diff options
Diffstat (limited to '409/CH24')
| -rwxr-xr-x | 409/CH24/EX24.1/Example24_1.sce | 44 | ||||
| -rwxr-xr-x | 409/CH24/EX24.2/Example24_2.sce | 36 | ||||
| -rwxr-xr-x | 409/CH24/EX24.3/Example24_3.sce | 27 | ||||
| -rwxr-xr-x | 409/CH24/EX24.4/Example24_4.sce | 36 | ||||
| -rwxr-xr-x | 409/CH24/EX24.5/Example24_5.sce | 72 | ||||
| -rwxr-xr-x | 409/CH24/EX24.6/Example24_6.sce | 37 | ||||
| -rwxr-xr-x | 409/CH24/EX24.7/Example24_7.sce | 33 | ||||
| -rwxr-xr-x | 409/CH24/EX24.8/Example24_8.sce | 32 | ||||
| -rwxr-xr-x | 409/CH24/EX24.9/Example24_9.sce | 43 |
9 files changed, 360 insertions, 0 deletions
diff --git a/409/CH24/EX24.1/Example24_1.sce b/409/CH24/EX24.1/Example24_1.sce new file mode 100755 index 000000000..5334f0d6d --- /dev/null +++ b/409/CH24/EX24.1/Example24_1.sce @@ -0,0 +1,44 @@ +clear ; +clc; +// Example 24.1 +printf('Example 24.1\n\n'); +//page no. 720 +// Solution Fig. E24.1 + +// Assumptions to be made in eqn. 24.1 in following segment +printf('Assumptions to be made in eqn. 24.1 in following segments are:\n'); +//(a)- 1 to 5 +printf('\n(a)- 1 to 5.\n'); +printf(' 1. Change in potential energy(del_PE) = 0(no change in level) .\n'); +printf(' 2. Probably change in kinetic energy(del_KE)=0 .\n'); +printf(' 3. Change in energy = 0 (process appears to be steady).\n'); +printf(' Result : Q + W = del_H.\n'); + +//(b) 4 to 5 +printf('\n\n(b) 4 to 5.\n'); +printf(' 1. Q = W = 0 \n'); +printf(' 2. Probably change in kinetic energy(del_KE)=0.\n'); +printf(' 3. Change in energy = 0 (process appears to be steady).\n'); +printf(' Result : del_H = -del_PE . \n'); + +//(c) 3 to 4 +printf('\n\n(c) 3 to 4.\n'); +printf(' 1. Q = W = 0 \n'); +printf(' 2. Probably change in kinetic energy(del_KE)=0.\n'); +printf(' 3. Change in energy = 0 (process appears to be steady).\n'); +printf(' Result : del_H = -del_PE . \n'); + +//(d) 3 to 5 +printf('\n\n(d) 3 to 5.\n'); +printf(' 1. Q = W = 0 \n'); +printf(' 2. Probably change in kinetic energy(del_KE)=0.\n'); +printf(' 3. Change in energy = 0 (process appears to be steady).\n'); +printf(' 4. Change in potential energy(del_PE) = 0(no change in level) .\n'); +printf(' Result : del_H = 0 . \n'); + +//(e)- 1 to 3 +printf('\n(e) 1 to 3.\n'); +printf(' 1. Change in potential energy(del_PE) = 0(no change in level) .\n'); +printf(' 2. Probably change in kinetic energy(del_KE)=0 .\n'); +printf(' 3. Change in energy = 0 (process appears to be steady).\n'); +printf(' Result : Q + W = del_H.\n');
\ No newline at end of file diff --git a/409/CH24/EX24.2/Example24_2.sce b/409/CH24/EX24.2/Example24_2.sce new file mode 100755 index 000000000..e23e35553 --- /dev/null +++ b/409/CH24/EX24.2/Example24_2.sce @@ -0,0 +1,36 @@ +clear ; +clc; +// Example 24.2 +printf('Example 24.2\n\n'); +//page no. 725 +// Solution + +printf('Table to carry out degree of freedom analysis:\n'); +// Number of variables involved +printf('\nI. Number of variables involved.