diff options
Diffstat (limited to '1943')
130 files changed, 3774 insertions, 0 deletions
diff --git a/1943/CH1/EX1.1/Ex1_1.sce b/1943/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..64f14893d --- /dev/null +++ b/1943/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,35 @@ + +clf()
+clc
+clear
+//Input data
+C=30//Capacity in MW
+M=70//Loads are taken above 70 MW
+t1=[0,6]//Time range in hours
+t2=[6,10]//Time range in hours
+t3=[10,12]//Time range in hours
+t4=[12,16]//Time range in hours
+t5=[16,20]//Time range in hours
+t6=[20,22]//Time range in hours
+t7=[22,24]//Time range in hours
+L=[30,70,90,60,100,80,60]//Load in MW
+
+//Calculations
+E=((L(1)*(t1(2)-t1(1)))+(L(2)*(t2(2)-t2(1)))+(L(3)*(t3(2)-t3(1)))+(L(4)*(t4(2)-t4(1)))+(L(5)*(t5(2)-t5(1)))+(L(6)*(t6(2)-t6(1)))+(L(7)*(t7(2)-t7(1))))//Energy generated in MWh
+AL=(E/24)//Average load in MW
+PL=max(L(1),L(2),L(3),L(4),L(5),L(6),L(7))//Peak load in MW
+LF=(AL/PL)//Load factor of the plant
+E1=((L(3)-M)*(t3(2)-t3(1)))+((L(5)-M)*(t5(2)-t5(1)))+((L(6)-M)*(t6(2)-t6(1)))//Energy generated if the load above 70 MW is supplied by a standby unit of 30 MW capacity in MWh
+T=(t3(2)-t3(1))+(t5(2)-t5(1))+(t6(2)-t6(1))//Time during which the standby unit remains in operation in h
+AL1=(E1/T)//Average load in MW
+LF1=(AL1/C)//Load factor
+U=(E1/(C*T))//Use factor
+
+//Output
+t=[0,0,6,6,10,10,12,12,16,16,20,20,22,22,24,24]//Time for plotting load curve in hours
+l=[0,30,30,70,70,90,90,60,60,100,100,80,80,60,60,0]//Load for plotting load curve in MW
+plot(t,l)//Load curve taking Time in hours on X-axis and Load in MW on Y- axis
+xtitle('Load Curve','Time hours','Load MW');
+
+
+printf('(a)Load factor of the plant is %3.2f\n(b)Load factor of a standby equipment of %3.0f capacity if it takes up all the loads above %3.0f MW is %3.2f\n(c)Use factor is %3.2f',LF,C,M,LF1,U)
diff --git a/1943/CH1/EX1.10/Ex1_10.sce b/1943/CH1/EX1.10/Ex1_10.sce new file mode 100755 index 000000000..f748d2aa6 --- /dev/null +++ b/1943/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,34 @@ + +clf()
+clc
+clear
+//Input data
+t1x=[0,6]//Time range in hours
+t2x=[6,12]//Time range in hours
+t3=[12,14]//Time range in hours
+t4=[14,18]//Time range in hours
+t5=[18,24]//Time range in hours
+L=[30,90,60,100,50]//Load in MW
+
+//Calculations
+t1=[0,6,6,12,12,14,14,18,18,24,24]//Time in hours for Load curve
+L1=[30,30,90,90,60,60,100,100,50,50,0]//Load in MW for Load curve
+t2=[0,4,4,10,10,12,12,18,18,24,24]//Time in hours for Load duration curve
+L2=[100,100,90,90,60,60,50,50,30,30,24]//Load in MW for Load duration curve
+E=((L(1)*(t1x(2)-t1x(1)))+(L(2)*(t2x(2)-t2x(1)))+(L(3)*(t3(2)-t3(1)))+(L(4)*(t4(2)-t4(1)))+(L(5)*(t5(2)-t5(1))))//Energy generated in MWh
+AL=E/24//Average load in MW
+MD=max(L(1),L(2),L(3),L(4),L(5))//Maximum demand in MW
+LF=(AL/MD)//Load factor
+Lx=[30,10]//Loads for selecting suitable generating units in MW
+tx=[24,18,10,4]//Time for selecting suitable generating units in hrs
+PC=(Lx(1)*tx(4)+Lx(2)*1)//Plant capacity in MW
+CF=(E/(PC*24))//Capacity factor
+
+//Output
+subplot(221)
+plot(t1,L1)//Load curve taking Time in hrs on X- axis and Load in MW on Y- axis
+xtitle('Load curve','Time hrs','Load MW')
+subplot(222)
+plot(t2,L2)//Load duration curve taking Time in hrs on X- axis and Load in MW on Y- axis
+xtitle('Load duration curve','Time hrs','Load MW')
+printf('(c)Suitable generating units to supply the load are\ni)One unit of %3.0f MW will run for %3.0f hours\nii)One unit of %3.0f MW will run for %3.0f hours\niii)One unit of %3.0f MW will run for %3.0f hours\niv)One unit of %3.0f MW will run for %3.0f hours\n\n(d)Load factor is %3.2f\n\n(e)Capacity of the plant is %3.0f MW and Capacity factor is %3.3f',Lx(1),tx(1),Lx(1),tx(2),Lx(1),tx(3),Lx(2),tx(4),LF,PC,CF)
diff --git a/1943/CH1/EX1.11/Ex1_11.sce b/1943/CH1/EX1.11/Ex1_11.sce new file mode 100755 index 000000000..a41e1df0c --- /dev/null +++ b/1943/CH1/EX1.11/Ex1_11.sce @@ -0,0 +1,30 @@ + +clc
+clear
+//Input data
+C=10//Capacity of generating unit in MW
+MD=[6,3.6,0.4]//Maximum demand for domestic consumers, industrial consumers and street-lighting load respectively in MW
+L=[0.2,0.5,0.3]//Load factor for domestic consumers, industrial consumers and street-lighting load respectively
+CC=10000//Capital cost of the plant per kW in Rs
+RC=3600000//Total rumming cost per year in Rs
+AID=10//Annual interest and depreciation on capital cost in percent
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+E=((MD(1)*L(1))+(MD(2)*L(2))+(MD(3)*L(3)))*Y*1000//Energy supplied per year to all three consumers in kWh
+OC=(RC/E)//Operating charges per kWh in Rs
+CCP=(C*1000*CC)//capital cost of the plant in Rs
+FCY=((AID/100)*CCP)//Fixed charges per year in Rs
+FCkW=(FCY/CC)//Fixed charges per kW in Rs
+//a) For domestic consumers
+TC1=((FCkW*MD(1)*1000)+(OC*MD(1)*L(1)*Y*1000))//Total chrges in Rs
+OC1=(TC1/(MD(1)*L(1)*Y*1000))*100//Overall cost per kWh in paise
+//b)For industrial consumers
+TC2=((FCkW*MD(2)*1000)+(OC*MD(2))*L(2)*Y*1000)//Total chrges in Rs
+OC2=(TC2/(MD(2)*L(2)*Y*1000))*100//Overall cost per kWh in paise
+//c) For street-lighting load
+TC3=((FCkW*MD(3)*1000)+(OC*MD(3))*L(3)*Y*1000)//Total chrges in Rs
+OC3=(TC3/(MD(3)*L(3)*Y*1000))*100//Overall cost per kWh in paise
+
+//Output
+printf('Overall cost of energy per kWh for:\n(a)Domestic consumers is %3.0f paise\n(b)Industrial consumers is %3.0f paise\n(c)Street-lighting load is %3.0f paise',OC1,OC2,OC3)
diff --git a/1943/CH1/EX1.12/Ex1_12.sce b/1943/CH1/EX1.12/Ex1_12.sce new file mode 100755 index 000000000..483b99e8f --- /dev/null +++ b/1943/CH1/EX1.12/Ex1_12.sce @@ -0,0 +1,15 @@ + +clc
+clear
+//Input data
+CC=(80*10^6)//Capital cost in Rs
+L=30//Useful life in years
+S=5//Salvage value of the capital cost in percent
+i=0.06//Yearly rate of compound interest
+
+//Calculations
+A=((100-S)/100)*CC//Difference of capital cost and salvage value in Rs
+P=((A*i)/((1+i)^L-1))//The amount of money to be saved annually in Rs
+
+//Output
+printf('The amount of money to be saved annually is Rs.%3.0f/-',P)
diff --git a/1943/CH1/EX1.13/Ex1_13.sce b/1943/CH1/EX1.13/Ex1_13.sce new file mode 100755 index 000000000..67df2dace --- /dev/null +++ b/1943/CH1/EX1.13/Ex1_13.sce @@ -0,0 +1,20 @@ + +clc
+clear
+//Input data
+i=4000//Initial investment in Rs crore
+Y=4//Period in years
+A=1200//Amount added in Rs crore
+B=400//Amount paid from 5th year onwards to the 12th year in Rs crore
+a=5//5th year
+b=12//12th year
+y=30//Period in years
+C=600//Salvage value in Rs crore
+I=0.1//Interest rate
+
+//Calculations
+X=(1/(1+I))//X value for calculations
+PW=(i+(A*X^Y)+((B/I)*X^b*((I+1)^b-1))-((B/I)*X^a*((I+1)^a-1))-(C*X^y))//Present worth of the payments at the time of commissioning in Rs. crores
+
+//Output
+printf('Present worth of the payments at the time of commissioning is Rs.%3.2f crores',PW)
diff --git a/1943/CH1/EX1.14/Ex1_14.sce b/1943/CH1/EX1.14/Ex1_14.sce new file mode 100755 index 000000000..bd064f64a --- /dev/null +++ b/1943/CH1/EX1.14/Ex1_14.sce @@ -0,0 +1,19 @@ + +clc
+clear
+//Input data
+O=1000//Combined output of two units in MW
+//Two coal generating units P and Q have the incremental heat rate defined by
+//(IR)P=0.4818*10^-7.LP^4 - 0.9089*10^-4.LP^3 + 0.6842*10^-1.LP^2 - 0.2106*10.LP + 9860
+//(IR)R=0.9592*10^-7.LQ^4 - 0.7811*10^-4.LQ^3 + 0.2625*10^-1.LQ^2 - 0.2189*10.LQ + 9003
+
+//Calculations
+//LP+LQ=1000
+//By making (IR)P=(IR)Q and solving the above three equations by a numerical methos such as Newton-Raphson algorithm, we get
+LP=732.5//Heat rate in MW
+LQ=(O-LP)//Heat rate in MW
+IR=0.4818*10^-7*LP^4 - 0.9089*10^-4*LP^3 + 0.6842*10^-1*LP^2 - 0.2106*100*LP + 9860
+IR1=0.9592*10^-7*LQ^4 - 0.7811*10^-4*LQ^3 + 0.2625*10^-1*LQ^2 - 0.2189*10*LQ + 9003
+
+//Output
+printf('Incremental heat transfer rate at which the combined output of the two units is %3.0f MW is IR = (IR)P = (IR)Q = %i kJ/kWh',O,IR)
diff --git a/1943/CH1/EX1.15/Ex1_15.sce b/1943/CH1/EX1.15/Ex1_15.sce new file mode 100755 index 000000000..06d2ae5d7 --- /dev/null +++ b/1943/CH1/EX1.15/Ex1_15.sce @@ -0,0 +1,30 @@ + +clc
+clear
+//Input data
+F=2700//Fixed cost of the thermal station per kW of installed capacity per year in Rs,
+FO=40//Fuel and operating costs per kWh generated in paise
+L=[100,75,50,25]//Load factors
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+FC=(F/Y)*100//Fixed costs per kW per hour in paise
+E1=(L(1)/100)//Energy produced in 1 hr with 1 kW plant in kWh
+FOC1=(E1*FO)//Fuel and operating cost in paise
+TC1=(FC+FOC1)//Total cost per hr in paise
+C1=(TC1/E1)//Cost per kWh in paise
+E2=(L(2)/100)//Energy produced in 1 hr with 1 kW plant in kWh
+FOC2=(E2*FO)//Fuel and operating cost in paise
+TC2=(FC+FOC2)//Total cost per hr in paise
+C2=(TC2/E2)//Cost per kWh in paise
+E3=(L(3)/100)//Energy produced in 1 hr with 1 kW plant in kWh
+FOC3=(E3*FO)//Fuel and operating cost in paise
+TC3=(FC+FOC3)//Total cost per hr in paise
+C3=(TC3/E3)//Cost per kWh in paise
+E4=(L(4)/100)//Energy produced in 1 hr with 1 kW plant in kWh
+FOC4=(E4*FO)//Fuel and operating cost in paise
+TC4=(FC+FOC4)//Total cost per hr in paise
+C4=(TC4/E4)//Cost per kWh in paise
+
+//Output
+printf('==============================================================================\nLoad Energy produced Fixed cost Fuel and Total cost Cost per\nfactor in 1hr with per hr operating cost per hr kWh\n(percent) 1kW plant(kWh) (paise) (paise) (paise) (paise)\n==============================================================================\n%3.0f %3.0f %3.0f %3.0f %3.0f %3.0f\n%3.0f %3.2f %3.0f %3.0f %3.0f %3.0f\n%3.0f %3.2f %3.0f %3.0f %3.0f %3.0f\n%3.0f %3.2f %3.0f %3.0f %3.0f %3.0f\n==============================================================================',L(1),E1,FC,FOC1,TC1,C1,L(2),E2,FC,FOC2,TC2,C2,L(3),E3,FC,FOC3,TC3,C3,L(4),E4,FC,FOC4,TC4,C4)
diff --git a/1943/CH1/EX1.2/Ex1_2.sce b/1943/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..b402f8ecf --- /dev/null +++ b/1943/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,18 @@ + +clc
+clear
+//Input data
+P=60//Peak load on power plant in MW
+L=[30,20,10,14]//Loads having maximum demands in MW
+C=80//Capacity of the power plant in MW
+A=0.5//Annual load factor
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+AL=(P*A)//Average load in MW
+E=(AL*1000*Y)/10^6//Energy supplied per year in kWh*10^6
+DF=(P/(L(1)+L(2)+L(3)+L(4)))//Demand factor
+DIF=((L(1)+L(2)+L(3)+L(4))/P)//Diversity factor
+
+//Output
+printf('(a) The average load on the power plant is %3.0f MW \n(b) The energy supplied per year is %3.1f *10^6 kWh \n(c) Demand factor is %3.3f \n(d) Diversity factor is %3.3f',AL,E,DF,DIF)
diff --git a/1943/CH1/EX1.3/Ex1_3.sce b/1943/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..ec1b5b1f1 --- /dev/null +++ b/1943/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,22 @@ + +clc
+clear
+//Input data
+C=210//Capacity of thermal power plant in MW
+P=160//Maximum load in MW
+L=0.6//Annual load factor
+m=1//Coal consumption per kWh of energy generated
+Rs=450//Cost of coal in Rs per tonne
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+AL=(L*P)//Average load in MW
+E=(AL*Y)//Energy generated per year in MWh
+CL=(E*1000)//Coal required per year in kg
+CY=(E*Rs)//Cost of coal per year
+CE=CL//Cost of energy sold in Rs
+RY=(CE-CY)/10^7//Revenue earned by the power plant per year in Rs crore
+CF=(AL/C)//Capacity factor
+
+//Output
+printf('(a) The annual revenue earned by the power plant is Rs %3.2f crore \n(b) Capacity factor is %3.3f',RY,CF)
diff --git a/1943/CH1/EX1.4/Ex1_4.sce b/1943/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..63fbed5a9 --- /dev/null +++ b/1943/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+L=0.75//Load factor
+C=0.60//Capacity factor
+U=0.65//Use factor
+M=60//Maximum power demand in MW
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+A=(L*M)//Average load in MW
+P=((A*1000)*Y)/10^6//Annual energy production in kWh *10^6
+PC=(A/C)//Plant capacity in MW
+R=(PC-M)//Reserve capacity in MW
+HIO=(P*1000/(U*PC))//Hours in operation in hrs
+NH=(Y-HIO)//Hours not in service in a year in hrs
+
+//Output
+printf('(a) Annual energy production is %3.1f * 10^6 kWh \n(b) Reserve capacity over and above the peak load is %3.0f MW \n(c) The hours during which the plant is not in service per year is %3.0f hrs',P,R,NH)
+
diff --git a/1943/CH1/EX1.5/Ex1_5.sce b/1943/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..e2de1847d --- /dev/null +++ b/1943/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,44 @@ + +clc
+clear
+//Input data
+D=500//Maximum demand in MW
+L=0.7//Load factor
+//1)Steam power plant 2)Hydroelectric power plant 3)Nuclear power plant
+CC=[3,4,5]//Capital cost per MW installed in Rs. crore
+I=[6,5,5]//Interest in percent
+D=[6,4,5]//Depreciation in percent
+OP=[30,5,15]//Operating cost (including fuel) per kWh
+TD=[2,3,2]//Transmission and distribution cost per kWh
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+//1)Steam power plant
+CCX=(CC(1)*D*10^7)//Capital cost in Rs
+IX=((I(1)/100)*CCX)//Interest in Rs
+DX=((D(1)/100)*CCX)//Depreciation in Rs
+AFCX=IX+DX//Annual fixed cost in Rs
+EX=(L*D*1000*Y)//Energy generated per year in kWh
+RX=(OP(1)+TD(1))//Running cost/kWh in paise
+OX=((AFCX/EX)+(RX/100))*100//Overall cost/kWh in paise
+
+//2)Hydroelectric Power plant
+CCY=(CC(2)*D*10^7)//Capital cost in Rs
+IY=((I(2)/100)*CCY)//Interest in Rs
+DY=((D(2)/100)*CCY)//Depreciation in Rs
+AFCY=IY+DY//Annual fixed cost in Rs
+EY=(L*D*1000*Y)//Energy generated per year in kWh
+RY=(OP(2)+TD(2))//Running cost/kWh in paise
+OY=((AFCY/EY)+(RY/100))*100//Overall cost/kWh in paise
+
+//3)Nuclear power plant
+CCZ=(CC(3)*D*10^7)//Capital cost in Rs
+IZ=((I(3)/100)*CCZ)//Interest in Rs
+DZ=((D(3)/100)*CCZ)//Depreciation in Rs
+AFCZ=IZ+DZ//Annual fixed cost in Rs
+EZ=(L*D*1000*Y)//Energy generated per year in kWh
+RZ=(OP(3)+TD(3))//Running cost/kWh in paise
+OZ=((AFCZ/EZ)+(RZ/100))*100//Overall cost/kWh in paise
+
+//Output
+printf('(i)Overall cost per kWh in Steam power plant is %3.0f paise \n(ii)Overall cost per kWh in Hydroelectric power plant is %3.0f paise \n(iii)Overall cost per kWh in Nuclear power plant is %3.0f paise',OX,OY,OZ)
diff --git a/1943/CH1/EX1.6/Ex1_6.sce b/1943/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..bda7572fe --- /dev/null +++ b/1943/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,25 @@ + +clc
+clear
+//Input data
+C=210//Capacity in MW
+ID=12//Interest and depreciation in percent
+CC=18000//Capital cost/kW installed in Rs
+L=0.6//Annual load factor
+AC=0.54//Annual capacity factor
+RC=(200*10^6)//Annual running charges in Rs
+E=6//Energy consumed by power plant auxiliaries in percent
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+MD=(C/L)*AC//Maximum demand in MW
