diff options
Diffstat (limited to '3831/CH11')
| -rw-r--r-- | 3831/CH11/EX11.10/Ex11_10.sce | 12 | ||||
| -rw-r--r-- | 3831/CH11/EX11.12/Ex11_12.sce | 21 | ||||
| -rw-r--r-- | 3831/CH11/EX11.13/Ex11_13.sce | 25 | ||||
| -rw-r--r-- | 3831/CH11/EX11.14/Ex11_14.sce | 24 | ||||
| -rw-r--r-- | 3831/CH11/EX11.15/Ex11_15.sce | 23 | ||||
| -rw-r--r-- | 3831/CH11/EX11.16/Ex11_16.sce | 54 | ||||
| -rw-r--r-- | 3831/CH11/EX11.17/Ex11_17.sce | 15 | ||||
| -rw-r--r-- | 3831/CH11/EX11.2/Ex11_2.sce | 14 | ||||
| -rw-r--r-- | 3831/CH11/EX11.3/Ex11_3.sce | 11 | ||||
| -rw-r--r-- | 3831/CH11/EX11.6/Ex11_6.sce | 11 | ||||
| -rw-r--r-- | 3831/CH11/EX11.7/Ex11_7.sce | 16 |
11 files changed, 226 insertions, 0 deletions
diff --git a/3831/CH11/EX11.10/Ex11_10.sce b/3831/CH11/EX11.10/Ex11_10.sce new file mode 100644 index 000000000..d0515bdbf --- /dev/null +++ b/3831/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,12 @@ +// Example 11_10
+clc;funcprot(0);
+// Given data
+T=20.0;// °C
+beta=0.207*10^-6;// K^-1
+k=45.9*10^-11;// m^2/N
+
+// Solution
+v_f=0.001002;// m^3/kg
+v=v_f;// m^3/kg
+c_pminusc_v=(((T+273.15)*beta^2*v)/k)*10^-3;// kJ/(kg.K)
+printf("\nThe difference between c_p and c_v for saturated liquid water,c_p-c_v=%1.2e kJ/(kg.K)",c_pminusc_v);
diff --git a/3831/CH11/EX11.12/Ex11_12.sce b/3831/CH11/EX11.12/Ex11_12.sce new file mode 100644 index 000000000..a3a969613 --- /dev/null +++ b/3831/CH11/EX11.12/Ex11_12.sce @@ -0,0 +1,21 @@ +// Example 11_12
+clc;funcprot(0);
+// Given data
+T=60.0;// °F
+p_1=14.7;// psia
+r_c=19.2:1;// Compression ratio
+
+// Solution
+// From Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics,we find that, at 60.0°F = 520.R,
+u_1=88.62;// Btu/lbm
+p_r1=1.2147;
+v_r1=158.58;
+v_2byv_1=1/19.2;
+v_r2=v_r1*v_2byv_1;
+// Scanning down the v_r column in Table C.16a, we find that vr = 8.26 at about
+T_2=1600-459.67;// °F
+u_2=286.06;// Btu/lbm
+p_r2=71.73;
+p_2=p_1*(p_r2/p_r1);// psia
+W_12bym=u_1-u_2;// Btu/lbm
+printf("\nThe final temperature and pressure of the air at the end of the compression stroke,T_2=%4.0f°F and p_2=%3.1f psia. \nThe work required per lbm of air present,1W2/m=%3.2f Btu/lbm",T_2,p_2,W_12bym);
diff --git a/3831/CH11/EX11.13/Ex11_13.sce b/3831/CH11/EX11.13/Ex11_13.sce new file mode 100644 index 000000000..31120dd5e --- /dev/null +++ b/3831/CH11/EX11.13/Ex11_13.sce @@ -0,0 +1,25 @@ +// Example 11_13
+clc;funcprot(0);
+// Given data
+m=8.20;// lbm
+V=1.00;// ft^3
+T=-78.0;// °F
+
+// Solution
+// From Table C.12a, we find that
+T_c=240;// R
+p_c=507;// psia
+v_c=1.49/28.011;// ft^3/lbm
+// Also, from Table C.13a, we find that
+R=0.0709;// Btu/lbm.R
+T_R=(T+460)/T_c;
+v=V/m;// ft^3/lbm
+// Assume 'a' instead of '
+v_ca=(R*(T_c*778.16))/(p_c*144);// ft^3/lbm
+v_Ra=v/v_ca;
+// Using T_R = T/Tc = 1.60 and v′R = v/v′c = 0:67, we find from Figure 11.6 that
+p_R=2.10;
+Z=0.850;
+p=p_c*p_R;// psia
+printf("\nThe pressure exerted by 8.20 lbm of the carbon monoxide,p=%4.0f psia",p);
+// The answer is different due to round off error.