\n'); +printf('\n For materials:\n'); +printf(' Hot gas : 4 component flows, T, and p 6 \n'); +printf(' Cool gas : 4 component flows, T, and p 6\n'); +printf(' Water in : 1 component flow, T, and p 3\n'); +printf(' Water out : 1 component flow, T, and p 3\n'); +printf('\n Energy:\n'); +printf(' Q and W 2 \n'); +printf(' H,KE and PE associated with each stream flow 12 \n'); +printf('\_______________________________________________________________________\n'); +printf(' Total 32\n'); +printf('\n\nII. Number of equations and specifications.\n'); +printf('\n Specified values:\n'); +printf(' Hot gas : 4 component flows, T, and p 6 \n'); +printf(' Cool gas : T, and p 2\n'); +printf(' Water in : T, and p 2\n'); +printf(' Water out : T, and p 2\n'); +printf('\n Specified in the energy balance:\n'); +printf(' Q and W 2 \n'); +printf(' KE and PE associated with each of 4 stream flow 8 \n'); +printf('\n Material balance: \n'); +printf(' 4 species balances plus water 5 \n'); +printf('\n Energy balance: 1\n'); +printf('\n H in each stream is a function of specified T and p 4\n'); +printf('\_______________________________________________________________________\n'); +printf(' Total 32\n'); +printf('\n Therefore, by analysing the above table it is clear that degrees of freedom of system is (32 - 32) = 0 \n');
\ No newline at end of file diff --git a/409/CH24/EX24.3/Example24_3.sce b/409/CH24/EX24.3/Example24_3.sce new file mode 100755 index 000000000..efcad7c20 --- /dev/null +++ b/409/CH24/EX24.3/Example24_3.sce @@ -0,0 +1,27 @@ +clear ; +clc; +// Example 24.3 +printf('Example 24.3\n\n'); +//page no. 728 +// Solution Fig. E24.3 + +// Given + m_CO2 = 10 ;// mass of CO2 - [lb] + Ti_CO2 = 80 ;// Initial temperature of CO2 - [degree F] + Vi = 4.0 ;// Initial volume of CO2-[cubic feet] + f_CO2 = 40/100 ;// Fraction of CO2 that convert to liquid finally + s_Vi = Vi /m_CO2 ;// Initial specific volume of CO2 - [cubic feet/lb] + s_Vf = s_Vi ;// Constant volume -[cubic feet/lb] +// Use the CO2 chart in Appendix J to necessary data, according to book it is +// CO2 is gas at start of process and reference state for the CO2 chart is -40 degree F , saturated liquid +// From chart +Pi = 300 ;// Intial pressure - [psia] +del_Hi = 160 ;// Intial change in specific enthalpy - [Btu/lb] +// Now again use chart to get fnal condition fixed by constant volume line and quality 0.6 , according to book it is +del_Hf = 81 ;// Final change in specific enthalpy - [Btu/lb] +Pf = 140 ;//Final pressure - [psia] +// Use conditions given in problem ( W= 0 ,since volume is constant ,therefore del_PE and del_KE =0 ),simplifing the energy balance equation we get Q = del_H - del_(PV) +// Analysing the given conditions dof of system = 0 , with 1 eqn. and 1 unknown Q +Q = ((del_Hf - del_Hi) - (Pf * s_Vf * 144/778.2 - Pi * s_Vi * 144/778.2))*m_CO2 ;// Heat removed from the extinguisher -[Btu] + + printf(' Heat removed from the extinguisher is %i Btu .\n',Q);
