What is Heat exchanger?
A heat exchanger is a system used to transfer heat between two or more fluids. Heat exchangers are used in both cooling and heating processes. The fluids may be separated by a solid wall to prevent mixing or they may be in direct contact. They are widely used in space heating, refrigeration, air conditioning, power stations, chemical plants, petrochemical plants, petroleum refineries, natural-gas processing, and sewage treatment. The classic example of a heat exchanger is found in an internal combustion engine in which a circulating fluid known as engine coolant flows through radiator coils and air flows past the coils, which cools the coolant and heats the incoming air. Another example is the heat sink, which is a passive heat exchanger that transfers the heat generated by an electronic or a mechanical device to a fluid medium, often air or a liquid coolant.
How to Calculate NTU relation of heat exchanger with one shell pass and 2, 4, 6 tube pass?
NTU relation of heat exchanger with one shell pass and 2, 4, 6 tube pass calculator uses Number of Transfer Units = -((1+Heat capacity ratio^2)^-0.5)*(ln(((2/Effectiveness of Heat Exchanger)-1-Heat capacity ratio-((1+Heat capacity ratio^2)^-0.5)))/((2/Effectiveness of Heat Exchanger)-1-Heat capacity ratio+((1+Heat capacity ratio^2)^-0.5))) to calculate the Number of Transfer Units, The NTU relation of heat exchanger with one shell pass and 2, 4, 6 tube pass formula is defined as the amount of energy units transferred between the fluids in the heat exchanger. Number of Transfer Units is denoted by NTU symbol.
How to calculate NTU relation of heat exchanger with one shell pass and 2, 4, 6 tube pass using this online calculator? To use this online calculator for NTU relation of heat exchanger with one shell pass and 2, 4, 6 tube pass, enter Heat capacity ratio (C) & Effectiveness of Heat Exchanger (ϵ) and hit the calculate button. Here is how the NTU relation of heat exchanger with one shell pass and 2, 4, 6 tube pass calculation can be explained with given input values -> -0.132276 = -((1+0.5^2)^-0.5)*(ln(((2/0.1)-1-0.5-((1+0.5^2)^-0.5)))/((2/0.1)-1-0.5+((1+0.5^2)^-0.5))).