diff --git a/3831/CH11/EX11.14/Ex11_14.sce b/3831/CH11/EX11.14/Ex11_14.sce new file mode 100644 index 000000000..f4286bc10 --- /dev/null +++ b/3831/CH11/EX11.14/Ex11_14.sce @@ -0,0 +1,24 @@ +// Example 11_14
+clc;funcprot(0);
+// Given data
+V=0.100;// m^3
+p=20.0;// MPa
+m=15.6;// kg
+T=1000;// °C
+
+// Solution
+// From Table C.12b, we find the critical state properties of methane to be
+T_c=191.1;// K
+p_c=4.64;// MPa
+v=V/m;// m^3/kg
+v_1=v;// m^3/kg
+v_2=v_1;// m^3/kg
+// Table C.13b, gives the gas constant for methane as
+R=0.518;// kJ/kg.K
+// Assume 'a' instead of '
+v_Ra=(v*p_c*10^3)/(R*T_c);
+T_R=(T+273.15)/T_c;
+p_R=32.0;
+p_2byp_c=p_R;
+p_2_worstcase=p_R*p_c;// MPa
+printf("\nThe maximum pressure in the CNG tank at this worst case temperature,(p_2)_worstcase=%3.0f MPa",p_2_worstcase);
diff --git a/3831/CH11/EX11.15/Ex11_15.sce b/3831/CH11/EX11.15/Ex11_15.sce new file mode 100644 index 000000000..088cdb68e --- /dev/null +++ b/3831/CH11/EX11.15/Ex11_15.sce @@ -0,0 +1,23 @@ +// Example 11_15
+clc;funcprot(0);
+// Given data
+p_1=20.0;// MPa
+T_1=150;// °C
+p_2=0.101;// MPa
+
+// Solution
+// From Table C.12b, we find the critical temperature and pressure for CO2 are
+T_c=304.2;// K
+p_c=7.39;// MPa
+M_CO2=44.01;// kg/kg mole
+c_p=0.845;// kJ/kg.K
+p_R1=p_1/p_c;
+T_R1=(T_1+273.15)/T_c;
+// Assume s_1=[(h*-h)/T_c]_1
+s_1=14.0;// kJ/kgmole·K
+p_R2=p_2/p_c;
+// Assume s_2=[(h*-h)/T_c]_2
+// h_2-h_1=0
+T_2=(T_1+273.15)-((s_1/c_p)*(T_c/M_CO2));// K
+T_2=T_2-273.15;// °C
+printf("\nThe exit temperature of the throttle,T_2=%2.1f°C",T_2);
diff --git a/3831/CH11/EX11.16/Ex11_16.sce b/3831/CH11/EX11.16/Ex11_16.sce new file mode 100644 index 000000000..6e3b97503 --- /dev/null +++ b/3831/CH11/EX11.16/Ex11_16.sce @@ -0,0 +1,54 @@ +// Example 11_16
+clc;funcprot(0);
+// Given data
+p_1=150;// psia
+p_2=15.0*10^3;// psia
+T_1=80.0;// °F
+T_2=T_1;// °F
+
+// Calculation
+// (a)
+// The properties of ethylene at its critical state and its molecular mass are found in Table C.12a as
+T_c=508.3;// R
+p_c=742;// psia
+M=28.05;// lbm/lbmoles
+p_R1=p_1/p_c;
+T_R1=(T_1+459.67)/T_c;
+p_R2=p_2/p_c;
+T_R2=T_R1;
+// Using p_R1 and T_R1, Figure 11.9 gives the enthalpy correction for state 1 as
+// Assume s_1=[(h*-hbar)/T_c]_1
+s_1=1.50;// kJ/kgmole.K
+s_1=s_1*(1/4.1865);// Btu/(lbmole.R)
+// Using p_R2 and T_R2, Figure 11.9 gives the enthalpy correction for state 2 as
+// Assume s_2=[(h*-hbar)/T_c]_2
+s_2=31.5;// kJ/kgmole.K
+s_2=s_2*(1/4.1865);// Btu/(lbmole.R)
+// h*2-h*1=0;
+// dh=h_2-h_1;
+dh=0-([s_2-s_1]*(T_c/M));// Btu/lbm
+// (b)
+p_R1=0.202;
+T_R1=1.06;
+Z_1=0.940;
+p_R2=20.2;
+T_R2=T_R2;
+Z_2=2.15;
+R=55.1;// ft.lbf/(lbm.R)
+v_1=(Z_1*R*(T_1+459.67))/(p_1*144);// ft^3/lbm
+v_2=(Z_2*R*(T_2+459.67))/(p_2*144);// ft^3/lbm
+du=dh-(((p_2*144)*v_2*(1/778.16))-((p_1*144)*v_1*(1/778.16)));// Btu/lbm
+// (c)
+// s*2-s*1=dS;
+dS=(c_p*log(T_2/T_1))-((R/778.16)*log(p_2/p_1));// Btu/lbm.R
+// Using p_R1 and T_R1, Figure 11.11 gives the entropy correction for state 1 as
+// Assume (s*bar-sbar)_1=S_1
+S_1=1.50;// kJ/kgmole.K
+S_1=S_1*(1/4.1865);// Btu/(lbmole.R)
+// Using p_R2 and T_R2, Figure 11.11 gives the entropy correction for state 2 as
+S_2=2.22;// kJ/kgmole.K
+S_2=S_2*(1/4.1865);// Btu/(lbmole.R)
+// d_s=S_1-S_2
+ds=dS-([S_2-S_1]*(1/M));// Btu/(lbm.R)