\ No newline at end of file diff --git a/409/CH24/EX24.4/Example24_4.sce b/409/CH24/EX24.4/Example24_4.sce new file mode 100755 index 000000000..9e5f581ca --- /dev/null +++ b/409/CH24/EX24.4/Example24_4.sce @@ -0,0 +1,36 @@ +clear ; +clc; +// Example 24.4 +printf('Example 24.4\n\n'); +//page no. 730 +// Solution + +// Pick the system as gas plus heater +// Given +Pi = 1.5 ;// Intial pressure - [Pa] +Vi = 2*10^-3 ;// Initial volume of gas - [cubic metre] +Ti = 300 ;// Initial temperature - [K] +W = 480 ;// Work done by heater on system +t = 5 ;// Time for which current is supplied -[ min] +m_ht = 12 ;// Mass of the heater - [g] +C_ht = 0.35 ;// Heat capacity of heater - [ J/gK] +R = 8.314 ;// Ideal gas constant - [(Pa*cubic metre)/(g mol* K)] + +// It is assumed that heat transfer across system boundary for this short time is negligible , therefore Q = 0 +// Using the above assumption the equation reduces to del_U = W, therefore +del_U = W ;// Change in nternal energy - [J] + +// Gas is assumed to be ideal, therefore get n by using pv = nRT +n = (Pi*Vi)/(R*Ti) ;// Number of moles of argon gas -[g mol] +Cp = (5/2)* R ;// Specific heat capacity of argon gas at constant pressure - [ J/gK] +Cv = Cp - R ;// Specific heat capacity of argon gas at constant volume - [ J/gK] +// del_Ug = n*Cv*(Tf - Ti) - change in internal energy of gas +// del_Uh = m_ht*C_ht*(Tf - Ti) - change in internal energy of gas +// get total change in internal energy = del_Ug + del_Uh , and solve it for Tf ( final temperature ) +deff('[y]=f(Tf)','y=m_ht*C_ht*(Tf - Ti) + n*Cv*(Tf - Ti) - del_U'); +Tf=fsolve(400,f) ;// Final temperature -[K] +funcprot(0); + printf(' Final temperature of gas is %.0f K .\n',Tf); + + Pf = (Tf/Ti)*Pi ;// Final pressure - [Pa] + printf(' Final pressure in chamber is %.2f Pa .\n',Pf);
\ No newline at end of file diff --git a/409/CH24/EX24.5/Example24_5.sce b/409/CH24/EX24.5/Example24_5.sce new file mode 100755 index 000000000..795119241 --- /dev/null +++ b/409/CH24/EX24.5/Example24_5.sce @@ -0,0 +1,72 @@ +clear ; +clc; +// Example 24.5 +printf('Example 24.5\n\n'); +//page no. 732 +// Solution Fig. E24.5 + +// Pick the system as shown in above figure of book +// Given +m_water = 10 ;// Mass of water - [lb] +T_water = 35 ;// Temperature of water - [degree F] +m_ice = 4 ;// Mass of ice - [lb] +T_ice = 32 ;// Temperature of ice - [degree F] +m_stm = 6 ;// Initial mass of steam -[lb] +T_stm = 250 ;// Temperature of stm - [degree F] +p = 20 ;// Pressure of system -[psia] + +m_total = m_water + m_ice + m_stm ;// Mass of H2O in three phases initially -[lb] +// By following conditions of book, Q = 0, W = 0 , change in PE and change in KE = 0, the energy balance reduces to del_U = 0 + +// According to book additional information is obtained from the steam table and CD at given conditions ,it is as follows +U_ice = -143.6 ;// Specific internal energy of ice -[Btu/lb] +U_water = 3.025 ;// Specific internal energy of water -[Btu/lb] +U_stm = 1092.25 ;// Specific internal energy of steam -[Btu/lb] +V_water = 0.0162 ;// Specific volume of water -[cubic feet/lb] +V_stm = 20.80 ;// Specific volume of steam -[cubic feet/lb] +V_total = m_stm*V_stm ;//Total volume of container ignoring volume of water and ice as they are neglgible + +V_sys = V_total/m_total ;// Specific volume of system -[cubic feet/lb] +U_sys =(m_water*U_water + m_ice*U_ice + m_stm*U_stm)/m_total ;// Final specific internal energy of system -[Btu/lb] + +// Trial and error method +// Assume two temperatures and find volume total