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Diffstat (limited to '3843/CH11/EX11.14')
| -rw-r--r-- | 3843/CH11/EX11.14/Ex11_14.sce | 28 |
1 files changed, 28 insertions, 0 deletions
diff --git a/3843/CH11/EX11.14/Ex11_14.sce b/3843/CH11/EX11.14/Ex11_14.sce new file mode 100644 index 000000000..eecb7c8b8 --- /dev/null +++ b/3843/CH11/EX11.14/Ex11_14.sce @@ -0,0 +1,28 @@ +// Example 11_14
+clc;funcprot(0);
+// Given data
+m_w3=10000;// kg/min
+T_ain=20;// The temperature of air at inlet in °C
+phi_1=50;// Humidity in %
+T_aout=32;// The temperature of air at exit in °C
+phi_2=98;// Humidity in %
+T_win=40;// The temperature of water at inlet in °C
+T_wout=25;// The temperature of water at exit in °C
+
+// Calculation
+// (a)
+// From the psychrometric chart we find
+h_1=37;// kJ/kg of dry air
+h_2=110;// kJ/kg of dry air
+w_1=0.0073;// kgH2O/kg dry air
+w_2=0.0302;// kgH2O/kg dry air
+// From steam tables
+h_3=167.5;// kJ/kg
+h_4=104.9;// kJ/kg
+m_a=(m_w3*(h_4-h_3))/(h_1-h_2+((w_2-w_1)*h_4));// kg/min
+// From the psychrometric chart we find
+v_1=0.84;// m^3/ kg dry air
+Vdot=m_a*v_1;// m^3/min
+// (b)
+m_4=m_w3-((w_2-w_1)*m_a);// kg/min
+printf("\n(a)The volume flow rate of air into the cooling tower,Vdot=%4.0f m^3/min \n(b)The mass flux of water,m_4=%4.0f kg/min",Vdot,m_4);
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