+RSC=(C-MD)//Reserve Capacity in MW
+AL=(L*MD)//Average load in MW
+EP=(AL*1000*Y)//Energy produced per year in kWh
+NE=((100-E)/100)*EP//Net energy delivered in kWh
+AID=((ID/100)*CC*C*1000)//Annual interest and depreciation in Rs
+T=(AID+RC)//Total annual cost in Rs
+CP=(T/NE)*100//Cost of power generation in paise
+
+//Output
+printf('(a) The cost of power generation per kWh is %3.0f paise \n(b) The reserve capacity is %3.0f MW',CP,RSC)
diff --git a/1943/CH1/EX1.7/Ex1_7.sce b/1943/CH1/EX1.7/Ex1_7.sce new file mode 100755 index 000000000..3d4fb3143 --- /dev/null +++ b/1943/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,25 @@ + +clc
+clear
+//Input data
+L=200//The total load supplied by the plants in MW
+//The incremental fuel costs for generating units a and b of power plant are given by
+//dFa/dPa=0.065Pa+25
+//dFb/dPb=0.08Pa+20
+
+//Calculations
+//Solving two equations
+//Pa+Pb=200
+//0.065Pa+25=0.08Pb+20
+A=[1 1
+ 0.065 -0.08]//Coefficient matrix
+B=[L
+ (20-25)]//Constant matrix
+X=inv(A)*B//Variable matrix
+P=100//If load is shared equally then Pa=Pb=100MW
+a=(((0.065*P^2)/2)+(25*P))-(((0.065*X(1)^2)/2)+(25*X(1)))//increase in fuel cost for unit a in Rs. per hour
+b=(((0.08*P^2)/2)+(20*P))-(((0.08*X(2)^2)/2)+(20*X(2)))//increase in fuel cost for unit a in Rs. per hour
+x=a+b//Net increase in fuel cost due to departure from economic distribution of load in Rs. per hour
+
+//Output
+printf('(a)The economic loading of two units when the total load supplied by the power plants is 200 MW are %3.2f MW and %3.2f MW\n(b)The loss in fuel cost per hour if the load is equally shared by both units is Rs.%3.2f per hour',X(1),X(2),x)
diff --git a/1943/CH1/EX1.8/Ex1_8.sce b/1943/CH1/EX1.8/Ex1_8.sce new file mode 100755 index 000000000..a950e16a4 --- /dev/null +++ b/1943/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,26 @@ + +clc
+clear
+//Input data
+C=200//Installed capacity of the plant in MW
+CC=400//Capital cost in Rs crores
+ID=12//Rate of interest and depreciation in percent
+AC=5//Annual cost of fuel, salaries and taxation in Rs. crores
+L=0.5//Load factor
+AL2=0.6//Raised Annual load
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+AvL=(C*L)//Average Load in MW
+E=(AvL*1000*Y)//Energy generated per year in kWh
+IDC=((ID/100)*CC*10^7)//Interest and depreciation (fixed cost) in Rs
+T=(IDC+(AC*10^7))//Total annual cost in Rs
+CP1=(T/E)*100//Cost per kWh in paise
+AvL2=(C*AL2)//Average Load in MW
+E2=(AvL2*1000*Y)//Energy generated per year in kWh
+CP2=(T/E2)*100//Cost per kWh in paise
+S=((CP1)-(CP2))//Saving in cost per kWh in paise
+S1=ceil(S)//Rounding off to next higher integer
+
+//Output
+printf('Cost of generation per kWh is %3.0f paise \n Saving in cost per kWh if the annual load factor is raised to 60 percent is %3.0f paise',CP1,S1)
diff --git a/1943/CH1/EX1.9/Ex1_9.sce b/1943/CH1/EX1.9/Ex1_9.sce new file mode 100755 index 000000000..5905c9beb --- /dev/null +++ b/1943/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,18 @@ + +clc
+clear
+//Input data
+C=300//Capacity of power plant in MW
+MXD=240//Maximum demand in MW in a year
+MND=180//Minimum demand in MW in a year
+//Assuming the load duration curve shown in Figure E1.9 on page no 30 to be straight line
+Y=8760//Number of hours in a year of 365 days
+
+//Calculations
+E=((MND*Y)+0.5*(MXD-MND)*Y)*1000//Energy supplied per year in kWh
+AL=(E/Y)//Average load in kW
+L=((AL/1000)/MXD)//Load factor
+CF=((AL/1000)*Y)/(C*Y)//Capacity factor
+
+//Output
+printf('(a) Load factor is %3.3f \n(b) Capacity factor is %3.2f',L,CF)
diff --git a/1943/CH10/EX10.1/Ex10_1.sce b/1943/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..6852fb019 --- /dev/null +++ b/1943/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,22 @@ + +clc
+clear
+//Input data
+P=4000//Power in kW
+N=400//Speed in r.p.m
+h=200//Head in m
+e=90//Efficiency in percent
+d=1.5//Diameter in m
+vd=10//Percentage decrease in velocity
+a=165//Angle with which jet is deflected in degrees
+
+//Calculations
+V1=sqrt(2*9.81*h*(e/100))//Velocity in m/s
+Vb=(3.14*d*N)/60//Velocity in m/s
+nn=((2*(1-((e/100)*cosd(a)))*(V1-Vb)*Vb)/V1^2)*100//Efficiency in percent
+p=(P/(nn/100))//Power developed in kW
+pj=(p/2)//Power developed per jet in kW
+dx=sqrt((pj*8)/(3.14*V1^3))//Diameter of each jet in m
+
+//Output
+printf('(a) the efficiency of the runner is %3.2f percent \n (b) the diameter of each jet is %3.4f m',nn,dx)
diff --git a/1943/CH10/EX10.10/Ex10_10.sce b/1943/CH10/EX10.10/Ex10_10.sce new file mode 100755 index 000000000..2a17ef557 --- /dev/null +++ b/1943/CH10/EX10.10/Ex10_10.sce @@ -0,0 +1,26 @@ + +clc
+clear
+//Input data
+gh=35//Gross head in m
+md=2//Mean diameter in m
+N=145//Speed in rpm
+a=30//Angle in degrees
+oa=28//Outlet angle in degrees
+x=7//Percentage of gross head lost
+y=8//Reduction in relative velocity in percent
+
+//Calculations
+H=((100-x)/100)*gh//Net haed in m
+V1=sqrt(2*9.81*H)//Velocity in m/s
+Vb=(3.14*md*N)/60//Velocity in m/s
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Angle in degrees
+Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s
+Vr2=((100-y)/100)*Vr1//Velocity in m/s
+Vw1=(V1*cosd(a))//Velocity in m/s
+Vw2=(Vb-(Vr2*cosd(oa)))//Velocity in m/s
+E=((Vb*(Vw1-Vw2))/9.81)//Workdone in kg.m/kg
+nb=(E/gh)*100//Hydraulic efficiency in percent
+
+//Output
+printf('Blade angle at inlet is %3.0f degrees \n Hydraulic efficiency is %3.0f percent',b1,nb)
diff --git a/1943/CH10/EX10.11/Ex10_11.sce b/1943/CH10/EX10.11/Ex10_11.sce new file mode 100755 index 000000000..d7a073ed2 --- /dev/null +++ b/1943/CH10/EX10.11/Ex10_11.sce @@ -0,0 +1,25 @@ + +clc
+clear
+//Input data
+P=10000//Power in kW
+h=12//Head in m
+Nr=2//Speed ratio
+Fr=0.65//Flow ratio
+x=0.3//Diameter of hub is 0.3 times the eternal diameter of the vane
+on=94//Overall efficiency in percent
+
+//Calculations
+Q=(P/(9.81*h*(on/100)))//Discharge in m^3/s
+Vr1=(Fr*sqrt(2*9.81*h))//Velocity in m/s
+Ab=(Q/Vr1)//Area of flow in m^2
+D=sqrt(((Ab*4)/3.14)/(1-x^2))//Diameter of runner in m
+Vb=(Nr*sqrt(2*9.81*h))//Velocity in m/s
+N=((Vb*60)/(3.14*D))//Speed in rpm
+f=50//Taking frequency as 50 Hz
+p=(120*50)/N//Number of poles
+N1=(120*f)/int(p)//Speed in rpm
+Ns=(N1*sqrt(P))/h^(5/4)//Specific speed
+
+//Output
+printf('(a) the speed is %3.1f rpm \n (b) the diameter of the runner is %3.2f m \n (c) the specific speed is %3.0f',N1,D,Ns)
diff --git a/1943/CH10/EX10.12/Ex10_12.sce b/1943/CH10/EX10.12/Ex10_12.sce new file mode 100755 index 000000000..bcb9c5b8f --- /dev/null +++ b/1943/CH10/EX10.12/Ex10_12.sce @@ -0,0 +1,16 @@ + +clc
+clear
+//Input data
+P=10000//Power in kW
+h=25//Head in m. In textbook it is given wrong as 2 m
+N=135//Speed in rpm
+h1=20//Head in m
+
+//Calculations
+Ns=((N*sqrt(P))/h^(5/4))//Specific speed
+N1=sqrt(h1/h)*N//Speed in rpm
+P2=P/(h/h1)^(3/2)//Power in kW
+
+//Output
+printf('Specific speed is %3.1f \n Normal speed is %3.1f rpm \n Output under a head of %i m is %3.0f kW',Ns,N1,h1,P2)
diff --git a/1943/CH10/EX10.13/Ex10_13.sce b/1943/CH10/EX10.13/Ex10_13.sce new file mode 100755 index 000000000..68eabc0cf --- /dev/null +++ b/1943/CH10/EX10.13/Ex10_13.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+Q=175//Discharge in m^3/s
+h=18//Head in meter
+N=150//Speed in rpm
+oe=82//Overall efficiency in percent
+Ns1=460//Maximum specific speed
+Ns2=350//Maximum specific speed
+d=1000//Density in kg/m^3
+
+//Calculations
+P=(d*Q*9.81*h*(oe/100)*10^-3)//power in kW
+P1=((Ns1*h^(5/4))/N)^2//Power in kW
+n1=P/P1//No.of turbains
+P2=((Ns2*h^(5/4))/N)^2//Power in kW
+n2=ceil(P/P2)//No.of turbains
+
+//Output
+printf('The number of turbines in \n (a) Francis turbine are%3.0f \n (b) Kaplan turbine are %i',n1,n2)
diff --git a/1943/CH10/EX10.14/Ex10_14.sce b/1943/CH10/EX10.14/Ex10_14.sce new file mode 100755 index 000000000..955b1ba8a --- /dev/null +++ b/1943/CH10/EX10.14/Ex10_14.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+Ns=210//Specific speed
+P=30//Power in MW
+N=180//Speed in rpm
+Q=0.6//Discharge in m^3/s
+h=4.5//Head in m
+e=88//Efficiency in percent
+d=1000//Density in kg/m^3
+
+//Calculations
+Pm=(d*Q*9.81*h*(e/100)*10^-3)//Power in kW
+Nm=(Ns*h^(5/4))/sqrt(Pm)//Speed in rpm
+Hp=((N*sqrt(P*1000))/Ns)^(4/5)//Head in m
+Dpm=(Nm/N)*sqrt(Hp/h)//Scale ratio
+Qp=(P*10^6)/(d*9.81*Hp*(e/100))//Discharge in m^3/s
+
+//Output
+printf('Speed is %3.0f rpm \n Power is %3.3f kW \n Scale ratio is %3.3f \n Flow through the turbine is %3.1f m^3/s',Nm,Pm,Dpm,Qp)
diff --git a/1943/CH10/EX10.15/Ex10_15.sce b/1943/CH10/EX10.15/Ex10_15.sce new file mode 100755 index 000000000..f1e755894 --- /dev/null +++ b/1943/CH10/EX10.15/Ex10_15.sce @@ -0,0 +1,17 @@ + +clc
+clear
+//Input data
+x=1/5//Scale model
+h=1.5//Head in m
+P=5//Power in kW
+N=450//Speed in rpm
+h1=30//Head in m
+
+//Calculations
+N1=(x*N)/sqrt(h/h1)//Speed in rpm
+Ns=(N*sqrt(P))/h^(5/4)//Specific speed
+P1=((Ns*h1^(5/4))/N1)^2//Power in kW
+
+//Output
+printf('Speed is %3.0f rpm \n Power is %3.0f kW',N1,P1)
diff --git a/1943/CH10/EX10.16/Ex10_16.sce b/1943/CH10/EX10.16/Ex10_16.sce new file mode 100755 index 000000000..cb34ba104 --- /dev/null +++ b/1943/CH10/EX10.16/Ex10_16.sce @@ -0,0 +1,15 @@ + +clc
+clear
+//Input data
+h=19//Head in m
+Q=3//Flow rate in m^3/s
+N=600//Speed in rpm
+h1=5//Head in m
+
+//Calculations
+N1=N/sqrt(h/h1)//Speed in rpm
+Q1=Q/sqrt(h/h1)//Discharge in m^3/s
+
+//Output
+printf('Speed of the turbine is %3.1f rpm \n Maximum flow rate is %3.1f m^3/s',N1,Q1)
diff --git a/1943/CH10/EX10.17/Ex10_17.sce b/1943/CH10/EX10.17/Ex10_17.sce new file mode 100755 index 000000000..c3dcd2909 --- /dev/null +++ b/1943/CH10/EX10.17/Ex10_17.sce @@ -0,0 +1,23 @@ + +clc
+clear
+//Input data
+Q=350//Discharge in m^3/s
+h=30//Head in m
+e=87//Turbine efficiency in percent
+f=50//Frequency in Hz
+p=24//Number of poles
+Ns1=300//Specific speed
+Ns2=820//Specific speed
+d=1000//Dnsity of water in kg/m^3
+
+//Calculations
+N=(120*f)/p//Speed in rpm
+P=d*Q*9.81*h*(e/100)*10^-3//Power in kW
+P1=((Ns1*h^(5/4))/N)^2//Power in kW
+n1=P/P1//No.of turbines
+P2=((Ns2*h^(5/4))/N)^2//Power in kW
+n2=ceil(P/P2)//No.of turbines
+
+//Output
+printf('Least number of machines required if using \n (a) Francis turbines are %3.0f \n (b) Kaplan turbines are %3.0f',n1,n2)
diff --git a/1943/CH10/EX10.18/Ex10_18.sce b/1943/CH10/EX10.18/Ex10_18.sce new file mode 100755 index 000000000..cb263a3a7 --- /dev/null +++ b/1943/CH10/EX10.18/Ex10_18.sce @@ -0,0 +1,23 @@ + +clc
+clear
+//Input data
+h=27//Head in m
+A=430//Area in sq.km
+R=150//Rainfall in cm/year
+pr=65//Percentage of rainfall utilised
+pe=95//Penstock efficiency in percent
+te=80//Turbine efficiency in percent
+ge=86//Generator efficiency in percent
+lf=0.45//Load factor
+d=1000//Density of water in kg/m^3
+
+//Calculations
+Q=A*10^6*(R/100)*(pr/100)//Discharge in m^3 per year
+Qs=(Q/(365*24*3600))//Quantity of water per second in m^3
+P=(pe/100)*(te/100)*(ge/100)*d*Qs*9.81*h*10^-3//Power in kW
+plc=(P/lf)//Peak load capacity in kW
+C=(plc/(2*(ge/100)))//Capacity of each unit in kW
+
+//Output
+printf('(a) Power developed is %3.0f kW \n (b) As the available head is low, Kaplan turbines are suggested.\n Two turbines each of 3000kW capacity may be installed.',P)
diff --git a/1943/CH10/EX10.19/Ex10_19.sce b/1943/CH10/EX10.19/Ex10_19.sce new file mode 100755 index 000000000..4b1d3f522 --- /dev/null +++ b/1943/CH10/EX10.19/Ex10_19.sce @@ -0,0 +1,24 @@ + +clc
+clear
+//Input data
+q=[30,25,20,0,010,50,80,100,110,65,45,30]//Mean discharge in millions of cu.m per month respectively
+h=90//Head in m
+n=86//Overall efficiency in percent
+
+//Calculations
+Qm=(q(1)+q(2)+q(3)+q(4)+q(5)+q(6)+q(7)+q(8)+q(9)+q(10)+q(11)+q(12))/12//Mean discharge in millions m^3/s
+Q=[30,30,25,25,20,20,0,0,10,10,50,50,80,80,100,100,110,110,65,65,45,45,30,30,0]//Discharge(million m^3/month) on y-axis
+y=[0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12]//Months on x-ais
+D=[110,100,90,80,70,60,50,40,30,25,20,10,0]//Discharge per month in million m^3
+pt=[8.3,16.7,25,25,25,33.3,41.7,50,66.7,75,83.3,91.7,100]//Percentage time
+Po=((Qm*10^6*9.81*h*(n/100))/(30*24*3600*1000))//Power developed in MW
+
+//Output
+subplot(121)
+plot(y,Q)//Graph Discharge(million m^3/month) vs Month
+xtitle('Discharge(million m^3/month) vs Month','Months','Discharge(million m^3/month)')
+subplot(122)
+plot(pt,D)//Graph percentage time vs Discharge(million m^3/month)
+xtitle('percentage time vs Discharge(million m^3/month)','percentage time','Discharge(million m^3/month)')
+printf('Power developed is %3.2f MW',Po)
diff --git a/1943/CH10/EX10.2/Ex10_2.sce b/1943/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..ed4e9740f --- /dev/null +++ b/1943/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,23 @@ + +clc
+clear
+//Input data
+P=6000//Power in kW
+h=300//Net head availabe in m
+N=550//Speed in r.p.m
+rd=(1/10)//Ratio of jet diameter to wheel diameter
+nh=0.85//Hydraulic efficiency
+Cv=0.98//Coefficient of velocity
+f=0.46//Speed ratio
+d=1000//Density in kg/m^3
+
+//Calculations
+V1=Cv*sqrt(2*9.81*h)//Velocity in m/s
+Vb=f*sqrt(2*9.81*h)//Velocity in m/s
+Q=((P*10^3)/(nh*d*9.81*h))//Discharge in m^3/s
+D=((Vb*60)/(3.14*N))//Diameter in m
+d=(D/10)//Diameter of jet in m
+n=(Q/((V1*(3.14/4)*d^2)))//Number of jets
+
+//Output
+printf('(a) the number of jets are%3.0f \n (b) diameter of each jet is %3.3f m \n (c) diameter of the wheel is %3.2f m \n (d) the quantity of water required is %3.1f m^3/s',n,d,D,Q)
diff --git a/1943/CH10/EX10.3/Ex10_3.sce b/1943/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..4c29ad4d6 --- /dev/null +++ b/1943/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+P=10//Capacity in MW
+h=500//Head in m
+Ns=10//Specific speed of the turbine
+on=80//Overall efficiency in percent
+Cv=0.98//Coefficient of velocity
+x=0.46//Speed of the bucket wheel to the velocity of jet
+da=1000//Density in kg/m^3
+
+//Calculations
+N=(Ns*h^(5/4))/sqrt(P*10^3)//Speed in r.p.m
+V=(Cv*sqrt(2*9.81*h))//Velocity in m/s
+Vb=(x*V)//Speed of bucket wheel in m/s
+D=((60*Vb)/(3.14*N))//Diameter in m
+d=sqrt((P*10^6)/((on/100)*(3.14/4)*da*V*9.81*h))//Diameter in m
+
+//Output
+printf('Diameter of jet is %3.3f m \n Diameter of bucket wheel is %3.2f m',d,D)
diff --git a/1943/CH10/EX10.4/Ex10_4.sce b/1943/CH10/EX10.4/Ex10_4.sce new file mode 100755 index 000000000..312867c29 --- /dev/null +++ b/1943/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,15 @@ + +clc
+clear
+//Input data
+Cv=0.97//Coefficient of velocity
+f=0.45//Friction coefficient
+h=0.85//Head in m
+d=1000//Density in kg/m^3
+n=1//For a single jet turbine
+
+//Calculations
+Ns=((60/3.14)*(f*sqrt(2*9.8))*sqrt(n*(3.14/4)*Cv*sqrt(2*9.8)*9.8*h))//Specific speed in terms of d/D
+
+//Output
