Heat Exchanger Device - Assignment Example

SUMMARY Heat exchangers are the devices that are designed to facilitate exchange of heat between two fluids with heat being transferred from the hot fluid to the colder fluid. In heat transfer of exchangers convection takes place in each of the fluids while conduction is involved in the walls that separate the fluids. In this report there is explanation of the comparison of parallel flow, counter flow and cross flow in the heat exchanger with conclusion on which type of flow is the most efficient. The design was such that hot water was flowing through the middle pipe with the flow being set at a constant rate. The heat transferred from the hot water to the cold water was calculated in each case by considering the in and out temperatures on both the hot and cold water. The calculation was used in finding out which flow was most efficient where the counter flow was found to be the most efficient while cross flow was the least efficient. The high efficiency of counter flow is attributed to the fact that in this flow temperature difference between the two fluids is kept constant during the whole process. As mentioned at the beginning heat transfer occurs due to conduction and convection and thus it is necessary to establish total heat transfer coefficient. Copper oxide which results from oxidation process of the copper metal is likely to have effect on the conduction part of the coefficient and fouling factor is introduced in order to take care of this. In the second part of this report there is explanation of the analysis by logarithmic mean temperature difference, which is an equivalent mean temperature difference between two fluids for the entire heat exchanger. This is applicable in he case of parallel and counter flow but for cross flow there is the introduction of the f factor. The effectiveness-NTU is the other method that is used for analysis where the NTU is derived from the overall heat transfer coefficient and the maximum heat capacity of the fluids and the effectiveness is then found from NTU. The heat transfer is found by multiplying effectiveness e by the maximum possible heat transfer that is obtained from the temperature difference at the inlet and at the outlet. NTU-e method is found to be more accurate than LMTD method, because in NTU-e there is explanation of the flow, thermal resistance and initial temperatures but for the case of LMTD it purely depends on thermal resistance and end temperatures. Aim The main aim of this exercise is to compare the three basic heat exchanging principles like Counter flow Parallel flow Cross flow Then estimate the heat transfer by Log mean temperature difference method Effectiveness –NTU method Basic structure of heat exchanger Two pumps are used for pumping cold water and hot water. The hot water which flows in the inner side is directed to a water heater for heating after circulating the pipe. The rate of hot water flow is kept constant. The design is such that the hot water passes through copper pipe that is located at the centre of the heat exchanger. The cold water which is pumped by the other pump also is channelled through copper pipe with the choice of material, being based on the high conductivity of copper. The manner in which the flow of the water flow with respect to each other determines whether there is parallel floe, counter flow or cross flow. Changing from parallel flow to counter flow can be achieved by simply swapping the inlet and outlet of the cold water pipes. The temperatures are measured by thermometers which are placed at the inlet and outlet of by the cold and hot water. For cross flow there are twos additional thermometer for measuring out let temperature. . Measurements The measurements which were made are in this practical are summarized in Table 1 2 and 3. Table 1 Types of flow Temperatures category Parallel flow Counter flow Cross flow Hot Water In Hot Water Out Cold Water In Cold Water Out Table 2 Mass flow rates All Hot flow Cold-Parallel flow Cold Counter flow Cold Cross flow 0.38 0.026 0.027 0.025 Table 3 Calculations: Calculation of actual heat (): Parallel flow ( From tables Specific heat of water Cp at 50 C =4.18 Heat transferred by hot water Using the formula heat transferred by hot water =( According to Yunus, 2009) Q=0.38 *4.18 *(50-48) =3.18kW Heat gained by cold water Q = Q=0.026 *418 *(41C-18C) =2.5kW Counter flow ( Heat lost by hot water Specific heat of water Cp at =4.180  at 50 Using the formula: Q = (Yunus, 2009) Q= 0.38 * 4.18 *(50-47) =4.8W Heat gained by cold water Q = Q= 0.027 x 4.18 x(42-17) =2.8kW Cross flow heat (: Heat transferred by hot water Specific heat of water Cp at 50=4.180 Q = = 0.38 * 4.180*(50-48) =3.18kW