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 --- 542/CH16/EX16.1/Example_16_1.sci | 30 ++++++++++++++++ 542/CH16/EX16.2/Example_16_2.sci | 27 ++++++++++++++ 542/CH16/EX16.3/Example_16_3.sci | 60 +++++++++++++++++++++++++++++++ 542/CH16/EX16.4/Example_16_4.sci | 40 +++++++++++++++++++++ 542/CH16/EX16.5/Example_16_5.sci | 77 ++++++++++++++++++++++++++++++++++++++++ 542/CH16/EX16.6/Example_16_6.sci | 41 +++++++++++++++++++++ 6 files changed, 275 insertions(+) create mode 100755 542/CH16/EX16.1/Example_16_1.sci create mode 100755 542/CH16/EX16.2/Example_16_2.sci create mode 100755 542/CH16/EX16.3/Example_16_3.sci create mode 100755 542/CH16/EX16.4/Example_16_4.sci create mode 100755 542/CH16/EX16.5/Example_16_5.sci create mode 100755 542/CH16/EX16.6/Example_16_6.sci (limited to '542/CH16') diff --git a/542/CH16/EX16.1/Example_16_1.sci b/542/CH16/EX16.1/Example_16_1.sci new file mode 100755 index 000000000..edef797b7 --- /dev/null +++ b/542/CH16/EX16.1/Example_16_1.sci @@ -0,0 +1,30 @@ +clear; +clc; +printf("\n Example 16.1"); +//For the first drying operation +w1 = 0.25; //in kg/kg +w = 0.10; //in kg/kg +wc = 0.15; //in kg/kg +we = 0.05; //in kg/kg +f1 = w1-we; //in kg/kg +fc = wc - we; //in kg/kg +f = w - we; //in kg/kg + +//x = mA +function[x]=TotalDryingTime(t) + x = (1/t)*[(f1-fc)/fc + log(fc/f)]; + funcprot(0); +endfunction +printf("\n mA = %.3f kg/sec",TotalDryingTime(15)); + +//For the second drying operation +w12 = 0.30; //in kg/kg +w2 = 0.08; //in kg/kg +wc2 = 0.15; //in kg/kg +we2 = 0.05; //in kg/kg +f12 = w12 - we2; //in kg/kg +fc2 = wc2 - we2; //in kg/kg +f2 = w2 - we2; //in kg/kg + +t1=(1/TotalDryingTime(15))*[(0.25-0.10)/0.10 + log(0.10/0.03)]; +printf("\n The total drying time is then %.1f ksec or %.2f hr ",t1,t1*1000/3600); diff --git a/542/CH16/EX16.2/Example_16_2.sci b/542/CH16/EX16.2/Example_16_2.sci new file mode 100755 index 000000000..4f0f1ba41 --- /dev/null +++ b/542/CH16/EX16.2/Example_16_2.sci @@ -0,0 +1,27 @@ +clear; +clc; +printf("\n Example 16.2"); +E = [1 0.64 0.49 0.38 0.295 0.22 0.14]; +J = [0 0.1 0.2 0.3 0.5 0.6 0.7]; +plot(J,E,rect=[0,1,0,1]); +xtitle("Plot for drying data","J = kt/L^2","E"); + +//For the 10 mm strips +mi = (0.28 - 0.07); //Initial free moisture content in kg/kg +mf = (0.13-0.07); //Final free moisture content in kg/kg +//at +t = 25; //time is in ksecs +E = (0.06/0.21); +//at E = 0.286 ,J = 0.52 from the plot of given data and J = kt/L^2 +k = poly([0],'k'); +k1 = roots(0.52 - (k*t)/(10/2)^2); +printf("\n k = %f",k1); + +//for the 60 mm planks +m1i = (0.22 - 0.07); //Initial free moisture content in kg/kg +m1f = (0.13 - 0.07); //Final free moisture content in kg/kg +E = (m1f/m1i); +//at E = 0.20 from yhe plot of the given data J = 0.63 and J = kt/L^2 +t1 = 0.63*(60/2)^(2)/k1; +printf("\n t1 = %d ksecs or %.1f days",t1,(t1*1000/(3600*24))); + diff --git a/542/CH16/EX16.3/Example_16_3.sci b/542/CH16/EX16.3/Example_16_3.sci new file mode 100755 index 000000000..cc88b6783 --- /dev/null +++ b/542/CH16/EX16.3/Example_16_3.sci @@ -0,0 +1,60 @@ +clear; +clc; +printf("\n example 16.3"); +//(a) Air +//G kg/s dry air enter with 0.006G kg/s water vapour and hence the heat content of this stream= +//[(1.00G) + (0.006G × 2.01)](385 − 273) = 113.35G kW + +//(b) Wet solid +//0.125 kg/s enter containing 0.40 kg water/kg wet solid, assuming the moisture is expressed on a wet basis. +flowWater = (0.125*0.40); //in kg/sec +flowDrySolid = (0.125-0.050); //in kg/sec +//Hence heat content of this stream +q = [(0.050*4.18)+(0.075*0.88)]*(295-273); +printf("\n The heat content of this stream = %.2f