diff options
Diffstat (limited to '572/CH9/EX9.4/c9_4.sce')
-rwxr-xr-x | 572/CH9/EX9.4/c9_4.sce | 59 |
1 files changed, 59 insertions, 0 deletions
diff --git a/572/CH9/EX9.4/c9_4.sce b/572/CH9/EX9.4/c9_4.sce new file mode 100755 index 000000000..755be5818 --- /dev/null +++ b/572/CH9/EX9.4/c9_4.sce @@ -0,0 +1,59 @@ +//(9.4) Air enters the compressor of an ideal air-standard Brayton cycle at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10. The turbine inlet temperature is 1400 K. Determine (a) the thermal efficiency of the cycle, (b) the back work ratio, (c) the net power developed, in kW.
+
+//solution
+
+//variable initialization
+T1 = 300 //in kelvin
+AV = 5 //volumetric flow rate in m^3/s
+p1 = 100 //in kpa
+pr = 10 //compressor pressure ratio
+T3 = 1400 //turbine inlet temperature in kelvin
+
+//analysis
+//At state 1, the temperature is 300 K. From Table A-22,
+h1 = 300.19 //in kj/kg
+pr1 = 1.386
+
+pr2 = pr*pr1
+//interpolating in Table A-22,
+h2 = 579.9 //in kj/kg
+//from Table A-22
+h3 = 1515.4 //in kj/kg
+pr3 = 450.5
+
+pr4 = pr3*1/pr
+//Interpolating in Table A-22, we get
+h4 = 808.5 //in kj/kg
+
+//part(a)
+eta = ((h3-h4)-(h2-h1))/(h3-h2) //thermal efficiency
+printf('the thermal efficiency is: %f',eta)
+
+//part(b)
+bwr = (h2-h1)/(h3-h4) //back work ratio
+printf('\nthe back work ratio is: %f',bwr)
+
+//part(c)
+R = 8.314 //universal gas constant, in SI units
+M = 28.97 //molar mass of air in grams
+mdot = AV*p1/((R/M)*T1) //mass flow rate in kg/s
+
+Wcycledot = mdot*((h3-h4)-(h2-h1)) //The net power developed
+printf('\n the net power developed, in kW is: %f',Wcycledot)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
|