It has 1 shell pass, 5 tubes each making 2 passes, each 5m long. External diameter of 22mm and internal diameter of 20mm with wall roghness of 1.5 micro m. Made of stainless steel with a thermal conductivity of 15W/mk.It cools oil using water.Water enters: 55 degrees C, k=0.681W/mK, Cp=4211J/kgK, Pr=1.7, density=958kg/m3, μ=0.000276Pa.s.Oil enters: 80 degrees C, exits: 55 degrees C, mass flow rate: 2kg/s, density=842, μ=0.0186, k=0.137, Cp=2206, Pr=300.1. How do you find the heat transfer rate from oil to water? I did Q=mCp(Tin-Tout) ---gt; 2*2.206(80-60) = 88.24Kw...is this right?2. How do you find the mass flow rate of the water to achieve no.1?
1by measuring the temperature and going from there 2is more complex. But cut and trail is a good method. The heat exchanger is a two cavities system separated by the heat transfer media. To make it easier to understand, take an automobile radiator. It is a series of tubes where the liquid passes in the tubes, and the tubes are thermally connected with fins to the air The liquid has a density of about 1000 times that of air. So ideally, you would need 1000 more surface area on the air side than the liquid side. The construction difference between a truck radiator and a car, both having about the same thermal load would be very different. How tight can you squeeze the fins, how thick does it have to be (how many layers), how much air pressure…life span, thermal cycling conditions... Hope this properly answers your question and helps you find an answer
Your first step seems correct however I am used to heat being expressed in BTUs or Calories. In order to determine the amount of water needed, you need to calculate the surface area of the exchanger. Next you have to calculate or estimate the over all heat transfer coefficient for the unit. Typically in coolers using water, the water is on the tube side of the exchanger. To calculate the overall heat transfer coefficient you have to assume a velocity for the water and determine the various film coefficients for the water side and the oil side of the unit. You will find that the heat transfer capability of the metal in the tubes has very little to do with the overall heat transfer coefficient. The viscosity of the oil and the velocity of the water have a much bigger impact. The relationship that you will be dealing with is this basic equation: Heat transferred equals the overall heat transfer coefficient times the log mean temperature difference between the two fluid times the surface area. This means you are going to have to use trial and error to balance, film coefficients, water flow rate and log mean temperature differences to find a solution. This is really not a simple problem if it is to be done by direct calculation. There are computer programs that are a great aid. Here is one site that might be of help:
