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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
commitb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
treeab291cffc65280e58ac82470ba63fbcca7805165 /3012/CH7
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Diffstat (limited to '3012/CH7')
-rwxr-xr-x3012/CH7/EX7.1/Ex7_1.sce19
-rwxr-xr-x3012/CH7/EX7.10/Ex7_10.sce49
-rwxr-xr-x3012/CH7/EX7.2/Ex7_2.sce52
-rwxr-xr-x3012/CH7/EX7.3/Ex7_3.sce47
-rwxr-xr-x3012/CH7/EX7.4/Ex7_4.sce16
-rwxr-xr-x3012/CH7/EX7.5/Ex7_5.sce27
-rwxr-xr-x3012/CH7/EX7.6/Ex7_6.sce49
-rwxr-xr-x3012/CH7/EX7.7/Ex7_7.sce35
-rwxr-xr-x3012/CH7/EX7.8/Ex7_8.sce57
-rwxr-xr-x3012/CH7/EX7.9/Ex7_9.sce22
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)