diff options
Diffstat (limited to '409/CH27')
| -rwxr-xr-x | 409/CH27/EX27.1/Example27_1.sce | 19 | ||||
| -rwxr-xr-x | 409/CH27/EX27.2/Example27_2.sce | 27 | ||||
| -rwxr-xr-x | 409/CH27/EX27.3/Example27_3.sce | 19 | ||||
| -rwxr-xr-x | 409/CH27/EX27.4/Example27_4.sce | 30 | ||||
| -rwxr-xr-x | 409/CH27/EX27.5/Example27_5.sce | 29 | ||||
| -rwxr-xr-x | 409/CH27/EX27.6/Example27_6.sce | 30 |
6 files changed, 154 insertions, 0 deletions
diff --git a/409/CH27/EX27.1/Example27_1.sce b/409/CH27/EX27.1/Example27_1.sce new file mode 100755 index 000000000..2f60d9937 --- /dev/null +++ b/409/CH27/EX27.1/Example27_1.sce @@ -0,0 +1,19 @@ +clear ; +clc; +// Example 27.1 +printf('Example 27.1\n\n'); +//page no. 838 +// Solution E27.1 + +// Given +V_w = 1 ;// Volume of given water -[L] +P_atm = 100 ;// Atmospheric pressure - [kPa] + +//W = -p*del_V +V_H2O = 0.001043 ;// Specific volume of water from steam table according to book- [cubic metre] +V_vap = 1.694 ;// Specific volume of vapour from steam table according to book- [cubic metre] +V1 = 0 ;// Initial volume of H2O in bag-[cubic metre] +V2 = (V_w*V_vap)/(1000*V_H2O) ;// Final volume of water vapour -[cubic metre] +W = -P_atm*(V2 -V1)* 1000 ;// Work done by saturated liquid water -[J] + +printf(' Work done by saturated liquid water is %.3e J.\n',W) ;
\ No newline at end of file diff --git a/409/CH27/EX27.2/Example27_2.sce b/409/CH27/EX27.2/Example27_2.sce new file mode 100755 index 000000000..81e508d7c --- /dev/null +++ b/409/CH27/EX27.2/Example27_2.sce @@ -0,0 +1,27 @@ +clear ; +clc; +// Example 27.2 +printf('Example 27.2\n\n'); +//page no. 840 +// Solution E27.2 + +// Given +m_N2 = 1 ;// Moles of N2 taken -[kg mol] +p = 1000;// Pressure of cylinder-[kPa] +T = 20 + 273 ;// Temperature of cylinder -[K] +a_pis = 6 ;// Area of piston - [square centimetre] +m_pis = 2 ;// Mass of pston - [kg] +R = 8.31 ;// Ideal gas constant - [(kPa*cubic metre)/(K * kgmol)] + +V = (R*T)/p ;// Specific volue of gas at initial stage -[cubic metre/kg mol] +V1 = V * m_N2 ;// Initial volume of gas - [cubic metre] +V2 = 2*V1 ;// Final volume of gas according to given condition -[cubic metre] + +// Assumed surrounding pressure constant = 1 atm +p_atm = 101.3 ;// Atmospheric pressure-[kPa] +del_Vsys = V2 -V1 ;// Change in volume of system -[cubic metre] +del_Vsurr = - del_Vsys ;// Change in volume of surrounding -[cubic metre] +W_surr = -p_atm*del_Vsurr ;// Work done by surrounding - [kJ] +W_sys = -W_surr ;// Work done by system - [kJ] + +printf(' Work done by gas(actually gas + piston system) is %.0f kJ.\n',W_sys) ;
\ No newline at end of file diff --git a/409/CH27/EX27.3/Example27_3.sce b/409/CH27/EX27.3/Example27_3.sce new file mode 100755 index 000000000..b639cb0be --- /dev/null +++ b/409/CH27/EX27.3/Example27_3.sce @@ -0,0 +1,19 @@ +clear ; +clc; +// Example 27.3 +printf('Example 27.3\n\n'); +//page no. 845 +// Solution + +// Given +p_plant = 20 ;// Power generated by plant-[MW] +h = 25 ;// Height of water level - [m] +V = 100 ;// Flow rate of water -[cubic metre/s] +d_water = 1000 ;// Density of water - [ 1000 kg / cubic metre] +g = 9.807 ;// Acceleration due to gravity-[m/square second] + +M_flow = V*d_water ;// Mass flow rate of water -[kg/s] +del_PE = M_flow*g*h ;// Potential energy change of water per second -[W] +eff = (p_plant*10^6) /(del_PE) ;// Efficiency of plant + +printf(' Efficiency of plant is %.2f .\n',eff) ;