+printf('The specific speed of a single jet Pelton wheel is about %3.0f (d/D) where d and D represent the jet and bucket wheel diameters respectively',Ns)
diff --git a/1943/CH10/EX10.5/Ex10_5.sce b/1943/CH10/EX10.5/Ex10_5.sce new file mode 100755 index 000000000..ce36206d8 --- /dev/null +++ b/1943/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,19 @@ + +clc
+clear
+//Input data
+n=4//Number of jets
+d=60//Diameter of each jet in mm
+a=165//Angle in degrees
+v=45//Speed of the bucket wheel in m/s
+de=1000//Density in kg/m^3
+
+//Calculations
+v1=(2*v)//Jet velocity in m/s
+Q=(3.14/4)*(d/1000)^2*v1//Discharge in m^3/s
+P=(1-cosd(a))*(v1^2/4)*Q*de*10^-3//Power developed in kW
+P4=(P*4)//For four jets in kW
+nd=((1-cosd(a))/2)*100//Maximum efficiency in percent
+
+//Output
+printf('Velocity of the jet for maximum efficiency is %3.0f m/s \n Power developed is %i kW \n Hydraulic efficiency is %3.1f percent',v1,P4,nd)
diff --git a/1943/CH10/EX10.6/Ex10_6.sce b/1943/CH10/EX10.6/Ex10_6.sce new file mode 100755 index 000000000..ced9a9bdd --- /dev/null +++ b/1943/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,19 @@ + +clc
+clear
+//Input data
+v=20//Peripheral velocity in m/s
+vw=17//Velocity of whirl in m/s
+vr=2//Radial velocity in m/s
+Q=0.7//Flow in m^3/s
+hn=80//Hydraulic efficiency in percent
+d=1000//Density in kg/m^3
+
+//Calculations
+H=((vw*v)/(9.81*(hn/100)))//Head on the wheel in m
+P=(d*Q*9.81*H*(hn/100)*10^-3)//Power generated in kW
+al=180-atand(vr/vw)//Angle of guide vanes in degrees
+bl=atand(vr/(v-vw))//Inlet blade angle in degrees
+
+//Output
+printf('Head on the wheel is %3.1f m \n The power generated by the turbine is %3.0f kW \n Eit angle of guide vanes is %3.2f degrees and Inlet blade angle is %3.1f degrees',H,P,al,bl)
diff --git a/1943/CH10/EX10.7/Ex10_7.sce b/1943/CH10/EX10.7/Ex10_7.sce new file mode 100755 index 000000000..2d0d4a10b --- /dev/null +++ b/1943/CH10/EX10.7/Ex10_7.sce @@ -0,0 +1,27 @@ + +clc
+clear
+//Input data
+od=1.5//Outer diameter in m
+id=0.75//Inner diameter in m
+h=150//Head in m
+P=14000//Power in kW
+Ns=120//Specific speed
+vw2=0//Velocity in m/s
+a=(11+(20/60))//Angle in degrees
+hn=92//Hydraulic efficiency in percent
+
+//Calculations
+N=(Ns*h^(5/4))/sqrt(P)//Speed in rpm
+vb1=(3.14*od*N)/60//velocity in m/s
+vw1=(((hn/100)*9.81*h)/vb1)//velocity in m/s
+vf1=(tand(a)*vw1)//Velocity in m/s
+vf2=vf1//Velocity in m/s
+b1=atand(vf1/(vb1-vw1))//Angle in degrees
+b1x=(b1-int(b1))*60//For output
+vb2=(vb1/2)//Velocity in m/s
+b2=atand(vf1/(vb2-vw2))//Angle in degrees
+b2x=(b2-int(b2))*60//For output
+
+//Output
+printf('Inlet blade angle is %3.0f degrees %3.0f minutes \n Outlet blade angle is %3.0f degrees %3.0f minute',b1,b1x,b2,b2x)
diff --git a/1943/CH10/EX10.8/Ex10_8.sce b/1943/CH10/EX10.8/Ex10_8.sce new file mode 100755 index 000000000..e2a597ba7 --- /dev/null +++ b/1943/CH10/EX10.8/Ex10_8.sce @@ -0,0 +1,27 @@ + +clc
+clear
+//Input data
+h=70//net head in m
+N=700//speed in rpm
+o=85//over all efficiency in %
+P=350//shaft power in kW
+he=92//hydraulic efficiency in %
+fr=.22//flow ratio
+b=.1//breadth ratio
+s=2//outer diameter in terms of inner diametre
+//Calculations
+vf1=fr*sqrt(2*9.81*h)//velocity in m/s
+q=(P/(9.81*h*(o/100)))//discharge in m^3/s
+d1=sqrt(q/(.94*b*vf1*3.14))//diameter in metre
+b1=d1*b//breadth in metre
+d2=d1/2//diametre in metre
+vb1=(3.14*d1*N)/60//velocity in m/s
+vw1=((he/100)*9.81*h)/vb1//velcity in m/s
+a=atand(vf1/vw1)//angle in degrees
+bet=atand(vf1/(vw1-vb1))//angle in degrees
+vb2=(d2/d1)*vb1//velocity in m/s
+bet2=atand(vf1/vb2)//angle in degrees
+
+//Output
+printf('(a)the guide vane angle is %3.1f degrees \n (b)the runner vane angle at inlet is %3.1f degrees and outlet is %3.2f degrees \n (c)the diametres of the runner at inlet is %3.1f metre and outlet is %3.2f metre\n (d)the width of the wheel at inlet is %3.2f metre',a,bet,bet2,d1,d2,b1)
diff --git a/1943/CH10/EX10.9/Ex10_9.sce b/1943/CH10/EX10.9/Ex10_9.sce new file mode 100755 index 000000000..12ff71e5e --- /dev/null +++ b/1943/CH10/EX10.9/Ex10_9.sce @@ -0,0 +1,40 @@ + +clc
+clear
+//Input data
+n=4//Number of units
+P=70000//Power in kVA
+f=50//Frequency in Hz
+p=10//No.of pair of poles
+h=505//Gross head in m
+tn=94//Transmission efficiency in percent
+po=260//Power in MW
+e=91//Efficiency in percent
+nn=0.98//Nozzle efficiency
+Cv=0.98//Coefficient of velocity
+x=0.48//Vb=0.48 V
+dd=25//Nozzle diameter is 25% bigger than jet diameter
+a=165//Angle of buckets in degrees
+de=99.75//Discharge efficiency in percent
+
+//Calculations
+N=(120*f)/(p*2)//Synchronous speed in r.p.m
+nh=((tn/100)*h)//Net head in m
+pt=(po*10^3)/n//Power developed per turbine in kW
+ip=(pt/(e/100))//Input water power in kW
+Q=(ip/(9.81*nh))//Discharge in m^3/s
+Qj=(Q/n)//Discharge per jet in m^3/s
+V1=Cv*sqrt(2*9.81*nh)//Velocity in m/s
+d=sqrt((4/3.14)*(Qj/V1))//Diameter of jet in m
+nd=(1+(dd/100))*d//Nozzle tip diameter in m
+Vb=(x*V1)//Velocity in m/s
+D=((Vb*60)/(3.14*N))//Pitch circle diameter of the wheel in m
+Ns=((N*sqrt(po*10^3))/nh^(5/4))//Specific speed
+jr=(D/d)//Jet ratio
+nob=(jr/2)+15//Number of buckets
+nobb=ceil(nob)//Rounding off to next integer
+W=((V1-Vb)*(1-(nn*cosd(a)))*Vb)/9.81//Workdone per kg in kg.m/kg
+nth=((W/nh)*de)//Hydraulic efficiency in percent
+
+//Output
+printf('(a) the discharge of the turbine is %3.2f m^3/s \n (b) the jet diameter is %3.3f m \n (c) the nozzle tip diameter is %3.3f m \n (d) the pitch circle diameter of the wheel is %3.2f m \n (e) the specific speed is %3.2f \n (f) the number of buckets on the wheel are %3.0f \n (g) the workdone per kg of water on the wheel is %3.2f kg.m/kg \n (h) the hydraulic efficiency is %3.0f percent',Q,d,nd,D,Ns,nobb,W,nth)
diff --git a/1943/CH11/EX11.1/Ex11_1.sce b/1943/CH11/EX11.1/Ex11_1.sce new file mode 100755 index 000000000..c10570883 --- /dev/null +++ b/1943/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,39 @@ + +clc
+clear
+//Input data
+C=3.5//Capacity in litres
+P=13.1//Indicated power in kW/m^3
+N=3600//Speed in rpm
+ve=82//Volumetric efficiency in percent
+p1=1.013//Pressure in bar
+T1=25+273//Temperature in K
+rp=1.75//Pressure ratio
+ie=70//Isentropic efficiency in percent
+me=80//Mechanical efficiency in percent
+g=1.4//Ratio of specific heats
+R=0.287//Gas constant in kJ/kg.K
+Cp=1.005//Specific heat in kJ/kg.K
+
+//Calculations
+EC=(C/1000)//Engine capacity in m^3
+Vs=(N/2)*EC//Swept volume in m^3
+Vui=(ve/100)*Vs//Unsupercharged induced volume in m^3/min
+dp=(rp*p1)//Blower delivery pressure in bar
+T2sT1=(rp)^((g-1)/g)//Ratio of temperatures
+T2s=(T2sT1*T1)//Temperature in K
+dT21=(T2s-T1)/(ie/100)//Difference in temperature in K
+T2=dT21+T1//Temperature in K
+EV=(Vs*dp*T1)/(p1*T2)//Equivalent volume in m^3/min
+iiv=EV-Vui//Increase in induced volume in m^3/min
+iip=(P*iiv)//Increase in indicated power in kW
+iipi=((dp-p1)*100*Vs)/60//Increase in induced power due to increase in induction pressure in kW
+tiip=iip+iipi//Total increase in indicated power in kW
+tibp=tiip*(me/100)//Total increase in brake power in kW
+ma=(dp*100*Vs)/(60*R*T2)//Mass of air in kg/s
+WI=(ma*Cp*(T2-T1))//Work input to heater in kW
+Pb=(WI/(me/100))//Power required in kW
+NI=tibp-Pb//Net increase in brake power in kW
+
+//Output
+printf('Net increase in brake power is %3.2f kW',NI)
diff --git a/1943/CH11/EX11.2/Ex11_2.sce b/1943/CH11/EX11.2/Ex11_2.sce new file mode 100755 index 000000000..081dcf9db --- /dev/null +++ b/1943/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,32 @@ + +clc
+clear
+//Input data
+p1=0.97//Pressure in bar
+t1=30+273//Temperature in K
+p2=2.1//Pressure in bar
+af=18//Air fuel ratio
+t3=580+273//Temperature in K
+p3=1.9//Pressure in bar
+p4=1.06//Pressure in bar
+iec=0.75//Isentropic efficiency of compressor
+iet=0.85//Isentropic efficiency of turbine
+cpa=1.01//Specific heat for air in kJ/kg.K
+ga=1.4//Ratio of specific heats
+cpex=1.15//Specific heat in kJ/kg.K
+gex=1.33//Ratio of specific heats
+
+//Calculations
+t2s=t1*(p2/p1)^((ga-1)/ga)//Tempeature in K
+t21=(t2s-t1)/iec//Temperature in K
+t2=t21+t1//Temperature in K
+T2=t2-273//Temperature in degree C
+t3t4s=(p3/p4)^((gex-1)/gex)//Ratio of temperatures
+t4s=(t3/t3t4s)//Temperature in K
+t4=t3-((t3-t4s)*iet)//Temperature in K
+T4=t4-273//Temperature in degree C
+mp=(((cpex*(1+(1/af))*(t3-t4))-(cpa*(t2-t1)))/(cpex*(1+(1/af))*(t3-t4)))*100//Percentage of mechanical power loss
+
+//Output
+printf('(a) the temperature of air leaving the compressor is %3.2f degree C \n (b) the temperature of gases leaving the turbine is %3.2f degree C \n (c) the mechanical power loss in the turbocharger as a percentage of the power generated in the turbine is %3.2f percent',T2,T4,mp)
+
diff --git a/1943/CH11/EX11.3/Ex11_3.sce b/1943/CH11/EX11.3/Ex11_3.sce new file mode 100755 index 000000000..92f75f60c --- /dev/null +++ b/1943/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,33 @@ + +clc
+clear
+//Input data
+a=215//Current in A
+v=210//Voltage in V
+e=85//Efficiency in percent
+q=11.8//Quantity of fuel supplied in kg/h
+cv=43//Calorific value in MJ/kg
+af=18//Air fuel ratio
+w=560//Water in litres/h
+tw=38//Temeparature in degree C
+te=97//Temeparature in degree C
+cp=1.04//Specific heat in kJ/kg.K
+ta=30//Temeparature in degree C
+l=32//Percentage lost
+sw=4.187//Specific heat in kJ/kg.K
+
+//Calculations
+P=(a*v)/1000//Power in kW
+BP=(P/(e/100))//Brake power in kW
+E=(q/3600)*cv*1000//Energy supplied in kW
+mg=(q/3600)*(1+af)//Rate of gases in kg/s
+he=(mg*cp*(te-ta))+((w/3600)*sw*tw)//Heat carried away by exhaust gases in kW
+hj=(l/100)*E//Heat lost to jacket cooling water in kW
+pBP=(BP/E)*100//Percentage
+pE=(E/E)*100//Percentage
+phe=(he/E)*100//Percenatge
+phj=(hj/E)*100//Percenatge
+
+//Output
+printf(' ENERGY BALANCE SHEET \n (in kW) (in percent)\n 1. Brake power %3.2f %3.2f \n 2. Heat carried away by exhaust gases %3.2f %3.2f \n 3. Heat lost to jacket cooling water %3.2f %3.2f \n 4. Heat loss unaccounted %3.2f %3.2f \n Total %3.2f %3.2f',BP,pBP,he,phe,hj,phj,(E-(BP+he+hj)),(((E-(BP+he+hj))/E)*100),E,(pBP+phe+phj+(((E-(BP+he+hj))/E)*100)))
+
diff --git a/1943/CH11/EX11.4/Ex11_4.sce b/1943/CH11/EX11.4/Ex11_4.sce new file mode 100755 index 000000000..44a474513 --- /dev/null +++ b/1943/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,45 @@ + +clc
+clear
+//Input data
+t=20//Trial time in minutes
+NL=680//Net brake load in N
+mep=3//Mean effective pressure in bar
+N=360//Speed in rpm
+Fc=1.56//Fuel consumption in kg
+cw=160//Cooling water in kg
+Tw=32//Temperature of water at inlet in degree C
+Wo=57//Water outlet temperature in degree C
+a=30//Air in kg
+Ta=27//Room temperature in degree C
+Te=310//Exhaust gas temperature in degree C
+d=210//Bore in mm
+l=290//Stroke in mm
+bd=1//Brake diameter in m
+cv=44//Calorific value in MJ/kg
+st=1.3//Steam formed in kg per kg fuel in the exhaust
+cp=2.093//Specific heat of steam in exhaust in kJ/kg.K
+cpx=1.01//Specific heat of dry exhaust gases in kJ/kg.K
+cpw=4.187//Specific heat of water in kJ/kg.K
+
+//Calculations
+ip=(mep*100*(l/1000)*(3.14/4)*(d/1000)^2*N)/60//Indicated Power in kW
+bp=((2*3.14*N*(NL*(1/2)))/60)/1000//Brake power in kW
+nm=(bp/ip)*100//Mechanical efficiency in percent
+qs=(Fc*cv*10^3)//Heat supplied in kJ
+qip=(ip*t*60)//Heat equivalent of ip in kJ
+qcw=(cw*cpw*(Wo-Tw))//Heat carried away by cooling water in kJ
+tm=(Fc*a)//Toatl mass of exhaust gas in kg
+ms=(st*Fc)//Mass of steam formed in kg
+mde=(tm-ms)//Mass of dry exhaust gas in kg
+Ed=(mde*cpx*(Te-Ta))//Energy carried away by dry exhaust gases in kJ
+Es=(ms*((cpw*(100-Ta))+2257.9+(cp*(Te-100))))//Energy carried away by steam in kJ
+TE=(Ed+Es)//Total energy carried away by exhaust gases in kJ
+ue=(qs-(qip+qcw+TE))//Unaccounted energy in kJ
+pqip=(qip/qs)*100//Percentage
+pqcw=(qcw/qs)*100//Percentage
+pTE=(TE/qs)*100//Percentage
+pue=(ue/qs)*100//Percentage
+
+//Output
+printf('Indicated power is %3.2f kW \n Brake power is %3.3f kW \n\n ENERGY BALANCE SHEET \n (in kJ) (in percent)\n 1. Energy equivalent in ip %3.0f %3.2f \n 2. Energy carried away by cooling water %3.0f %3.2f \n 3. Energy carried away by exhaust gases %3.0f %3.2f \n 4. Unaccounted for energy loss %3.0f %3.2f \n Total %3.0f %3.2f',ip,bp,qip,pqip,qcw,pqcw,TE,pTE,ue,pue,qs,(pqip+pqcw+pTE+pue))
diff --git a/1943/CH11/EX11.5/Ex11_5.sce b/1943/CH11/EX11.5/Ex11_5.sce new file mode 100755 index 000000000..c5833f410 --- /dev/null +++ b/1943/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,38 @@ + +clc
+clear
+//Input data
+Vbm=360//Blade velocity in m/s
+b1=20//Blade angle at inlet in degrees
+a2=b1//Angle in degrees
+b2=52//Blade angle at exit in degrees
+a1=b2//Angle in degrees
+R=50//Degree of reaction in percent
+dm=0.45//Mean diameter of the blade in m
+bh=0.08//Mean blade height in m
+
+//Calculations
+Vf=(Vbm/(tand(b2)-tand(b1)))//Velocity in m/s
+rt=(dm/2)+(bh/2)//Mean radius in m
+Vbt=(Vbm*(rt/(dm/2)))//Velocity in m/s
+Vw1m=Vf*tand(a1)//Velocity in m/s
+Vw1t=(Vw1m*((dm/2)/rt))//Velocity in m/s
+dVw1=(Vf*(tand(b1)+tand(b2))*Vbm)/Vbt//Velocity in m/s
+rr=(dm/2)-(bh/2)//Radius in m
+Vbr=(Vbm*(rr/(dm/2)))//Velocity in m/s
+Vw1r=(Vw1m*((dm/2)/rr))//Velocity in m/s
+Vr2=Vf/cosd(b2)//Velocity in m/s
+dVwr=((Vw1m+((Vr2*sind(b2))-Vbm))*Vbm)/Vbr//Velocity in m/s
+a1r=atand(Vw1r/Vf)//Angle in degrees
+a2r=atand((dVwr-Vw1r)/Vf)//Angle in degrees
+b1r=atand((Vw1r-Vbr)/Vf)//Angle in degrees
+b2r=atand((Vbr+(Vf*tand(a2r)))/Vf)//Angle in degrees
+a1t=atand(Vw1t/Vf)//Angle in degrees
+a2t=atand((dVw1-Vw1t)/Vf)//Angle in degrees
+b1t=atand((Vw1t-Vbt)/Vf)//Angle in degrees
+b2t=atand((Vbt+(Vf*tand(a2t)))/Vf)//Angle in degrees
+Rt=((Vf*(tand(b2t)-tand(b1t)))/(2*Vbt))*100//Degree of reaction at the tip in percent
+Rr=((Vf*(tand(b2r)-tand(b1r)))/(2*Vbr))*100//Degree of reaction at the root in percent
+
+//Output
+printf('(a)The flow velocity is %3.0f m/s \n (b) The blade angles at the tip are : \n Fixed blades (root) are %3.2f degrees and %3.2f degrees \n Moving blades (root) are %3.2f degrees and %3.2f degrees \n Fixed blades (tip) are %3.2f degrees and %3.2f degrees \n Moving blades (tip) are %3.2f degrees and %3.2f degrees \n (c) The degree of reaction at : \n the tip is %3.0f percent \n the root is %3.0f percent',Vf,a1r,a2r,b1r,b2r,a1t,a2t,b1t,b2t,Rt,Rr)
diff --git a/1943/CH11/EX11.6/Ex11_6.sce b/1943/CH11/EX11.6/Ex11_6.sce new file mode 100755 index 000000000..f91afb86d --- /dev/null +++ b/1943/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,28 @@ + +clc
+clear
+//Input data
+N=16000//Speed in rpm
+T1=17+273//Temperature in K
+rp=4//Pressure ratio
+in=82//Isentropic efficiency in percent
+s=0.85//Slip factor
+a=20//Angle in degrees
+d=200//Diameter in mm
+V=120//Velocity in m/s
+cp=1.005//Specific heat in kJ/kg.K
+g=1.4//Ratio of specific heats
+
+//Calculations
+T2sT1=(rp)^((g-1)/g)//Temperature ratio
