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10 files changed, 157 insertions, 0 deletions

diff --git a/3831/CH6/EX6.1/Ex6_1.sce b/3831/CH6/EX6.1/Ex6_1.sce
new file mode 100644
index 000000000..b8fa4fa12
--- /dev/null
+++ b/3831/CH6/EX6.1/Ex6_1.sce
@@ -0,0 +1,22 @@
+// Example 6_1

+clc;funcprot(0);

+// Given data

+V=300;// ft/s

+D=6/12;// ft

+R=D/2;// ft

+Z=15;// ft

+g=32.174;// ft/s^2

+g_c=32.174;// lbm.ft/lbf.s^2

+

+// Calculation

+// From the superheated steam table, Table C.3a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at 100. psia and 500.°F,

+v=5.587;// ft^3/lbm

+h=1279.1;// Btu/lbm

+A=%pi*(3/12)^2;// ft^2

+mdot=(A*V)/v;// lbm/s

+ke=(V^2)/(2*g_c);// ft.lbf/lbm

+ke=ke*(1/778.16);// Btu/lbm

+pe=(g*Z)/g_c;// // ft.lbf/lbm

+pe=pe*(1/778.16);// Btu/lbm

+E_mf=-[mdot*(h+ke+pe)];// Btu/s

+printf("\nThe mass flow energy transport rate of steam,E_mass flow=%1.2e Btu/s",E_mf);

diff --git a/3831/CH6/EX6.10/Ex6_10.sce b/3831/CH6/EX6.10/Ex6_10.sce
new file mode 100644
index 000000000..7831668a3
--- /dev/null
+++ b/3831/CH6/EX6.10/Ex6_10.sce
@@ -0,0 +1,15 @@
+// Example 6_10

+clc;funcprot(0);

+// Given data

+p_1=2000;// psig

+T_1=200+459.67;// R

+T_T=70.0+459.67;// R

+m_R=0.500;// lbm/s

+W_c=-3.00;// hp

+k=1.4;// The specific heat ratio of nitrogen

+

+// Calculation

+m_Rbym_D=(k-1)/[(k*(T_1/T_T))-1];// The ratio of recycled mass flow rate to discharge mass flow rate

+c_p=0.248;// Btu/(lbm.R)

+Q_H=(m_R*c_p*(T_1-T_T))+[(W_c)*550*(1/778)];// Btu/s

+printf("\nThe rate of recycle heat transfer required,Q_H=%2.1f Btu/s",Q_H);

diff --git a/3831/CH6/EX6.2/Ex6_2.sce b/3831/CH6/EX6.2/Ex6_2.sce
new file mode 100644
index 000000000..8fecb1a65
--- /dev/null
+++ b/3831/CH6/EX6.2/Ex6_2.sce
@@ -0,0 +1,18 @@
+// Example 6_2

+clc;funcprot(0);

+// Given data

+D=1.00;// inch

+T=60.0;// °F

+p=80.0;// psig

+mdot=0.800;// lbm/s

+v=0.01603;// ft^3/lbm

+g_c=32.174;// lbm.ft/lbf.s^2

+g=32.174;// ft/s^2

+

+// Calculation

+V_in=(4*mdot*v)/(%pi*D^2*(1/12)^2);// ft/s

+p_in=94.7;// psia

+p_out=14.7;// psia

+V_out=[(V_in^2)+(2*g_c*v*(p_in-p_out)*144)]^(1/2);// ft/s

+Z_out=V_out^2/(2*g);// ft

+printf("\n(a)The outlet velocity from the nozzle,(V_out)_a=%3.0f ft/s \n(b)The height to which the stream of water rises above the nozzle outlet when the nozzle is pointed straight up,(Z_out)_b=%3.0f ft.",V_out,Z_out)

diff --git a/3831/CH6/EX6.3/Ex6_3.sce b/3831/CH6/EX6.3/Ex6_3.sce
new file mode 100644
index 000000000..b0ddc7201
--- /dev/null
+++ b/3831/CH6/EX6.3/Ex6_3.sce
@@ -0,0 +1,18 @@
+// Example 6_3

+clc;funcprot(0);

+// Given data

+p_1=2.00;// MPa

+p_2=0.100;// MPa

+T_2=150;// °C

+h_1=2776.4;// kJ/kg

+h_2=2776.4;// kJ/kg

+

+// Calculation

+h_f1=908.8;// kJ/kg

+h_fg1=1890.7;// kJ/kg

+h_g1=2799.5;// kJ/kg

+x_1=(h_1-h_f1)/h_fg1;// The quality of steam

+x_1=x_1*100;// The quality of steam in %

+T_1=212.4;// °C

+mu_J=(T_1-T_2)/(p_1-p_2);// °C/MPa

+printf("\nThe quality of the wet steam in the pipe,x=%2.1f percentage \nJoule-Thomson coefficient,mu_J=%2.1f°C/MPa",x_1,mu_J);

diff --git a/3831/CH6/EX6.4/Ex6_4.sce b/3831/CH6/EX6.4/Ex6_4.sce
new file mode 100644
index 000000000..3686f743a
--- /dev/null
+++ b/3831/CH6/EX6.4/Ex6_4.sce
@@ -0,0 +1,17 @@
+// Example 6_4