\ No newline at end of file diff --git a/409/CH27/EX27.4/Example27_4.sce b/409/CH27/EX27.4/Example27_4.sce new file mode 100755 index 000000000..b61cdfcb9 --- /dev/null +++ b/409/CH27/EX27.4/Example27_4.sce @@ -0,0 +1,30 @@ +clear ; +clc; +// Example 27.4 +printf('Example 27.4\n\n'); +//page no. 845 +// Solution Fig.E27.4 + +// Given +LHV = 36654 ;// LHV value of fuel - [kJ/ cubic metre] +Q1 = 16 ;//- [kJ/ cubic metre] +Q2 = 0 ;//- [kJ/ cubic metre] +Q3 = 2432 ;//- [kJ/ cubic metre] +Q4 = 32114 ;//- [kJ/ cubic metre] +Q41 = 6988 ;//- [kJ/ cubic metre] +Q8 = 1948 ;//- [kJ/ cubic metre] +Q9 = 2643 ;//- [kJ/ cubic metre] +Q81 = 2352 - Q8 ;// - [kJ/ cubic metre] +Q567 = 9092 ;// Sum of Q5, Q6 and Q7- [kJ/ cubic metre] + +//(a) +G_ef = (LHV+ Q1 +Q2 + Q3 - Q9)/(LHV) ;// Gross efficiency +printf('(a) Gross efficiency is %.3f .\n',G_ef) ; + +//(b) +T_ef = (Q567+Q8)/(LHV+ Q1 +Q2 + Q3) ;//Thermal efficiency +printf(' (b) Thermal efficiency is %.3f .\n',T_ef) ; + +//(c) +C_ef = Q4/(Q4 + Q41) ;// Combustion efficiency +printf(' (c) Combustion efficiency is %.3f .\n',C_ef) ;
\ No newline at end of file diff --git a/409/CH27/EX27.5/Example27_5.sce b/409/CH27/EX27.5/Example27_5.sce new file mode 100755 index 000000000..da9833b0d --- /dev/null +++ b/409/CH27/EX27.5/Example27_5.sce @@ -0,0 +1,29 @@ +clear ; +clc; +// Example 27.5 +printf('Example 27.5\n\n'); +//page no. 850 +// Solution + +// Given +V1 = 5 ;// Volume of gas initially - [cubic feet] +P1 = 1 ;// Initial pressure - [atm] +P2 = 10 ;// Final pressure - [atm] +T1 = 100 + 460 ;// initial temperature - [degree Rankine] +R = 0.7302 ;// Ideal gas constant -[(cubic feet*atm)/(lb mol)*(R)] +//Equation of state pV^1.4 = constant + +//(a) +//Energy balance reduces to del_E = del_U = del_W +V2 = V1*(P1/P2)^(1/1.4) ;// Final volume - [cubic feet] +W1_rev = integrate('-(P1)*(V1/V)^(1.4)','V',V1,V2) ;// Reversible work done in compresion in a horizontal cylinder with piston -[cubic feet *atm] +W1 = W1_rev *1.987/.7302 ;// Conversion to Btu -[Btu] + +printf('\n (a)Reversible work done in compression in a horizontal cylinder with piston is %.1f Btu .\n ',W1); + +//(b) +n1 = (P1*V1)/(R*T1) ;// Number of moles of gas +W2_rev = integrate('(V1)*(P1/P)^(1/1.4)','P',P1,P2) ;// Reversible work done in compresion in a rotary compressor -[cubic feet *atm] +W2 = W2_rev *1.987/.7302 ;// Conversion to Btu -[Btu] + +printf('\n (b)Reversible work done in a rotary compressor is %.1f Btu .\n ',W2);
\ No newline at end of file diff --git a/409/CH27/EX27.6/Example27_6.sce b/409/CH27/EX27.6/Example27_6.sce new file mode 100755 index 000000000..650f46728 --- /dev/null +++ b/409/CH27/EX27.6/Example27_6.sce @@ -0,0 +1,30 @@ +clear ; +clc; +// Example 27.6 +printf('Example 27.6\n\n'); +//page no. 853 +// Solution + +// Given +m_water = 1 ;// Mass flow rate of water -[lb/min] +P1 = 100 ;// Initial pressure - [psia] +P2 = 1000 ;// Final pressure - [psia] +T1 = 80 + 460 ;// initial temperature - [degree Rankine] +T2 = 100 + 460 ;// final temperature - [degree Rankine] +h = 10 ;// Difference in water level between entry and exit of stream-[ft] +g = 32.2 ;// Accleration due to gravity - [ft/ square second] +gc = 32.2 ;//[(ft*lbm)/(lbf*square second)] + +// The mechanical energy balance reduces to W = PV_work + del_PE ....(A) +// From steam table , specific volume of liquid water at 80 and 100 degree F is noted , according to book it is as follows- +v1 = .01607 ;// specific volume of liquid water at 80 degree F -[cubic feet/lbm] +v2 = .01613 ;// specific volume of liquid water at 100 degree F -[cubic feet/lbm] +// But for pratical purposes wwater is taken to be incompressible and specific volume can be taken as v, ith following value +v= 0.0161 ;// -[cubic feet/lbm] + +del_PE = (h*g)/(gc*778) ;// Change in potential energy - [Btu/lbm] +PV_work = integrate('(v)*(12^2/778)','P',P1,P2) ;// PV work done -[Btu/lbm] +//From eqn. (A) +W = PV_work + del_PE ;// Work per minute required to pump 1 lb water per minute - [Btu/lbm] + +printf('\n Work per minute required to pump 1 lb water per minute is %.2f Btu/lbm .\n ',W);
\ No newline at end of file |