diff options
Diffstat (limited to '3831/CH16')
| -rw-r--r-- | 3831/CH16/EX16.1/Ex16_1.sce | 12 | ||||
| -rw-r--r-- | 3831/CH16/EX16.10/Ex16_10.sce | 16 | ||||
| -rw-r--r-- | 3831/CH16/EX16.11/Ex16_11.sce | 16 | ||||
| -rw-r--r-- | 3831/CH16/EX16.12/Ex16_12.sce | 45 | ||||
| -rw-r--r-- | 3831/CH16/EX16.13/Ex16_13.sce | 28 | ||||
| -rw-r--r-- | 3831/CH16/EX16.14/Ex16_14.sce | 30 | ||||
| -rw-r--r-- | 3831/CH16/EX16.15/Ex16_15.sce | 15 | ||||
| -rw-r--r-- | 3831/CH16/EX16.2/Ex16_2.sce | 16 | ||||
| -rw-r--r-- | 3831/CH16/EX16.3/Ex16_3.sce | 23 | ||||
| -rw-r--r-- | 3831/CH16/EX16.4/Ex16_4.sce | 14 | ||||
| -rw-r--r-- | 3831/CH16/EX16.5/Ex16_5.sce | 17 | ||||
| -rw-r--r-- | 3831/CH16/EX16.6/Ex16_6.sce | 24 | ||||
| -rw-r--r-- | 3831/CH16/EX16.7/Ex16_7.sce | 19 | ||||
| -rw-r--r-- | 3831/CH16/EX16.8/Ex16_8.sce | 21 |
14 files changed, 296 insertions, 0 deletions
diff --git a/3831/CH16/EX16.1/Ex16_1.sce b/3831/CH16/EX16.1/Ex16_1.sce new file mode 100644 index 000000000..0ba509586 --- /dev/null +++ b/3831/CH16/EX16.1/Ex16_1.sce @@ -0,0 +1,12 @@ +// Example 16_1
+clc;funcprot(0);
+// Given data
+T=20+273.15;// K
+V=90.0;// km/h
+g_c=1;// The gravitational constant
+c_p=1.004;// kJ/kg.K
+
+// Solution
+T_0=T*(1+(((V*10^3/(3600*1000))^2)/(2*g_c*c_p*T)));// K
+T_0=T_0-273.15;// °C
+printf("\nThe stagnation temperature,T_0=%2.1f°C",T_0)
diff --git a/3831/CH16/EX16.10/Ex16_10.sce b/3831/CH16/EX16.10/Ex16_10.sce new file mode 100644 index 000000000..2d64a15c2 --- /dev/null +++ b/3831/CH16/EX16.10/Ex16_10.sce @@ -0,0 +1,16 @@ +// Example 16_10
+clc;funcprot(0);
+// Given data
+m=5.00*10^-3;// kg
+T=20.0+273.15;// K
+p=101.3*10^3;// kg/(m.s^2)
+R=286;// m^2/(s^2.K)
+D=3.00*10^-3;// m
+g=9.81;// m/s^2
+g_c=1;// The gravitational constant
+
+// Calculation
+W=(m*g)/g_c;// N
+rho=p/(R*T);// kg/m^3
+V_in=((4*g_c*W)/(rho*%pi*D^2))^(1/2);// m/s
+printf("\nThe velocity of the jet,V_in=%2.1f m/s",V_in);
diff --git a/3831/CH16/EX16.11/Ex16_11.sce b/3831/CH16/EX16.11/Ex16_11.sce new file mode 100644 index 000000000..d6b365224 --- /dev/null +++ b/3831/CH16/EX16.11/Ex16_11.sce @@ -0,0 +1,16 @@ +// Example 16_11
+clc;funcprot(0);
+// Given data
+M_x=5.50;// The Mach number
+p_x=14.7;// lbf/in^2
+T_x=70.0+459.67;// °F
+k=1.4;// The specific heat ratio
+R=53.34;// ft.lbf/lbm.R
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+M_y=((((k-1)*M_x^2)+2)/((2*k*M_x^2)+1-k))^(1/2);// The Mach number
+T_y=T_x*[(1+(((k-1)/2)*M_x^2))/(1+(((k-1)/2)*M_y^2))];// R
+p_y=p_x*(M_x/M_y)*(T_y/T_x)^(1/2);// lbf/in^2
+V_wind=(M_x*sqrt(k*g_c*R*T_x))-(M_y*sqrt(k*g_c*R*T_y));// ft/s
+printf("\nThe pressure directly behind the shock wave,p_y=%3.0f lbf/in^2 \nThe temperature directly behind the shock wave,T_y=%4.0f R \nThe wind velocity directly behind the shock wave,V_wind=%1.0e ft/s",p_y,T_y,V_wind);
