameters. 2. An example problem is included to demonstrate calculating the rate of condensation, overall heat transfer coefficient, NTU, and effectiveness for a steam condenser heat exchanger. 3. Key results from the example problem are that the rate of steam condensation is 0.0714 kg/s, the overall heat transfer coefficient is 255.9 W/m2C, the NTU is 0.628, and the effectiveness is 0.47.
Effectiveness and number of transfer units for Parallel flow
- 1. Gandhinagar institute of technology(012) ALA Subject:- HT(2151909) Topic name:- Effectiveness and number of transfer units for “Parallel flow” Prepared by:- jani Parth u. (150120119051) guided by :-Prof. Nikita gupta 1
- 2. content 1.Introduction 2.NTU (Number of transfer unit) method 3.effectiveness for parallel method 4.Example 2
- 3. 1. introduction Heat exchanger may be defined as an equipment which transfer the energy from a hot fluid to cold fluid, with maximum rate and minimum investment Type of heat exchanger :- classification according nature of heat exchanger, relative direction of fluid motion ,design and construction features. 1. Nature of heat exchange process i. Direct contact heat exchanger ii. Indirect contact heat exchanger (A) Regenerators (B) Recuperators 3
- 4. ... 2. Relative direction of fluid motion i. Parallel flow ii. Counter flow iii. Cross flow 3. Design and construction features i. Concentric tubes ii. Shell and tubes 4. Physical state of fluid (A) Condensers (B) Evaporators 4
- 5. 2. NTU method (Number of transfer unit) The LMTD Is used in equation if the inlet and output temperatures of the two fluids are known When the problem is determine the inlet or exit temperature for particular heat exchanger are not known, the analysis is performed more easily, by using a method based on effectiveness of the heat exchanger and number of transfer unit. Heat exchanger effectiveness is used which does not involve any of the outlet temperatures. UdAQ 5
- 6. 3. Heat exchanger effectiveness The heat transfer effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer. ε = 𝑎𝑐𝑡𝑢𝑎𝑙 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 ℎ𝑒𝑎𝑡 𝑟𝑎𝑛𝑠𝑓𝑒𝑟 = 𝑄 𝑄 𝑀𝐴𝑋 )( )( 11min 21 ch hhh ttc ttC )( )( 11min 12 ch ccc ttC ttC dtccmdQ dthcmdQ UdAdQ cc hh 6
- 7. … )( )( 12min 12 ch ccc ttC ttC 1 11min 2 )( c C c t C tcthc t h ch hh C ttC tt )( 11min 12 )( )( 11min 21 ch hhh ttc ttC 7
- 8. ... By compiling non- dimensional grouping, ε can be expressed as function of three non dimensional parameter ,this is know as NTU method. ch ch cc dQdtdt 11 ch chch cc ttUdAttd 11 )()( c h hch ch chch ch C C C UA tt tt cc UdA tt ttd 1ln 11)( 11 22 8
- 9. … c h hch ch C C C UA tt tt 1exp 11 22 c h hch chch ch C C C UA cc ttCtt tt 1exp 11 )( 1 11min11 11 9 ch c h CC C C C C UA 11 1exp1 min min
- 10. Example 1 Steam condenses at atmospheric pressure on the external surface of the tubes of the steam condenser. The tubes are 12 in number and each is 30mm in diameter and 10m long .the inlet and outlet temperatures of cooling water flowing inside the tubes are 25 c and 60 c respectively.if the flow rate is 1.1 kg/s, calculate following: 1. The rate of condensation of steam 2. The Mean overall heat transfer coefficient based on the inner surface area 3. The NTU 4. The effectiveness of he condenser SOLUTION :- given : N = 12; 𝑑𝑖= 30 mm = 0.03 m ; L=10 m ; 𝑡 𝑐1=25 c, 𝑡 𝑐2=60 c,𝑡ℎ1=𝑡ℎ2= 100 c , 𝑚 𝑤=𝑚 𝑐=1.1 kg/s i. The rate of condensation of steam, 𝑚 𝑠 (= 𝑚ℎ) 10
- 11. … heat lost by steam = heat gained by water 𝑚 𝑠 × ℎ 𝑓𝑔= 𝑚 𝑐 × 𝑐 𝑝𝑐 (𝑡 𝑐2 - 𝑡 𝑐1) 𝑚 𝑠 × 2257= 1.1× 4.187 (60-25) 𝑚 𝑠 = 0.0714 kg/s ii. The mean overall heat transfer coefficient total heat transfer rate is given by Q = 𝑚 𝑐× 𝑐 𝑝𝑐× (𝑡 𝑐2- 𝑡 𝑐1) =1.1× 4187× (60-25) =161199.5 j/s ϴ 𝑚= ϴ1 − ϴ2 ln ϴ1 ϴ2 = 55.68 c 11
- 12. ... Substituting the value in the above equation , we get 191199.5= U× 11.31 ×55.68 U= 255.9 w/𝑚2 c iii. The number of transfer units in condenser 𝑐 𝑚𝑎𝑥 refer to the hot fluids which remains temperatures. Therefore,𝑐 𝑚𝑖𝑛 refers to water, 𝑐 𝑚𝑖𝑛refers to water; 𝑐 𝑚𝑖𝑛= m×𝑐 𝑝𝑐 = 1.1 × 4187 = 4605.7 𝑤/𝑐 NTU = 𝑈𝐴 𝐶 𝑚𝑖𝑛 = 255.9×11.31 4605.7 =0.628 12
- 13. … iv. the effectiveness of the condenser ε = 1 - exp (NTU) ε = 1 - exp (-0.628) = 0.47 13
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