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+//Chemical Engineering Thermodynamics

+//Chapter 3

+//First Law of Thermodynamics

+

+//Example 3.7

+clear;

+clc;

+

+//Given

+m = 5000;//Amount of steam recived per hour in Kg

+H1 = 666;//Specific enthalpy when steam entered in the turbine in Kcal/Kg

+H2 = 540;//Specific enthalpy when steam left the turbine in Kcal/Kg

+u1 = 3000/60;//velocity at which steam entered in m/sec

+u2 = 600/60;//velocity at which steam left in m/sec

+Z1 = 5;//height at which steam entered in m

+Z2 = 1;//height at which steam left in m

+Q = -4000;//heat lost in Kcal

+g = 9.81;

+

+//To calculate the horsepuwer output of the turbine

+delH = H2-H1;//change in enthalpy in Kcal

+delKE = ((u2^2)-(u1^2)/(2*g))/(9.8065*427);//change in kinetic energy in Kcal; 1kgf = 9.8065 N

+delPE = ((Z2-Z1)*g)/(9.8065*427);//change in potential energy in Kcal

+W = -(m*(delH+delKE+delPE))+Q;//work delivered in Kcal/hr

+W1 = W*(427/(3600*75));//work delivered by turbine in hp

+mprintf('Work delivered by turbine is %f hp',W1);

+//end
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