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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /572/CH7 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '572/CH7')
| -rwxr-xr-x | 572/CH7/EX7.1/c7_1.sce | 21 | ||||
| -rwxr-xr-x | 572/CH7/EX7.10/c7_10.sce | 62 | ||||
| -rwxr-xr-x | 572/CH7/EX7.2/c7_2.sce | 51 | ||||
| -rwxr-xr-x | 572/CH7/EX7.3/c7_3.sce | 57 | ||||
| -rwxr-xr-x | 572/CH7/EX7.4/c7_4.sce | 20 | ||||
| -rwxr-xr-x | 572/CH7/EX7.5/c7_5.sce | 38 | ||||
| -rwxr-xr-x | 572/CH7/EX7.6/c7_6.sce | 65 | ||||
| -rwxr-xr-x | 572/CH7/EX7.7/c7_7.sce | 45 | ||||
| -rwxr-xr-x | 572/CH7/EX7.8/c7_8.sce | 67 | ||||
| -rwxr-xr-x | 572/CH7/EX7.9/c7_9.sce | 24 |
10 files changed, 450 insertions, 0 deletions
diff --git a/572/CH7/EX7.1/c7_1.sce b/572/CH7/EX7.1/c7_1.sce new file mode 100755 index 000000000..23015cfec --- /dev/null +++ b/572/CH7/EX7.1/c7_1.sce @@ -0,0 +1,21 @@ +// (7.1) A cylinder of an internal combustion engine contains 2450 cm3 of gaseous combustion products at a pressure of 7 bar and a temperature of 867C just before the exhaust valve opens. Determine the specific exergy of the gas, in kJ/kg. Ignore the effects of motion and gravity, and model the combustion products as air as an ideal gas. Take T0 = 300 K (27C) and p0= 1.013 bar.
+
+
+//solution
+
+
+//variable initialization
+v = 2450 //volume of gaseous products in cm^3
+P = 7 //pressure of gaseous product in bar
+T = 867 //temperature of gaseous product in degree celcius
+T0 = 300 //in kelvin
+P0 = 1.013 //in bar
+
+//from table A-22
+u = 880.35 //in kj/kg
+u0 = 214.07 //in kj/kg
+s0(T) = 3.11883 //in kj/kg.k
+s0(T0) = 1.70203 //in kj/kg.k
+
+e = (u-u0) + (P0*(8.314/28.97)*[((T+273)/P)-(T0/P0)]) - T0*[s0(T)-s0(T0)-(8.314/28.97)*log(P/P0)] //in kj/kg
+printf('the specific exergy of the gas, in kJ/kg is \n\t e = %f',e)
\ No newline at end of file diff --git a/572/CH7/EX7.10/c7_10.sce b/572/CH7/EX7.10/c7_10.sce new file mode 100755 index 000000000..1647d151e --- /dev/null +++ b/572/CH7/EX7.10/c7_10.sce @@ -0,0 +1,62 @@ +//(7.10) A cogeneration system consists of a natural gas-fueled boiler and a steam turbine that develops power and provides steam for an industrial process. At steady state, fuel enters the boiler with an exergy rate of 100 MW. Steam exits the boiler at 50 bar, 466C with an exergy rate of 35 MW. Steam exits the turbine at 5 bar, 205C and a mass flow rate of 26.15 kg/s. The unit cost of the fuel is 1.44 cents per kw.h of exergy. The costs of owning and operating the boiler and turbine are, respectively, dollar 1080/h and dollar 92/h. The feedwater and combustion air enter with negligible exergy and cost. The combustion products are discharged directly to the surroundings with negligible cost. Heat transfer with the surroundings and kinetic and potential energy effects are negligible. Let T0 = 298 K. (a) For the turbine, determine the power and the rate exergy exits with the steam, each in MW. (b) Determine the unit costs of the steam exiting the boiler, the steam exiting the turbine, and the power, each in cents per kw.h of exergy. (c) Determine the cost rates of the steam exiting the turbine and the power, each in $/h.