diff --git a/3831/CH16/EX16.12/Ex16_12.sce b/3831/CH16/EX16.12/Ex16_12.sce new file mode 100644 index 000000000..07c18c308 --- /dev/null +++ b/3831/CH16/EX16.12/Ex16_12.sce @@ -0,0 +1,45 @@ +// Example 16_12
+clc;funcprot(0);
+// Given data
+p_os=7.00;// MPa
+T_os=2000;// °C
+D_t=0.0200;// m
+D_e=0.100;// m
+k=1.40;// The specific heat ratio
+R=286;// m^2/(s^2.K)
+g_c=1;// The gravitational constant
+
+// Calculation
+// (a)
+A_t=(%pi*D_t^2)/4;// m^2
+mdot=(0.0404*(p_os*10^6)*A_t)/sqrt(T_os+273.15);// kg/s
+// (b)
+A_r=(D_e/D_t)^2;// (A_r=A_exit/A*)
+M_e=5.00;// Mach number at exit
+// Assume p_exit/p_os=p_r
+p_r=1.89*10^-3;// Pressure ratio
+// Assume T_exit/T_os=T_r
+T_r=0.16667;// Temperature ratio
+p_e=p_r*p_os*10^3;// The exit pressure in kN/m^2
+T_exit=T_r*(T_os+273.15);// K
+c_e=sqrt(k*g_c*R*T_exit);// The velocity of sound at the exit in m/s
+V_exit=c_e*M_e;// m/s
+// (c)
+M_x=5.0;// The Mach number
+p_x=13.23;// kN/m^2
+T_x=378.8;// K
+// Table C.19 is a tabular version of these equations, and at Mx = 5.0, we again have a direct entry
+M_y=0.415;// The Mach number
+// Assume p_osy/p_osx=p_ros
+p_ros=0.06172;
+// Assume p_y/p_x=p_rxy
+p_rxy=29.00;
+// Assume p_osy/p_x=p_rosyx
+p_rosyx=32.654;
+// Assume T_y/T_x=T_yx
+T_yx=5.800;
+p_osx=p_os*10^3;// kN/m^2
+p_B=p_ros*p_osx;// The required back pressure in kN/m^2
+// Alternatively
+p_B=p_rosyx*p_x;// The required back pressure in kN/m^2
+printf("\n(a)The mass flow rate required for supersonic flow in the diverging section,mdot=%1.2f kg/s \n(b)The Mach number, pressure,temperature and velocity at the exit of the diverging section with this massflow rate,M_exit=%1.2f,p_exit=%2.1f kN/m^2,T_exit=%3.1f K,V_exit=%4.0f m/s \n(c)The outside back pressure required to produce a standing normal shock wave at the exit of the diverging section,p_B=%3.0f kN/m^2",mdot,M_e,p_e,T_exit,V_exit,p_B);
diff --git a/3831/CH16/EX16.13/Ex16_13.sce b/3831/CH16/EX16.13/Ex16_13.sce new file mode 100644 index 000000000..59fc65793 --- /dev/null +++ b/3831/CH16/EX16.13/Ex16_13.sce @@ -0,0 +1,28 @@ +// Example 16_13
+clc;funcprot(0);
+// Given data
+p_os=3.00;// atm
+T_os=20.0;// °C
+p_B=1.00;// atm
+A_r=2.0;// The exit to throat area ratio fo r the nozzle
+k=1.4;// The specific heat ratio
+R=286;// m^2/(s^2.K)
+g_c=1;// The gravitational constant
+
+// Calculation
+p_a=p_os*(2/(k+1))^(k/(k-1));// atm
+// Since we are given Aexit/A* = A_E/A*= 2.00, we can find ME by inverting Eq. (16.23b).However, in this case, it is again much easier to use Table C.18 for this area ratio and read (approximately),
+M_E=2.20;// The Mach number at exit
+// Assume p_rEos=p_E/p_os
+p_rEos=0.09352;
+p_E=p_rEos*p_os;// atm
+// Assume p_r=p_osy/p_osx