so as to bracket value of U_sys, Here e take T1 = 190 and T2 = 200 degree F +// Obtain necessary data from steam table at corresponding temperatures + +T1 = 190 ;// assumed temperature +U1 = [157.17 1071.83] ;//specific internal energy of liquid and vapour respetively -[Btu/lb] +V1 = [0.0165 41.01] ;// Specific volume of liquid and vapour respetively -[cubic feet/lb] +x1 = V_sys/V1(2) ;// Quality of vapour +U1_sys = (1-x1)*U1(1) + x1*U1(2); // Specific internal energy of system at T1-[Btu/lb] + +T2 = 200 ;// assumed temperature +U2 = [168.11 1073.96];// specific internal energy of liquid and vapour respetively -[Btu/lb] +V2 = [0.017 33.601] ;// Specific volume of liquid and vapour respetively -[cubic feet/lb] +x2 = V_sys/V2(2) ; // Quality of vapour +U2_sys = (1-x2)*U2(1) + x2*U2(2) ;// Specific internal energy of system at T2-[Btu/lb] + +// Check whether assumption is right +if (U_sys > U1_sys ) + if ( U_sys <U2_sys) + printf('Assumption is right, now find exact temperature by interpolation between 2 assumed temperatures.\n'); + else + printf('Assumption is wrong, assume different T2.\n'); + end + else + printf('Assumption is wrong,assume different T1.\n'); + end + +// Interpolation, to get final temperature corresponding to U_sys +T_sys = T1 + ((T2 - T1)*(U_sys - U1_sys))/(U2_sys - U1_sys); + + printf(' The final temperature obtained by interpolation between 2 assumed temperatures is %.2f degree F.\n',T_sys); + +// Now obtain specific volume of vapour data at final temperature from steam table and use it to calculate x(quality) , according to book it is +V_vap = 39.35 ;//specific volume of vapour data at final temperature -[cubic feet/lb] +x = V_sys /V_vap ;// Quality of gas at final temperature + + //Final state +Vap = m_total*x ;// Mass of vapour at final state - [lb] +stm_con = m_stm - Vap ;// Mass of steam condenses - [lb] + +printf('\nTherefore, mass of steam condenses is %.2f lb.\n',stm_con);
\ No newline at end of file diff --git a/409/CH24/EX24.6/Example24_6.sce b/409/CH24/EX24.6/Example24_6.sce new file mode 100755 index 000000000..4bb77a796 --- /dev/null +++ b/409/CH24/EX24.6/Example24_6.sce @@ -0,0 +1,37 @@ +clear ; +clc; +// Example 24.6 +printf('Example 24.6\n\n'); +//page no. 736 +// Solution Fig. E24.6 + +// Pick the system as shown in above figure of book +// Given +h1 = -15 ;// Initial level of water from ground level -[ft] +h2 = 165 ;//Final level of water from ground level -[ft] +V_rate = 200 ;// Volume flow rate of water - [gal/hr] +Q1 = 30000 ;// Heat input by heater - [Btu/hr] +Q2 = 25000 ;// Heat lost by system -[Btu/hr] +T1 = 35 ;// Initial temperature of water - [degree F] +g = 32.2 ;// Acceleration due to gravity - [ft/ square second] +p_pump = 2 ;// Power of pump -[hp] +f_w = 55/100 ;// Fraction of rated horsepower that i used in pumping water +Cp = 1 ;// Specific heat capacity of water - [Btu/lb*F] + +// Use following conditions to simplify the energy balance +// 1. Proces is in steady state , so change in energy = 0 +// 2. m1 = m2 = m +// 3. change in KE = 0 , because we will assume that v1 = v2 = 0 +// The energy balance reduce to Q + W = del_(H*m + PE*m) + +m = V_rate * 8.33 ;// Total mass of water pumped -[lb] +del_PE = (m* g *(h2 - h1))/(32.2*778) ;// Change in PE - [Btu/hr] +Q = Q1 - Q2 ;// Net heat exchange -[Btu/hr] +W = 2* f_w * 60 * 33000/778 ;// Work on system - [Btu/hr] +del_H = Q + W - del_PE ;// By using reduced energy balance - [Btu/hr] +// Also del_H = m* Cp * (T2 - T1), all is known except T2 , solve for T2 +deff('[y] = f(T2)','y = m*Cp*(T2-T1) - del_H'); +T2 = fsolve(40,f) ;// Boiling point temperature +funcprot(0); + + printf(' Final temperature of water that enters storage tank is %.1f degree F .\n',T2);
\ No newline at end of file diff --git a/409/CH24/EX24.7/Example24_7.sce b/409/CH24/EX24.7/Example24_7.sce new file mode 100755 index 000000000..f8ce840f7 --- /dev/null +++ b/409/CH24/EX24.7/Example24_7.sce @@ -0,0 +1,33 @@ +clear ; +clc; +// Example 24.7 +printf('Example 24.7\n\n'); +//page no. 738 +// Solution Fig. E24.7 + +// Pick the system as shown in above figure of book +// Given +T_stm = 250 + 273 ;// Temperature of entering steam - [K ] +Q_loss = -1.5 ;// Rate of heat loss from system - [kJ/s ] +T_mi = 20 + 273 ;//Temperature of entering material -[K ] +T_mf = 100 + 273 ;// Temperature of material after heating - [K] +m_m = 150 ;// Mass of charged material - [kg] +Cp_m = 3.26 ;// Average heat capacity of material - [ J/(g*K)] + +// Use following conditions to simplify the energy balance +// 1. Proces is not in steady state , so change in energy not equals 0. +// 2. Assume del_KE and del_PE = 0. +// 3. Assume del_KE and del_PE = 0, for entering and exiting material . +// 4. W = 0. +// 5. Assume m1 = m2 = m_stm +// The energy balance reduce to del_E = del_U = Q - del_(H*m) .... eqn. (b) + +del_U = m_m*Cp_m*(T_mf - T_mi) ;// Change in enthalpy of system , because del_(pV) = 0 for liquid and solid charge -[kJ] +Q_loss_total = Q_loss * 3600; // Total heat loss by system n 1 hour - [kJ] +// We need the value of specific change in enthalpy value of saturated steam(del_H_steam), according to book we can obtain this value from steam table, it's value is -1701 kJ/kg +del_H_steam = -1701 ;// Specific change in enthalpy value of saturated steam -[kJ/kg] +// Get mass of steam per kg charge from reduced energy balance(eqn. (b)) +m_stm_total = (del_U - Q_loss_total)/(-del_H_steam) ;// Total mass of stea used - [kg] +m_stm = m_stm_total/m_m ;// Mass of steam used per kg of charge - [kg] + + printf(' Mass of steam used per kg of charge is %.3f kg .\n',m_stm);
\ No newline at end of file diff --git a/409/CH24/EX24.8/Example24_8.sce b/409/CH24/EX24.8/Example24_8.sce new file mode 100755 index 000000000..3a5978de5 --- /dev/null +++ b/409/CH24/EX24.8/Example24_8.sce @@ -0,0 +1,32 @@ +clear ; +clc; +// Example 24.8 +printf('Example 24.8\n\n'); +//page no. 741 +// Solution Fig. E24.8 + +// Pick the system of whole process as shown in above figure of book +// Given +Q = 1.63 ;// Heat loss from the process - [ kW ] +m_bm = 150 ;// Mass flow rate of biological media into the sterlizer -[kg/min] +T_bm = 50 +273 ;// Temperature of biological media into the sterlizer - [K] +T_sm = 75 + 273 ;// Temperature of sterlize media out of