+
+
+//solution
+
+//variable initialization
+EfFdot = 100 //exergy rate of fuel entering the boiler in MW
+cF = 1.44 //unit cost of fuel in cents per kw.h
+Zbdot = 1080 //the cost of owning and operating boiler in dollars per hour
+Ef1dot = 35 //exergy rate of exiting steam from the boiler in MW
+p1 = 50 //pressure of exiting steam from the boiler in bar
+T1 = 466 //temperature of exiting steam from the boiler in degree celcius
+Ztdot = 92 //the cost of owning and operating turbine in dollars per hour
+p2 = 5 //pressure of exiting steam from the turbine in bars
+T2 = 205 //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 //in kelvin
+
+
+//part(a)
+//from table A-4,
+h1 = 3353.54 //in kj/kg
+h2 = 2865.96 //in kj/kg
+//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
+
+//from table A-4
+s1 = 6.8773 //in kj/kg.k
+s2 = 7.0806 //in kj/kg.k
+
+Ef2dot = Ef1dot+m2dot*(h2-h1-T0*(s2-s1))/1000 //the rate exergy exits with the steam in MW
+printf('for the turbine,the power in MW is:\t%f',Wedot)
+printf('\nfor the turbine,the rate exergy exits with the steam in MW is:\t%f',Ef2dot)
+
+//part(b)
+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
+
+printf('\n\nthe unit costs of the steam exiting the boiler in cents per kw.h of exergy is:\t%f',c1)
+printf('\nthe unit costs of the steam exiting the turbine in cents per kw.h of exergy is:\t%f',c2)
+printf('\nunit cost of power in cents per kw.h is:\t%f',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
+
+printf('\n\nthe cost rate of the steam exiting the turbine in dollars per hour is: %f',C2dot)
+printf('\nthe cost rate of the power in dollars per hour is: %f',Cedot)
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/572/CH7/EX7.2/c7_2.sce b/572/CH7/EX7.2/c7_2.sce new file mode 100755 index 000000000..a20526380 --- /dev/null +++ b/572/CH7/EX7.2/c7_2.sce @@ -0,0 +1,51 @@ +//(7.2) Refrigerant 134a, initially a saturated vapor at -28C, is contained in a rigid, insulated vessel. The vessel is fitted with a paddle wheel connected to a pulley from which a mass is suspended. As the mass descends a certain distance, the refrigerant is stirred until it attains a state where the pressure is 1.4 bar. The only significant changes of state are experienced by the suspended mass and the refrigerant. The mass of refrigerant is 1.11 kg. Determine (a) the initial exergy, final exergy, and change in exergy of the refrigerant, each in kJ. (b) the change in exergy of the suspended mass, in kJ. (c) the change in exergy of an isolated system of the vessel and pulley–mass assembly, in kJ. Discuss the results obtained, and compare with the respective energy changes. Let T0 = 293 K (20C), p0 = 1 bar.
+
+//solution
+
+//variable initialization
+mR = 1.11 //mass of the refrigerant in kg
+T1 = -28 //initial temperature of the saturated vapor in degree celcius
+P2 = 1.4 //final pressure of the refrigerant in bar
+T0 = 293 //in kelvin
+P0 = 1 //in bar
+
+//part (a)
+//from table A-10
+u1 = 211.29 //in kj/kg
+v1 = .2052 //in m^3/kg
+s1 = .9411 //in kj/kg.k
+//from table A-12
+u0 = 246.67 //in kj/kg
+v0 = .23349 //in m^3/kg
+s0 = 1.0829 //in kj/kg.k
+
+E1 = mR*[(u1-u0) + P0*10^5*(v1-v0)*10^(-3)-T0*(s1-s0)]
+
+//from table A-12
+u2 = 300.16 //in kj/kg
+s2 = 1.2369 //in kj/kg.k
+v2 = v1
+
+E2 = mR*[(u2-u0) + P0*10^5*(v2-v0)*10^(-3)-T0*(s2-s0)]
+
+printf('part(a)the initial exergy in kj is :\n\t E1 = %f',E1)
+printf('\nthe final exergy in kj is :\n\t E2 = %f',E2)
+printf('\nthe change in exergy of the refrigerant in kj is \n\tdeltaE = %f',E2-E1)
+
+
+//part (b)
+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
+printf('\n\n\npart(b)the change in exergy of the suspended mass, in kJ is :\n\t deltaE = %f',deltaE)
+
+
+//part(c)
+deltaEiso = (E2-E1) + deltaE
+printf('\n\n\npart(c)the change in exergy of an isolated system of the vessel and pulley–mass assembly, in kJ is :\n\t deltaEiso = %f',deltaEiso)
+
+
+
+
diff --git a/572/CH7/EX7.3/c7_3.sce b/572/CH7/EX7.3/c7_3.sce new file mode 100755 index 000000000..8506d92c5 --- /dev/null +++ b/572/CH7/EX7.3/c7_3.sce @@ -0,0 +1,57 @@ +//(7.3) Water initially a saturated liquid at 100C is contained in a piston–cylinder assembly. The water undergoes a process to the corresponding saturated vapor state, during which the piston moves freely in the cylinder. For each of the two processes described below, determine on a unit of mass basis the change in exergy, the exergy transfer accompanying work, the exergy transfer accompanying heat, and the exergy destruction, each in kJ/kg. Let T0 = 20C, p0 = 1.014 bar. (a) The change in state is brought about by heating the water as it undergoes an internally reversible process at constant temperature and pressure. (b) The change in state is brought about adiabatically by the stirring action of a paddle wheel.