+p_r=1.00/3.00;
+// From Table C.19 at p_osy/p_osx=0.333
+M_x=2.98;// The Mach number
+M_y=0.476;// The Mach number
+T_e=0.50813*(T_os+273.15);// K
+c_exit=sqrt(k*g_c*R*T_e);// m/s
+M_exit=M_E;// The Mach number at exit
+V_exit=M_exit*c_exit;// m/s
+printf("\nThe exit pressure,p_E=%0.3f atm\nThe exit temperature,T_exit=%3.2f K \nThe exit velocity,V_exit=%3.0f m/s",p_E,T_e,V_exit);
diff --git a/3831/CH16/EX16.14/Ex16_14.sce b/3831/CH16/EX16.14/Ex16_14.sce new file mode 100644 index 000000000..69adc41ae --- /dev/null +++ b/3831/CH16/EX16.14/Ex16_14.sce @@ -0,0 +1,30 @@ +// Example 16_14
+clc;funcprot(0);
+// Given data
+p_inlet=456.2;// kN/m^2
+T_inlet=283.7;// K
+p_exit=370.4;// kN/m^2
+T_exit=260.1;// K
+V_exit=474.8;// m/s
+k=1.67;// The specific heat ratio for helium
+R=2077.0;// m^2/(s^2.K)
+g_c=1;// The gravitational constant
+
+// Calculation
+// (a)
+c_osi=sqrt(k*g_c*R*T_inlet);// m/s
+c_inlet=c_osi;// m/s
+n_N=((((k-1)/2)*(V_exit/c_inlet)^2)/(1-((p_exit/p_inlet)^((k-1)/k))));// The nozzle’s efficiency
+// (b)
+C_v=sqrt(n_N);// The nozzle’s velocity coefficient
+// (c)
+R=2.077;// kJ/kg.K
+rho_e=p_exit/(R*T_exit);// kg/m^3
+M_exit=1.0;// The exit Mach number
+T_os=T_inlet;// K
+p_os=p_inlet;// kN/m^2
+T_es=T_os*(2/(k+1));// K
+rho_es=(p_os/(R*T_os))*[2/(k+1)]^(1/(k-1));// kg/m^3
+V_es=sqrt(k*g_c*R*10^3*T_es);// m/s
+C_d=(rho_e*V_exit)/(rho_es*V_es);// The nozzle’s discharge coefficient
+printf("\n(a)The nozzle’s efficiency,n_N=%0.3f \n(b)The nozzle’s velocity coefficient,C_v=%0.3f \n(c)The nozzle’s discharge coefficient,C_d=%0.3f",n_N,C_v,C_d);
diff --git a/3831/CH16/EX16.15/Ex16_15.sce b/3831/CH16/EX16.15/Ex16_15.sce new file mode 100644 index 000000000..6872faec3 --- /dev/null +++ b/3831/CH16/EX16.15/Ex16_15.sce @@ -0,0 +1,15 @@ +// Example 16_15
+clc;funcprot(0);
+// Given data
+M_in=0.890;// The inlet Mach number
+p_osi=314.7;// kPa
+p_ose=249.3;// kPa
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+n_D=(((((1+((((k-1)/2)*M_in^2)))*(p_ose/p_osi)^((k-1)/k)))-1)/(((k-1)*M_in^2)/2))*100;// %
+// (b)
+p_i=p_osi/((1+(((k-1)/2)*M_in^2))^(k/(k-1)));// kPa
+C_p=(p_ose-p_i)/(p_osi-p_i);// The diffuser’s pressure recovery coefficient
+printf("\n(a)The diffuser’s efficiency,n_D=%2.1f percentage \n(b)The diffuser’s pressure recovery coefficient,C_p=%0.3f",n_D,C_p);
diff --git a/3831/CH16/EX16.2/Ex16_2.sce b/3831/CH16/EX16.2/Ex16_2.sce new file mode 100644 index 000000000..7f40f256f --- /dev/null +++ b/3831/CH16/EX16.2/Ex16_2.sce @@ -0,0 +1,16 @@ +// Example 16_2
+clc;funcprot(0);
+// Given data
+T=20+273.15;// K
+V=25.0;// m/s
+k=1.40;// The specific heat ratio
+p=0.101;// MPa
+g_c=1;// The gravitational constant
+c_p=1.004;// kJ/kg.K
+R=0.286;// kJ/kg.K
+
+// Solution