kW",q); + +//Heat out +//(a) Air +//Heat in exit air = [(1.00 G) + (0.006 G × 2.01)](310 − 273) = 37.45G kW. +fd = 0.075; //mass flow rate of dry solids in kg/sec +w = 0.05*0.075/(1+0.05); //water in the dried solids leaving in kg/secs +we = (0.050 - w); //The water evaporated into gas stream in kg/secs + +//Assuming evaporation takes place at 295 K,then: +qout = 0.0464*[2.01*(310-295)+2449+4.18*(295-273)]; +printf("\n Heat in the water vapour = %.1f kW",qout); + +//the total heat in this stream = (119.30 + 37.45G) kW. +//(b) Dried solids + + + //The dried solids contain 0.0036 kg/s water and hence heat conten t of this stream is: + q1 = [(0.075*0.88)+(0.0036*4.18)/(305-273)]; + printf("\n The dried solids contain 0.0036 kg/s water and hence heat content of this stream is: %.2f kW",q1); + + +//(c) Losses +//These amount to 20 kJ/kg dry air or 20m kW. +//Heat Balance +G = poly([0],'G'); +G1 = roots(113.35*G + 6.05 - 119.30 - 37.45*G - 2.59 - 20*G); +printf("\n G = %.2f kg/secs",G1); +printf("\n Water in the outlet stream %.4f kg/secs",0.006*2.07+0.0464); +printf("\n The humidity H = %.4f kg/kg dry air",0.0588/2.07); + + + + + + + + + + + + + + + + + diff --git a/542/CH16/EX16.4/Example_16_4.sci b/542/CH16/EX16.4/Example_16_4.sci new file mode 100755 index 000000000..0df32648e --- /dev/null +++ b/542/CH16/EX16.4/Example_16_4.sci @@ -0,0 +1,40 @@ +clear; +clc; +printf("\n Example 16.4"); +//In 100 kg of feed + + //mass of water = + mw = 100*30/100; //mass of water in kg + //mass of dry solids = + md = 100-30; //mass of dry solids + +//and: +//For b kg water in the dried solids: 100b/(b + 70) = 15.5 +b = poly([0],'b'); +b1 = roots(100*b - 15.5*(b+70)); +printf("\n water in the product ,b = %.1f kg",b1); + + //Initial water content + w1 = 30/70; //Intial moisture content in kg/kg dry solids + //Final moisture content + w2 = (12.8/70); //Final moisture content in kg/kg dry solids + //water to be removed + w3 = (30-12.8); //water to be removed in kg + +//Surface of drying +S = (0.03*70); //Surface for drying in m^2 +rate = (0.0007*2.1); //Rate of drying in kg/sec + +//As the final moisture content is above the critical value, all the drying is at this constant rate and the time of drying is: +t = 17.2/0.00147; +printf("\n As the final moisture content is above the critical value, all the drying is at this constant rate and the time of drying is: %d ksecs or %.2f hr",t,t/3600); + + + + + + + + + + diff --git a/542/CH16/EX16.5/Example_16_5.sci b/542/CH16/EX16.5/Example_16_5.sci new file mode 100755 index 000000000..a7e03099a --- /dev/null +++ b/542/CH16/EX16.5/Example_16_5.sci @@ -0,0 +1,77 @@ +clear; +clc; +printf("\n Example 16.5"); +//Inlet air temperature +Tair = 400; //Inlet air temperature in kelvins +H = 0.01; //Humidity is in kg/kg dry air +//*therefore wet bulb temperature = +Twetbulb = 312; //Inlet wet-bulb temperature +NTU = 1.5; //Number of transfer units + +//Then for adiabatic drying the outlet air temperature,To is given by +To = poly([0],'To'); +To1 = roots(exp(1.5)*(To-312)-(400-312)); +printf("\n For adiabatic drying the outlet air temperature = %.1f K",To1); + +//Solids outlet temperature will be taken to be maximum allowable,325K +//From the steam tables in the Appendix, the latent heat of vaporisation of water at 312 K is2410 kJ/kg. Again from steam tables, the specific heat capacity of water vapour = 1.88 kJ/kg K and that of the solids will be taken as 2.18 kJ/kg K. + +//For a mass flow of solids of 0.35 kg/s and inlet and outlet moisture contents of 0.15 and 0.005 kg/kg dry solids respectively, the mass of water evaporated = 