+
+
+//solution
+
+//variable initialization
+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
+
+deltae = ug-uf + P0*10^5*(vg-vf)/(10^3)-T0*(sg-sf)
+
+//exergy 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
+
+printf('part(a)the change in exergy in kj/kg is:\n\t deltae = %f ',deltae)
+printf('\nthe exergy transfer accompanying work in kj/kg is:\n\t etaw = %f',etaw)
+printf('\nthe exergy transfer accompanying heat in kj/kg is:\n\t etah = %f',etah)
+printf('\nthe exergy destruction in kj/kg is:\n\t ed = %f',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
+
+printf('\n\n\npart(b)the change in exergy in kj/kg is:\n\t Deltae = %f ',Deltae)
+printf('\nthe exergy transfer accompanying work in kj/kg is:\n\t Etaw = %f',Etaw)
+printf('\nthe exergy transfer accompanying heat in kj/kg is:\n\t Etah = %f',Etah)
+printf('\nthe exergy destruction in kj/kg is:\n\t Ed = %f',Ed)
+
+
+
+
+
+
diff --git a/572/CH7/EX7.4/c7_4.sce b/572/CH7/EX7.4/c7_4.sce new file mode 100755 index 000000000..762084987 --- /dev/null +++ b/572/CH7/EX7.4/c7_4.sce @@ -0,0 +1,20 @@ +//(7.4) For the gearbox of Examples 2.4 and 6.4(a), develop a full exergy accounting of the power input. Let T0 = 293 K.
+
+
+//solution
+
+//Since the gearbox volume is constant, the rate of exergy transfer accompanying power reduces to the power itself. Accordingly, exergy is transferred into the gearbox via the high-speed shaft at a rate equal to the power input, 60 kW, and exergy is transferred out via the low-speed shaft at a rate equal to the power output, 58.8 kW. Additionally, exergy is transferred out accompanying heat transfer and destroyed by irreversibilities within the gearbox.
+
+T0 = 293 //in kelvin
+Qdot = -1.2 //in KW, from example 6.4a
+Tb = 300 //temperature at the outer surface of the gearbox in kelvin from example 6.4a
+sigmadot = 4e-3 //rate of entropy production in KW/k from example 6.4a
+
+R = (1-T0/Tb)*Qdot //time rate of exergy transfer accompanying heat
+Eddot = T0*sigmadot //rate of exergy destruction
+
+printf('balance sheet')
+printf('\nrate of exergy in:\n high speed shaft\t\t 60Kw')
+printf('\nDisposition of the exergy:\n• Rate of exergy out\nlow-speed shaft\t\t 58.8Kw')
+printf('\nheat transfer in kw\t\t%f',norm(R))
+printf('\n• Rate of exergy destruction in kw\t\t%f',Eddot)
diff --git a/572/CH7/EX7.5/c7_5.sce b/572/CH7/EX7.5/c7_5.sce new file mode 100755 index 000000000..20e742b33 --- /dev/null +++ b/572/CH7/EX7.5/c7_5.sce @@ -0,0 +1,38 @@ +//Superheated water vapor enters a valve at 3.0 MPa, 320C and exits at a pressure of 0.5 MPa. The expansion is a throttling process. Determine the specific flow exergy at the inlet and exit and the exergy destruction per unit of mass flowing, each in kJ/kg. Let T0 = 25C, p0= 1 atm.
