The Fundamentals of a Heat Exchanger - Lab Report Example

A PRACTICAL REPORT ON HEAT TRANSFER (HEAT EXCHANGER) Name Institutional Affiliation Date Table of Contents Introduction 2 Aim of the experiment 4 The fundamentals of a heat exchanger 4 Discussion and Analysis 5 Calculations 7 Results 25 Conclusion 30 Reference 30 Introduction Heat exchanger is an equipment that helps in heat transfer between to fluids acting as medium of transfer (Çengel & Turner, 2001). The transfer of heat is two-fold, with the first one being conduction whereby heat is transferred across a wall separating the two fluids (medium). The second mode of heat transfer is convection that involves the transfer of heat within each fluid. This report discusses and analyses three different types of flow and suggests which one among them is the most efficient and effective. The types of heat flow that will be analysed in this report include cross flow, parallel flow and counter flow . In each flow process, hot water at constant mass flow rate is allowed to pass through the middle pipe made of copper. The temperature of hot and cold water that flows out and in respectively in the heat exchanger are measured to help in the computation of the quantity of heat transferred in each of the three scenarios (Çengel & Turner, 2001). The figure obtained is divided by the flow rate of the cold water in each case. This helps in determining which flow process among the three is efficient. When this is carried out, it is observed that the counter flow process maintains a uniform difference in temperature throughout the entire flow process. Because of this, both cross flow and counter flow process display higher efficiency compared to parallel flow process. As has been mentioned above, heat exchange arises due to convection and conduction heat transfer processes. This makes it imperative to incorporate all the potential coefficients of heat transfer. There are impediments that affect the rate of heat transfer through the copper tubes. One of these impediments is the formation of an oxide layer copper tube’s surface. The analysis of the results using logarithmic mean as explained in part two of the report. This formula helps to determine the equivalent average difference in temperature for the whole exchanger between two mediums. Because of this, cross flow must consider a correction factor unlike the remaining two in this situation. Further heat exchange analysis requires the application of NTU technique which measures effectiveness of the heat exchange process. This value is obtained from the overall coefficient of heat exchange and minimum capacity of the fluid’s heat. The transfer of heat is the product of maximum attainable amount of heat transfer and heat exchanger effectiveness obtained from NTU. It is worth appreciating that the NTU-e method portrays much accuracy compared to LMTD method. This is because the former incorporates thermal resistance, the flow and the original fluid and ambient temperatures but the latter incorporates only original temperatures and thermal resistance. Aim of the experiment The objective of this experiment is to compare the three basic principles of heat exchange, namely; Parallel flow Cross flow Counter flow From the data tabulated above, heat transfer is estimated by applying two basic principles; Effectiveness-NTU principle Logarithmic mean temperature difference principle The fundamentals of a heat exchanger Heat exchanger device has two water pumps, one pumps water into the exchanger while the other takes care of the fluid that exits the heat exchanger. When the water enters the heat exchanger, it gets heated up as it passes through the copper pipe that is situated in the middle of the heat exchanger device at a constant mass flow rate (Çengel & Turner, 2001). Cold water is pumped into the outside of the copper tube and at this time, the flow type in terms of direction is established as either counter flow, cross flow, or parallel flow. These flow processes are possible when the water tubes are swapped. The outlet and inlet temperatures of the hot and cold water are measured by the thermometers which are situated at the ends of the heat exchanger device. For the cross flow, another thermometer is introduced right in the middle of the exchanger to measure the third temperature so that an average outlet temperature is computed. Discussion and Analysis Computation of the Actual heat transfer The table below summarises the parameters which will help in computation of the three flows. Measurements data: Flow types Temperatures Parallel flow Counter flow Cross flow Hot Water In Hot Water Out Cold Water In Cold Water Out The length of heat exchanger=5.3m Mass-flow Rates of flows All Hot flow Cold-Parallel flow Cold Counter flow Cold Cross flow 0.38 0.026 0.027 0.025 Calculations Calculating actual heat(): Parallel flow (: Cp (specific heat of water) at = Using