+p_os=p*(1+((V^2/1000)/(2*g_c*c_p*T)))^(k/(k-1));// The isentropic stagnation pressure in MPa
+rho=(p*10^3)/(R*T);// kg/m^3
+rho_os=rho*(1+((V^2/1000)/(2*g_c*c_p*T)))^(1/(k-1));// The isentropic stagnation density in kg/m^3
+printf("\nThe isentropic stagnation pressure,p_os=%0.4f MPa \nThe isentropic stagnation density,rho_os=%1.4f kg/m^3",p_os,rho_os);
diff --git a/3831/CH16/EX16.3/Ex16_3.sce b/3831/CH16/EX16.3/Ex16_3.sce new file mode 100644 index 000000000..c18a212f8 --- /dev/null +++ b/3831/CH16/EX16.3/Ex16_3.sce @@ -0,0 +1,23 @@ +// Example 16_3
+clc;funcprot(0);
+// Given data
+p_1=14.7;// psia
+T_1=1000;// °F
+V_1=1612;// ft/s
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+// Station 1
+p_1=14.7;// psia
+T_1=1000;// °F
+h_1=1534.4;// Btu/lbm
+s_1=2.1332;// Btu/lbm.R
+// Station os
+s_os=s_1;// Btu/lbm.R
+h_os=h_1+(V_1^2/(2*g_c));// Btu/lbm
+//Table C.3a, in Thermodynamic Tables to accompany Modern Engineering Thermodynamics a Mollier diagram for steam
+p_os=20.0;// psia
+T_os=1100;// °F
+v_os=46.4;// ft^3/lbm
+rho_os=1/v_os;// lbm/ft^3;
+printf("\nThe isentropic stagnation temperature,T_0=%4.0f°F \nThe isentropic stagnation pressure,p_os=%2.1f psia \nThe isentropic stagnation density,rho_os=%0.3f lbm/ft^3",T_os,p_os,rho_os);
diff --git a/3831/CH16/EX16.4/Ex16_4.sce b/3831/CH16/EX16.4/Ex16_4.sce new file mode 100644 index 000000000..3c15fa3f2 --- /dev/null +++ b/3831/CH16/EX16.4/Ex16_4.sce @@ -0,0 +1,14 @@ +// Example 16_4
+clc;funcprot(0);
+// Given data
+T=35+273.15;// K
+V=300;// m/s
+
+// Solution
+// Using Table C.13b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics for the values of the specific heat ratio and the gas constant for methane, we get
+k_methane=1.30;// The specific heat ratio
+g_c=1;// The gravitational constant
+R_methane=518;// J/kg.K
+c_methane=sqrt(k_methane*g_c*R_methane*T);// m/s
+M_methane=V/c_methane;//The Mach number
+printf("\nThe Mach number of the methane,M_methane=%0.3f",M_methane);
diff --git a/3831/CH16/EX16.5/Ex16_5.sce b/3831/CH16/EX16.5/Ex16_5.sce new file mode 100644 index 000000000..4f5044cec --- /dev/null +++ b/3831/CH16/EX16.5/Ex16_5.sce @@ -0,0 +1,17 @@ +// Example 16_5
+clc;funcprot(0);
+// Given data
+T=-20.0+273.15;// K
+p=0.500;// atm
+M=0.850;// The Mach number
+k=1.40;// The specific heat ratio
+R=286;// J/kg.K
+g_c=1;// The gravitational constant
+
+// Solution
+V=M*sqrt(k*g_c*R*T);// m/s
+T_os=T*(1+(((k-1)*M^2)/2));// K
+T_os=T_os-273.15;// °C
+p_os=p*(1+(((k-1)*M^2)/2))^(k/(k-1));// atm
+p_os=p_os*1.013*10^2;// kPa
+printf("\nThe aircraft’s velocity,V=%3.0f m/s \nThe isentropic stagnation temperature,T_os=%2.1f°C \nThe isentropic stagnation pressure,p_os=%2.1f KPa",V,T_os,p_os);
