diff options
Diffstat (limited to '3843/CH2')
| -rw-r--r-- | 3843/CH2/EX2.1/Ex2_1.sce | 21 | ||||
| -rw-r--r-- | 3843/CH2/EX2.2/Ex2_2.sce | 22 | ||||
| -rw-r--r-- | 3843/CH2/EX2.3/Ex2_3.sce | 18 | ||||
| -rw-r--r-- | 3843/CH2/EX2.4/Ex2_4.sce | 15 | ||||
| -rw-r--r-- | 3843/CH2/EX2.5/Ex2_5.sce | 13 | ||||
| -rw-r--r-- | 3843/CH2/EX2.6/Ex2_6.sce | 14 | ||||
| -rw-r--r-- | 3843/CH2/EX2.7/Ex2_7.sce | 37 |
7 files changed, 140 insertions, 0 deletions
diff --git a/3843/CH2/EX2.1/Ex2_1.sce b/3843/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..4bab098ec --- /dev/null +++ b/3843/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,21 @@ +// Example 2_1
+clc;funcprot(0);
+// Given data
+P_a=1;// kPa
+P_b=100;// kPa
+P_c=10000;// kPa
+v_ga=129.2;// m^3/kg
+v_fa=0.001;// m^3/kg
+v_gb=1.694;// m^3/kg
+v_fb=0.001;// m^3/kg
+v_gc=0.01803;// m^3/kg
+v_fc=0.00145;// m^3/kg
+
+// Calculation
+// (a)
+v_fga=v_ga-v_fa;// m^3/kg
+// (b)
+v_fgb=v_gb-v_fb;// m^3/kg
+// (c)
+v_fgc=v_gc-v_fc;// m^3/kg
+printf("\n(a)v_fg=%3.1f m^3/kg \n(b)v_fg=%1.3f m^3/kg \n(c)v_fg=%0.5f m^3/kg",v_fga,v_fgb,v_fgc);
diff --git a/3843/CH2/EX2.2/Ex2_2.sce b/3843/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..0d619aa53 --- /dev/null +++ b/3843/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,22 @@ +// Example 2_2
+clc;funcprot(0);
+// Given data
+m=4;// The mass of water in kg
+V=1;// m^3
+T=150;// °C
+v=0.3928;// m^3/kg
+
+// Calculation
+// Table C-1 is used
+V=m*v;// m^3
+// (a)
+P=475.8;// The pressure in kPa
+// (b)
+v=1/4;// m^3/kg
+v_f=0.00109;// m^3/kg
+v_g=0.3928;// m^3/kg
+x=(v-v_f)/(v_g-v_f);// The quality of steam
+m_g=m*x;// The mass of vapor in kg
+// (c)
+V_g=v_g*m_g;// The volume of the vapor in m^3
+printf("\n(a)The pressure,P=%3.1f kPa \n(b)The mass of vapor,m_g=%1.3f kg \n(c)The volume of the vapor,V_g=%0.4f m^3",P,m_g,V_g);
diff --git a/3843/CH2/EX2.3/Ex2_3.sce b/3843/CH2/EX2.3/Ex2_3.sce new file mode 100644 index 000000000..52e7dd4f3 --- /dev/null +++ b/3843/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,18 @@ +// Example 2_3
+clc;funcprot(0);
+// Given data
+m=4;// The mass of water in kg
+P=220;// kPa
+x=0.8;// The quality of steam
+
+// Calculation
+// Use Table C-2.To determine the appropriate numbers at 220 kPa we linearly interpolate between 0.2 and 0.3 MPa.
+P_1=0.2*10^3;// kPa
+P_2=0.3*10^3;// kPa
+v_g1=0.8857;// m^3/kg
+v_g2=0.6058;// m^3/kg
+v_g=(((P-P_1)/(P_2-P_1))*(v_g2-v_g1))+v_g1;// m^3/kg
+v_f=0.0011;// m^3/kg
+v=v_f+(x*(v_g-v_f));// m^3/kg
+V=m*v;// The total volume occupied in m^3
+printf("\nThe final volume occupied by the mixture,V=%1.3f m^3",V);
diff --git a/3843/CH2/EX2.4/Ex2_4.sce b/3843/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..ea49bd74f --- /dev/null +++ b/3843/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,15 @@ +// Example 2_4
+clc;funcprot(0);
+// Given data
+m=2;// The mass of water in lb
+P=540;// psia
+T=700;// °F
+
+// Calculation
+// Use Table C-3E.
