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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /3012/CH7 | |
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initial commit / add all books
Diffstat (limited to '3012/CH7')
-rwxr-xr-x | 3012/CH7/EX7.1/Ex7_1.sce | 19 | ||||
-rwxr-xr-x | 3012/CH7/EX7.10/Ex7_10.sce | 49 | ||||
-rwxr-xr-x | 3012/CH7/EX7.2/Ex7_2.sce | 52 | ||||
-rwxr-xr-x | 3012/CH7/EX7.3/Ex7_3.sce | 47 | ||||
-rwxr-xr-x | 3012/CH7/EX7.4/Ex7_4.sce | 16 | ||||
-rwxr-xr-x | 3012/CH7/EX7.5/Ex7_5.sce | 27 | ||||
-rwxr-xr-x | 3012/CH7/EX7.6/Ex7_6.sce | 49 | ||||
-rwxr-xr-x | 3012/CH7/EX7.7/Ex7_7.sce | 35 | ||||
-rwxr-xr-x | 3012/CH7/EX7.8/Ex7_8.sce | 57 | ||||
-rwxr-xr-x | 3012/CH7/EX7.9/Ex7_9.sce | 22 |
10 files changed, 373 insertions, 0 deletions
diff --git a/3012/CH7/EX7.1/Ex7_1.sce b/3012/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..77763035a --- /dev/null +++ b/3012/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,19 @@ +// Given:- +v = 2450.00 // volume of gaseous products in cm^3 +P = 7.00 // pressure of gaseous product in bar +T = 867.00 // temperature of gaseous product in degree celcius +T0 = 300.00 // in kelvin +P0 = 1.013 // in bar + +// From table A-22 +u = 880.35 // in kj/kg +u0 = 214.07 // in kj/kg +s0T = 3.11883 // in kj/kg.k +s0T0 = 1.70203 // in kj/kg.k + +// Calculations + +e = (u-u0) + (P0*(8.314/28.97)*(((T+273)/P)-(T0/P0))) - T0*(s0T-s0T0-(8.314/28.97)*log(P/P0)) // kj/kg + +// Results +printf( ' The specific exergy of the gas is %.3f kJ/kg.',e) diff --git a/3012/CH7/EX7.10/Ex7_10.sce b/3012/CH7/EX7.10/Ex7_10.sce new file mode 100755 index 000000000..9d703feb2 --- /dev/null +++ b/3012/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,49 @@ +// Given:- +EfFdot = 100.00 // exergy rate of fuel entering the boiler in MW +cF = 1.44 // unit cost of fuel in cents per kw.h +Zbdot = 1080.00 // the cost of owning and operating boiler in dollars per hour +Ef1dot = 35.00 // exergy rate of exiting steam from the boiler in MW +p1 = 50.00 // pressure of exiting steam from the boiler in bar +T1 = 466.00 // temperature of exiting steam from the boiler in degree celcius +Ztdot = 92.00 // the cost of owning and operating turbine in dollars per hour +p2 = 5.00 // pressure of exiting steam from the turbine in bars +T2 = 205.00 // temperature of exiting steam from the turbine in degree celcius +m2dot = 26.15 // mass flow rate of exiting steam from the turbine in kg/s +T0 = 298.00 // in kelvin + + +// Part(a) +// From table A-4, +h1 = 3353.54 // in kj/kg +h2 = 2865.96 // in kj/kg +s1 = 6.8773 // in kj/kg.k +s2 = 7.0806 // in kj/kg.k + +// Calculations +// From assumption,For each control volume,Qcvdot = 0 and kinetic and potential energy effects are negligible,the mass and energy rate +// balances for a control volume enclosing the turbine reduce at steady state to give +Wedot = m2dot *(h1-h2)/1000 // power in MW +Ef2dot = Ef1dot+m2dot*(h2-h1-T0*(s2-s1))/1000 // the rate exergy exits with the steam in MW + +// Results +printf( ' For the turbine,the power is %.2f MW.',Wedot) +printf( ' For the turbine,the rate exergy exits with the steam is %.2f MW.',Ef2dot) + +// Part(b) +// Calculations +c1 = cF*(EfFdot/Ef1dot) + ((Zbdot/Ef1dot)/10**3)*100 // unit cost of exiting steam from boiler in cents/Kw.h +c2 = c1 // Assigning the same unit cost to the steam entering and exiting the turbine +ce = c1*((Ef1dot-Ef2dot)/Wedot) + ((Ztdot/Wedot)/10**3)*100 // unit cost of power in cents/kw.h + +// Results +printf('The unit costs of the steam exiting the boiler of exergy is: %.2f cents per kw.h.',c1) +printf('The unit costs of the steam exiting the turbine of exergy is: %.2f cents per kw.h.',c2) +printf('Unit cost of power is: %f cents per kw.h.',ce) + +// Part(c) +C2dot = (c2*Ef2dot*10**3)/100 // cost rate for low-pressure steam in dollars per hour +Cedot = (ce*Wedot*10**3)/100 // cost rate for power in dollars per hour + +// Results +printf( ' The cost rate of the steam exiting the turbine is: %.2f dollars per hour.',C2dot) +printf( ' The cost rate of the power is: %.2f dollars per hour.',Cedot) diff --git a/3012/CH7/EX7.2/Ex7_2.sce b/3012/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..af9546a57 --- /dev/null +++ b/3012/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,52 @@ +// Given:- +mR = 1.11 // mass of the refrigerant in kg +T1 = -28.00 // initial temperature of the saturated vapor in degree celcius +P2 = 1.4 // final pressure of the refrigerant in bar +T0 = 293.00 // in kelvin +P0 = 1.00 // in bar + +// Part (a) +// From table A-10 +u1 = 211.29 // in kj/kg +v1 = 0.2052 // in m^3/kg +s1 = 0.9411 // in kj/kg.k +// From table A-12 +u0 = 246.67 // in kj/kg +v0 = 0.23349 // in m^3/kg +s0 = 1.0829 // in kj/kg.k + +// From table A-12 +u2 = 300.16 // in kj/kg +s2 = 1.2369 // in kj/kg.k +v2 = v1 + +// Calculations +E1 = mR*((u1-u0) + P0*(10**5)*(v1-v0)*(10**(-3))-T0*(s1-s0)) +E2 = mR*((u2-u0) + P0*(10**5)*(v2-v0)*(10**(-3))-T0*(s2-s0)) + +// Results for Part A +printf( ' Part(a) The initial exergy is %.2f kJ.',E1) +printf( ' The final exergy is %.2f kJ.',E2) +printf( ' The change in exergy of the refrigerant is %.2f kj',E2-E1) + + +// Part (b) +// Calculations +deltaU = mR*(u2-u1) +// From energy balance +deltaPE = -deltaU +// With the assumption::The only significant changes of state are experienced by the refrigerant and the suspended mass. For the refrigerant, +// there is no change in kinetic or potential energy. For the suspended mass, there is no change in kinetic or internal energy. Elevation is +// the only intensive property of the suspended mass that changes +deltaE = deltaPE + +// Results for part b +printf( ' Part(b)The change in exergy of the suspended mass is %.3f kJ',deltaE) + + +// Part(c) +// Calculations +deltaEiso = (E2-E1) + deltaE + +// Results +printf( ' Part(c)The change in exergy of an isolated system of the vessel and pulley–mass assembly is %.2f kJ',deltaEiso) diff --git a/3012/CH7/EX7.3/Ex7_3.sce b/3012/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..1055014ad --- /dev/null +++ b/3012/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,47 @@ +// Given :- +T = 373.15 // initial temperature of saturated liquid in kelvin +T0 = 293.15 // in kelvin +P0 = 1.014 // in bar + +// Part(a) +// From table A-2 +ug = 2506.5 // in kj/kg +uf = 418.94 // in kj/kg +vg = 1.673 // in m^3/kg +vf = 1.0435*(10**(-3)) // in m^3/kg +sg = 7.3549 // in kj/kg.k +sf = 1.3069 // in kj/kg.k + + +// Calculations +// Energy transfer accompanying work +etaw = 0 // since p = p0 +// Exergy transfer accompanying heat +Q = 2257 // in kj/kg,obtained from example 6.1 +etah = (1-(T0/T))*Q + +// Exergy destruction +ed = 0 // since the process is accomplished without any irreversibilities +deltae = ug-uf + P0*(10**5)*(vg-vf)/(10**3)-T0*(sg-sf) + +// Results +printf( ' Part(a)the change in exergy is %.2f kJ/kg.',deltae) +printf( ' The exergy transfer accompanying work is %.2f kJ/kg.',etaw) +printf( ' The exergy transfer accompanying heat is %.2f kJ/kg',etah) +printf( ' The exergy destruction is %.2f kJ/kg.',ed) + + +// Part(b) +Deltae = deltae // since the end states are same +Etah = 0 // since process is adiabatic +// Exergy transfer along work +W = -2087.56 // in kj/kg from example 6.2 +Etaw = W- P0*(10**5)*(vg-vf)/(10**3) +// Exergy destruction +Ed = -(Deltae+Etaw) + +// Results +printf( ' Part(b)the change in exergy is %.2f kJ/kg.',Deltae) +printf( ' The exergy transfer accompanying work is %.2f kJ/kg.',Etaw) +printf( ' The exergy transfer accompanying heat is %.2f kJ/kg.',Etah) +printf( ' The exergy destruction is %.2f kJ/kg.',Ed) diff --git a/3012/CH7/EX7.4/Ex7_4.sce b/3012/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..20098082a --- /dev/null +++ b/3012/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,16 @@ +// Given:- +T0 = 293.00 // in kelvin +Qdot = -1.2 // in KW, from example 6.4a +Tb = 300.00 // temperature at the outer surface of the gearbox in kelvin from example 6.4a +sigmadot = 0.004 // rate of entropy production in KW/k from example 6.4a + +// Calculations +R = -(1-T0/Tb)*Qdot // time rate of exergy transfer accompanying heat +Eddot = T0*sigmadot // rate of exergy destruction + +// Results +printf( ' Balance sheet'); +printf( '\n Rate of exergy in high speed shaft 60Kw' ) +printf( '\n Disposition of the exergy: Rate of exergy out low-speed shaft %.1f Kw',58.8 ) +printf( '\n Heat transfer is %.3f kw.',R) +printf( '\n Rate of exergy destruction is %.3f kw',Eddot) diff --git a/3012/CH7/EX7.5/Ex7_5.sce b/3012/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 000000000..e6639a76e --- /dev/null +++ b/3012/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,27 @@ +// Given:- +p1 = 3.0 // entry pressure in Mpa +p2 = 0.5 // exit pressure in Mpa +T1 = 320.0 // entry temperature in degree celcius +T0 = 25.0 // in degree celcius +p0 = 1.0 // in atm + +// From table A-4 +h1 = 3043.4 // in kj/kg +s1 = 6.6245 // in kj/kg.k +h2 = h1 // from reduction of the steady-state mass and energy rate balances +s2 = 7.4223 // Interpolating at a pressure of 0.5 MPa with h2 = h1, units in kj/kg.k + +// From table A-2 +h0 = 104.89 // in kj/kg +s0 = 0.3674 // in kj/kg.k + +// Calculations +ef1 = h1-h0-(T0+273)*(s1-s0) // flow exergy at the inlet +ef2 = h2-h0-(T0+273)*(s2-s0) // flow exergy at the exit +// From the steady-state form of the exergy rate balance +Ed = ef1-ef2 // the exergy destruction per unit of mass flowing is + +// Results +printf( ' The specific flow exergy at the inlet is %.2f kJ/kg.',ef1) +printf( ' The specific flow exergy at the exit is %.2f kJ/kg.',ef2) +printf( ' The exergy destruction per unit of mass flowing is %.2f kJ/kg.',Ed) diff --git a/3012/CH7/EX7.6/Ex7_6.sce b/3012/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..2820cf276 --- /dev/null +++ b/3012/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,49 @@ +// Given:- +T1 = 610.0 // temperature of the air entering heat exchanger in kelvin +p1 = 10.0 // pressure of the air entering heat exchanger in bar +T2 = 860.0 // temperature of the air exiting the heat exchanger in kelvin +p2 = 9.70 // pressure of the air exiting the heat exchanger in bar +T3 = 1020.0 // temperature of entering hot combustion gas in kelvin +p3 = 1.10 // pressure of entering hot combustion gas in bar +p4 = 1.0 // pressure of exiting hot combustion gas in bar +mdot = 90.0 // mass flow rate in kg/s +T0 = 300.0 // in kelvin +p0 = 1.0 // in bar + +// Part (a) +// From table A-22 +h1 = 617.53 // in kj/kg +h2 = 888.27 // in kj/kg +h3 = 1068.89 // in kj/kg + +// Calculations +h4 = h3+h1-h2 + +// Using interpolation in table A-22 gives +T4 = 778 // in kelvin + +// Results +printf( ' The exit temperature of the combustion gas is %f kelvin.',T4); + +// Part(b) +// From table A-22 +s2 = 2.79783 // in kj/kg.k +s1 = 2.42644 // in kj/kg.k +s4 = 2.68769 // in kj/kg.k +s3 = 2.99034 // in kj/kg.k + +// Calculations for part b + +deltaR = (mdot*((h2-h1)-T0*(s2-s1-(8.314/28.97)*log(p2/p1))))/1000 +deltRc = mdot*((h4-h3)-T0*(s4-s3-(8.314/28.97)*log(p4/p3)))/1000 + +// Results for part b +printf( ' The net change in the flow exergy rate from inlet to exit of compressed gas is %.3f MW.',deltaR) +printf( ' The net change in the flow exergy rate from inlet to exit of hot combustion gas is %.3f MW.',deltRc) + +// Part(c) +//From an exergy rate balance +Eddot = -deltaR-deltRc + +// Results +printf( ' The rate exergy destroyed, is %.3f MW.',Eddot) diff --git a/3012/CH7/EX7.7/Ex7_7.sce b/3012/CH7/EX7.7/Ex7_7.sce new file mode 100755 index 000000000..5793d53e0 --- /dev/null +++ b/3012/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,35 @@ +// Given:- +p1 = 30.0 // pressure of entering steam in bar +t1 = 400.0 // temperature of entering steam in degree celcius +v1 = 160.0 // velocity of entering steam in m/s +t2 = 100.0 // temperature of exiting saturated vapor in degree celcius +v2 = 100.0 // velocity of exiting saturated vapor in m/s +W = 540.0 // rate of work developed in kj per kg of steam +Tb = 350.0 // the temperature on the boundary where heat transfer occurs in kelvin +T0 = 25.0 // in degree celcius +p0 = 1.0 // in atm + +// From table A-4 +h1 = 3230.9 // in kj/kg +s1 = 6.9212 // in kj/kg.k +// From table A-2 +h2 = 2676.1 // in kj/kg +s2 = 7.3549 // in kj/kg.k +// From example 6.6 +Q = -22.6 // in kj/kg + +// Calculations +DELTAef = (h1-h2)-(T0+273)*(s1-s2)+(v1**2-v2**2)/(2*1000) +// The net exergy carried in per unit mass of steam flowing in kj/kg +Eq = (1-(T0+273)/Tb)*(Q) // exergy transfer accompanying heat in kj/kg +Ed = ((1-(T0+273)/Tb)*(Q))-W+(DELTAef) // The exergy destruction determined by rearranging the steady-state form of the exergy + // rate balance + +// Results +printf( ' Balance sheet') +printf( ' Net rate of exergy %f kJ/kg,',DELTAef) +printf( ' Disposition of the exergy:') +printf( '* Rate of exergy out') +printf( ' Work %f kJ/kg.',W) +printf( ' Heat transfer %f',-Eq) +printf( '• Rate of exergy destruction %f kJ/kg.',Ed) diff --git a/3012/CH7/EX7.8/Ex7_8.sce b/3012/CH7/EX7.8/Ex7_8.sce new file mode 100755 index 000000000..5cbf01eed --- /dev/null +++ b/3012/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,57 @@ +// Given:- +clc; +m1dot = 69.78 // in kg/s +p1 = 1.0 // in bar +T1 = 478.0 // in kelvin +T2 = 400.0 // in kelvin +p2 = 1.0 // in bar +p3 = 0.275 // in Mpa +T3 = 38.9 // in degree celcius +m3dot = 2.08 // in kg/s +T4 = 180.0 // in degree celcius +p4 = 0.275 // in Mpa +p5 = 0.07 // in bar +x5 = 0.93 +Wcvdot = 876.8 // in kW +T0 = 298.0 // in kelvin + + +// Part(a) +// From table A-22 +h1 = 480.35 // in kj/kg +h2 = 400.97 // in kj/kg +s1 = 2.173 // in kj/kg +s2 = 1.992 // in kj/kg + +// From table A-2E +h3 = 162.82 // in kj/kg +s3 = 0.5598 // in kj/kg.k +// Using saturation data at 0.07 bars from Table A-3 +h5 = 2403.27 // in kj/kg +s5 = 7.739 // in kj/kg.k +//The net rate exergy carried out by the water stream + +// From table A-4 +h4 = 2825.0 // in kj/kg +s4 = 7.2196 // in kj/kg.k +// Calculations +netRE = m1dot*(h1-h2-T0*(s1-s2-(8.314/28.97)*log(p1/p2))) // the net rate exergy carried into the control volume +netREout = m3dot*(h5-h3-T0*(s5-s3)) +// From an exergy rate balance applied to a control volume enclosing the steam generator +Eddot = netRE + m3dot*(h3-h4-T0*(s3-s4)) // the rate exergy is destroyed in the heat-recovery steam generator + +// From an exergy rate balance applied to a control volume enclosing the turbine +EdDot = -Wcvdot + m3dot*(h4-h5-T0*(s4-s5)) // the rate exergy is destroyed in the tpurbine + +// Results +printf( '\n balance sheet') +printf( '\n- Net rate of exergy in: %f kJ/kg.',netRE) +printf( '\n Disposition of the exergy:') +printf( '\n• Rate of exergy out') +printf( '\n power developed %f kJ/kg.',netRE-netREout-Eddot-EdDot) +printf( '\n water stream %f',netREout) +printf( '\n• Rate of exergy destruction') +printf( '\n heat-recovery steam generator %f kJ/kg',Eddot) +printf( '\n turbine %f',EdDot) + +// note : answer is slightly different because of rounding off error. diff --git a/3012/CH7/EX7.9/Ex7_9.sce b/3012/CH7/EX7.9/Ex7_9.sce new file mode 100755 index 000000000..5dcf9b32e --- /dev/null +++ b/3012/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,22 @@ +// Given:- +T0 = 273.00 // in kelvin +pricerate = 0.08 // exergy value at $0.08 per kw.h + +// From example 6.8 +sigmadotComp = 17.5e-4 // in kw/k +sigmadotValve = 9.94e-4 // in kw/k +sigmadotcond = 7.95e-4 // in kw/k + +// Calculations +// The rates of exergy destruction +EddotComp = T0*sigmadotComp // in kw +EddotValve = T0*sigmadotValve // in kw +Eddotcond = T0*sigmadotcond // in kw + +mCP = 3.11 // From the solution to Example 6.14, the magnitude of the compressor power in kW + +// Results +printf( ' Daily cost in dollars of exergy destruction due to compressor irreversibilities = %.3f',EddotComp*pricerate*24) +printf( ' Daily cost in dollars of exergy destruction due to irreversibilities in the throttling valve = %.3f',EddotValve*pricerate*24) +printf( ' Daily cost in dollars of exergy destruction due to irreversibilities in the condenser = %.3f ',Eddotcond*pricerate*24) +printf( ' Daily cost in dollars of electricity to operate compressor = %.3f',mCP*pricerate*24) |