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diff --git a/1309/CH9/EX9.7/Result9_7.pdf b/1309/CH9/EX9.7/Result9_7.pdf Binary files differnew file mode 100755 index 000000000..8c9dbdd87 --- /dev/null +++ b/1309/CH9/EX9.7/Result9_7.pdf diff --git a/1309/CH9/EX9.7/ch9_7.sce b/1309/CH9/EX9.7/ch9_7.sce new file mode 100755 index 000000000..2503fc00f --- /dev/null +++ b/1309/CH9/EX9.7/ch9_7.sce @@ -0,0 +1,72 @@ +clc; +clear; +printf("\t\t\tChapter9_example7\n\n\n"); +// (a) Determine the UA product for the exchanger. (b) Calculate the exit temperatures for the exchanger, assuming that only the inlet temperatures are known +// properties of engine oil at (190 + 158)/2 = 174°F = 176 degree F from appendix table C4 +rou_1= 0.852*62.4; // density in lbm/ft^3 +cp_1=0.509; // specific heat BTU/(lbm-degree Rankine) +v_1= 0.404e-3; // viscosity in ft^2/s +kf_1 = 0.08; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_1 = 2.98e-3; // diffusivity in ft^2/hr +Pr_1 = 490; // Prandtl Number +m_1=39.8; // mass flow rate in lbm/min +// temperatures in degree F +T1=190; +T2=158; +// properties of air at (126 + 166)/2 = 146°F = 606 degree R from appendix table D1 +rou_2= 0.0653; // density in lbm/ft^3 +cp_2=0.241; // specific heat BTU/(lbm-degree Rankine) +v_2= 20.98e-5; // viscosity in ft^2/s +kf_2 = 0.01677 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_2 = 1.066; // diffusivity in ft^2/hr +Pr_2 = 0.706; // Prandtl Number +m_2=67; // mass flow rate in lbm/min +// temperatures in degree F +t1=126; +t2=166; +// Heat Balance +q_air=m_2*cp_2*60*(t2-t1); +q_oil=m_1*cp_1*60*(T1-T2); +printf("\nThe heat gained by air is %.2e BTU/hr",q_air); +printf("\nThe heat lost by oil is %.2e BTU/hr",q_oil); +// for counterflow +LMTD=((T1-t2)-(T2-t1))/(log((T1-t2)/(T2-t1))); +printf("\nThe LMTD for counter flow configuration is %.1f degree F",LMTD); +// Frontal Areas for Each Fluid Stream +Area_air=(9.82*8)/144; +Area_oil=(3.25*9.82)/144; +printf("\nThe Core frontal area on the air side is %.3f sq.ft\nThe Core frontal area on the oil side is %.3f sq.ft ",Area_air,Area_oil); +// Correction Factors (parameters calculated first) +S=(t2-t1)/(T1-t1); +R=(T1-T2)/(t2-t1); +F=0.87; //value of correction factor from figure 9.21a corresponding to above calculated values of S and R +// Overall Coefficient (q = U*A*F*LMTD) +UA=q_air/(F*LMTD); +printf("\nThe Overall Coefficient is %.2e BTU/(hr. degree R)",UA); +// determining the capacitances +mcp_air=m_2*cp_2*60; +mcp_oil=m_1*cp_1*60; +printf("\nThe capacitance value of air is %d BTU/(hr. degree R)",mcp_air); +printf("\nThe capacitance value of engine oil is %d BTU/(hr. degree R)",mcp_oil); +if mcp_air>mcp_oil then + mcp_max=mcp_air; + mcp_min=mcp_oil; + printf("\nEngine Oil has minimum capacitance"); + else mcp_max=mcp_oil; + mcp_min=mcp_air; + printf("\nAir has minimum capacitance"); +end +// determination of parameters for determining effectiveness +mcp_min_max=mcp_min/mcp_max; +NTU=(UA/mcp_min); +printf("\nThe required parameters are mcp_min/mcp_max=%.3f and (UoAo/mcp_min)=%.2f",mcp_min_max,NTU); +effectiveness=0.62; //value of effectiveness from figure 9.21b corresponding to the above calculated values of capacitance ratio and (UoAo/mcp_min):'); +t2_c=(T1-t1)*effectiveness+t1; +T2_c=T1-(mcp_min_max)*(t2_c-t1); +printf("\n\t\t\tSummary of Requested Information\n"); +printf("\n(a) UA = %.2e BTU/(hr. degree R)",UA); +printf("\n(b) The Outlet temperatures (degree F)"); +printf("\n\tCalculated\tGiven in Problem Statement"); +printf("\nAir\t\t%d\t%d",t2_c,t2); +printf("\nEngine Oil\t%d\t%d",T2_c,T2); + |