+v_f=1.3040;// ft^3/lbm
+v_g=1.0727;// ft^3/lbm
+x=0.4;// The quality of steam
+v=v_f+(x*(v_g-v_f));// ft^3/lbm
+V=m*v;// The final volume of the container in ft^3
+printf("\nThe final volume of the container,V=%1.3f ft^3.",V);
diff --git a/3843/CH2/EX2.5/Ex2_5.sce b/3843/CH2/EX2.5/Ex2_5.sce new file mode 100644 index 000000000..a0924cd1a --- /dev/null +++ b/3843/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,13 @@ +// Example 2_5
+clc;funcprot(0);
+// Given data
+V=0.6;// m^3
+P_gage=200;// The gage pressure in kPa
+T=20+273;// K
+P_atm=100;// kPa
+R=287;// N.m/kg.K
+
+// Calculation
+P=P_gage+P_atm;// The absolute pressure in kPa
+m=(P*10^3*V)/(R*T);// The mass of air in kg
+printf("\nThe mass of air,m=%1.2f kg",m);
diff --git a/3843/CH2/EX2.6/Ex2_6.sce b/3843/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..9ae0c98bb --- /dev/null +++ b/3843/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,14 @@ +// Example 2_6
+clc;funcprot(0);
+// Given data
+// T(z)=15-0.00651°C
+z_0=0;// m
+P=101;// m
+z_1=3000;// m
+
+// Calculation
+// Using the given equation for T(z) we have
+// dP=P/(29.3)*(288-0.00651 z)
+// By solving this equation,we get
+P=101*exp(-0.368);// kPa
+printf("\nThe pressure at an elevation of 3000 m,P=%2.1f kPa.",P);
diff --git a/3843/CH2/EX2.7/Ex2_7.sce b/3843/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..7055758c5 --- /dev/null +++ b/3843/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,37 @@ +// Example 2_7
+clc;funcprot(0);
+// Given data
+T=500+273;// K
+rho=24;// The density in kg/m^3
+R=0.462;// kJ/kg.K
+v=1/rho;// m^3/kg
+
+// Calculation
+// (a)
+P=rho*R*T;// kPa
+printf("\n(a)Using the ideal gas equation,The pressure of steam(P)=%4.0f kPa.",P);
+// (b)
+// Using values for a and b from Table B-8,the vander Waals equation provides
+a=1.703;
+b=0.00169;
+P=((R*T)/(v-b))-(a/v^2);// kPa
+printf("\n(b)Using the vander Waals equation,the pressure of steam(P)=%4.0f kPa.",P);
+// (c)
+// Using values for a and b from Table B-8,the Redlich-Kwong equation provides
+a=43.9;
+b=0.00117;
+P=((R*T)/(v-b))-(a/(v*(v+b)*sqrt(T)));// kPa
+printf("\n(c)Using the Redlich-Kwong equation,the pressure of steam(P)=%4.0f kPa.",P);
+// (d)
+T_c=647.4;// The critical temperature in K
+T_R=T/T_c;// The reduced temperature
+P_c=8000;// The critical pressure in kPa
+P_R=P/P_c;// The reduced pressure
+// By using the reduced temperature and the reduced pressure
+Z=0.93;// The compressibilty factor
+P=(Z*R*T)/v;// kPa
+printf("\n(d)By using the compressibilty factor,the pressure of steam(P)=%4.0f kPa.",P);
+// (e)
+// By using the steam tables,
+P=8000;// kPa
+printf("\n(e)By using the steam tables,the pressure of steam(P)=%4.0f kPa.",P);
|