0.35(0.15 − 0.005) = 0.0508 kg/s. + +//For unit mass of solids ,the heat duty includes: + //Heat to the solids + qsolids = 2.18*(325-300); //heat to solids in kJ/kg + //Heat to raise the moisture to the dew point + qdew = (0.15*4.187*(312-300)); //in kJ/kg + //Heat of vaporisation + qvap = 2410*(0.15-0.005); //in kJ/kg + //Heat to raise remaining moisture to solids outlet temperature + qremaining = (0.05*4.187)*(325-312); + //Heat to raise evaporated moisture to the air outlet temp. + qevapo = (0.15-0.005)*1.88*(331.5-312); + qtotal = qsolids + qdew + qvap + qremaining + qevapo; + printf("\n Total heat = %.1f kJ/kg or %d kW",qtotal,qtotal*0.35); + +//The humid heat of entering air is 1.03 kJ/kg K + //G1 (1 +H1) = Q/Cp1(T1 − T2) + //where: G1 (kg/s) is the mass flowrate of inlet air, + //H1 (kg/kg) is the humidity of inlet air, + //Q (kW) is the heat duty, + //Cp1 (kJ/kg K) is the humid heat of inlet air + //and: T1 and T2 (K) are the inlet and outlet air temperatures respectively. + G1 = poly([0],'G1'); + G = roots(G1*(1+0.01)-146/(1.03*(400-331.5))); + printf("\n Mass flow rate of inlet air = %.2f kg/secs",G); + printf("\n Mass flow rate of dry air ,Ga = %.2f kg/secs",G/1.01); + printf("\n the humidity of the outlet air H2 = %.4f kg/kg",0.01+0.0508/2.05); + +//At a dry bulb temperature of 331.5 K, with a humidity of 0.0347 kg/kg, the wet-bulb temperatureof the outlet air, from Figure 13.4 in Volume 1, is 312 K, the same as the inlet, which is the case for adiabatic drying. + +//The dryer diameter is then found from the allowable mass velocity of the air and the entering air flow and for a mass velocity of 0.95 kg/m^2.secs, the cross sectional area of the dryer is +printf("\n The cross sectional area of the dryer is %.2f m^2",2.07/0.95); +printf("\n The equivalent diameter of the dryer = %.2f m",[(4*2.18)/%pi]^0.5); + +//With a constant drying temperature of 312 K: + //at the inlet + deltaT1 = 400-312; //inlet temperature is in deg K + //at the outlet + deltaT2 = 331.5-312; //outlet temperature is in deg K + Tlogmean = (deltaT1 - deltaT2)/log(deltaT1/deltaT2); + printf("\n Logarithmic mean temperature difference = %.1f deg K",Tlogmean); +//The length of the dryer, L is then: L = Q/(0.0625πDG^(0.67)*Tm) + //where D (m) is the diameter + //and G(kg/m^2.secs) is the air mass velocity. + L = 146/[0.0625*(%pi)*1.67*(0.95)^(0.67)*45.5]; + printf("\n The length of the dryer = %.1f m",L); + printf("\n Length/diameter ratio = %d ",10.1/1.67); + + + + + + + + + + + diff --git a/542/CH16/EX16.6/Example_16_6.sci b/542/CH16/EX16.6/Example_16_6.sci new file mode 100755 index 000000000..f55ccd97b --- /dev/null +++ b/542/CH16/EX16.6/Example_16_6.sci @@ -0,0 +1,41 @@ +clear; +clc; +printf("\n Example 16.6"); +H = 0.036; //Humidity is in kg/kg at 811 K +//Taking R as 90 per cent and P as 101.3 kN/m2, then, for assumed values of Tb of 321, 333 and 344 K + //Pw = 13,20 and 32 kN/m2, respectively + //G = 27.8, 12.9 and 6.02 kg/s, respectively. + +//for Tb = 321, 333 and 344 K, +//G = 7.16, 7.8 and 7.54 kg/s respectively. +Tb = [321 333 344]; +G1 = [27.8 12.9 6.02]; //Temperature is in kelvins +G = [7.16 7.8 7.54]; //flow rate in kg/secs +plot2d(Tb,G,style=3); +plot2d(Tb,G1,style=2); +xtitle("Temperature vs Flow rate","Temperature Tb(K)","Flow rate G(kg/secs)"); + + +//Plotting G against Tb for each equation on the same axis, then +Go = 8.3; //Gas flow rate is in kg/secs +Tb = 340; //temperature is in Kelvins +uf = 0.61; //velocity is in m/secs + +D = sqrt(340*(8.3+(1.58*1.26))/(278*0.61)); +printf("\n D = %.2f m",D); + + + + + + + + + + + + + + + + -- cgit