### Heat exchanger

Transcript: HEAT TRANSFER GROUP 2 Photography credit : ABB engineering and consulting Problem statement Problem statement In a factory , heat exchanger is used to preheat the mixure of alcohol ( Ethanol and Methanol) in order to be used for the distilation process. From the distilation process fusel oil is produced , it is then used as the hotstream in the heat exchanger. In a 2 shell and tube the hot stream the mass flow rate of 3-Methyl-1Butanol (cp=2.586kJ/kg°C) is fed into the tube side of 1kg/s at 100°C. The mixture of alcohol flows (cp=3.237kJ/kg°C) through the shell of 2kg/s (mole fraction of methanol is 0.45 and ethanol is 0.55) at room temperature (30°C). Determine the exit temperature of the cold and hot stream. Given : U=159.8 w/m^2*K A= 3.5m^2 1 TEMA type : BFM Material : SS304 Shell pass: 2 Tube pass: 4 Shell ID/OD: 175 mm / 180.54mm Tube OD : 19.05mm Length : 2.5m Pitch :25.4mm Number if baffle : 30 Type of baffle : single segmental Heat exchanger specification Heat exchanger specification 2 Design Aspen Exchanger design 3 Console Geometry The optimum ratio of baffle spacing to shell size diameter is normally between 0.3 to 0.6 4 Process data & Geometry Fouling resistance is less than 0.001 which means that the prefered tube size for light tube side fouling service is 0.75 inch O.D. 5 Exchanger Geometry Baffle cuts between 20% and 35% be employed. Horizontal baffle cut is recommended for single phase fluid on the shell side for minimizes accumulation of the deposit at the buttom of the shell and also prevent stratification. 6 Tube layout 7 Analyzing the Graph Analyzing the Graph T exit of fusel oil 67.62°C T exit of alcohols 46.56°C For liquid-liquid HEX : temperature difference 11-24°C obtained : 21.06°C } ΔT 8 Warnings Warnings 9 HTRI PROGRAM HTRI PROGRAM 10 3D Picture 3D Picture 11 HEAT EXCHANGER Veeranuch Sirisook 5810755370 Suphasin Perunavin 5810751775 Chanyanuch lertjaruwong 5810751544 Chanyanuch Chanunan 5710753038 Suphawit keakultanes 5810750843 Jiroj Sariddipanthawat 5810750082 Hand Calculation T h,out =? Tc,out =? Cold stream (inlet) : Ethanol+Methanol mC = 1.5 kg/s Tc,in = 30°C P= 2 bar Hot stream (inlet) : 3-Methyl-1-Butanol mh=1 kg/s Th,in= 100°C P= 2 bar Mole fraction of substances : Ethanol : 0.55 Methanol: 0.45 Fusel oil (3-Methyl-1Butanol) : 1 12 parameters The NTU method is used to determine the outlet temperature of both streams . NOTE : we know all the inlet parameters (Area , U , Ltube and Dtube) Given parameters: Tube OD : 19.05mm Length : 2.5m Surface area (A) = 3.5 m2 U = 519.8 W/m2.C 13 Calculations Method 1. Obtain HEX data (known : surface area , U ,Length , Diameters etc.) 2.Properties of fluids : to obtain specific heat capacities 3. Find maximum heat transfer possible : Qmax=CminΔTmax 4.Determine Heat transfer effectiveness "ε" ε (C,NTU) - can use graph or formula - C= Cmin/Cmax - NTU equation 5. Calculate Q actual = εQmaxห 6. Obtain the outlet temperatures 14 Assumption 1.Steady state 2. neglect ΔKE and ΔPE of the fluids 3. Fluid properties are constant 4.Heat loss to surrounding is neligible as the heat exchanger is new and well insulated. 15 Property table Ethanol Methanol FROM : Mole fraction of the mixture : Methanol = 0.45 ; Ethanol = 0.55 Cp,mixture=(mole fraction methanol)( Cp,methanol) + (mole fraction ethanol)(Cp,ethanol) = (0.45)(3.499 kg/J.K) + (0.55)(3.086 kg/J.K) = 3.271 kg/J.K 16 Property table Now we have : Cp,mixture : 3.271 kg/J.K Cp,fusel oil : 2.586 kg/J.K 17 Finding minimum heat capacity (Cmin) Given : ṁc =1.5 kg/s ṁh = 1 kg/s from the equation --> [C,rate = mass flowrate * liquid specific heat] Cc =mc Cp,cold=1.5x3.271= 4.9078 kg/J.K Ch=ṁh Cp,hot =1x2.586 = 2.586 kg/J.K The one with the least value is Cmin and higher one is Cmax Calculations Cmax Cmin 18 Find ∆Tmax : From the equation ---> ∆Tmax= Th,in-Tc,in ∆Tmax= Th,in-Tc,in= 100 – 30 = 70 °C Plugging ∆Tmax and Cmin in to this equation ----> Qmax=CminΔTmax Qmax=CminΔTmax=(2.586kg/J.K )( 70C)=181 kW Find Heat transfer effectiveness "ε" C = Cmin/Cmax = 2.586/4.9078 = 0.526 NTU = 3.5x519.8/(2.586x〖10〗^3 ) = 0.704 From graph ε=0.45 19 Find actual Q : Q ̇= εQmax〖=0.45x181=81.45 kW〗 This Q is used to obtain the outlet temperatures For cold side Q ̇=Cc*∆Tc 81.45kW =(4.9078 kg/J.K )(Tc,out-30) Tc,out=46.596° C For hot side Q ̇=Ch *∆Th 81.45 =(2.586 kg/J.K )(〖100-T〗h,out ) Th,out=68.5 °C 20 EXTRA : CORRECTION FACTORS ***chose the graph that is suitable for the type of HEX you're using*** P=(t2-t1)/(T1-t1)= (68.5-100)/(30-100)= 0.45 R =(T1-T2)/(t2-t1)= (30-48.6)/(30-1068.5-100)=0.60 From graph correction factor is 0.99 which is good as log mean temperature correction factor should be greater than 0.75 to avoid temperature approach problems. 21 Result summary Result summary Comparision between Aspen EDR and hand calculation 22 TEMA SHEET 23 Comparison chart Comparison chart 24