Heat gained by cold water Q = = 0.025 * 4.180 *(38-16) =2.3kW Calculating velocity of the flows inside the pipes Velocity  Where density of water  Velocity of hot water for parallel, counter &cross flow: A = Parallel Cold water flow's velocity:  0.026kg/s Counter Cold water flow's velocity: A = 0.027 Cross flow cold water flow's velocity:  0.025kg/s Cold flow area in cross flow is calculated using the equation: A= (D-d)×L D=0.0635 d =0.0127m L=5.3m A =  Finding Reynolds numbers and Nuselt numbers: hot water in parallel, counter and cross flow: With Re being greater than 40000 but less than 400000 it qualifies to be turbulent flow Thus we apply  Using table A-15 it is found 3.55 Convection heat transfer coefficient: for Cold outside flow:  for the case of parallel flow Using  d =0.0508m Referring to   Re is in the range of between 40-4000 depicting laminar because Referring to table A-15 at   (Yunus, 2009) Convection heat transfer coefficient:  Counter flow: d =0.0508m 40C Re range is between 40-4000 thus flow is laminar Referring to table A-15 at   (Yunus, 2009) Convection heat transfer coefficient: Convection heat transfer coefficient  We know  d =0.0508m The flow is laminar because Re 40 Referring to table A-15 at   (Yunus, 2009) Convection heat transfer coefficient: Finding the product of overall heat transfer coefficient and area (UA)  Where  is the thermal resistance Inner area of the pipe: Outer area of the pipe Calculating predicted heat transfer by LMTD method for parallel flow: = 0.1 , where  represents fouling factor Therefore UA= Heat predicted LMTD (Q): UA*LMTD Where LMTD Counter flow heat transfer heat transfer prediction LMTD method = 0.1 , where  is the fouling factor = 0.1 , where  represents fouling factor Therefore UA= Heat predicted by LMTD (Q): UA*LMTD Where LMTD Calculating predicted heat transfer by LMTD method for cross flow = 0.1 , where  represents fouling factor Therefore UA= Heat predicted by LMTD (Q): F*UA*LMTD Where F represents correction factor From the thermal fluid chart (c) for cross flow page 942 F=0.95 Calculation of predicted heat transfer by NTU method: Parallel flow Where  is the maximum possible heat transfer given by T max Ratio of heat capacity: (Number of transfer units) NTU=UA/ Effectiveness: T max =  * For Counter flow: * T max Ratio of heat capacity: 2 (Number of transfer units) NTU=UA/ Effectiveness: T max =  * Cross flow: * T max Ratio of heat capacity: (Number of transfer units) NTU=UA/ Effectiveness: =0.086 T max =  * DISCUSSION Compare the actual heat transferred in three cases. From the experiment it can be seen that counter flow is the most effect followed by parallel flow while parallel flow is the least effective with their temperature difference in the cold water outlet and inlet being 25, 23 and 22 respectively. It can be seen that in all the three types of flow thermodynamics laws have been obeyed where the temperature of the cold water does not rise to be above the initial temperature of the hot water. Actual heat transfer and predicted methods.   heat lost by heat gained by LMTD method NTU-e, Method   hot water cold water     (kwatt) (kwatt) (kwatt) kwatt           counter flow 4.8kW 2.8kW           parrallel flow 3.18kW 2.5kW           cross flow 3.18kW 2.3kW 0. Even though there is insulation of pipes from the results it can be seen that that the heat gained by cold water is less than the heat lost by the hot water in all the three types of flows. This can be attributed to the following Some of the heat is retained by the pipes The insulation cannot be 100% by any mean There is a lot of time required for flow to be developed fully and thus there is no full length flow When the NTU-e method and LMTD method are compared it can be seen that the former is more accurate than the later. The explanation for this is LMTD method is depend on mean of end temperatures as well as the thermal resistance. In NTU-e method there consideration of fluid flow including the Reynolds number For flow to develop full it needs time but in this practical there was no computer simulation of properties of flow which was necessary in the analysis of prediction of heat transfer. CONCLUSION Counter flow has shown he highest efficiency while cross flow exhibited the least efficiency. There can only be the 100% of heat transfer from hot water to cold water in ideal situation but practically it is not possible NTU-e, prediction sho higher accuracy than LMTD method prediction and this can be attributed the it deals with fluid flow while in the later there is dependence on thermal resistance and end temperatures. REFERENCE Unus A. Cengal, Fundamentals of thermal-fluid sciences-second edition published by Mc Graw Hill Read More

The hot water which flows in the inner side is directed to a water heater for heating after circulating the pipe. The rate of hot water flow is kept constant. The design is such that the hot water passes through copper pipe that is located at the centre of the heat exchanger. The cold water which is pumped by the other pump also is channelled through copper pipe with the choice of material, being based on the high conductivity of copper. The manner in which the flow of the water flow with respect to each other determines whether there is parallel floe, counter flow or cross flow.

Changing from parallel flow to counter flow can be achieved by simply swapping the inlet and outlet of the cold water pipes. The temperatures are measured by thermometers which are placed at the inlet and outlet of by the cold and hot water. For cross flow there are twos additional thermometer for measuring out let temperature. . Measurements The measurements which were made are in this practical are summarized in Table 1 2 and 3. Table 1 Types of flow Temperatures category Parallel flow Counter flow Cross flow Hot Water In Hot Water Out Cold Water In Cold Water Out Table 2 Mass flow rates All Hot flow Cold-Parallel flow Cold Counter flow Cold Cross flow 0.38 0.026 0.027 0.025 Table 3 Calculations: Calculation of actual heat (): Parallel flow ( From tables Specific heat of water Cp at 50 C =4.

18 Heat transferred by hot water Using the formula heat transferred by hot water =( According to Yunus, 2009) Q=0.38 *4.18 *(50-48) =3.18kW Heat gained by cold water Q = Q=0.026 *418 *(41C-18C) =2.5kW Counter flow ( Heat lost by hot water Specific heat of water Cp at =4.180  at 50 Using the formula: Q = (Yunus, 2009) Q= 0.38 * 4.18 *(50-47) =4.8W Heat gained by cold water Q = Q= 0.027 x 4.18 x(42-17) =2.8kW Cross flow heat (: Heat transferred by hot water Specific heat of water Cp at 50=4.

180 Q = = 0.38 * 4.180*(50-48) =3.18kW Heat gained by cold water Q = = 0.025 * 4.180 *(38-16) =2.3kW Calculating velocity of the flows inside the pipes Velocity  Where density of water  Velocity of hot water for parallel, counter &cross flow: A = Parallel Cold water flow's velocity:  0.026kg/s Counter Cold water flow's velocity: A = 0.027 Cross flow cold water flow's velocity:  0.025kg/s Cold flow area in cross flow is calculated using the equation: A= (D-d)×L D=0.0635 d =0.0127m L=5.

3m A =  Finding Reynolds numbers and Nuselt numbers: hot water in parallel, counter and cross flow: With Re being greater than 40000 but less than 400000 it qualifies to be turbulent flow Thus we apply  Using table A-15 it is found 3.55 Convection heat transfer coefficient: for Cold outside flow:  for the case of parallel flow Using  d =0.0508m Referring to   Re is in the range of between 40-4000 depicting laminar because Referring to table A-15 at   (Yunus, 2009) Convection heat transfer coefficient:  Counter flow: d =0.

0508m 40C Re range is between 40-4000 thus flow is laminar Referring to table A-15 at   (Yunus, 2009) Convection heat transfer coefficient: Convection heat transfer coefficient  We know  d =0.0508m The flow is laminar because Re 40 Referring to table A-15 at   (Yunus, 2009) Convection heat transfer coefficient: Finding the product of overall heat transfer coefficient and area (UA)  Where  is the thermal resistance Inner area of the pipe: Outer area of the pipe Calculating predicted heat transfer by LMTD method for parallel flow: = 0.

1 , where  represents fouling factor Therefore UA= Heat predicted LMTD (Q): UA*LMTD Where LMTD Counter flow heat transfer heat transfer prediction LMTD method = 0.1 , where  is the fouling factor = 0.1 , where  represents fouling factor Therefore UA= Heat predicted by LMTD (Q): UA*LMTD Where LMTD Calculating predicted heat transfer by LMTD method for cross flow = 0.

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