From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 2510/CH22/EX22.9/Ex22_9.sce | 71 +++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 71 insertions(+) create mode 100755 2510/CH22/EX22.9/Ex22_9.sce (limited to '2510/CH22/EX22.9/Ex22_9.sce') diff --git a/2510/CH22/EX22.9/Ex22_9.sce b/2510/CH22/EX22.9/Ex22_9.sce new file mode 100755 index 000000000..13a64b8e8 --- /dev/null +++ b/2510/CH22/EX22.9/Ex22_9.sce @@ -0,0 +1,71 @@ +//Variable declaration: +po = 53*16.0185 //Density of oil (kg/m^3) +co = 0.46*4186.7 //Heat capacity of oil (J/kg. C) +pi = %pi +muo = 150/1000 //Dynamic viscosity of oil (kg/m.s) +ko = 0.11*1.7303 //Thermal conductivity of oil (W/m. C) +qo = 28830*4.381*10**-8 //Volumetric flowrate of oil (m^3/s) +pw = 964 //Density of water (kg/m^3) +cw = 4204 //Heat capacity of water (J/kg. C) +muw = 0.7/3600*1.4881 //Dynamic viscosity of water (kg/m.s) +kw = 0.678 //Thermal conductivity of water (W/m. C) +qw = 8406*4.381*10**-8 //Volumetric flowrate of water (m^3/s) +t1 = 23.5 //Initial temperature of oil ( C) +t2 = 27 //Final temperature of oil ( C) +T1 = 93 //Water heating temperature of water ( C) +syms T2 //Minimum temperature of heating water ( C) +syms A //Heat transfer area (m^2) +Uc = 35.4 //Clean heat transfer coefficient (W/m^2.K) +Rf = 0.0007 //Thermal resistance (m^2.K/W) +D = 6*0.0254 //Inside diameter of pipe (m) + +//Calculation: +vo = muo/po //Kinematic viscosity of oil (m^2/s) +mo = po*qo //Mass flowrate of oil (kg/s) +vw = muw/pw //Kinematic viscosity of (m^2/s) +mw = pw*qw //Masss flow rate of water (kg/s) +Q1 = mo*co*(t2-t1) //Duty of exchanger of oil (W) +T2m = t1 //Lowest possible temperature of the water ( C) (part 1) +Qmw = mw*cw*(T1-T2m) //Maximum duty of exchanger of water (W) (part 2) +Q2 = mw*cw*(T1-T2) //Duty of exchanger of water in terms of T2 (W) +x = eval(solve(Q1-Q2,T2)) //Solving value for T2 ( C) +T3 = x; //Minimum temperature of heating water ( C) +DT1 = T3-t1 //Inlet temperature difference ( C) +DT2 = T1-t2 //Outlet temperature difference ( C) +DTlm = (DT1-DT2)/log(DT1/DT2) //Log mean temperature difference ( C) +Ud1 = 1/Uc+Rf //Dirty heat transfer coefficient (W/m^2.K) (part 3) +Ud2 = 34.6 //Dirty heat transfer coefficient (W/m^2. C) +Q3 = Ud2*A*DTlm //Duty of exchanger (W) (part 4) +y = eval(solve(Q1-Q3,A)) //Heat transfer area (m^2) +A1 = y //Required heat transfer area (m^2) +L = A1/(pi*D) //Required heat transfer length (m) +Qmo = mo*co*(T1-t1) //Maximum duty of exchanger of oil (W) (part 5) +Qm = Qmw //Maximum duty of exchanger (W) +E = Q1/Qm*100 //Effectiveness (%) +NTU = Ud2*A1/(mw*cw) //Number of transfer units + +//Result: +disp("1. The lowest possible temperature of the water is :") +disp(T2m) +disp(" C .") + +disp("2. The log mean temperature difference is : ") +disp (DTlm) +disp(" C .") + +disp("3. The overall heat transfer coefficient for the new clean exchanger is : ") +disp (Ud2) +disp ("W/m^2. C .") + +disp("4. The length of the double pipe heat exchanger is : ") +disp(L) +disp (" m .") + +disp("5. The effectiveness of the exchanger is : ") +disp(E) +disp("%") + +disp("The NTU of the exchanger is : ") +disp(NTU) + +// Answers are correct. Please calculate manually. -- cgit