the sterlizer - [K] +P_ss = 300 ;// Pressure of satureted steam entering the steam heater - [kPa] +P_sc = 300 ;// Pressure of satureted condensate exiting the steam heater - [kPa] + +// Additional data of change in enthalpy is obtained from the steam table, according to book the data are +H_w1 = 207.5 ;// Change in specific enthalpy of water at 50 degree C - [kJ/kg] +H_w2 = 310.3 ;// Change in specific enthalpy of water at 75 degree C - [kJ/kg] +H_ss = 2724.9 ;//Change in specific enthalpy of satureted steam entering the steam heater at 300 kPa - [kJ/kg] +H_sc = 561.2 ;//Change in specific enthalpy of satureted condensate exiting the steam heater at 300 kPa - [kJ/kg] + +// Use following conditions to simplify the energy balance +// 1. Proces is in steady state , so change in energy = 0. +// 2. Assume del_KE and del_PE = 0. +// 3. W = 0. +// 4. Assume m1 = m2 = m_stm +// The energy balance reduce to Q = H_out - H_in , using it +m_sm = m_bm ;// By material balance -[kg/min] +m_stm = (Q*60 - m_sm*H_w2 + m_bm * H_w1 )/(H_sc - H_ss ) ;// Mass flow rate of steam entering the steam heater - [kg/min] + + printf(' Mass flow rate of steam entering the steam heater is %.2f kg steam/min .\n',m_stm);
\ No newline at end of file diff --git a/409/CH24/EX24.9/Example24_9.sce b/409/CH24/EX24.9/Example24_9.sce new file mode 100755 index 000000000..847303e67 --- /dev/null +++ b/409/CH24/EX24.9/Example24_9.sce @@ -0,0 +1,43 @@ +clear ; +clc; +// Example 24.9 +printf('Example 24.9\n\n'); +//page no. 742 +// Solution Fig. E24.9a and Fig. E24.9b + +// Given + +// For material balance +F = 20000 ;// Feed rate of saturated liquid - [kg/h] +F_Bz = 0.5 ;// Fraction of benzene in feed +F_Tol = 0.5 ;// Fraction of toluene in feed +D_Bz = 0.98 ;// Fraction of benzene in distillate +D_Tol = 0.02 ;// Fraction of toluene in distillate +B_Bz = 0.04 ;// Fraction of benzene in bottoms +B_Tol = 0.96 ;// Fraction of toluene in bottoms +R_by_D = 4.0 ;// Recycle ratio +// Analysing the condition for material balance , degree of freedom is 0. +// Solve equations obtained by material balances , simultaneously to get B and D +a = [1 1;B_Bz D_Bz] ;// Matrix formed by coefficients of unknown +b = [ F ; F_Bz*F ] ;// Matrix formed by contants +x = a\b ;// Matrix of solutions +B = x(1) ;// Bottoms - [kg/h] +D = x(2) ;//Distillate - [kg/h] +R = D * R_by_D ;// Recycle - [kg/h] +V = R + D ;// Overhead vapour - [kg/h] + +// For energy balance +// According to book additional data obtained from the fig.E24.9b are +H_F = 165 ;// Change in enthalpy of F - [kJ/kg] +H_B = 205 ;// Change in enthalpy of B - [kJ/kg] +H_D = 100 ;// Change in enthalpy of D - [kJ/kg] +H_R = 100 ;// Change in enthalpy of R - [kJ/kg] +H_V = 540 ;// Change in enthalpy of V - [kJ/kg] + +Qc = R*H_R + D*H_D - V*H_V ;// The heat duty in the condenser - [kJ] +Qr = D*H_D + B*H_B - F*H_F - Qc;// The heat duty to the reboiler - [kJ] + +printf(' Ditillate (D) %.2e kg/h.\n',D); +printf(' Bottoms (B) %.2e kg/h.\n',B); +printf(' The heat duty in the condenser (Qc) %.2e kJ/h.\n',Qc); +printf(' The heat duty to the reboiler (Qr) %.2e kJ/h.\n',Qr);
\ No newline at end of file |