diff options
Diffstat (limited to '3831/CH19')
-rw-r--r-- | 3831/CH19/EX19.1/Ex19_1.sce | 14 | ||||
-rw-r--r-- | 3831/CH19/EX19.2/Ex19_2.sce | 11 | ||||
-rw-r--r-- | 3831/CH19/EX19.3/Ex19_3.sce | 20 | ||||
-rw-r--r-- | 3831/CH19/EX19.4/Ex19_4.sce | 12 | ||||
-rw-r--r-- | 3831/CH19/EX19.5/Ex19_5.sce | 22 | ||||
-rw-r--r-- | 3831/CH19/EX19.6/Ex19_6.sce | 17 |
6 files changed, 96 insertions, 0 deletions
diff --git a/3831/CH19/EX19.1/Ex19_1.sce b/3831/CH19/EX19.1/Ex19_1.sce new file mode 100644 index 000000000..b73f02a26 --- /dev/null +++ b/3831/CH19/EX19.1/Ex19_1.sce @@ -0,0 +1,14 @@ +// Example 19_1
+clc;funcprot(0);
+// Given data
+T=20.0+273.16;// K
+d=0.0100;// m
+alpha_cu=3.50*10^-6;// V/K
+rho_e=5.00*10^-9;// ohm m
+dphibydx=1.00;// Voltage gradient in V/m
+
+// Solution
+A=(%pi/4)*d^2;// m^2
+I=(A/rho_e)*dphibydx;// A
+Q_P=alpha_cu*T*I;// W
+printf('\nThe Peltier heat flow,Q_P=%2.1f W',Q_P);
diff --git a/3831/CH19/EX19.2/Ex19_2.sce b/3831/CH19/EX19.2/Ex19_2.sce new file mode 100644 index 000000000..2b57c3e9e --- /dev/null +++ b/3831/CH19/EX19.2/Ex19_2.sce @@ -0,0 +1,11 @@ +// Example 19_2
+clc;funcprot(0);
+// Given data
+T=100.0;// °C
+
+// Solution
+// (a)
+alpha_fecu=-(-13.4+(0.028*T)+(0.00039*T^2))*10^-6;// V/K
+// (b)
+pi_fecu=(T+273.16)*alpha_fecu;// V
+printf('\n(a)The relative Seebeck coefficient,alpha_fecu=%1.2e V/K \n(b)The relative Peltier coefficient,pi_fecu=%1.2e V',alpha_fecu,pi_fecu);
diff --git a/3831/CH19/EX19.3/Ex19_3.sce b/3831/CH19/EX19.3/Ex19_3.sce new file mode 100644 index 000000000..2897ca7f9 --- /dev/null +++ b/3831/CH19/EX19.3/Ex19_3.sce @@ -0,0 +1,20 @@ +// Example 19_3
+clc;funcprot(0);
+// Given data
+T_H=100;// °C
+T_C=0;// °C
+alpha_ch=23.0*10^-6;// V/K
+alpha_al=-18.0*10^-6;// V/K
+
+// Solution
+// (a)
+alpha_chal=alpha_ch-alpha_al;// V/K
+phi_alch=alpha_chal*(T_H-T_C);// V
+// (b)
+pi_ch1=alpha_ch*(T_C+273.15);// V
+pi_al1=alpha_al*(T_C+273.15);// V
+pi_chal1=pi_ch1-pi_al1;// V
+pi_ch2=alpha_ch*(T_H+273.15);// V
+pi_al2=alpha_al*(T_H+273.15);// V
+pi_chal2=pi_ch2-pi_al2;// V
+printf('\n(a)alpha_ch-al=%2.0e V/K \n phi_al-ch=%1.1e V \n(b)At the 0.00°C = 273.15 K junction, \npi_ch=%1.2e V \npi_al=%1.2e V \npi_ch-al=%2.1e V \nAt the 100.°C = 373.15 K junction,\npi_ch=%1.2e V \npi_al=%1.2e V \npi_ch-al=%2.1e V ',alpha_chal,phi_alch,pi_ch1,pi_al1,pi_chal1,pi_ch2,pi_al2,pi_chal2);
diff --git a/3831/CH19/EX19.4/Ex19_4.sce b/3831/CH19/EX19.4/Ex19_4.sce new file mode 100644 index 000000000..7b33c8310 --- /dev/null +++ b/3831/CH19/EX19.4/Ex19_4.sce @@ -0,0 +1,12 @@ +// Example 19_4
+clc;funcprot(0);
+// Given data
+mu=1.50*10^-5;// The viscosity of the CO_2 in kg/(m.s)
+T_1=300;// K
+T_2=305;// K
+k_p=1.00*10^-6;// m^2
+k_o=2.00*10^4;// The osmotic heat conductivity in m^2/s
+
+// Solution
+dp=-((mu*k_o)/k_p)*log(T_2/T_1);// N/m^2
+printf('\nThe steady state thermomolecular pressure difference across the membrane,p_2-p_1=%4.0f N/m^2',dp);
diff --git a/3831/CH19/EX19.5/Ex19_5.sce b/3831/CH19/EX19.5/Ex19_5.sce new file mode 100644 index 000000000..d0febe114 --- /dev/null +++ b/3831/CH19/EX19.5/Ex19_5.sce @@ -0,0 +1,22 @@ +// Example 19_5
+clc;funcprot(0);
+// Given data
+T_1=30+273.15;// K
+T_2=T_1;// K
+dp=10.0;// kPa
+d=0.0100;// m
+rho=996;// kg/m^3
+k_p=1.00*10^-12;// m^2
+mu=891*10^-6;// kg/(s.m)
+dx=0.100;// m
+Q=15.0;// The isothermal energy transport rate in this system in J/s
+
+// Solution
+// (a)
+A=(%pi/4)*d^2;// m^2
+m=-((rho*A*k_p)/mu)*((dp*10^3)/dx);// kg/s
+// (b)
+k_o=-(Q/A)/((-dp*10^3)/dx);// m^2/s
+// (c)
+S_i=Q/T_1;// J/(s.K)
+printf('\n(a)The thermomechanical mass flow rate between the vessels,m=%1.2e kg/s \n(b)The osmotic heat conductivity coefficient,k_o=%1.2f m^2/s \n(c)The isothermal entropy transport rate induced by the thermomechanical mass flow rate,S_i=%0.4f J/(s.K)',m,k_o,S_i);
diff --git a/3831/CH19/EX19.6/Ex19_6.sce b/3831/CH19/EX19.6/Ex19_6.sce new file mode 100644 index 000000000..56b8330e9 --- /dev/null +++ b/3831/CH19/EX19.6/Ex19_6.sce @@ -0,0 +1,17 @@ +// Example 19_6
+clc;funcprot(0);
+// Given data
+rho=996;// kg/m^3
+Q=8.70;// J/s
+T=30+273;// K
+k_t=0.610;// J/(s.K.m)
+k_o=1.91;// m^2/s
+k_p=1.00*10^-12;// m^2
+mu=891*10^-6;// kg/(s.m)
+dx=0.100;// m
+
+// Solution
+m=(rho*Q)/((T*(k_t/k_o))+(mu*(k_o/k_p)));// kg/s
+dTbydx=-(T*m)/(rho*k_o);// K/m
+dT=dTbydx*dx;// K
+printf('\nThe induced isobaric mass flow rate,m=%1.2e kg/s \nThe resulting temperature difference between the vessels,dT=%1.2e K',m,dT);
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