+
+//solution
+
+//variable initialization
+p1 = 3 //entry pressure in Mpa
+p2 = .5 //exit pressure in Mpa
+T1 = 320 //entry temperature in degree celcius
+T0 = 25 //in degree celcius
+p0 = 1 //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
+
+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
+
+printf(' the specific flow exergy at the inlet in kj/kg is :\n\t ef1 =%f',ef1)
+printf('\nthe specific flow exergy at the exit in kj/kg is:\n\t ef2 = %f', ef2)
+printf('\nthe exergy destruction per unit of mass flowing in kj/kg is:\n\t = %f',Ed)
+
+
+
+
+
diff --git a/572/CH7/EX7.6/c7_6.sce b/572/CH7/EX7.6/c7_6.sce new file mode 100755 index 000000000..057e2c91d --- /dev/null +++ b/572/CH7/EX7.6/c7_6.sce @@ -0,0 +1,65 @@ +//Compressed air enters a counterflow heat exchanger operating at steady state at 610 K, 10 bar and exits at 860 K, 9.7 bar. Hot combustion gas enters as a separate stream at 1020 K, 1.1 bar and exits at 1 bar. Each stream has a mass flow rate of 90 kg/s. Heat transfer between the outer surface of the heat exchanger and the surroundings can be ignored. Kinetic and potential energy effects are negligible. Assuming the combustion gas stream has the properties of air, and using the ideal gas model for both streams, determine for the heat exchanger(a) the exit temperature of the combustion gas, in K. (b) the net change in the flow exergy rate from inlet to exit of each stream, in MW. (c) the rate exergy is destroyed, in MW. Let T0 = 300 K, p0 = 1 bar.
+
+
+//solution
+
+//variable initialization
+T1 = 610 //temperature of the air entering heat exchanger in kelvin
+p1 = 10 //pressure of the air entering heat exchanger in bar
+T2 = 860 //temperature of the air exiting the heat exchanger in kelvin
+p2 = 9.7 //pressure of the air exiting the heat exchanger in bar
+T3 = 1020 //temperature of entering hot combustion gas in kelvin
+p3 = 1.1 //pressure of entering hot combustion gas in bar
+p4 = 1 //pressure of exiting hot combustion gas in bar
+mdot = 90 //mass flow rate in kg/s
+T0 = 300 //in kelvin
+p0 = 1 //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
+
+//from reduction of mass and energy rate balances for the control volume at steady state
+h4 = h3+h1-h2
+
+//using interpolation in table A-22 gives
+T4 = 778 //in kelvin
+printf('the exit temperature of the combustion gas in kelvin is:\n\tT4 = %f',T4)
+
+
+//part(b)
+//from table A-22
+s2 = 2.79783 //in kj/kg.k
+s1 = 2.42644 //in kj/kg.k
+
+deltaR = (mdot*((h2-h1)-T0*(s2-s1-(8.314/28.97)*log(p2/p1))))/1000
+
+//from table A-22
+s4 = 2.68769 //in kj/kg.k
+s3 = 2.99034 //in kj/kg.k
+
+deltRc = mdot*((h4-h3)-T0*(s4-s3-(8.314/28.97)*log(p4/p3)))/1000
+
+printf('\nthe net change in the flow exergy rate from inlet to exit of compressed gas in MW is:\n\t deltaR = %f',deltaR)
+printf('\nthe net change in the flow exergy rate from inlet to exit of hot combustion gas in MW is:\n\tdeltRc =%f ',deltRc)
+
+//part(c)
+//from an exergy rate balance
+Eddot = -deltaR-deltRc
+
+printf('\nthe rate exergy destroyed, in MW is:Eddot = %f',Eddot)
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/572/CH7/EX7.7/c7_7.sce b/572/CH7/EX7.7/c7_7.sce new file mode 100755 index 000000000..cac0d1a39 --- /dev/null +++ b/572/CH7/EX7.7/c7_7.sce @@ -0,0 +1,45 @@ +//Steam enters a turbine with a pressure of 30 bar, a temperature of 400C, a velocity of 160 m/s. Steam exits as saturated vapor at 100C with a velocity of 100 m/s. At steady state, the turbine develops work at a rate of 540 kJ per kg of steam flowing through the turbine. Heat transfer between the turbine and its surroundings occurs at an average outer surface temperature of 350 K. Develop a full accounting of the net exergy carried in by the steam, per unit mass of steam flowing. Neglect the change in potential energy between inlet and exit. Let T0 = 25C, p0= 1 atm.