+T2s=T1*T2sT1//Temeprature in K
+dTs=(T2s-T1)//Temperature difference in K
+dT=dTs/(in/100)//Temperature difference in K
+Wc=(cp*dT)//Power input in kJ/kg
+Vb1=(3.14*(d/1000)*N)/60//Velocity in m/s
+Vw1=(V*sind(a))//Pre-whirl velocity in m/s
+Vb2=sqrt(((Wc*1000)+(Vb1*Vw1))/s)//Velocity in m/s
+d2=((Vb2*60)/(3.14*N))*1000//Tip diameter in mm
+
+//Output
+printf('Impeller tip diameter is %3.0f mm',d2)
diff --git a/1943/CH11/EX11.7/Ex11_7.sce b/1943/CH11/EX11.7/Ex11_7.sce new file mode 100755 index 000000000..89a575331 --- /dev/null +++ b/1943/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,30 @@ + +clc
+clear
+//Input data
+T1=25+273//Temperature in K
+rp=6//Pressure ratio
+Vb=220//Mean velocity in m/s
+b1=45//Angle in degrees
+a2=b1//Angle in degrees
+b2=15//Angle in degrees
+a1=b2//Angle in degrees
+R=50//Degree of reaction in percent
+n=10//Number of stages
+in=83//Isentropic efficiency in percent
+cp=1.005//Specific heat in kJ/kg.K
+g=1.4//Ratio of specific heats
+
+//Calculations
+V1=(Vb/(sind(b2)+(cosd(a1)*tand(a2))))//Velocity in m/s
+V2=(V1*cosd(b2))/cosd(b1)//Velocity in m/s
+dVw=(V2*sind(a2))-(V1*sind(a1))//Velocity in m/s.Textbook answer is wrong. Correct answer is 127 m/s
+T2sT1=rp^((g-1)/g)//Temperature ratio
+T2s=(T2sT1*T1)//Temperature in K
+dTs=(T2s-T1)//Temperature difference in K
+dT=(dTs/(in/100))//Temperature difference in K
+W=(cp*dT)//Workdone in kJ/kg
+w=(W*10^3)/(Vb*dVw*n)//Work done factor
+
+//Output
+printf('Workdone factor of the compressor is %3.2f',w)
diff --git a/1943/CH11/EX11.8/Ex11_8.sce b/1943/CH11/EX11.8/Ex11_8.sce new file mode 100755 index 000000000..3cfa76979 --- /dev/null +++ b/1943/CH11/EX11.8/Ex11_8.sce @@ -0,0 +1,43 @@ + +clc
+clear
+//Input data
+p1=1//Pressure in bar
+T1=20+273//Temperature in K
+Tm=900+273//Maximum temperature in K
+rp=6//Pressure ratio
+e=0.7//Effectiveness of regenerator
+ma=210//Rate of air flow in kg/s
+CV=40800//Calorific value in kJ/kg
+ic=0.82//Isentropic efficiencies of both the compressors
+it=0.92//Isentropic efficiencies of both the turbine
+cn=0.95//Combustion efficiency
+mn=0.96//Mechanical efficiency
+gn=0.95//Generator efficiency
+cp=1.005//Specific heat of air in kJ/kg.K
+cpg=1.08//Specific heat of gas in kJ/kg.K
+g1=1.4//Ratio of specific heats for air
+g=1.33//Ratio of specific heats for gas
+
+//Calculations
+pi=sqrt(p1*rp)//Intermediate pressure in bar
+T2sT1=(pi/p1)^((g1-1)/g1)//Temperature ratio
+T2s=(T2sT1*T1)//temperature in K
+T2=((T2s-T1)/ic)+T1//Temperature in K
+T4s=(T1*(rp/pi)^((g1-1)/g1))//Temperature in K
+T4=((T4s-T1)/ic)+T1//Temperature in K
+T7s=(Tm/(rp/p1)^((g-1)/g))//Temperature in K
+T7=Tm-(it*(Tm-T7s))//Temperature in K
+T5=(e*(T7-T4))+T4//Temperature in K
+mf=1/((cp*(Tm-T5))/((CV*cn)-(cp*(Tm-T5))))//Air fuel ratio
+Wgt=((1+(1/mf))*cpg*(Tm-T7))//Workdone by turbine in kJ/kg of air
+Wc=(cp*((T2-T1)+(T4-T1)))//Workdone by compressor in kJ/kg of air
+Wnet=(Wgt-Wc)//Net workdone in kJ/kg of air
+Q=(CV*cn)/mf//Heat supplied in kJ/kg of air
+ncy=(Wnet/Q)*100//Cycle efficiency in percent
+PO=(Wnet*ma*mn*gn)/10^3//Power output in MW
+Fc=(ma*3600*(1/mf))//Fuel consumption per hour in kg
+SFC=(Fc/(PO*10^3))//Specific fuel consumption in kg/kW.h
+
+//Output
+printf('(a) the air fuel ratio is %3.2f \n (b) the cycle efficiency is %3.1f percent \n (c) the power supplied by the plant is %3.0f MW \n (d) the specific fuel consumption of the plant is %3.3f kg/kW.h and the fuel consumption per hour is %3.2f kg',mf,ncy,PO,SFC,Fc)
diff --git a/1943/CH12/EX12.1/Ex12_1.sce b/1943/CH12/EX12.1/Ex12_1.sce new file mode 100755 index 000000000..ac0483950 --- /dev/null +++ b/1943/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,29 @@ + +clc
+clear
+//Input data
+V=64000//Volume in m^3
+Q=8300//Discharge in m^3/hr
+p1=1//Pressure in bar
+T1=20+273//Temperature in K
+p2=100//Pressure in bar
+pn=70//Polytropic efficiency in percent
+pt=60//Peaking turbine efficiency in percent
+g=1.4//Ratio of specific heats
+cp=1.005//Specific heat in kJ/kg.K
+R=0.287//Gas constant in kJ/kg.K
+
+//Calculations
+T2sT1=(p2/p1)^((g-1)/g)//Temperature ratio
+T2s=(T1*T2sT1)//Temperature in K
+T21=(T2s-T1)/(pn/100)//Difference in Temperatures in K
+T2=(T21+T1)-273//Temperature in degree C
+v=(R*T1)/(p2*100)//Specific volume in m^3/kg
+mf=(Q/(v*3600))//Mass flow rate in kg/s
+E=(mf*cp*T21)/1000//Rate of energy storage in MW
+t=(V/Q)//Storage time in hour
+tE=(E*t)//Total energy storage in MWh
+Ed=(tE*(pt/100))//Total energy delivered by the peaking turbine in MWh
+
+//Output
+printf('(a) the compressed sir temperature is %3.0f degree C \n (b) the storage time is %3.2f hour \n (c) the total energy storage is %3.0f MWh \n (d) the total energy delivered by the peaking turbine is %3.0f MWh',T2,t,tE,Ed)
diff --git a/1943/CH12/EX12.2/Ex12_2.sce b/1943/CH12/EX12.2/Ex12_2.sce new file mode 100755 index 000000000..7baf60e2b --- /dev/null +++ b/1943/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,32 @@ + +clc
+clear
+//Input data
+V=175000//Volume in m^3
+d=4//Diameter in m
+U=1.5//Overall heat transfer coefficient in W/m^2.K
+p2=2//Pressure in bar
+p1=20//Pressure in bar
+Ta=20//Ambient temperature in degree C
+cp=4.35//Specific heta of water in kJ/kg.K
+e=96//Efficiency in percent
+ppe=25//Peaking plant efficiency in percent
+
+//Calculations
+//At 20 bar
+T1=212.37//Saturation temperature in degree C
+hf1=908.5//Enthalpy in kJ/kg
+vf1=0.0011766//Specific volume in m^3/kg
+//At 2 bar
+T2=120.23//Saturation temperature in degree C
+hf2=504.8//Enthalpy in kJ/kg
+vf2=0.0010605//Specific volume in m^3/kg
+ad=(1/2)*((1/vf1)+(1/vf2))//Average density of water in kg/m^3
+tc=(d*ad*cp*1000)/(4*U*3600)//Time constant in h
+ts=(log(1/(1-((1-((e/100)))/((T1-Ta)/(T1-T2))))))*tc//Storage time in h
+m=(V/vf1)//Mass of water needed in kg
+E=(m*(hf1-hf2))/(3600*10^3)//Total energy stored in MWh
+Ed=(E*(e/100)*(ppe/100))//Energy delivered in MWh
+
+//Output
+printf('(a) the storage time is %3.3f h \n (b) the total energy stored in the accumulator is %3.1f MWh \n (c) the total energy that can be delivered by the peaking turbine is %3.2f MWh',ts,E,Ed)
diff --git a/1943/CH2/EX2.1/Ex2_1.sce b/1943/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..bff6595c3 --- /dev/null +++ b/1943/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,37 @@ + +clc
+clear
+//Input data
+p1=40//Initial pressure of steam in bar
+T1=500//Initial temperature of steam in degree C
+m1=5500//Rate of steam in kg/h
+p2=2//Pressure of steam after expansion in bar
+n1=0.83//Isentropic efficiency
+q=0.87//Quality
+m2=2700//Mass flow rate in kg/h
+p3=0.1//Pressure of steam after expansion in l.p turbine in bar
+n2=0.78//Isentropic efficiency
+
+//Calculations
+h1=3445.3//Enthalpy in kJ/kg
+s1=7.0901//Entropy in kJ/kg.K which is 1.5301+x2s*5.5970
+x2s=(5.5600/5.5970)//dryness fraction
+h2s=(504.7+(x2s*2201.9))//Enthalpy in kJ/kg
+h2=h1-(n1*(h1-h2s))//Enthalpy in kJ/kg
+h3=(504.7+(q*2201.9))//Enthalpy in kJ/kg
+h4=((m2*h3+m1*h2)/(m1+m2))//Enthalpy in kJ/kg
+x4=(2183.78/2201.9)//dryness fraction
+s4=(1.5301+x4*5.5970)//Entropy in kJ/kg.K
+x5s=0.8574//dryness fraction
+h5s=(191.84+x5s*2392.5)//Enthalpy in kJ/kg
+dh4h5=(n2*(h4-h5s))//Difference in enthalpy (h4-h5) in kJ/kg
+h6=191.83//Enthalpy in kJ/kg
+W1=((m1*(h1-h2))+((m1+m2)*dh4h5))/3600//Power output of the plant in kW
+Q1=(m1*(h1-h6))/3600//Heat input in kW
+n1=(W1/Q1)*100//Efficiency in percent
+WT=(m1*(h1-h2))/3600//Power output without the geothermal heat supply in kW
+Q2=(m1*(h1-h6))/3600//Heat input without the geothermal heat supply in kW
+n2=(WT/Q2)*100//Efficiency of the cycle without the geothermal heat supply in percent
+
+//Output
+printf('(a)Power output of the cycle is %3.1f kW \n Efficiency of the cycle is %3.1f percent \n\n (b)Without geothermal heat supply \n Power output of the cycle is %3.2f kW \n Efficiency of the cycle is %3.2f percent',W1,n1,WT,n2)
diff --git a/1943/CH2/EX2.2/Ex2_2.sce b/1943/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..7acb57e2b --- /dev/null +++ b/1943/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,46 @@ + +clc
+clear
+//Input data
+p1=90//Initial pressure of steam in bar
+T1=500//Initial temperature of steam in degree C
+O=(500*1000)//Output in kW
+T2=40//Condensation temperature in degree C
+nhp=0.92//Efficiency of h.p turbine
+nlp=0.9//Efficiency of l.p turbine
+np=0.75//Isentropic efficiency of the pump
+TTD=-1.6//Temperature in degree C
+
+//Calculations
+p2=(0.2*p1)//Optimum reheat pressure in bar
+h1=3386.1//Enthalpy in kJ/kg
+s1=6.6576//Entropy in kJ/kg.K
+s2s=s1//Entropy in kJ/kg.K
+h2s=2915//Enthalpy in kJ/kg
+h3=3469.8//Enthalpy in kJ/kg
+s3=7.4825//Entropy in kJ/kg.K
+x4s=(s3-0.5725)/7.6845//Dryness fraction
+h4s=(167.57+x4s*2406.7)//Enthalpy in kJ/kg
+h5=167.57//Enthalpy in kJ/kg
+h7=883.42//Enthalpy in kJ/kg
+Wps=(0.001008*p1*10)//Workdone by the pump in kJ/kg
+h6s=176.64//Enthalpy in kJ/kg
+dh1h2=(nhp*(h1-h2s))//Difference in enthalpy (h1-h2) in kJ/kg
+h2=h1-dh1h2//Enthalpy in kJ/kg
+dh3h4=(nlp*(h3-h4s))//Difference in enthalpy (h3-h4) in kJ/kg
+h4=h3-dh3h4//Enthalpy in kJ/kg
+Wp=(Wps/np)//Workdone by the pump in kJ/kg
+h6=(Wp+h5)//Enthalpy in kJ/kg
+tsat=207.15//Saturation temperature at 18 bar in degree C
+t9=(tsat-TTD)//Temperature in degree C
+h9=875//Enthalpy in kJ/kg
+m=((h9-h6)/(h2-h7))//Mass of steam in kg
+WT=(dh1h2+(1-m)*dh3h4)//Workdone by the turbine in kJ/kg
+Wnet=(WT-Wp)//Net workdone in kJ/kg
+ws=(O/Wnet)//Mass flow rate of steam at turbine inlet in kg/s
+Q1=((h1-h9)+(1-m)*(h3-h2))//Heat input in kJ/kg
+n=(Wnet/Q1)*100//Efficiency of the cycle in percent
+Wr=(Wnet/WT)//Work ratio
+
+//Output
+printf('(a)Mass flow rate of steam at turbine inlet is %3.0f kg/s \n (b)The cycle efficiency is %3.2f percent \n (c)Work ratio is %3.3f',ws,n,Wr)
diff --git a/1943/CH2/EX2.3/Ex2_3.sce b/1943/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..6a736cc07 --- /dev/null +++ b/1943/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,29 @@ + +clc
+clear
+//Input data
+p1=70//Pressure at which an ideal seam power plant operates in bar
+T1=550//Temperature at which an ideal seam power plant operates in degrees C
+p2=0.075//Pressure at which an ideal seam power plant operates in bar
+
+//Calculations
+TB=285.9//Saturation temperature at 70 bar in degree C
+TC=40.3//Saturation temperature at 0.075 bar in degree C
+Tr=(TB-TC)/(7+1)//Temperature rise per heater for maximum cycle efficiency in degree C
+t1=(TB-Tr)//Temperature at heater 1 in degree C
+P1=4.33//Pressure at heater 1 in MPa
+t2=(t1-Tr)//Temperature at heater 2 in degree C
+P2=2.5318//Pressure at heater 2 in MPa
+t3=(t2-Tr)//Temperature at heater 3 in degree C
+P3=1.367//Pressure at heater 3 in MPa
+t4=(t3-Tr)//Temperature at heater 4 in degree C
+P4=0.6714//Pressure at heater 4 in MPa
+t5=(t4-Tr)//Temperature at heater 5 in degree C
+P5=0.2906//Pressure at heater 5 in MPa
+t6=(t5-Tr)//Temperature at heater 6 in degree C
+P6=0.108//Pressure at heater 6 in MPa
+t7=(t6-Tr)//Temperature at heater 7 in degree C
+P7=32.65//Pressure at heater 7 in kPa
+
+//Output
+printf('The optimum pressure and temperature at different heaters are: \n Heater 1: t1 = %3.1f degree C and p1 = %3.2f MPa\n Heater 2: t2 = %3.1f degree C and p2 = %3.4f MPa\n Heater 3: t3 = %3.1f degree C and p3 = %3.3f MPa\n Heater 4: t4 = %3.1f degree C and p4 = %3.4f MPa\n Heater 5: t5 = %3.1f degree C and p5 = %3.4f MPa\n Heater 6: t6 = %3.1f degree C and p6 = %3.3f MPa\n Heater 7: t7 = %3.1f degree C and p7 = %3.2f kPa',t1,P1,t2,P2,t3,P3,t4,P4,t5,P5,t6,P6,t7,P7)
diff --git a/1943/CH2/EX2.4/Ex2_4.sce b/1943/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..bc29b9e30 --- /dev/null +++ b/1943/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,16 @@ + +clc
+clear
+//Input data
+ng=0.97//Efficiency of electric generator
+nt=0.95//Efficiency of turbine
+nb=0.92//Efficiency of boiler
+nc=0.42//Efficiency of cycle
+no=0.33//Efficiency of overall plant
+
+//Calculations
+na=(no/(ng*nt*nb*nc))//Efficiency of auxiliaries
+n=(1-na)*100//Percentage of total electricity generated which is consumed in running the auxiliaries
+
+//Output
+printf('Percentage of total electricity generated which is consumed in running the auxiliaries is %3.2f percent',n)
diff --git a/1943/CH2/EX2.5/Ex2_5.sce b/1943/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..9be045e01 --- /dev/null +++ b/1943/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+T1=140//Temperature with which feed water enters into economiser in degree C
+T2=[25,250]//Temperature from air is preheated to in degree C
+P1=60//Pressure with which steam leaves the drum in bar
+x1=0.98//Dryness fraction
+T3=450//Temperature with which steam leaves the superheater in degree C
+cc=25.2//Calorific value of coal in MJ/kg
+r=8.5//Rate of evaporation of steam per kg coal
+wf=1//Mass of coal in kg
+R=15//Air fuel ratio by mass
+Cpa=1.005//Specific heat of air at constant pressure in kJ/kg.K
+Cpw=4.2//Specific heat of water at constant pressure in kJ/kg.K
+
+//Calculations
+h1=(T1*Cpw)//Enthalpy in kJ/kg
+hf=1213.35//Enthalpy in kJ/kg
+h2=hf//Enthalpy in kJ/kg
+hfg=1571//Enthalpy in kJ/kg
+h4=3301.8//Enthalpy in kJ/kg
+h3=(hf+x1*hfg)//Enthalpy in kJ/kg
+n=((r*(h4-h1))/(wf*cc*1000))*100//Efficiency
+he=(r*(h2-h1))/wf*10^-3//Heat transfer in the economiser in MJ/kg
+hb=(r*(h3-h2))/wf*10^-3//Heat transfer in the boiler in MJ/kg
+hs=(r*(h4-h3))/wf*10^-3//Heat transfer in the superheater in MJ/kg
+ha=(R*Cpa*(T2(2)-T2(1)))/wf*10^-3//Heat transfer in the air preheater in MJ/kg
+pe=((h2-h1)/(h4-h1))*100//Percentage of total heat absorbed in the economiser in percent
+pb=((h3-h2)/(h4-h1))*100//Percentage of total heat absorbed in the boiler in percent
+ps=((h4-h3)/(h4-h1))*100//Percentage of total heat absorbed in the superheater in percent
+
+//Output
+printf('Efficiency of steam generator is %3.2f percent \n\n Heat transfer per kg fuel in \n (i)economiser is %3.4f MJ/kg \n (ii)boiler is %3.3f MJ/kg \n (iii)superheater is %3.3f MJ/kg \n (iv)air pre-heater is %3.3f MJ/kg \n\n Percentage of total heat absorption taking place in \n (i)economiser is %3.2f percent \n (ii)boiler is %3.2f percent \n (iii)superheater is %3.2f percent',n,he,hb,hs,ha,pe,pb,ps)
diff --git a/1943/CH2/EX2.6/Ex2_6.sce b/1943/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..150d63c6d --- /dev/null +++ b/1943/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,73 @@ + +clc
+clear
+//Input data
+p1=150//Pressure of inlet steam in bar
+T1=550//Temperature of steam in degree C
+p2=20//Pressure after expansion in bar
+T2=500//Reheat temperature in degree C
+pc=0.075//Condenser pressure in bar
+php=50//Pressure of steam in h.p turbine in bar
+pip=[10,5,3]//Pressure of steam in i.p turbines in bar
+plp=1.5//Pressure of steam in l.p turbine in bar
