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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /839/CH16/EX16.1/Example_16_1.sce | |
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diff --git a/839/CH16/EX16.1/Example_16_1.sce b/839/CH16/EX16.1/Example_16_1.sce new file mode 100755 index 000000000..c25334b10 --- /dev/null +++ b/839/CH16/EX16.1/Example_16_1.sce @@ -0,0 +1,52 @@ +//clear//
+clear;
+clc;
+
+//Example 16.1
+//Given
+mdot = 20000; //[lb/h]
+xin = 0.20;
+xout = 0.50;
+Pg = 20; //[lbf/in.^2]
+Pabs = 1.93; //[lbf/in.^2]
+U = 250; //[Btu/ft^2-h-F]
+Tf = 100; //[F]
+
+//Solution
+//the amount of water in feed and thick liquor, from material balance
+w_feed = 80/20; //[lb/per pound of solid]
+w_liquor = 50/50; //[lb/per pound of solid]
+//water evaporated
+w_eva = w_feed-w_liquor; //[lb/per pound of solid]
+//or
+w_eva = w_eva*mdot*xin; //[lb/h]
+//Flow raye of thick liquor is
+ml_dot = mdot-w_eva //[lb/h]
+
+//Steam consumed
+//Since with strong solutions of NaOH the heat of dilution is not negligible,
+//the rate of heat transfer is found from Eq.(16.4) and Fig. 16.8.
+//The vaporiztion temperature of the 50 percent solution at a pressure of 100 mmHg
+//is found as follows
+Tb_w = 124; //[F], at 100 mmHg, from Appendix 7
+Tb_s = 197; //[F], from Fig. 16.8
+BPE = Tb_s-Tb_w; //[F]
+//From Fig. 16.8, the enthalpies of the feed and thick liquor are found
+Hf = 55; //[Btu/lb], 20% solids, 100 [F]
+H = 221; //[Btu/lb], 50% solids, 197 [F]
+//Enthalpy of the leaving water vapor is found from the steam table
+Hv = 1149; //[Btu/lb], At 197 [F] and 1.93 [lbf/in.^2]
+//Enthalpy of the vapor leaving the evaporator
+lambda_s = 939;//[Btu/lb], At 20 [lbf/in.^2], from Appendix 7
+//Using Eq.(16.4), the rate of heat transfer and steam consumption
+q = (mdot-ml_dot)*Hv + ml_dot*H - mdot*Hf; //[Btu/h]
+ms_dot = q/lambda_s; //[lb/h]
+disp('lb/h',ms_dot,'steam consumed is')
+//Economy
+Economy = ml_dot/ms_dot
+disp(Economy,'Economy')
+//Heating Surface
+//The condensation temperature of the steam is 259 [F], the heating area required is
+
+A = q/(U*(259-197)) //[ft^2]
+disp('ft^2',A,'heating area required is')
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