+
+//solution
+
+//variable initialization
+p1 = 30 //pressure of entering steam in bar
+t1 = 400 //temperature of entering steam in degree celcius
+v1 = 160 //velocity of entering steam in m/s
+t2 = 100 //temperature of exiting saturated vapor in degree celcius
+v2 = 100 //velocity of exiting saturated vapor in m/s
+W = 540 //rate of work developed in kj per kg of steam
+Tb = 350 //the temperature on the boundary where heat transfer occurs in kelvin
+T0 = 25 //in degree celcius
+p0 = 1 //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
+
+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
+
+//from example 6.6
+Q = -22.6 //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 balanceff
+
+printf('balance sheet')
+printf('\nNet rate of exergy in:\t%f',DELTAef)
+printf('\nDisposition of the exergy:')
+printf('\n• Rate of exergy out')
+printf('\nwork\t%f',W)
+printf('\nheat transfer\t%f',-Eq)
+printf('\n• Rate of exergy destruction\t%f',Ed)
+
+
+
+
+
+
+
+
diff --git a/572/CH7/EX7.8/c7_8.sce b/572/CH7/EX7.8/c7_8.sce new file mode 100755 index 000000000..3df7082ba --- /dev/null +++ b/572/CH7/EX7.8/c7_8.sce @@ -0,0 +1,67 @@ +//(7.8) Suppose the system of Example 4.10 is one option under consideration for utilizing the combustion products discharged from an industrial process. (a) Develop a full accounting of the net exergy carried in by the combustion products. (b) Discuss the design implications of the results.
+
+
+//solution
+
+//variable initialization
+m1dot = 69.78 //in kg/s
+p1 = 1 //in bar
+T1 = 478 //in kelvin
+T2 = 400 //in kelvin
+p2 = 1 //in bar
+p3 = .275 //in Mpa
+T3 = 38.9 //in degree celcius
+m3dot = 2.08 //in kg/s
+T4 = 180 //in degree celcius
+p4 = .275 //in Mpa
+p5 = .07 //in bar
+x5 = .93
+Wcvdot = 876.8 //in kW
+T0 = 298 //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
+
+netRE = m1dot*(h1-h2-T0*(s1-s2-(8.314/28.97)*log(p1/p2))) //the net rate exergy carried into the control volume
+
+
+//from table A-2E
+h3 = 162.82 //in kj/kg
+s3 = .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
+
+netREout = m3dot*(h5-h3-T0*(s5-s3)) //the net rate exergy carried out by the water stream
+
+//from table A-4
+h4 = 2825 //in kj/kg
+s4 = 7.2196 //in kj/kg.k
+
+//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
+
+
+printf('balance sheet')
+printf('\nNet rate of exergy in:\t%f',netRE)
+printf('\nDisposition of the exergy:')
+printf('\n• Rate of exergy out')
+printf('\npower developed\t%f',1772.8-netREout-Eddot-EdDot)
+printf('\nwater stream\t%f',netREout)
+printf('\n• Rate of exergy destruction')
+printf('\nheat-recovery steam generator\t%f',Eddot)
+printf('\nturbine\t%f',EdDot)
+
+
+
+
+
+
diff --git a/572/CH7/EX7.9/c7_9.sce b/572/CH7/EX7.9/c7_9.sce new file mode 100755 index 000000000..0d9a5040a --- /dev/null +++ b/572/CH7/EX7.9/c7_9.sce @@ -0,0 +1,24 @@ +//(7.9) For the heat pump of Examples 6.8 and 6.14, determine the exergy destruction rates, each in kW, for the compressor, condenser, and throttling valve. If exergy is valued at $0.08 per kw.h, determine the daily cost of electricity to operate the compressor and the daily cost of exergy destruction in each component. Let T0 = 273 K (0C), which corresponds to the temperature of the outside ai.
+
+
+//solution
+
+T0 = 273 //in kelvin
+pricerate = .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
+
+//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
+
+printf('Daily cost in dollars of exergy destruction due to compressor irreversibilities =\t %f',EddotComp*pricerate*24)
+printf('\naDaily cost in dollars of exergy destruction due to irreversibilities in the throttling valve =\t %f',EddotValve*pricerate*24)
+printf('\naDaily cost in dollars of exergy destruction due to irreversibilities in the condenser =\t %f',Eddotcond*pricerate*24)
+printf('\naDaily cost in dollars of electricity to operate compressor =\t %f',mCP*pricerate*24)
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