+printf("\n(a)The change in specific enthalpy,h_2-h_1=%3.0f Btu/lbm \n(b)The change in specific internal energy,u_2-u_1=%3.0f Btu/lbm \n(c)The change in specific entropy of the ethylene,s_2-s_1=%0.3f Btu/lbm.R",dh,du,ds);
+// The answer vary due to round off error
diff --git a/3831/CH11/EX11.17/Ex11_17.sce b/3831/CH11/EX11.17/Ex11_17.sce new file mode 100644 index 000000000..2b34f4c8d --- /dev/null +++ b/3831/CH11/EX11.17/Ex11_17.sce @@ -0,0 +1,15 @@ +// Example 11_17
+clc;funcprot(0);
+// Given data
+m_A=1.00;// lbm
+p_A=1.00;// psia
+T_A=200;// °F
+m_B=5.00;// lbm
+p_B=5.00;// psia
+T_B=400;// °F
+
+// Calculation
+T_2=((m_A*(T_A+459.67))+(m_B*(T_B+459.67)))/(m_A+m_B);// R
+T_2=T_2-459.67;// °F
+p_2=((m_A+m_B)*(T_2+459.67))/(((m_A*(T_A+459.67))/p_A)+((m_B*(T_B+459.67))/p_B));// psia
+printf("\nThe final temperature,T_2=%3.0f°F \nThe final pressure,p_2=%1.2f psia",T_2,p_2);
diff --git a/3831/CH11/EX11.2/Ex11_2.sce b/3831/CH11/EX11.2/Ex11_2.sce new file mode 100644 index 000000000..5456fc8ea --- /dev/null +++ b/3831/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,14 @@ +// Example 11_2
+clc;funcprot(0);
+// Given data
+p=200;// psia
+T=400;// °F
+
+// Solution
+// From Table C.3a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at this state,
+u=1123.5;// Btu/lbm
+h=1210.8;// Btu/lbm
+s=1.5602;// Btu/lbm.R
+f=u-((T+459.67)*s);// Btu/lbm
+g=h-((T+459.67)*s);// Btu/lbm
+printf("\nThe value of the specific Helmholtz function for superheated water vapor,f=%3.0f Btu/lbm \nThe value of the specific Gibbs function for superheated water vapor,g=%3.0f Btu/lbm",f,g);
diff --git a/3831/CH11/EX11.3/Ex11_3.sce b/3831/CH11/EX11.3/Ex11_3.sce new file mode 100644 index 000000000..a1fb34060 --- /dev/null +++ b/3831/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,11 @@ +// Example 11_3
+clc;funcprot(0);
+// Given data
+p=1.00;// MPa
+
+// Solution
+// From Table C.2b at p = 1.00 MPa, we find that,
+h_fg=2015.3;// kJ/kg
+T_sat=179.90;// °C
+s_fg=h_fg/(T_sat+273.15);// kJ/kg .K
+printf("\nThe phase change entropy for water,s_fg=%1.4f kJ/kg.K",s_fg);
diff --git a/3831/CH11/EX11.6/Ex11_6.sce b/3831/CH11/EX11.6/Ex11_6.sce new file mode 100644 index 000000000..e3108f37c --- /dev/null +++ b/3831/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,11 @@ +// Example 11_6
+clc;funcprot(0);
+// Given data
+L_0=0.0700;// m
+L=0.200;// m
+T=20.0;// °C
+K=0.150;// N/K
+
+// Solution
+Q_12=(-K*(T+273.15)*L_0*((L/L_0)-1)^3)/3;// N.m
+printf("\n(c)The required heat transfer,Q_12=%1.2f N.m",Q_12);
diff --git a/3831/CH11/EX11.7/Ex11_7.sce b/3831/CH11/EX11.7/Ex11_7.sce new file mode 100644 index 000000000..3e172a8d2 --- /dev/null +++ b/3831/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,16 @@ +// Example 11_7
+clc;funcprot(0);
+// Given data
+// ln p_sat=14.05-(6289.78/T_sat)-(913998.2/T_sat^2);
+// T_sat=°F + 461.2
+T=212;// °F
+R=0.1102;// Btu/(lbm.R)
+
+// Solution
+T_sat=T+461.2;// R
+// From Table C.13a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics
+h_fg=[(6289.78)+((1827997.8)/(T_sat))]*R;// Btu/lbm
+// Table C.1a gives
+h_fg_212F=970.4;// Btu/lbm
+p_error=((h_fg-h_fg_212F)/h_fg_212F)*100;// %
+printf("\nThus, the value obtained from Rankine’s equation is in error by only %1.2f percentage.",p_error);
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