diff options
Diffstat (limited to '409/CH22')
| -rwxr-xr-x | 409/CH22/EX22.1/Example22_1.sce | 24 | ||||
| -rwxr-xr-x | 409/CH22/EX22.2/Example22_2.sce | 31 | ||||
| -rwxr-xr-x | 409/CH22/EX22.3/Example22_3.sce | 29 | ||||
| -rwxr-xr-x | 409/CH22/EX22.4/Example22_4.sce | 32 | ||||
| -rwxr-xr-x | 409/CH22/EX22.5/Example22_5.sce | 29 |
5 files changed, 145 insertions, 0 deletions
diff --git a/409/CH22/EX22.1/Example22_1.sce b/409/CH22/EX22.1/Example22_1.sce new file mode 100755 index 000000000..9d4ba7844 --- /dev/null +++ b/409/CH22/EX22.1/Example22_1.sce @@ -0,0 +1,24 @@ +clear; +clc; +// Example 22.1 +printf('Example 22.1\n\n'); +//page no. 651 +// Solution + +//Assume that properties of water can be used to substitute properties of solution +// Given +V = 1.673 ;// Volume of closed vessel-[cubic metre] +m = 1 ;// mass of saturated liquid vaporized-[kg] +Pi = 1 ;// Initial pressure -[atm] +Ti = 10 ;// Initial temperature -[degree C] +Pf = 1 ;// final pressure -[atm] +Tf = 100 ;// final temperature -[degree C] + +// Use steam table to obtain additional information at given condition +Ui = 35 ;// Initial enthalpy-[kJ/kg] +Uf = 2506.0 ;// Final enthalpy -[kJ/kg] + +// Use eqn. 22.2 after modifiying it using given conditions(W = 0,del_KE = 0 and del_PE = 0 ) +Q = m*(Uf - Ui) ;// Heat transferred to the vessel - [kJ] + +printf('\nHeat transferred to the vessel is %.1f kJ .\n ',Q);
\ No newline at end of file diff --git a/409/CH22/EX22.2/Example22_2.sce b/409/CH22/EX22.2/Example22_2.sce new file mode 100755 index 000000000..70cbd545f --- /dev/null +++ b/409/CH22/EX22.2/Example22_2.sce @@ -0,0 +1,31 @@ +clear ; +clc; +// Example 22.2 +printf('Example 22.2\n\n'); +//page no. 652 +// Solution + +// Given +T1 = 80 ;// Initial temperature -[degree F] +T1 = 40 ;// final temperature -[degree F] + +// Additional data obtained from steam table at given temperatures and corresponding vapour pressures +p1 = 0.5067 ;// Initial saturation pressure-[psia] +p2 = 0.1217 ;// Final saturation pressure-[psia] +V1 = 0.01607 ;// Initial specific volume - [cubic feet/lb] +V2 = 0.01602 ;// Final specific volume - [cubic feet/lb] +H1 = 48.02 ;//Initial specific enthalpy -[Btu/lb] +H2 = 8.05 ;// Final specific enthalpy -[Btu/lb] + +del_P = p2 - p1 ;// Change in pressure -[psia] +del_V = V2 - V1 ;// Change in specific volume -[cubic feet/lb] +del_H = H2 - H1 ;// Change in specific enthalpy -[Btu/lb] +del_pV = p2*144*V2/778 - p1*144*V1/778 ;// Change in pv-[Btu] +del_U = del_H - del_pV ;// Change in specific internal energy - [Btu/lb] +del_E = del_U ;// Change in specific total energy(since KE=0,PE=0 and W=0) -[Btu/lb] + +printf('\nChange in pressure is %.3f psia .\n ',del_P); +printf('\nChange in specific volume is %.5f cubic feet/lb (negligible value) .\n ',del_V); +printf('\nChange in specific enthalpy is %.2f Btu/lb .\n ',del_H); +printf('\nChange in specific internal energy is %.2f Btu/lb .\n ',del_U); +printf('\nChange in specific total energy is %.2f Btu/lb .\n ',del_E);
\ No newline at end of file diff --git a/409/CH22/EX22.3/Example22_3.sce b/409/CH22/EX22.3/Example22_3.sce new file mode 100755 index 000000000..75f37e64c --- /dev/null +++ b/409/CH22/EX22.3/Example22_3.sce @@ -0,0 +1,29 @@ +clear ; +clc; +// Example 22.3 +printf('Example 22.3\n\n'); +//page no. 662 +// Solution fig.E22.3a + +//Lets take tank to be system +// Given +T = 600 ; // Temperature of steam -[K] +P = 1000 ;// Pressure of steam -[kPa] + +// Additional data for steam obtained from CD database at T and P +U = 2837.73 ;// Specific internal energy-[kJ/kg] +H = 3109.44 ;// Specific enthalpy -[kJ/kg] +V = 0.271 ;// Specific volume -[cubic metre/kg] + +// Use eqn. 22.6 to get change in specific internal energy,by simplifing it with following assumption: +//1. Change in KE and PE of system = 0, therefore change