+clc;funcprot(0);

+// Given data

+Q=0;// kW

+W=0;// kW

+m_s=12.0;// kg/min

+p_1=1.00;// MPa

+T_1=500;// °C

+T_3=15;// °C

+T_4=20;// °C

+

+// Calculation

+h_1=3478.4;// kJ/kg

+h_2=762.8;// kJ/kg

+c_w=4.2;// kJ/kg.K

+m_w=m_s*(h_1-h_2)/[c_w*(T_4-T_3)];// kg/min

+printf("\nThe flow rate of cooling water taken from a local river,m_w=%4.0f kg/min",m_w);

diff --git a/3831/CH6/EX6.5/Ex6_5.sce b/3831/CH6/EX6.5/Ex6_5.sce
new file mode 100644
index 000000000..0e152bf5d
--- /dev/null
+++ b/3831/CH6/EX6.5/Ex6_5.sce
@@ -0,0 +1,16 @@
+// Example 6_5

+clc;funcprot(0);

+// Given data

+p_1=85.0;// psig

+p_2=10.0;// psig

+t=8.00;// hour

+m=20.0;// gal

+

+// Calculation

+mv=20.0/8.00;// gal/h

+mv=mv*0.13368*(1/3600);// ft^3/s

+W_shaft=mv*(p_1-p_2)*144;// ft.lbf/s

+W_shaft=W_shaft*(1/550);// hp

+W_shaft=W_shaft*746;// W

+W_shaft_ins=W_shaft*5*60*(1/2.50);// W

+printf("\nThe hydraulic power produced,(W_shaft)_instantaneous=%3.0f W",W_shaft_ins);

diff --git a/3831/CH6/EX6.6/Ex6_6.sce b/3831/CH6/EX6.6/Ex6_6.sce
new file mode 100644
index 000000000..f546890e8
--- /dev/null
+++ b/3831/CH6/EX6.6/Ex6_6.sce
@@ -0,0 +1,17 @@
+// Example 6_6

+clc;funcprot(0);

+// Given data

+p_1=2.00;// MPa

+T_1=800;// °C

+p_2=1.00;// MPa

+Wbymdot=2000;// kJ/kg

+

+// Calculation

+h_1=4150.4;// kJ/kg

+h_f2=29.30;// kJ/kg

+h_fg2=2484.9;// kJ/kg

+h_g2=2514.2;// kJ/kg

+h_2=h_1-Wbymdot;// kJ/kg

+x_2=(h_2-h_f2)/h_fg2;// The quality of steam

+x_2=x_2*100;// % vapor at the turbine’s outlet

+printf("\nThe quality of the steam at the outlet of an insulated steam turbine,x_2=%2.1f percentage.",x_2);

diff --git a/3831/CH6/EX6.7/Ex6_7.sce b/3831/CH6/EX6.7/Ex6_7.sce
new file mode 100644
index 000000000..78e861d2b
--- /dev/null
+++ b/3831/CH6/EX6.7/Ex6_7.sce
@@ -0,0 +1,12 @@
+// Example 6_7

+clc;funcprot(0);

+// Given data

+T_in=20.0;// °C

+p_in=50.0;// MPa

+c=4.126;// kN.m/kg.K

+

+// Calculation

+v_f=0.001002;// m^3/kg

+v=0.0009804;// m^3/kg

+T_finalfilled=T_in+((v*(p_in*10^3))/c);// °C

+printf("\nThe final temperature of the water in the tank,T_final filled=%2.1f°C",T_finalfilled);

diff --git a/3831/CH6/EX6.8/Ex6_8.sce b/3831/CH6/EX6.8/Ex6_8.sce
new file mode 100644
index 000000000..7118053eb
--- /dev/null
+++ b/3831/CH6/EX6.8/Ex6_8.sce
@@ -0,0 +1,11 @@
+// Example 6_8

+clc;funcprot(0);

+// Given data

+T_in=20.0;// °C

+p_in=1.40;// MPa

+k=1.40;// The specific heat ratio

+

+// Calculation

+T_finalfilling=k*(T_in+273.15);// K

+T_finalfilling=T_finalfilling-273.15;// °C

+printf("\nThe final temperature of the air in the tank,T_final filling=%3.0f°C",T_finalfilling);

diff --git a/3831/CH6/EX6.9/Ex6_9.sce b/3831/CH6/EX6.9/Ex6_9.sce
new file mode 100644
index 000000000..721e20bdb
--- /dev/null
+++ b/3831/CH6/EX6.9/Ex6_9.sce
@@ -0,0 +1,11 @@
+// Example 6_9

+clc;funcprot(0);

+// Given data

+// From  Example 6_8

+T_initial=137+273.15;// K

+k=1.4;// The specific heat ratio

+

+// Calculation

+T_finalemptying=T_initial*((2/k)-1);// K

+T_finalemptying=T_finalemptying-273.15;// °C

+printf("\nThe final temperature inside the tank immediately after the tank is empty,T_final emptying=%2.1f°C.",T_finalemptying);