+m=300*1000//Steam flow rate in kg/h
+
+//Calculations
+h1=3448.6//Enthalpy in kJ/kg
+h4=3467.6//Enthalpy in kJ/kg
+s1=6.5119//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+s3=s1//Entropy in kJ/kg.K
+s4=7.4317//Entropy in kJ/kg.K
+s5=s4//Entropy in kJ/kg.K
+s6=s5//Entropy in kJ/kg.K
+s7=s6//Entropy in kJ/kg.K
+s8=s7//Entropy in kJ/kg.K
+s9=s8//Entropy in kJ/kg.K
+t2=370//Temperature in degree C
+t3=245//Temperature in degree C
+t5=400//Temperature in degree C
+t6=300//Temperature in degree C
+t7=225//Temperature in degree C
+t8=160//Temperature in degree C
+h2=3112//Enthalpy in kJ/kg
+h3=2890//Enthalpy in kJ/kg
+h5=3250//Enthalpy in kJ/kg
+h6=3050//Enthalpy in kJ/kg
+h7=2930//Enthalpy in kJ/kg
+h8=2790//Enthalpy in kJ/kg
+x9=(s9-0.5764)/7.6751//Dryness fraction
+h9=168.79+x9*2406////Enthalpy in kJ/kg
+h10=168.79//Enthalpy in kJ/kg
+h11=h10+0.001*pip(2)*100//Enthalpy in kJ/kg
+h12=467.11//Enthalpy in kJ/kg
+t14=111.37//Temperature in degree C
+h14=467//Enthalpy in kJ/kg
+h13=h12//Enthalpy in kJ/kg
+h14=h13//Enthalpy in kJ/kg
+h15=h14//Enthalpy in kJ/kg
+h16=561.47//Enthalpy in kJ/kg
+h17=h16//Enthalpy in kJ/kg
+h18=640.23//Enthalpy in kJ/kg
+h19=h18+0.001*(p1-pip(2))*100//Enthalpy in kJ/kg
+h20=762.8//Enthalpy in kJ/kg
+h21=h20//Enthalpy in kJ/kg
+h22=1154.23//Enthalpy in kJ/kg
+h23=h22//Enthalpy in kJ/kg
+m1=((h23-h21)/(h2-h22))//Mass in kg
+m2=((h21-h19)-(m1*(h22-h20)))/(h5-h20)//Mass in kg
+m3=(((1-m1-m2)*(h18-h17))-((m1+m2)*(h20-h18)))/(h6-h18+h18-h17)//Mass in kg
+m4=((1-m1-m2-m3)*(h17-h15))/(h7-h16)//Mass in kg
+m5=(((1-m1-m2-m3-m4)*(h14-h11))-(m4*(h16-h12)))/(h8-h12+h14-h11)//Mass in kg
+WT=(h1-h2)+(1-m1)*(h2-h3)+(1-m1)*(h4-h5)+(1-m1-m2)*(h5-h6)+(1-m1-m2-m3)*(h6-h7)+(1-m1-m2-m3-m4)*(h7-h8)+(1-m1-m2-m3-m4-m5)*(h8-h9)//Workdone by turbine in kJ/kg
+Wp=(0.5+14.5+0.15)//Workdone in kJ/kg
+Wnet=(WT-Wp)//Net workdone in kJ/kg
+Q1=(h1-h23)+(1-m1)*(h4-h3)//Heat supplied in kJ/kg
+ncy=(Wnet/Q1)*100//Cycle efficiency in percent
+t23=264//Temperature in degree C
+sr=(3600/Wnet)//Steam rate in kJ/kWh
+hr=((Q1/Wnet)*3600)//Heat rate in kJ/kWh
+P=((Wnet*m)/3600)/10^3//Power output in MW
+
+//Output
+printf('(a) The cycle efficiency is %3.2f percent \n (b) The feedwater temperature is %i degree C \n (c) The steam rate is %3.2f kJ/kWh \n (d) The heat rate is %3.0f kJ/kWh \n (e) The quality of steam at turbine exhaust is %3.4f \n (f) The power output is %3.2f MW',ncy,t23,sr,hr,x9,P)
diff --git a/1943/CH2/EX2.7/Ex2_7.sce b/1943/CH2/EX2.7/Ex2_7.sce new file mode 100755 index 000000000..e927af656 --- /dev/null +++ b/1943/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,23 @@ + +clc
+clear
+//Input data
+m=10000//Mass flow rate of steam in kg/h
+p=3//Pressure of steam in bar
+P=1000//Power in kW
+n=0.7//Internal efficiency of turbine
+
+//Calculations
+dh=(P*3600)/m//Change in enthalpy in kJ/kg
+h2=2725.3//Enthalpy in kJ/kg from Fig. E2.7
+h1=dh+h2//Enthalpy in kJ/kg
+dh1h2s=dh/n//Change in enthalpy in kJ/kg
+h2s=h1-dh1h2s//Enthalpy in kJ/kg
+x2s=(h2s-561.47)/2163.8//Dryness fraction
+s2s=1.6718+x2s*(6.999-1.6718)//Entropy in kJ/kg.K
+s1=s2s//Entropy in kJ/kg.K
+p1=37.3//Pressure in bar from Mollier diagram
+t1=344//Temperature in degree C
+
+//Output
+printf('The steam condition required at inlet of the turbine: \n Enthalpy is %3.1f kJ/kg \n Entropy is %3.4f kJ/kg.K \n Pressure is %3.1f bar \n Temperature is %i degree C',h1,s1,p1,t1)
diff --git a/1943/CH2/EX2.8/Ex2_8.sce b/1943/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..2786b5b6b --- /dev/null +++ b/1943/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,37 @@ + +clc
+clear
+//Input data
+Pl=5.6//Power load in MW
+Hl=1.163//Heat load in MW
+p1=40//Pressure in bar
+T1=500+273//Temperature in K
+p2=0.06//Pressure in bar
+p3=2//Pressure in bar
+CV=25//Calorific value in MJ/kg
+n=88//Boiler efficiency in percent
+T=6//Temperature rise in degree C
+
+//Calculations
+h1=3445.3//Enthalpy in kJ/kg
+s1=7.0901//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+s3=s1//Entropy in kJ/kg.K
+x2=(s2-1.5301)/5.5970//Dryness fraction
+h2=2706.7//Enthalpy in kJ/kg
+h26=2201.9//Difference in enthalpy in kJ/kg
+w=(Hl*10^3)/h26//Rate of steam extraction in kg/h
+x3=(s1-0.52)/7.815//Dryness fraction
+h3=(149.79+x3*2416)//Enthalpy in kJ/kg
+h4=149.79//Enthalpy in kJ/kg
+ws=((Pl*10^3+(w*(h2-h3)))/((h1-h2)+(h2-h3)))//Steam generation capacity in kg/s
+ws1=(ws*3600)/1000//Steam generation capacity in t/h
+h7=(504.7+(1.061*10^-3*(p1-p3)*100))//Enthalpy in kJ/kg
+h5=(149.79+(1.006*100*p1*10^-3))//Enthalpy in kJ/kg
+Q1=(((ws-w)*(h1-h5))+(w*(h1-h7)))//Heat input in kW
+wf=((Q1/1000)/((n/100)*CV))*(3600/1000)//Fuel burning rate in t/h
+Q2=((ws-w)*(h3-h4))//Heat rejected to the condensor in kW
+wc=(Q2/(4.187*T))/1000//Rate of flow of cooling water in m^3/s
+
+//Output
+printf('(a) the steam generation capacity of the bolier is %3.2f t/h \n (b) the heat input to the boiler is %3.1f kW \n (c) the fuel burning rate of the bolier is %3.3f t/h \n (d) the heat rejected to the condensor is %3.0f kW \n (e) the rate of flow of cooling water in the condensor is %3.3f m^3/s',ws1,Q1,wf,Q2,wc)
diff --git a/1943/CH2/EX2.9/Ex2_9.sce b/1943/CH2/EX2.9/Ex2_9.sce new file mode 100755 index 000000000..4659c210b --- /dev/null +++ b/1943/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+m=21000//Steam rate in kg/h
+p1=17//Pressure in bar
+T1=230+273//Temperature in K
+P=132.56//Power in kW
+x2=0.957//Dryness fraction
+p2=3.5//Pressure in bar
+Pl=1337.5//Power in l.p turbine in kW
+p3=0.3//Pressure in bar
+x3=0.912//Dryness fraction
+
+//Calculations
+h1=2869.7//Enthalpy in kJ/kg
+s1=6.5408//Entropy in kJ/kg.K
+h2=(870.44+x2*1924.7)//Enthalpy in kJ/kg
+h3=h2//Enthalpy in kJ/kg
+h56=(Pl*3600)/m//Difference in Enthalpy in kJ/kg
+h6=(289.23+x3*2336.1)//Enthalpy in kJ/kg
+h5=2649.04//Enthalpy in kJ/kg
+s4s=s1//Entropy in kJ/kg.K
+x4s=(s4s-1.7275)/5.2130//Dryness fraction
+h4s=584.33+x4s*2148.1//Enthalpy in kJ/kg
+w=(P/(h1-h2))//Flow rate in kg/s
+ws=(m/3600)//Steam flow rate in kg/s
+h4=((ws*h5)-(w*h3))/(ws-w)//Enthalpy in kJ/kg
+x4=(h4-584.33)/2148.1//Dryness fraction
+W=(ws-w)*(h1-h4)//Power developed by h.p turbine in kW
+n=((h1-h4)/(h1-h4s))*100//Isentropic efficiency in percent
+
+//Output
+printf('(a) the steam quality at the exhaust of the h.p turbine is %3.3f \n (b) the power developed by the h.p turbine is %3.2f kW \n (c) the isentropic efficiency of the h.p turbine is %3.2f percent',x4,W,n)
diff --git a/1943/CH3/EX3.1/Ex3_1.sce b/1943/CH3/EX3.1/Ex3_1.sce new file mode 100755 index 000000000..540215eb9 --- /dev/null +++ b/1943/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+p=40//Pressure in bar
+T1=400+273//Temperature in K
+T2=40+273//Temperature in K
+x=[10,515.5,72.23,363.0,0.1478,0.5167,80.9*10^-6,0.0333]//Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m^3/kg)
+y=[0.2,277.3,38.35,336.55,0.0967,0.6385,77.4*10^-6,1.163]//Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m^3/kg)
+
+//Calculations
+h1=3216//Enthalpy in kJ/kg
+s1=6.7690//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+x2=(s2-0.5725)/(8.2570-0.5725)//Dryness fraction
+h2=167.57+x2*2406.7//Enthalpy in kJ/kg
+h3=167.57//Enthalpy in kJ/kg
+h4=(167.57+p*100*1.008*10^-3)//Enthalpy in kJ/kg
+h5=1087.31//Enthalpy in kJ/kg
+h6=2801.4//Enthalpy in kJ/kg
+ha=x(4)//Enthalpy in kJ/kg
+sa=x(6)//Entropy in kJ/kg.K
+sb=sa//Entropy in kJ/kg.K
+xb=(sb-y(5))/(y(6)-y(5))//Dryness fraction
+hb=(y(3)+xb*(y(4)-y(3)))//Enthalpy in kJ/kg
+hc=y(3)//Enthalpy in kJ/kg
+hd=hc//Enthalpy in kJ/kg
+m=(h6-h5)/(hb-hc)//Mass of mercury circulated per kg of steam
+Q1=m*(ha-hd)+(h1-h6)+(h5-h4)//Heat supplied in kJ/kg
+Q2=(h2-h3)//Heat rejected in kJ/kg
+nc=(1-(Q2/Q1))*100//Efficiency in percent
+
+//Output
+printf('(a) The amount of mercury circulated per kg of water is %3.4f kg \n (b) The efficiency of the combined cycle is %3.1f percent',m,nc)
diff --git a/1943/CH3/EX3.2/Ex3_2.sce b/1943/CH3/EX3.2/Ex3_2.sce new file mode 100755 index 000000000..f2783e253 --- /dev/null +++ b/1943/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,40 @@ + +clc
+clear
+//Input data
+m=5//Mass flow rate in kg/s
+p1=40//Pressure in bar
+T1=440+273//Temperature in K
+p2=1.5//Pressure in bar
+p3=1//Pressure in bar
+T3=60+273//Temperature in K
+p4=16//Pressure in bar
+T4=100+273//Temperature in K
+p5=9//Pressure in bar
+
+//Calculations
+h1=3307.1//Enthalpy in kJ/kg
+s1=6.9041//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+h2=2570.8//Enthalpy in kJ/kg
+h3=417.46//Enthalpy in kJ/kg
+h6=(251.13+(1.0172*10^-3)*(p3-0.1994)*100)//Enthalpy in kJ/kg
+m3=(m/2)//Mass flow rate in kg/s
+m6=m3//Mass flow rate in kg/s
+h4=(m3*h3+m6*h6)/m//Enthalpy in kJ/kg
+h5=(h4+(1.0291*10^-3)*(p1-p3)*100)//Enthalpy in kJ/kg
+ha=241.58//Enthalpy in kJ/kg
+sa=0.7656//Entropy in kJ/kg.K
+sb=sa//Entropy in kJ/kg.K
+hb=229.43//Enthalpy in kJ/kg
+hc=71.93//Enthalpy in kJ/kg
+hd=hc+(0.7914*10^-3*(p4-p5)*100)//Enthalpy in kJ/kg
+Q1=(m*(h1-h5))/1000//Heat supplied in kW
+Wnets=(m*((h1-h2)-(h5-h4)))//Net workdone by steam in kW
+mR12=(m3*(h2-h3))/(ha-hd)//Mass of R12 in kg/s
+WnetR=(mR12*((ha-hb)-(hd-hc)))//Net workdone by R12 in kW
+T=Wnets+WnetR//Total output in kW
+Qh=(m6*(h2-h6))//Heat rejected in kW
+
+//Output
+printf('(a) Rate of heat transfer in the steam generator is %3.3f kW \n (b) The net power output of the binary cycle is %i kW \n (c) The rate of heat transfer to the industrial process is %3.0f kW',Q1,T,Qh)
diff --git a/1943/CH3/EX3.3/Ex3_3.sce b/1943/CH3/EX3.3/Ex3_3.sce new file mode 100755 index 000000000..7594f5463 --- /dev/null +++ b/1943/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,41 @@ + +clc
+clear
+//Input data
+rp=7.5//Pressure ratio
+T1=15+273//Inlet air temperature in K
+T3=750+273//Maximum temperature in K
+T6=100+273//Temperature in K
+p1=50//Pressure in bar
+T7=600+273//Temperature in K
+p2=0.1//Pressure in bar
+P=200//Total power in MW
+CV=43.3//calorific value in MJ/kg
+cpg=1.11//Specific heat for gas in kJ/kg.K
+g=1.33//Ratio of specific heats for gas
+cpa=1.005//Specific heat for air in kJ/kg.K
+g1=1.4//Ratio of specific heats for air
+
+//Calculations
+T2=(T1*rp^((g1-1)/g1))//Temperature in K
+T4=(T3/rp^((g-1)/g))//Temperature in K
+ha=3670//Enthalpy in kJ/kg
+hb=2305//Enthalpy in kJ/kg
+hc=192//Enthalpy in kJ/kg
+hd=hc//Enthalpy in kJ/kg
+//ma*cpg*(T3-T6)=ms*(ha-hd)
+//ma*cpg*(T3-T4)-ma*cpa*(T2-T1)+ms*(ha-hb)=P*1000
+//Solving these two equations
+A=[cpg*(T3-T6) (hd-ha)
+ cpg*(T3-T4)-cpa*(T2-T1) (ha-hb)]//Coefficient matrix
+B=[0
+ (P*10^3)]//Constant matrix
+X=inv(A)*B//Variable matrix
+Wgt=(cpg*(T3-T4)-cpa*(T2-T1))*X(1)*10^-3//Net workdone by Gas turbine in MW
+Wst=(P-Wgt)//Net workdone by steam turbine in MW
+Q1=(X(1)*cpg*(T3-T2+T3-T4))//Heat supplied in MW
+nth=(P/(Q1*10^-3))*100//Thermal efficiency in percent
+af=(CV*10^3)/(cpg*(T3-T2+T3-T4))//Air fuel ratio
+
+//Output
+printf('(a) The flow rate of air is %3.2f kg/s and steam is %3.2f kg/s \n (b) The power outputs of the gas turbine is %3.2f MW and steam turbine is %3.2f MW \n (c) The thermal efficiency of the combined plant is %3.0f percent \n (d) The air fuel ratio is %3.1f',X(1),X(2),Wgt,Wst,nth,af)
diff --git a/1943/CH3/EX3.4/Ex3_4.sce b/1943/CH3/EX3.4/Ex3_4.sce new file mode 100755 index 000000000..77a3b0487 --- /dev/null +++ b/1943/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,90 @@ + +clc
+clear
+//Input data
+p1=1//Pressure in bar
+T1=25+273//Temperature in K
+rp=8//Pressure ratio of compressor
+Tm=900+273//Maximum temperature in K
+pd=3//pressure drop in combustion chamber in percent
+nc=0.88//Efficiency of compressor
+nt=0.88//Efficiency of turbine
+CV=44.43//Calorific value of fuel in MJ/kg
+cpa=1.006//Specific heat of air in kJ/kg.K
+cpg=1.148//Specific heat of gas in kJ/kg.K
+g1=1.333//Specific heat ratio of gas
+g=1.4//Specific heat ratio of air
+T3=425+273//Temperature in K
+p2=40//Pressure in bar
+p3=0.04//Condensor pressure in bar
+Th=170.4+273//Temperature of feed water to the HRSG in K
+nst=0.82//Efficiency of steam turbine
+pdh=5//Pressure drop in HRSG in kPa
+m=29.235//Steam flow rate in kg/s
+A=1.0401//si=1.0401+0.1728*(h/c)
+B=0.1728//si=1.0401+0.1728*(h/c)
+
+//Calculations
+//Gas turbine plant
+T2=(rp^((g-1)/(g*nt)))*T1//Temperature in K
+//Combustor
+pc=((pd/100)*rp)//Pressure loss in bar
+pcx=(rp-pc)//Pressure in bar
+f=((cpg*(Tm-T1))-(cpa*(T2-T1)))/((CV*10^3)-(cpa*(T2-T1)))//Fuel flow rate in kg/s
+af=(1-f)/f//Air fuel ratio
+//C8H18+12.5O2->8CO2+9H2O
+afc=(12.5*32)/(0.232*114)//Air fuel ratio for stoichiometric combustion
+ea=((af-afc)/afc)*100//Excess air in percent
+//Gas turbine
+p4=p1+0.05//Pressure in bar
+T4=(Tm/(pcx/p4)^(((g1-1)*nt)/g1))//Temperature in K
+//HRSG
+T5=250+30//Temeprature in K
+ha=3272//Enthalpy in kJ/kg
+hf=1087.31//Enthalpy in kJ/kg
+ws=(cpg*((T4-273)-T5))/(ha-hf)//Flow rate in kg/s
+he=721.1//Enthalpy in kJ/kg
+T6=(T4-273)-((ws*(ha-he))/cpg)//Temperature in degree C
+//Power output
+sa=6.853//Entropy in kJ/kg.K
+sbs=sa//Entropy in kJ/kg.K
+xbs=(sbs-0.4266)/8.052//Dryness fraction
+hbs=(121.46+xbs*2432.9)//ENthalpy in kJ/kg
+Wst=(m*(ha-hbs)*nst)//Workdone in kW
+wg=(m/ws)//gas flow rate in kg/s
+wa=(1-f)*wg//Air flow rate entering the compressor in kg/s
+Wgt=(wg*cpg*(Tm-T4))-(wa*cpa*(T2-T1))//Power output of gas turbine in kW
+TO=Wst+Wgt//Total power output in kW
+wf1=(f*wa)//Fuel mass flow rate in kg/s
+wf=4.466//Rounding off of wf1 for exact answers
+no=(TO/(wf*(CV*10^3)))*100//Overall efficiency of the combined plant in percent
+ns=((ha-hbs)/(ha-he))*nst//Efficiency of steam plant
+ngtp=(Wgt/(wf*(CV*10^3)))//Efficiency of the GT plant
+xL=((wg*cpg*(T6-(T1-273)))/(wf*(CV*10^3)))//Lost heat coefficient
+nov=(ns+ngtp-ns*ngtp-ngtp*xL)//The overall efficiency
+//Energy fluxes and irreversibilities
+si=(A+B*((18*1)/(8*12)))//si for octane C8H18
+dHo=(wf*CV*10^3)//Power in kW
+dGo=(si*dHo)//Power in kW
+TS=(dGo-dHo)//Power in kW
+//Compressor
+dS=(cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))//change in entropy in kJ/kg.K
+Ic=(wa*T1*dS)//power in kW
+Icx=((wg*T1*((cpg*log(Tm/T1))-(((cpg*(g1-1))/g1)*log(pcx))))-(wa*T1*((cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))))+TS)//Compressor in kW
+Icg=(-cpg*log(Tm/T4))-(((cpg*(g1-1))/g1)*log(p4/pcx))//Difference in entropy in kJ/kg.K
+IGT=(Icg*T1*wg)//Gas turbine in kW
+se=2.046//Enntropy in kJ/kg.K
+sae=(sa-se)//Difference in entropy in kJ/kg.K
+s64=(cpg*log((T6+273)/T4))-(((cpg*(g1-1))/g1)*log(p4/p1))//Difference in entropy in kJ/kg.K
+Ih=(T1*m*sae)+(wg*T1*s64)//For HRSG in kW
+hb=(ha-(nst*(ha-hbs)))//Enthalpy in kJ/kg
+xb=(hb-121.46)/2432.9//Dryness Fraction
+sb=(0.4226+xb*8.052)//Entropy in kJ/kg.K
+Ist=(m*(sb-sa)*T1)//For steam turbine in kW
+Iexh=(wg*cpg*((T6-(T1-273))-(T1*log((T6+273)/T1))))//For exhaust in kW
+Tl=Icx+Icg+IGT+Ih+Ist+Iexh//Exergy losses in kW