in total energy = change in internal energy +//2. W = 0,work done by or on the system +//3. Q = 0 , system is well insulated +//4. Change in KE and PE of entering steam = 0 +//5. H_out = 0, no stream exits the system +//6. Ut1 = 0, initially no mass exists in the system + +// By the reduced equation +Ut2 = H ;// Internal energy at final temperature-[kJ/kg] + +printf('\nThe specific internal energy at final temperature is %.2f kJ/kg. \nNow use two properties of the steam (P = %i kPa and Ut2 = %.2f kJ/kg) to find final temperature (T) from steam table. \nFrom steam table we get T = 764 K.',Ut2,P,Ut2);
\ No newline at end of file diff --git a/409/CH22/EX22.4/Example22_4.sce b/409/CH22/EX22.4/Example22_4.sce new file mode 100755 index 000000000..fba795617 --- /dev/null +++ b/409/CH22/EX22.4/Example22_4.sce @@ -0,0 +1,32 @@ +clear; +clc; +// Example 22.4 +printf('Example 22.4\n\n'); +//page no. 669 +// Solution + +// Take milk plus water in tank to be system +// Given +T1_water = 70 ;// Temperature of entering water -[degree C] +T2_water = 35 ;// Temperature of exiting water -[degree C] +T1_milk = 15 ;//Temperature of entering milk -[degree C] +T2_milk = 25 ;//Temperature of exiting milk -[degree C] + +// Get additional data from steam table for water and milk,assuming milk to have same properties as that of water. +H_15 = 62.01 ;//Change in specific internal energy-[kJ/kg] +H_25 = 103.86 ;//Change in specific internal energy-[kJ/kg] +H_35 = 146.69 ;//Change in specific internal energy-[kJ/kg] +H_70 = 293.10 ;//Change in specific internal energy-[kJ/kg] + +// Assumptions to simplify Equation 22.8 are: +printf('\nAssumptions to simplify Equation 22.8 are:\n'); +printf('1. Change in KE and PE of system = 0.\n'); +printf('2. Q = 0 ,because of way we picked the system,it is is well insulated.\n'); +printf('3. W = 0,work done by or on the system.\n'); + +//Basis m_milk = 1 kg/min , to directly get the answer . +m_milk = 1 ;// Mass flow rate of milk-[kg/min] +// By applying above assumtions eqn. 22.8 reduces to del_H = 0 .Using it get m_water- +m_water = (m_milk*(H_15 - H_25))/(H_35 - H_70) ; // Mass flow rate of water-[kg/min] +m_ratio = m_water/m_milk ;// Mass flow rate of water per kg/min of milk-[kg/min] +printf('\nMass flow rate of water per kg/min of milk is %.2f (kg water/min )/(kg milk/min).',m_ratio);
\ No newline at end of file diff --git a/409/CH22/EX22.5/Example22_5.sce b/409/CH22/EX22.5/Example22_5.sce new file mode 100755 index 000000000..7d686fce5 --- /dev/null +++ b/409/CH22/EX22.5/Example22_5.sce @@ -0,0 +1,29 @@ +clear ; +clc; +// Example 22.5 +printf('Example 22.5\n\n'); +//page no. 670 +// Solution + +// Take pipe between initial and final level of water +// Given +h_in = -20 ;// Depth of water below ground-[ft] +h_out = 5 ;// Height of water level above ground-[ft] +h = h_out - h_in ;// Total height to which water is pumped-[ft] +V = 0.50 ;// Volume flow rate of water - [cubic feet/s] +ef = 100; // Efficiency of pump - [%] +g = 32.2; // Acceleration due to gravity -[ft/square second] +gc = 32.2 ;//[(ft*lbm)/(second square*lbf)] + +M = V * 62.4 ;// mass flow rate - [lbm/s] +PE_in = 0 ;// Treating initial water level to be reference level +PE_out = (M*g*h*1.055)/(gc*778.2) ;// PE of discharged water -[lbm*(square feet/square second)] + +// Assumptions to simplify Equation 22.8 are: +//1. Change in KE = 0. +//2. Q = 0 -By given assumption +//3. Let us assume that temperature of water is same in well and when it is discharged, therefore del_H = 0 +// Reduced equation is W = del_PE, using this: +W = PE_out - PE_in ;//Work done on system = power delivered by pump, (since we are using mass flow rate and pump efficiency is 100 % , so W = Power) -[kW] + +printf('The electric power required by the pump is %.2f kW. \n', W);
\ No newline at end of file |