diff --git a/3831/CH16/EX16.6/Ex16_6.sce b/3831/CH16/EX16.6/Ex16_6.sce new file mode 100644 index 000000000..771b7dc2b --- /dev/null +++ b/3831/CH16/EX16.6/Ex16_6.sce @@ -0,0 +1,24 @@ +// Example 16_6
+clc;funcprot(0);
+// Given data
+p_os=1.00;// MPa
+T_os=20.0+273.15;// K
+k=1.40;// The specific heat ratio
+p=0.1013;// MPa
+g_c=1;// The gravitational constant
+R=286;// J/kg.K
+
+// Solution
+// (a)
+p_r=p/p_os;// The pressure ratio
+M=((2/(k-1))*(((p_os/p)^((k-1)/k))-1))^(1/2);// The exit Mach number
+// (b)
+T=(T_os/(1+(((k-1)*M^2)/2)))-273.15;// The exit temperature in °C
+// (c)
+V=M*sqrt(k*g_c*R*(T+273.15));// The exit velocity in m/s
+// (d)
+p_throat=p_os*[2/(k+1)]^(k/(k-1));// The pressure at the throat of the nozzle in MPa
+// (e)
+T_throat=T_os*[2/(k+1)];// The temperature at the throat of the nozzle in K
+T_throat=T_throat-273.15;// The temperature at the throat of the nozzle in °C
+printf("\n(a)The exit Mach number,M=%1.2f \n(b)The exit temperature,T=%3.0f°C \n(c)The exit velocity,V=%3.0f m/s \n(d)The pressure at the throat of the nozzle,p_throat=%0.3f MPa \n(e)The temperature at the throat of the nozzle,T_throat=%2.1f°C",M,T,V,p_throat,T_throat);
diff --git a/3831/CH16/EX16.7/Ex16_7.sce b/3831/CH16/EX16.7/Ex16_7.sce new file mode 100644 index 000000000..5d16670ca --- /dev/null +++ b/3831/CH16/EX16.7/Ex16_7.sce @@ -0,0 +1,19 @@ +// Example 16_7
+clc;funcprot(0);
+// Given data
+D_bag=3.00;// ft
+t_fill=30;// milliseconds
+p_air=15.00;// psia
+p_os=1500;// psia
+T_os=70.0+459.67;// R
+k=1.40;// The specific heat ratio
+R_air=53.34;// ft.lbf/lbm.R
+
+// Solution
+V_bag=(%pi*D_bag^3)/6;// ft^3
+T_air=T_os*(2/(k+1));// R
+rho_air=(p_air*144)/(R_air*T_air);// lbm/ft^3
+m_avg=(rho_air*V_bag)/(t_fill*10^-3);// lbm/s
+D_tube=[(4*m_avg*sqrt(T_os+459.67))/(0.532*%pi*p_os)]^(1/2);// in
+printf("\nThe minimum tube diameter,D_tube=%1.2f in",D_tube);
+// The answer vary due to round off error
diff --git a/3831/CH16/EX16.8/Ex16_8.sce b/3831/CH16/EX16.8/Ex16_8.sce new file mode 100644 index 000000000..3cf90b793 --- /dev/null +++ b/3831/CH16/EX16.8/Ex16_8.sce @@ -0,0 +1,21 @@ +// Example 16_8
+clc;funcprot(0);
+// Given data
+D_exit=0.0938;// in
+T_os=70.0;// °F
+p_osi=50.0;// psia
+V_T=1.00;// ft^3
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+p_r1=(2/(k+1))^(k/(k-1));// The pressure ratio
+p_exit=14.7;// psia
+p_exitbyp_os=p_exit/p_osi;// The pressure ratio
+// (b)
+p_os=p_exit/p_r1;// psia
+p_os=p_os*0.472;// psig
+// (c)
+A_a=(%pi*D_exit^2)/(4*144);// ft^2
+tau=31.95*log(p_osi/(p_os/0.472));// s
+printf("\n(a)p_exit/p_os=%0.3f,which is <0.528 therefore, initially, the flow is choked.\n(b)The flow remains choked until the tire deflates to a pressure of p_os=%2.1f psig \n(c)The valve stem unchokes at time,tau=%2.1f s",p_exitbyp_os,p_os,tau);
|