+T=Tl+Wgt+Wst//Total exergy output and exergy destruction in kW
+ee=((Wst+Wgt)/T)*100//Exergy efficiency in percent
+
+//Output
+printf('(a) Total power output is %3.2f kW and overall efficiency is %3.2f percent lost heat coefficient is %3.3f\n Exergy efficiency is %3.0f percent \n\n Input is %3.0f kW \n Total Output is %3.0f kW \n Total losses is %3.0f kW \n Exergy outut + exergy destruction = %3.0f kW which is 1.3 percent gretter than the exergy input',TO,no,xL,ee,dGo,(Wgt+Wst),Tl,T)
diff --git a/1943/CH3/EX3.5/Ex3_5.sce b/1943/CH3/EX3.5/Ex3_5.sce new file mode 100755 index 000000000..b8a2dfab8 --- /dev/null +++ b/1943/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,13 @@ + +clc
+clear
+//Input data
+n1=0.5//Efficiency of mercury
+n2=0.4//Efficiency of steam
+n3=0.25//Efficiency of composite cycle
+
+//Calculations
+n=(1-(1-n1)*(1-n2)*(1-n3))*100//Overall efficiency of the combined cycle in percent
+
+//Output
+printf('The overall efficiency of the combined cycle is %3.1f percent',n)
diff --git a/1943/CH3/EX3.6/Ex3_6.sce b/1943/CH3/EX3.6/Ex3_6.sce new file mode 100755 index 000000000..f45a7ccc9 --- /dev/null +++ b/1943/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,12 @@ + +clc
+clear
+//Input data
+z=30//Percentage of total energy of fuel
+n=40//Cycle efficiency in percent
+
+//Calculations
+on=((z/100)+(1-(z/100))*(n/100))*100//Overall efficiency in percent
+
+//Output
+printf('The overall efficiency of the combined plant is %3.0f percent',on)
diff --git a/1943/CH3/EX3.7/Ex3_7.sce b/1943/CH3/EX3.7/Ex3_7.sce new file mode 100755 index 000000000..60ca65b42 --- /dev/null +++ b/1943/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+Tc=1250+273//Cathode temperature in K
+Ta=500+273//Anode temperature in K
+e=1.602*10^-19//Charge in coloumb
+K=1.38*10^-23//Boltzmann constant in J/molecule.K
+b=18//Constant
+
+//Calculations
+Va=((b*K*Ta)/e)//Voltage of anode in V
+Vc=((b*K*Tc)/e)//Voltage of cathode in V
+Vo=Vc-Va//Output voltage in V
+Ja=(120*Ta^2*exp(-b))//Current density in Cathode in A/cm^2
+Jc=(120*Tc^2*exp(-b))//Current density in Anode in A/cm^2
+P=Vo*(Jc-Ja)//Power output per unit area in /cm^2
+nth=(((Tc-Ta)/Tc)*(b/(b+2)))*100//Thermal efficiency in percent
+
+//Output
+printf('(a) The output voltage is %3.4f V \n (b) The current density in the cathode is %3.3f A/cm^2 and anode is %3.3f A/cm^2 \n (c) Power output per unit area is %3.2f W/cm^2 \n (d) Thermal efficiency is %3.1f percent',Vo,Jc,Ja,P,nth)
diff --git a/1943/CH3/EX3.8/Ex3_8.sce b/1943/CH3/EX3.8/Ex3_8.sce new file mode 100755 index 000000000..70a6aa650 --- /dev/null +++ b/1943/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,37 @@ + +clc
+clear
+//Input data
+P=100//Power in kW
+V=115//Voltage in V
+To=1500//Outer temperature in K
+Te=1000//Exit temperature in K
+Ta=350//Ambient temperature in K
+nth=30//Thermal efficiency in percent
+nge=92//Generator efficiency in percent
+//Properties of thermoelectrons
+a=0.0012//At 1250K in V/K
+kp=0.02//In W/cm.K
+kn=0.03//In W/cm.K
+dp=0.01//In ohm.cm
+dn=0.012//In ohm.cm
+J=20//Current density in A/cm^2
+
+//Calculations
+zmax=(a^2/(sqrt(dp*kp)+sqrt(dn*kn))^2)//Maximum value of figure of merit in K^-1
+mo=sqrt(1+(zmax*((To+Te)/2)))//Optimum value of the resistance ratio
+nmax=(((To-Te)/To)*((mo-1)/(mo+(Te/To))))*100//Maximum thermal efficiency in percent
+Vl=(a*(To-Te)*(mo/(mo+1)))//Voltage per couple in V
+nc=(V/Vl)//Number of couples in series
+L=((a*(To-Te))/((1+mo)*(dp+dn)))/J//Length in cm
+A=((P*Te)/V)/J//Area in cm^2
+I=(J*A)//Current in A
+Vo=(a*(To-Te))//Voltage in V
+Q1=((a*I*To)-((1/2)*(L/A)*I^2*(dp+dn))+((A/L)*(kp+kn)*(To-Te)))/1000//Heat input to the thermoelectric generator in kW
+Q2=((a*I*Te)+((A/L)*(kp+kn)*(To-Te))+P)/1000//Heat rejected at full load in kW
+Q1n=(((A/L)*(kp+kn)*(To-Te)))/1000//At no load heat input in kW
+Q2n=Q1n//At no load heat rejected in kW
+no=((nmax/100)+(1-(nmax/100))*(nth/100)*(nge/100))*100//Overall efficiency in percent
+
+//Output
+printf('(a) The thermal efficiency of thermocouple generator is %3.1f percent \n (b) The number of thermo couples in series is %i \n (c) The lenght of the thermal elements is %3.3f cm and area is %3.2f cm^2 \n (d) The output open-circuit voltage is %3.1f V \n (e) At full load: \n The heat input is %3.3f kW \n The heat rejected is %3.3f kW \n At no load: \n The heat input is %3.3f kW \n The heat rejected is %3.3f kW \n (f) The overall efficiency of the combined thermo-electric steam power plant is %3.2f percent',nmax,nc,L,A,Vo,Q1,Q2,Q1n,Q2n,no)
diff --git a/1943/CH4/EX4.1/Ex4_1.sce b/1943/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..7c59f44cd --- /dev/null +++ b/1943/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,39 @@ + +clc
+clear
+//Input data
+C=84;//The mass of carbon present in the fuel in %
+H=10;//The mass of hydrogen present in the fuel in %
+S=3.2;//The mass of sulphur present in the fuel in %
+O=1.6;//The mass of oxygen present in the fuel in %
+I=1.2;//The mass of incombustible in the fuel in %
+X=15.72;//The flue gas of combined CO2 and SO2 by volume in %
+Og=1;//The flue gas of O2 by volume in %
+Y=100;//Let us consider the fuel oil in kg
+C1=12;//Molecular weight of Carbon
+H1=2;//Molecular weight of hydrogen
+S1=32;//Molecular weight of sulphur
+O1=32;//Molecular weight of oxygen
+Co2=44;//Molecular weight of carbondioxide
+So2=64;//Molecular weight of sulphurdioxide
+N1=28;//Molecular weight of nitrogen
+H2O=18;//Molecular weight of water
+
+//Calculations
+b=C/C1;//Equating coefficients of the carbon from equation
+g=H/H1;//Equating coefficients of the hydrogen from equation
+d=S/S1;//Equating coefficients of the sulphur from the equation
+e=(b+d)/(X/Og);//By volumetric analysis
+x=b+d+e+(g/2)-(O/O1);//Moles of oxygen are supplied for combustion
+f=3.76*x;//Equating coefficients of the nitrogen from equation
+Mo=x*O1;//Mass of oxygen supplied in kg
+Ma=Mo/0.232;//Mass of air supplied for 100 kg of fuel in kg
+Wa=Ma/100;//Mass of air supplied for 1 kg fuel in kg
+Wrh={(11.5*C)+(34.5*[(H)-(O/8)])+(4.3*S)}/100;//Theoretical air required per kg of fuel in kg
+E=[(Wa-Wrh)/Wrh]*100;//Percentage of excess air in %
+D=(b*Co2)+(d*So2)+(e*O1)+(f*N1);//Mass of dry flue gas formed for 100 kg fuel in kg
+dfg=D/100;//Mass of dry flue gas formed per kg of fuel in kg
+Mw=(g*H2O)/100;//Mass of water vapour formed per kg of fuel
+
+//Output
+printf('(a) Mass of air supplied WA = %3.2f kg \n (b)Percentage excess air supplied = %3.2f percentage \n (c)mass of dry flue gas formed = %3.2f kg \n (d) Mass of water vapour formed = %3.2f kg',Wa,E,dfg,Mw)
diff --git a/1943/CH4/EX4.10/Ex4_10.sce b/1943/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..62f265176 --- /dev/null +++ b/1943/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,20 @@ + +clc
+clear
+//Input data
+tg=180;//The gas temperature in degree centigrade
+p=250;//The draught produced by the ID fan in mm
+nf=0.52;//The efficiency of the fan
+Va=0.858;//Volume flow rate of air in m^3/kg
+g=9.812;//gravitational force constant in m/s^2
+Wf=10;//Coal rate in t/h
+Wa=12.9;//Actual air required per kg fuel in kg air/kg fuel
+ta=30;//Room temperature in degree centigrade
+
+//Calculations
+Wfg=[(Wf+(Wa*10))*1000]/3600;//Fuel gas required in kg/s
+Vfg=[Va*(tg+273)]/(ta+273);//Volume flow rate of fuel gas in m^3/kg
+ID=((Wfg*Vfg*p*g)/(nf))/1000;//ID fan motor capacity in kW
+
+//Output
+printf('The motor capacity of the ID fan is %3.2f kW ',ID)
diff --git a/1943/CH4/EX4.11/Ex4_11.sce b/1943/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 000000000..0004d7fb9 --- /dev/null +++ b/1943/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,39 @@ + +clc
+clear
+//Input data
+CO2=13.2;//The volume of carbondioxide present in the partial analysis of dry flue gas in %
+O2=3.2;//The volume of oxygen present in the partial analysis of dry flue gas in %
+C=88;//The mass of carbon present in the coal according to coal analysis on mass basis in %
+H=4.4;//The mass of hydrogen present in the coal according to coal analysis on mass basis in %
+A=7.6;//The mass of ash present in the coal according to coal analysis on mass basis in %
+M=0;//Moisture present in the fuel was nil
+Mc=12;//Molecular weight of the carbon
+Mh=2;//Molecular weight of the hydrogen
+Mo=32;//Molecular weight of the oxygen
+Mho=18;//Molecular weight of water
+p=101.325;//Atmospheric pressure in kPa
+
+//Calculations
+c=C/Mc;//Equating coefficients of the carbon from the equation
+g=H/Mh;//Equating coefficients of the hydrogen from the equation
+x=(CO2/100)/(O2/100);//From dry fuel gas analysis (dfg)
+d=[[(CO2/100)*(47.5)]-7.333]/[[(CO2/100)*(3.032)]-1];//Coefficient of the carbonmonoxide in the equations product side
+b=c-d;//Coefficient of the carbondioxide in the equation product side
+a=10.21-(0.742*d);//Coefficient of the oxygen in the reactant side of the equation
+e=b/x;//Coefficient of the oxygen in the product side of the equation
+f=3.76*a;//Equating coefficients of the nitrogen from the equation
+ma=(a*Mo)/0.232;//Mass of air supplied for 100 kg coal in kg
+ma1=ma/100;//Mass of air supplied per kg coal in kg
+T=b+d+e+f;//Total number of moles of dry flue gas (dfg)
+CO21=(b/T)*100;//Carbondioxide by volume in percentage
+O21=(e/T)*100;//Oxygen by volume in percentage
+CO1=(d/T)*100;//Carbonmonoxide by volume in percentage
+N21=(f/T)*100;//Nitrogen by volume in percentage
+Mwv=(g*Mho)/100;//Mass of watervapour formed per kg coal in kg
+Mf=(g)/(b+d+e+f+g);//Mole fraction of water vapour in flue gas
+P=Mf*p;//Partial pressure of water vapour in kPa
+D=32.9;//Dew point temperature from steam tables in degree centigrade
+
+//Output
+printf('(a)The complete volumetric composition of the dry flue gas is \n Carbondioxide by volume = %3.2f percentage \n Oxygen by volume = %3.2f percentage \n Carbonmonoxide by volume = %3.2f percentage \n Nitrogen by volume = %3.2f percentage \n (b) The actual amount of air supplied per kg coal = %3.2f kg \n (c) Mass of water vapour formed per kg coal = %3.2f kg \n (d) The dew point temperature of the flue gas = %3.2f degree centigrade ',CO21,O21,CO1,N21,ma1,Mwv,D)
diff --git a/1943/CH4/EX4.12/Ex4_12.sce b/1943/CH4/EX4.12/Ex4_12.sce new file mode 100755 index 000000000..87adcf53a --- /dev/null +++ b/1943/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,24 @@ + +clc
+clear
+//Input data
+H=200;//Height of the stack in m
+D=4;//Diameter of the stack in m
+m=1000;//Mass flow rate of gas in kg/s
+Ts=100;//Stack exit gas temperature in degree centigrade
+Ta=5;//Ambient air temperature in degree centigrade
+Vw=50;//Wind velocity in Km/h
+Cp=1.005;//Specific heat of the gas in kJ/kgK
+pi=3.142;//Mathematical constant the value of pi
+
+//Calculations
+Vw1=(50*1000)/(60*60);//Wind velocity in m/s
+Qe=m*Cp*(Ts-Ta);//Heat emission from plume in kW
+Qe1=Qe/1000;//Heat emission from the plume in MW
+p=(101.325)/(0.287*373);//Density of the gas in kg/m^3
+A=(pi*D^2)/4;//Area of the stack in m^2
+Vs=m/(p*A);//Stack gas exict velocity in m/s
+H1=[(2.62*(Qe1^(1/2))*1000)/Vw1]-[(0.029*Vs*D)/Vw1];//The height of the gas plume in m
+
+//Output
+printf('The height of the gas plume is H = %3.1f m ',H1)
diff --git a/1943/CH4/EX4.13/Ex4_13.sce b/1943/CH4/EX4.13/Ex4_13.sce new file mode 100755 index 000000000..b6bd4e83b --- /dev/null +++ b/1943/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,47 @@ + +clc
+clear
+//Input data
+CV=20;//Calorific value of the fuel in MJ/kg
+C=65;//The amount of carbon present in the fuel according to gravimetric analysis in %
+H=25;//The amount of hydrogen present in the fuel according to gravimetric analysis in %
+O=10;//The amount of oxygen present in the fuel according to gravimetric analysis in %
+p1=1;//Pressure at the inlet of the compressor in bar
+t1=27;//Temperature at the inlet of the compressor in degree centigrade
+p2=4;//The pressure which compressor compresses it isentropically in bar
+Re=78;//The regenerator effectiveness in %
+CO2=6;//The amount of carbondioxide according to the analysis of dry exhaust gas in %
+CO=1.5;//The amount of carbonmonoxide according to the analysis of dry exhaust gas in %
+Cp=1.005;//Specific heat capacity of the air in kJ/kgK
+i=1.44;//Isentropic index for the air
+Cp1=1.15;//Specific heat capacity of the air in kJ/kgK
+i1=1.33;//Isentropic index for the combustion products
+Mc=12;//Molecular weight of the carbon
+Mh=2;//Molecular weight of the hydrogen
+Mo=32;//Molecular weight of the oxygen
+Mho=18;//Molecular weight of water
+T0=288;//Datum temperature in K (Assumed)
+
+//Calculations
+h=(C/100)/(Mc);//Equating coefficients of the carbon from the equation
+e=(H/100)/Mh;//Equating coefficients of the hydrogen from the equation
+y=(CO/100)/(CO2/100);//From dry exhaust gas analysis for solving
+a=h/(1+y);//The coefficient of the carbondioxide in the product side of the equation
+b=h-a;//The coefficient of the carbonmonoxide in the product side of the equation
+z=b/(CO/100);//The sum of coefficients of the product side of the equation
+x=z-(b/2)+(e/2);//Mol of air supplied in kmol
+wa=x*28.96;//Air supplied in kg/kg fuel
+wf=1;//Assuming 1 kg of fuel supplied
+T2=(t1+273)*(p2/p1)^((i-1)/i);//Temperature at the outlet of the compressor in K
+T3=[[(wa*Cp*(T2-T0))+(wf*CV*1000)]/[(wa+wf)*(Cp1)]]+T0;//Maximum temperature of the cycle in K
+T4=T3/[(4)^((i1-1)/i1)];//Temperature at point of the cycle in K
+T5=[(Re/100)*(T4-T2)]+T2;//Temperature at point of the cycle in K
+Wc=wa*Cp*(T2-(t1+273));//Work done by the compressor in kW
+Wt=23.54*Cp1*(T3-T4);//Work done by the turbine in kW
+Q1=23.54*Cp1*(T3-T5);//Total work done by the system in kW
+nc=(Wt-Wc)/Q1;//Efficiency of the cycle
+nc1=nc*100;//Efficiency of the cycle in %
+spc=3600/(Wt-Wc);//Specific fuel consumption in kg/kWh
+
+//Output
+printf('(a) The maximum temperature of the cycle T3 = %3.0f K \n (b)Thermal efficiency of the plant = %3.3f or %3.2f percentage\n (c) Specific fuel consumption = %3.3f kg/kWh ',T3,nc,nc1,spc)
diff --git a/1943/CH4/EX4.2/Ex4_2.sce b/1943/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..d54483c56 --- /dev/null +++ b/1943/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,18 @@ + +clc
+clear
+//Input data
+CO2=11.5;//Percentage of carbondioxide present in combustion in %
+O2=2.7;//Percentage of oxygen present in the combustion in %
+CO=0.7;//Percentage of carbonmonoxide present in the combuston in %
+
+//Calculations
+a=85.1/3.76;//Equating moles for nitrogen from the equation
+x=(CO2+CO)/3;//Equating moles for carbon from the equation
+b=[a-CO2-(CO/2)-O2]*2;//Equating moles for oxygen from the equation
+y=a/x;//Moles of oxygen supplied for one mole of propane gas
+z=5;//Theoretically 5 moles of oxygen are required for reacting
+E=[(y-z)/z]*100;//The excess of air supplied in %
+
+//Output