the formula: Heat transferred by hot water So, (Yunus, 2009) Heat gained by cold water Counter flow(: Heat transferred by hot water Cp (specific heat of water) at = Using the formula: (Yunus, 2009) Heat gained by cold water Cross flow heat (: Heat transferred by hot water Cp (specific heat of water) at = Heat gained by cold water Calculating velocity of the flows inside the pipes: Velocity for hot water in parallel, counter &cross flow: Parallel Cold water flow's velocity: Counter Cold water flow's velocity: Cross Cold water flow's velocity: 0.025kg/s Cold flow area in cross flow can be calculated by this equation: A= (D-d)×L D=0.0635 d =0.0127m L=5.3m Finding Reynolds numbers and Nusselt numbers: Hot parallel, counter and cross flow: This is a Turbulent flow because Re greater than 40000 and less than 400000. (Thermal-fluid p788) So, From table A-15 Convection heat transfer coefficient: Convection heat transfer coefficient …….Thermal conductivity table A-15 For Cold outside flow: Parallel flow: d =0.0508m 40C So, The flow is laminar because Re rang is between 40 and 4000. From A-15 at 40˚C (Yunus, 2009) Convection heat transfer coefficient: Counter flow: 40C So, The flow is laminar because Re rang is between 40 and 4000. (From A-15)(Yunus, 2009) Convection heat transfer coefficient: cross flow: d =0.0508m 40˚C So,= 7 The flow is laminar because Re 40 (From A-15)(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 is the fouling factor So, UA= Heat predicted LMTD (Q): [UA][LMTD] LMTD Calculating predicted heat transfer by LMTD method for counter flow: = 0.1 , where is the fouling factor So, Heat predicted LMTD (Q): [UA][LMTD] Calculating predicted heat transfer by LMTD method for cross flow: , where is the fouling factor So, Heat predicted LMTD (Q): F x UA x LMTD Where F is the correction factor From chart (c) cross flow page 942(thermal fluid) F=0.95 Calculating predicted heat transfer by NTU method: Parallel flow T max Ratio of heat capacity: (Number of transfer units) NTU=UA/ Effectiveness: Counter flow: T max Ratio of heat capacity: 2 (Number of transfer units) NTU=UA/ Effectiveness: Cross flow: T max Ratio of heat capacity: (Number of transfer units) NTU=UA/ Effectiveness: =0.086 T max = A plot of temperature v distance (counter flow) and temperature vs. distance (parallel flow) is as shown in the graphs bellow. Results From the calculations, the amount of energy output can be summarized as follows for counter flow, parallel flow and cross flow respectively. Counter flow Heat lost by hot water Heat gained by cold water Heat required by LMTD method kw/C Heat required by NTU-Effectiveness method ε*Q_maxpossible=0.104*3.75kW=0.4kW Parallel flow Heat lost by hot water Heat gained by cold water Heat required by LMTD method Q_prLMTD=0.0125kw/C*17.5C=218.75W Heat required by NTU-Effectiveness method Q_prNTU=ε*Q_maxpossible=0.1*3.5kW=0.35kW Cross flow Heat lost by hot water Heat gained by cold water Heat required by LMTD method Q_prLMTD=0.95*0.01kw/C*20.4C=194W Heat required by NTU-Effectiveness method Q_prNTU=ε*Q_maxpossible=0.086*3kW=0.3kW Compare the actual transferred heat in the three cases From the data, parallel flow is not as effective as the other because the maximum temperature for hot and cold inlets in the same. Therefore it violates the law of dynamics. Counter flow is the most effective of the three. Cross flow has got both parallel and counter flows even though it is perpendicular to the hot water. The table below shows the comparison of the three types of flow in terms of heat loss by the hot water, heat gained by cold water, LTMD method and NTU-e method in regards to the energy output.   heat lost by heat gained by LMTD method NTU-ɛ, Method   hot water cold water       4.8kW  2.8kW  210.8W  0.4kW COUNTER FLOW   3.18kW  2.5kW  218.75kW  0.35kW  PARALLEL FLOW   3.18kW  2.3kW      CROSS FLOW The results obtained show that heat gained by cold water is not the same as heat lost by hot water despite perfect insulation. This is due to; Differences in mass flow rates Differences in diameter between hot water pipe and cold water pipe There can never be perfect lagging Flow takes time to develop in order to get the length of full flow From the predicted calculations, it can also be shown that; LMTD technique relies on mean of thermal resistance and temperature NTU-ɛ method considers the fluid flow and therefore considers Reynolds number Flow takes time to develop. Conclusion From the results, it can be seen that heat gain by cold water is not the same as that lost by hot water despite perfect insulation. This is due to difference in mass flow rate, non-efficient insulation and differences in diameters. Prediction methods also reveal that NTU-ɛ is more reliable than LMTD technique since it is more accurate than the latter. Reference Çengel, Y.A., and Turner, R.H. (2012). Fundamentals of thermal-fluid sciences. Boston: McGraw-Hill. Read More