+printf('The percentage excess air used is = %3.1f percentage',E)
diff --git a/1943/CH4/EX4.3/Ex4_3.sce b/1943/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..aba54e17b --- /dev/null +++ b/1943/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,23 @@ + +clc
+clear
+//Input data
+CO2=12.1;//The amount of carbondioxide released from the combustion in %
+O2=3.8;//The amount of oxygen released from the combustion in %
+CO=0.9;//The amount of carbonmonoxide released from the combustion in %
+MO=32;//Molecular weight of Oxygen
+
+//Calculations
+a=83.2/3.76;//Equating moles for nitrogen from the equation
+b=(2*a)-(2*CO2)-(2*O2)-CO;//Equating moles for oxygen from the equation
+x=CO2+CO;//Equating moles for carbon from the equation
+y=2*b;//Equating moles for hydrogen from the equation
+z=18.75;//Moles of Oxygen from the stoichiometric equation
+z1=a;//Moles of Oxygen from the combustion equation
+E=[(z1-z)/z]*100;//Percentage of excess air in%
+A=(a*MO)/0.232;//Actual air supplied per mole of C13H23
+Mc=179;//Molecular weight of C13H23
+Af=A/Mc;//Air fuel ratio during the test
+
+//Output
+printf('(a) The air fuel ratio during the test = %3.2f \n (b) The excess or deficiency of air used = %3.0f Percentage of excess air used',Af,E)
diff --git a/1943/CH4/EX4.4/Ex4_4.sce b/1943/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..d4098f1d9 --- /dev/null +++ b/1943/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,51 @@ + +clc
+clear
+//Input data
+C=61;//The mass of carbon present in the coal according to coal analysis on mass basis in %
+H=4;//The mass of hydrogen present in the coal according to coal analysis on mass basis in %
+O=3;//The mass of oxygen present in the coal according to coal analysis on mass basis in %
+N=2;//The mass of nitrogen present in the coal according to coal analysis on mass basis in %
+S=1;//The mass of sulphur present in the coal according to coal analysis on mass basis in %
+M=4;//The mass of moisture present in the coal according to coal analysis on mass basis in %
+A=25;//The mass of ash present in the coal according to coal analysis on mass basis in %
+HHV=24.3;//The high heating value of the coal i.e energy released by complete combustion of 1 kg fuel in MJ/kg
+CO2=12;//The amount of carbondioxide by volume according to dry flue gas analysis in %
+CO=1.5;//The amount of carbonmonoxide by volume according to dry flue gas analysis in %
+O2=7;//The amount of oxygen by volume according to dry flue gas analysis in %
+N2=79.5;//The amount of nitrogen by volume according to dry flue gas analysis in %
+Te=170;//Exhaust gas temperature in degree centigrade
+L=0.03;//Energy loss other than dry exhaust loss and incomplete combustion is 3% of HHV
+R=150;//Steam generation rate in t/h
+Po=100;//Steam condition at boiler outlet in bar
+To=500;//Steam condition at boiler outlet in degree centigrade
+Ti=160;//Feed water inlet temperature in degree centigrade
+HCO2=33083;//Heat of reaction in kJ/kg carbon
+HCO=9500;//Heat of reaction in kJ/kg carbon
+cp=1.05;//Heat capacity of dry flue gas (dfg) in kJ/kgK
+Ta=30;//The ambient temperature of air in degree centigrade
+Mc=44;//Molecular weight of Carbondioxide
+Mco=28;//Molecular weight of carbonmonoxide
+Mo=32;//Molecular weight of oxygen
+Mn=28;//Molecular weight of nitrogen
+Mx=12;//Molecular weight of carbon
+h1=3373.7;//Enthalpy at 100 bar and 500 degree centigrade in kJ/kg
+hf=675.55;//Enthalpy at 160 degree centigrade in kJ/kg
+hg=2724.7;//Enthalpy at 100 bar in kJ/kg
+
+//Calculations
+Mdfg=([(C/100)*[(Mc*(CO2/100))+(Mco*(CO/100))+(Mo*(N2/100))]]/[Mx*[(CO2/100)+(CO/100)]]);//Mass of dry flue gas produced per kg of fuel in kg
+Ed=Mdfg*cp*(Te-Ta);//Energy loss due to dry exhaust gas in kJ/kg fuel
+//Since Mdfg is 11.73kg through sciab calculation, there is a variation in Ed value and Ei value
+Ei=[(Mdfg)*(HCO2-HCO)*(Mx/Mco)]*[(Mco*(CO/100))/[(Mc*(CO2/100))+(Mco*(CO/100))+(Mo*(O2/100))+(Mco*(N2/100))]];//Energy loss due to incomplete combustion in kJ/kg fuel
+El=L*HHV;//Energy loss other than dry exhaust loss and incomplete combustion loss in MJ/kg fuel
+TEl=(Ed/1000)+(Ei/1000)+El;//Total energy loss in MJ/kg fuel
+Be=[(HHV-TEl)/(HHV)]*100;//Boiler efficiency in %
+Wf=([(R*1000)*(h1-hf)]/((Be/100)*HHV*1000))/3600;//The fuel burning rate in kg/s
+Wth=(11.5*(C/100))+(34.5*[(H/100)-(O/800)])+(4.3*(S/100));//Thearetical air required per kg of fuel in kg
+WA=[[(3.04*(N2/100)*(C/100))]/[(CO2/100)+(CO/100)]]-[(N/100)*(1/0.768)];//Actual air supplied per kg of fuel in kg
+per=[(WA-Wth)/Wth]*100;//Percentage excess air used in %
+pea=[(h1-hg)/(h1-hf)]*100;//Percentage of energy absorbed in the superheater
+
+//Output
+printf('(a)The amount of dry flue gas produced per kg fuel = %3.2f kg \n (b)The dry exhaust loss = %3.1f kJ/kg fuel and incomplete combustion loss per kg fuel = %3.2f kJ/kg fuel \n (c)The boiler efficiency = %3.2f percentage \n (d) THe fuel burning rate = %3.3f kg/s \n (e)The percentage of excess air used = %3.2f percentage \n (f) The percentage of energy absorbed in the superheater = %3.2f percentage',Mdfg,Ed,Ei,Be,Wf,per,pea)
diff --git a/1943/CH4/EX4.5/Ex4_5.sce b/1943/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..d809fa7d9 --- /dev/null +++ b/1943/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+C=83.7;//The amount of carbon present in the fuel oil according to ultimate analysis of a fuel oil in %
+H=12.7;//The amount of hydrogen present in the fuel oil according to ultimate analysis of a fuel oil in %
+O=1.2;//The amount of oxygen present in the fuel oil according to ultimate analysis of a fuel oil in %
+N=1.7;//The amount of nitrogen present in the fuel oil according to ultimate analysis of a fuel oil in %
+S=0.7;//The amount of sulphur present in the fuel oil according to ultimate analysis of a fuel oil in %
+td=27;//The dry bulb temperature of combustion air in degree centigrade
+tw=21;//The wet bulb temperature of combustion air in degree centigrade
+E=0.3;//Excess air and assuming complete combustion in %
+t=200;//Temperature to find total volume of combustion products in degree centigrade
+p=1.013;//Pressure to find total volume of combustion procucts in bar
+
+//Calculations
+Wth=(11.5*(C/100))+[34.5*((H/100)-(O/100)*(1/8))]+(4.3*(S/100));//Theoretical air required per kg of fuel in kg
+WA=(1+E)*Wth;//Actual air required per kg of fuel in kg/kg fuel
+sh=0.0132;//Specific humidity at DBT and WBT in kg moisture/kg dry air
+W=WA*sh;//Water vapour entering with air per kg fuel in kg vap/kg fuel
+Tw=(9*(H/100))+WA;//Total water vapour formed per kg fuel in kg
+CO2=(44/12)*(C/100);//mass of carbondioxide gas per kg of fuel
+O2=0.232*E*Wth;//Mass of oxygen gas per kg of fuel
+N2=0.768*(1+E)*Wth+(N/100);//Mass of nitrogen gas per kg of fuel
+SO2=(64/32)*(S/100);//Mass of nitrogen gas per kg of fuel
+H2O=1.383;//Mass of water per kg of fuel
+M=(CO2/44)+(O2/32)+(N2/28)+(SO2/64)+(H2O/18);//Moles of combustion gases formed per kg fuel
+VG=M*22.4*[(273+t)/273]*(1.013/1.013);//Volume of flue gases at 200 degree centigrade and 1.013 bar per kg fuel
+CO21=((CO2/44)/[(CO2/44)+(O2/32)+(N2/28)])*100;//Composition of dry flue gas CO2 by volume
+O21=((O2/32)/[(CO2/44)+(O2/32)+(N2/28)])*100;//Composition of dry flue gas O2 by volume
+N21=((N2/28)/[(CO2/44)+(O2/32)+(N2/28)])*100;//Composition of dry flue gas N2 by volume
+
+//Output
+printf('(a)The total volume of combustion products at 200 degee centigrade and 1.013 bar = %3.2f m^3 \n (b)The dry flue gas analysis based on carbondioxide,oxygen and nitrogen is \n Carbondioxide = %3.2f percent \n Oxygen = %3.2f percent \n Nitrogen = %3.2f percent',VG,CO21,O21,N21)
diff --git a/1943/CH4/EX4.6/Ex4_6.sce b/1943/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 000000000..dfacec0b8 --- /dev/null +++ b/1943/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,33 @@ + +clc
+clear
+//Input data
+C2H6=22.6;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %
+CH4=73.6;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %
+CO2=2.4;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %
+N2=1.4;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %
+E=0.25;//Assuming combustion air to be dry and in excess
+t=260;//The temperature for the total gas volume for complete combustion in degree centigrade
+p=1.013;//The pressure for the total gas volume for complete combustion in bar
+Mch=30;//Molecular weight of C2H6
+Mc=16;//Molecular weight of CH4
+Mco=44;//Molecular weight of CO2
+Mn=28;//Molecular weight of N2
+Mo=32;//Molecular weight of O2
+Mh=18;//Molecular weight of H2O
+
+//Calculations
+x=100;//Assuming 100 moles of fuel gas
+Mf=[(C2H6/100)*Mch]+[(CH4/100)*Mc]+[(N2/100)*Mn]+[(CO2/100)*Mco];//Molecular weight of fuel gas
+Ma=[(226.3*(Mo+(3.76*Mn))*(1+E))]/28.96;//Moles of air supplied
+Mc=1440;//Moles of combustion gas from the equation
+Mr=x+Ma+Mc;//Total reaction molecules
+Mwc=[(121.2*Mco)+(215*Mh)+(56.6*Mo)+(1065.4*Mn)]/Mc;//Molecular weight of combustion gas in kg/kgmol
+Mt=Mc/(x*20);//Total number of moles of combustion gas per kg fuel gas
+VG=Mt*22.4*[(273+t)/273];//Volume of combustion products per kg fuel gas
+CO21=(121.2/(121.2+56.6+1065.4))*100;//Gas analysis of CO2 by volume
+O21=(56.6/1243.2)*100;//Gas analysis of O2 by volume
+N2=(1065.4/1243.2)*100;//Gas analysis of N2 by volume
+
+//Output
+printf('(a)The molecular weight of the combustion products M = %3.2f kg/kg mol \n (b) The total gas volume for complete combustion at 260 degree centigrade and 1.013 bar is %3.2f m^3/kg fuel \n (c)The dry flue gas analysis on \n carbondioxide = %3.1f percent \n oxygen = %3.1f percent \n nitrogen = %3.1f percent ',Mwc,VG,CO21,O21,N2)
diff --git a/1943/CH4/EX4.7/Ex4_7.sce b/1943/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..cc1cdd3d6 --- /dev/null +++ b/1943/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,19 @@ + +clc
+clear
+//Input data
+CO21=9.7;//Carbondioxide gas analysis before the air preheater
+CO22=9.2;//Carbondioxide gas analysis after the air preheater
+O21=4.0;//Oxygen gas analysis before the air preheater
+O22=4.9;//Oxygen gas analysis after the air preheater
+N21=86.3;//Nitrogen gas analysis before the air preheater
+N22=85.9;//Nitrogen gas analysis after the air preheater
+C=72;//The coal used shows the carbon percentage by mass in %
+
+//Calculations
+W1=[(3.04)*(N21/100)*(C/100)]/[(CO21/100)];//Before air preheater in kg air/kg fuel
+W2=[(3.04)*(N22/100)*(C/100)]/[(CO22/100)];//After air preheater in kg air/kg fuel
+A=W2-W1;//Air leakage in kg air/kg fuel
+
+//Output
+printf('The air leakage into the air preheater per kg of coal fired is %3.0f kg air/kg fuel',A)
diff --git a/1943/CH4/EX4.8/Ex4_8.sce b/1943/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..65f3383a4 --- /dev/null +++ b/1943/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,31 @@ + +clc
+clear
+//Input data
+n=6;//Total lancashire boilers in a textile factory
+Ws=6;//Each boiler supplying steam in t/h
+p=16;//Pressure at which steam is supplied in bar
+t=250;//Temperature at which steam is supplied in degree centigrade
+CV=43960;//Calorific value of the fuel oil in kJ/kg
+no=75;//Overall efficiency of the boiler in %
+a=16;//The amount of air required for efficient burning of the fuel inkg
+H=20;//Drought of water gauge required at the base of chimney in mm
+tf=320;//The flue gases leave the boiler in degree centigrade
+ts=300;//The average temperature of the gases in the stack in degree centigrade
+ta=30;//The atmospheric temperature in degree centigrade
+R=0.287;//Real Gas constant in kJ/kgK
+h1=2919.2;//enthalpy at the entrance of the boiler in kJ/kgK
+hf=125.8;//Enthalpy at the feed in kJ/kgK
+pi=3.1412;//Mathematical constant
+g=9.81;//gravitational fore constant in m/s^2
+P=1.013;//Atmospheric pressure in bar
+
+//Calculations
+H1=[(H*R*(273+ta)*(273+ts))]/[P*100*[(273+ts)-(273+ta)]];//The draught produced in m
+Wf=[[(Ws*1000)*6*(h1-hf)]/((no/100)*CV)]/3600;//Air fuel ratio in kg/s
+Wa=a*Wf;//Actual air fuel ratio in kg/s
+Wfg=17*Wf;//Air fuel ratio in kg/s
+D=[[(Wfg*R*(273+ts)*(4/pi))]/[(101.3)*(2*g*H1)^(1/2)]]^(1/2);//Diameter at its base in m
+
+//Output
+printf('(a)The height of the stack H = %3.2f m \n (b)The diameter at its base D = %3.2f m ',H1,D)
diff --git a/1943/CH4/EX4.9/Ex4_9.sce b/1943/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..83570f703 --- /dev/null +++ b/1943/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,27 @@ + +clc
+clear
+//Input data
+Wf=10;//Coal rate in t/h
+C=78;//The mass of carbon present in the coal according to coal analysis on mass basis in %
+H=3;//The mass of hydrogen present in the coal according to coal analysis on mass basis in %
+O=3;//The mass of oxygen present in the coal according to coal analysis on mass basis in %
+S=1;//The mass of sulphur present in the coal according to coal analysis on mass basis in %
+M=7;//The mass of moisture present in the coal according to coal analysis on mass basis in %
+A=8;//The mass of ash present in the coal according to coal analysis on mass basis in %
+E=0.3;//Excess air in percentage
+p=180;//Plenum chamber pressure in mm water gauge
+nm=0.6;//Mechanical efficiency of the fan
+ta=30;//Room temperature in degree centigrade
+R=0.287;//Real gas constant
+P=101.325;//Atmospheric pressure in kPa
+g=9.812;//gravitational force constant m/s^2
+
+//Calculations
+Wth=(11.5*(C/100))+(34.5*[(H/100)-(O/(8*100))])+(4.3*(S/100));//Theoretical air required per kg fuel in kg air/kg fuel
+WA=Wth*(1+0.3);//Actual air required per kg fuel in kg air/kg fuel
+Va=(R*(273+ta))/P;//Volume flow rate of air in m^3/kg
+FD=((WA*Wf*1000*Va*p*g)/(3600*nm))/1000;//FD fan motor capacity in kW
+
+//Output
+printf('The required motor capacity needed for the FD fan is %3.2f kW ',FD)
diff --git a/1943/CH5/EX5.1/Ex5_1.sce b/1943/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..8caf9211e --- /dev/null +++ b/1943/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,14 @@ + +clc
+clear
+//Input data
+Vs=2500;//The mass of a bed of solid particles in kg
+p=2650;//The density of the solid in kg/m^3
+d=800*10^-6;//The mean particle size in m
+s=0.84;//The sphericity of the particle
+
+//Calculations
+As=(6*Vs)/(p*d*s);//The total surface area of the particles in the bed
+
+//Output
+printf(' The total surface area of the particles in the bed As = %3.0f m^2 ',As)
diff --git a/1943/CH5/EX5.2/Ex5_2.sce b/1943/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..c04c4f6dd --- /dev/null +++ b/1943/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+d=427*10^-6;//The mean particle size in m
+pg=1.21;//The density of air in kg/m^3
+v=1.82*10^-5;//The viscosity of air in kg/ms
+pl=1620;//The density of the loosely packed bed in kg/m^3
+ps=2780;//The density of the solids in kg/m^3
+c1=27.2;//(Grace,1982)constant value.
+c2=0.0408;//(Grace,1982)constant value
+g=9.812;//Gravitational forc constant in m/s^2
+
+//Calculations
+E=1-(pl/ps);//The voidage of the bed
+Ar=[(pg)*(ps-pg)*g*(d^3)]/v^2;//Archimedes number
+Re=[c1^2+(c2*Ar)]^(0.5)-c1;//Reynolds number
+Umf=Re*v/(pg*d);//Minimum superficial velocity in m/s
+
+//Output
+printf('(a) The voidage of the bed = %3.3f \n (b) The minimum fluidization velocity Umf = %3.3f m/s ',E,Umf)
diff --git a/1943/CH5/EX5.3/Ex5_3.sce b/1943/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..52d1b6f52 --- /dev/null +++ b/1943/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,24 @@ + +clc
+clear
+//Input data
+d=427*10^-6;//The mean particle size in m
+pg=1.21;//The density of air in kg/m^3
+v=1.82*10^-5;//The viscosity of air in kg/ms
+Umf=0.14;//Minimum superficial velocity in m/s
+Ar=7753;//Archimedes number from previous example problem
+
+//Calculations
+Re=(Umf*pg*d)/v;//Reynolds number
+function[f] = F(x);//function definition
+ f = 7753*x^2- 381.1*x -4793;
+endfunction
+x = 100;//Initial guss
+function[z] = D(x)//Derivative
+ z= 3*x^2 - 3;
+endfunction
+y = fsolve(x,F, D);
+
+//Output
+printf('The sphericity of particles is = %3.3f ',y)
+
diff --git a/1943/CH5/EX5.4/Ex5_4.sce b/1943/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..be366b709 --- /dev/null +++ b/1943/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,22 @@ + +clc
+clear
+//Input data
+O=35;//The output of the fluidized bed combustion system in MW
+n=0.80;//Efficiency of the fluidized bed combustion system
+H=26;//The heating value of coal in MJ/kg
+S=3.6;//Sulphur content in the coal in %
+C=3;//The calcium sulphur ratio
+Ca=85;//The amount of calcium carbonate in the limestone in %
+CaCO3=100;//The molecular weight of CaCO3
+
+//Calculations
+Cb=O/(n*H);//Coal burning rate in kg/s
+Cb1=Cb*3600;//Coal burning rate in kg/h
+Sf=(Cb1*(S/100))/32;//Flow rate of sulphur in Kmol/h
+Cf=Sf*C;//The flow rate of calcium in Kmol/h
+Caf=Cf*CaCO3;//Mass flow rate of CaCO3 in kg/h
+L=Caf/(Ca/100);//Mass flow rate of limestone in kg/h
+
+//Output
+printf('The required flow rate of limestone is %3.1f kg/h ',L)
diff --git a/1943/CH5/EX5.5/Ex5_5.sce b/1943/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..7cfceff4e --- /dev/null +++ b/1943/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,20 @@ + +clc
+clear
+//Input data
+CV=24;//The calorific value of the fuel in MJ/kg
+C=0.65;//The amount of calorific value released in the bed in %
+to=850;//Temperature at which products leave in degree centigrade
+ti=30;//The inlet temperature in degree centigrade
+tb=850;//The bed temperature in degree centigrade
+A=14.5;//The air fuel ratio by mass
+Cp=1.035;//The specific heat of the products leaving the bed surface in kJ/kgK
+B=7000;//The burning rate of coal in kg/h
+
+//Calculations
+H=(C*CV*1000)-(A*Cp*(to-ti));//Heat removal from the bed per kg fuel in kJ/kg fuel
+Hr=(H*B)/3600;//Rate of heat removal from the bed in kW
+Hb=(B/3600)*(1-C)*CV*1000;//The rate of heat removal from the above bed zone in kW
+
+//Output
+printf('(a) The rate of heat removal from the bed = %3.0f kW \n (b) The rate of heat removal from the above bed zone = %3.0f kW ',Hr,Hb)
diff --git a/1943/CH5/EX5.6/Ex5_6.sce b/1943/CH5/EX5.6/Ex5_6.sce new file mode 100755 index 000000000..12ab49288 --- /dev/null +++ b/1943/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,22 @@ + +clc
+clear
+//Input data
+tb=850;//The bed temperature in degree centigrade
+CV=25;//The calorific value of the fuel in MJ/kg
+A=9.5;//The stoichiometric air fuel ratio by mass
+E=20;//The amount of excess air used in %
+F=4.8;//The total fueling rate in MW
+p=0.3145;//The density of air at bed temperature in kg/m^3
+f=2;//The firing rate in MW/m^2
+v=2.7;//The fluidizing velocity in m/s
+
+//Calculations
+P=F/f;//Planform area in m^2
+m=(F*1000)/(CV*1000);//Fuel burning rate in kg/s
+ma=A*(1+(E/100))*m;//Mass flow rate of air in kg/s
+Pa=ma/(p*v);//Planform area in m^2
+
+//Output
+printf('(a) The planform area = %3.1f m^2 \n (b) Fuel burning rate = %3.3f kg/s \n Air flow rate = %3.4f kg/s \n Planform area = %3.2f m^2 ',P,m,ma,Pa)
+
diff --git a/1943/CH6/EX6.1/Ex6_1.sce b/1943/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..b70022200 --- /dev/null +++ b/1943/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,36 @@ + +clc
+clear
+//Input data
+H=18;//The length of furnace wall riser in m
+O=76.2;//The outer diameter of the furnace wall riser in mm
+T=6.1;//The thickness of the furnace wall riser in mm
+P=80;//Pressure at which saturated water is recieved in bar
+V=1.5;//The velocity of the saturated water in m/s
+CR=12.5;//Assuming circulation ratio
+S=1.2;//Assuming slip ratio
+g=9.81;//Gravitational force constant in m/s^2
+pi=3.142;//Mathematical constant
+
+//Calculations
+xt=1/CR;//The quality of steam at the top of the riser
+vf=0.001384;//Specific volume of saturated liquid at 80 bar in m^3/kg
+vfg=0.02214;//Specific volume of Evaporation gas at 80 bar in m^3/kg
+vg=0.02352;//Specific volume of saturated gas at 80 bar in m^3/kg
+pf=1/vf;//Density of the saturated liquid at 80 bar in kg/m^3
+vt=vf+(xt*vfg);//Specific volume of the steam at the top of the riser in m^3/kg
+pt=1/vt;//Density of steam at the top of the riser in kg/m^3
+pm=(pt+pf)/2;//Mean density in kg/m^3
+Ph=[H*g*(pf-pm)]/1000;//The pressure head developed in kPa
+C=(vf/vg)*S;//The part of calculation for the void fraction
+VF=1/[1+((1-xt)*C)/xt];//The void fraction at riser exit
+hfg=1441.3;//Enthalpy of the evaporation in kJ/kg
+di=O-12.2;//Inner diameter of the furnace wall riser in mm
+A=(pi*di^2)/4;//Inner area in m^2
+w=pf*A*V;//Mass flow rate of saturated water entering the riser in kg/s
+ws=xt*w;//The rate of steam formation in the riser tube in kg/s
+h=[(ws*hfg)/(O*H)]/1000;//Heat transfer rate per unit projected area in kW/m^2
+
+//output
+printf('(a) The pressure head developed = %3.1f kPa \n (b)The void fraction at riser exit = %3.4f \n (c) The heat transfer rate per unit projected area = %3.1f kW/m^2 ',Ph,VF,h)
+
diff --git a/1943/CH6/EX6.2/Ex6_2.sce b/1943/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..fb9cd8c51 --- /dev/null +++ b/1943/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,19 @@ + +clc
+clear
+//Input data
+t=60;//The temperature of water while supplying it to desuperheater in degree centigrade
+ws=200;//The amount of steam carrying in a steam line in t/h
+p=35;//The pressure of steam in bar
+ts=400;//The temperature to be maintained by the steam in degree centigrade
+to=450;//The outlet temperature of the steam from boiler in degree centigrade
+h1=3337.2;//The enthalpy of steam at 450 degree C in kJ/kg
+h2=252;//The enthalpy of water at 60 degree C in kJ/kg
+h3=3222.3;//The enthalpy of steam at 400 degree C in kJ/kg
+
+//Calculations
+w=(ws*(h1-h3))/(h3-h2);//Mass flow rate of water in t/h
+w1=w*(1000/3600);//Mass flow rate of water in kg/s
+
+//Output
+printf('The amount of water that must be sprayed is %3.3f t/h or %3.3f kg/s ',w,w1)
diff --git a/1943/CH6/EX6.3/Ex6_3.sce b/1943/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..cd844c174 --- /dev/null +++ b/1943/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,23 @@ + +clc
+clear
+//Input data
+H=15;//The high of downcomer riser circuit in m
+P=160;//The pressure at which downcomer riser circuit operates in bar
+xe=0.5;//The exit quality of the steam
+S=1.2;//Slip factor
+vf=0.001711;//Specific volume of saturated liquid in m^3/kg
+vg=0.009306;//Specific volume of saturated gas in m^3/kg
+g=9.806;//Gravitational force constant in m/s^2
+
+//Calculations
+C=S*(vf/vg);//The part of calculation for the void fraction
+VF=1/[1+((1-xe)*C)/xe];//The void fraction at riser exit
+pf=1/vf;//Density of the saturated liquid in kg/m^3
+pg=1/vg;//Density of the saturated gas in kg/m^3
+pm=pf-[[(pf-pg)/(1-C)]*[1-{(1/((VF)*(1-C)))-1}*log(1/(1-(VF*(1-C))))]];//The average mixture density in the riser in kg/m^3
+P1=g*(pf-pm)*H;//Pressure head developed due to natural circulation in N/m^2
+P2=P1/1000;//ressure head developed due to natural circulation in kPa
+
+//Output
+printf('The pressure head developed due to natural circulation is %3.0f N/m^2 or %3.3f kPa',P1,P2)
diff --git a/1943/CH6/EX6.4/Ex6_4.sce b/1943/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..5e320b124 --- /dev/null +++ b/1943/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,48 @@ + +clc
+clear
+//Input data
+W=120;//The amount of electricity produced in the power plant in MW
+po=100;//The pressure of the steam at the outlet of boiler in bar
+to=500;//The temperature of steam at the outlet of boiler in degree centigrade
+p=0.1;//The condenser pressure in bar
+nb=0.9;//The efficiency of the boiler
+CV=25.7;//The calorific value of the coal in MJ/kg
+ti=160;//The feed water temperature at boiler inlet in degree centigrade
+H=40;//The high of the risers in the furnace wall in m
+xt=0.08;//The quality of the steam at the top of the riser
+v=2;//The exit velocity of the riser and entering the drum in m/s
+Do=60;//The outer diameter of the risers in mm
+T=3;//The thickness of the wall in mm
+pi=3.142;//Mathematical constant
+g=9.806;//Gravitational force constant in m/s^2
+
+//Calculations
+h1=3374.8;//Enthalpy at point 1 in kJ/kg
+s1=6.6011;//Entropy at point 1 in kJ/kgK
+sf=0.6479;//Entropy of the saturated liquid at point 1 in kJ/kgK
+sg=7.5055;//Entropy of the Saturated vapour at point 1 in kJ/kgK
+x2=(s1-sf)/sg;//The quality of the steam
+h2=191.46+(x2*2393.29);//Enthalpy at point 2 in kJ/kg
+h3=191.46;//Enthalpy at point 3 in kJ/kg
+h5=675.5;//Enthalpy at point 5 in kJ/kg
+ws=(W*1000)/(h1-h2);//Mass flow rate of steam in kg/s
+wf=(ws*(h1-h5))/(nb*CV*1000);//Mass flow rate of fuel in kg/s
+E=ws/wf;//Evaporation factor
+vf=0.0014523;//The specific volume of saturated liquid in m^3/kg
+vg=0.0165884;//The specific volume of saturated vapour in m^3/kg
+vt=vf+(xt*vg);//Specific volume at the top in m^3/kg
+pt=1/vt;//Density of the steam at the top in kg/m^3
+pf=1/vf;//The density of the steam in kg/m^3
+pm=(pf+pt)/2;//The average mixture density in kg/m^3
+H1=[g*H*(pf-pm)]/10^5;//Pressure head available for natural circulation in bar
+CR=1/xt;//Circulation ratio
+di=(Do-(2*T))/1000;//The inner diameter of the riser in m
+A=(pi*di^2)/4;//Area for the inner diameter in m^2
+w=(A*pt*v*xt);//The rate of steam formation in the riser in kg/s
+Nr=ceil(ws)/w;//The number of risers
+hfg=1319.8;//Enthalpy of the evaporation in kJ/kg
+Ha=(w*hfg)/((Do/1000)*H);//Heat absorption rate per unit projected area of the riser in kW/m^2
+
+//Output
+printf('(a)The steam generation rate = %3.3f kg/s \n (b) The fuel burning rate = %3.3f kg/s \n (c) The evaporation factor = %3.2f \n (d) The pressure head available for natural circulation = %3.4f bar \n (e) The circulation ratio = %3.1f \n (f)The number of risers required = %3.0f \n (g) The heat absorbtion rate per unit projected area of the riser = %3.2f kW/m^2',ws,wf,E,H1,CR,Nr,Ha)
diff --git a/1943/CH6/EX6.5/Ex6_5.sce b/1943/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..612c5bd29 --- /dev/null +++ b/1943/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,26 @@ + +clc
+clear
+//Input data
+ws=64;//The steam flow rate in kg/s
+p=60;//The pressure at which steam leaves the boiler in bar
+m=0.02;//Moisture contant in the steam
+wf=62;//The feedwater flow rate in kg/s
+Pf=3;//concentration of feedwater in ppm
+wm=2;//The flow rate of makeup water
+Pm=50;//concentration of makeup water in ppm
+Ps=5;//Leaving the drum water in ppm
+Pw=1000;//The concentration in the drum water in ppm
+mf=7;//The fuel burning rate in kg/m
+CV=23;//The heating value in MJ/kg
+ta=30;//The room temperature in degree centigrade
+hf=1213.35;//Enthalpy of saturated liquid at 60 bar in kJ/kg
+ha=125.79;//Enthalpy at ambient temperature in kJ/kg
+
+//Calculations
+BD=[(wf*Pf)+(wm*Pm)-(m*ws*Ps)]/1000;//The rate of blowdown in kg/s
+E=[(BD*(hf-ha))/(mf*CV*1000)]*100;//The energy loss in blowdown in percentage
+S=m*ws*Ps*10^-6*3600*24;//Scale deposition in superheater tubes
+
+//Output
+printf('(a)The blowdown required = %3.4f kg/s \n (b) Heat loss in blowdown as a percentage of total heat released in the furnace = %3.2f percentage \n (c) The deposition of scale in superheater tube = %3.3f kg/day ',BD,E,S)
diff --git a/1943/CH6/EX6.6/Ex6_6.sce b/1943/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..95db5326b --- /dev/null +++ b/1943/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,37 @@ + +clc
+clear
+//Input data
+ws=600;//Mass flow rate of feedwater in kg/s
+p=140;//The inlet pressure of the feedwater in bar
+t=170;//The inlet temperature of the feedwater in degree centigrade
+wg=1250;//The mass flow rate of flue gases in kg/s
+tg2=450;//The temperature at which flue gases leave the economisers coils in degree centigrade
+Vf=12;//The velocity of the flue gas in m/s
+Vw=1.2;//The velocity of the water leaving the coil in m/s
+Do=0.07;//The outer diameter of the tube in m
+Di=0.06;//The inner diameter of the tube in m
+U=70;//The overall heat transfer coefficient in W/m^2K
+Cp=1.12;//The specific heat capacity of the flue gases in kJ/kgK
+V=0.08;//The vertical pitch of the coil in m
+B=4.8;//The width of the duct in m
+C=0.005;//The clearence on the both sides of the duct in m
+pi=3.142;//Mathematical constant
+
+//Calculations
+hf=1571.1;//The enthalpy of the saturated liquid at 140 bar in kJ/Kg
+ts=336.75;//The saturated temperature at 140 bar in degree centigrade
+vf=0.001611;//The specific volume of the saturated liquid at 140 bar in m^3/kg
+hf1=719.21;//The enthalpy of the saturated liquid at 170 degree C in kJ/kg
+vf1=0.001114;//The specific volume of the saturated liquid at 170 degree C in m^3/kg
+tg1=[(ws*(hf-hf1))/(wg*Cp)]+tg2;//The temperature at which flue gases enters the economisers coils in degree centigrade
+t1m=(478.25-280)/(log(478.25/280));//The mean temperature for inlet and exit temperature in degree centigrade
+Q=ws*(hf-hf1);//The rate of heat transfer in the economiser in kW
+Ao=[Q/(U*t1m)]*10^3;//The outer area in m^2
+n=[(ws*(vf/Vw)*(4/pi)*(1/Di^2))];//The number of coils needed in the economiser
+l=Ao/(n*pi*Do);//The length of one coil in m
+nt=l/(B-(2*C));//The number of turns in on ecoil
+VH=nt*V;//The vertical height of the duct occupied by the economiser coils
+
+//Output
+printf('(a) The number of coils needed in the economiser = %3.0f \n (b) The length of one coil = %3.1f m \n (c) The verticle height of the duct occupied by the economiser coils = %3.2f m ',n,l,VH)
diff --git a/1943/CH6/EX6.7/Ex6_7.sce b/1943/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 000000000..eca79643c --- /dev/null +++ b/1943/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,31 @@ + +clc
+clear
+//Input data
+tg2=160;//The temperature to which the flue gases are cooled in degree centigrade
+ta1=35;//The ambient temperature of the air in degree centigrade
+wa=1167;//The mass flow rate of air in kg/s
+Vg=13;//The inlet velocity of the flue gases in m/s
+U=30;//The overall heat transfer coefficient in W/m^2K
+Cpg=1.10;//The specific heat of the flue gas in kJ/kgK
+Cpa=1.005;//The specific heat of the air in kJ/kgK
+R=0.287;//Real gas constant in kJ/kgK
+wg=1250;//The mass flow rate of gas in kg/s
+tg1=450;//The temperature at the inlet of flue gas in degree centigrade
+P=101.325;//Atmospheric temperature in kPa
+pi=3.1414;//Mathematical constant
+Di=0.06;//The inner diameter of the tube in m
+Do=0.065;//The outer diameter of the tube in m
+
+//Calculations
+vg1=(R*(273+tg1))/P;//Specific volume of the gas in m^3/kg
+ta2=[(wg*Cpg*(tg1-tg2))/(wa*Cpa)]+ta1;//The temperature of the heated air in degree centigrade
+t1m=(75-125)/log(75/125);//The mean temperature of the inlet and exit temperature in degree centigrade
+Q=wg*Cpg*(tg1-tg2);//The rate of heat transfer in the economiser in kW
+Ao=[Q/(U*t1m)]*10^3;//The outer area in m^2
+n=[(wg*(vg1/Vg)*(4/pi)*(1/Di^2))];//The number of coils needed in the economiser
+l=Ao/(n*pi*Do);//The length of one coil in m
+
+//Output
+printf('(a)The length of the tubes = %3.2f m\n (b) The number of tubes = %3.0f ',l,n)
+
diff --git a/1943/CH6/EX6.8/Ex6_8.sce b/1943/CH6/EX6.8/Ex6_8.sce new file mode 100755 index 000000000..dcef6838b --- /dev/null +++ b/1943/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,25 @@ + +clc
+clear
+//Input data
+di=0.05;//The inner diameter of the superheater coil in m
+T=0.005;//The thickness of the coil in m
+p=60;//The pressure of the steam at the exit in bar
+t=500;//The temperature of the steam at the exit in degree centigrade
+V2=10;//The velocity of the steam at the exit in m/s
+ws=80;//The mass flow rate of steam in kg/s
+H=140;//The heat flux in the super heated coils in kW/m^2
+pi=3.142;//Mathematical constant
+Do=0.06;//The outer diameter of the tube in m
+
+//Calculations
+h1=2784.3;//The enthalpy of the saturated gas at 60 bar in kJ/kg
+h2=3422.2;//The enthalpy of the saturated gas at 500 degreeC in kJ/kg
+v2=0.05665;//The specific volume of gas at 500 degreeC in m^3/kg
+Q=ws*(h2-h1);//Heat absorption rate in superheater coil in kW
+Ao=Q/H;//Surface area required in m^2
+n=[(ws*(v2/V2)*(4/pi)*(1/di^2))];//The number of coils needed in the economiser
+l=Ao/(n*pi*Do);//The length of one coil in m
+
+//Output
+printf('(a)The length of the one coil = %3.2f m\n (b) The number of coils = %3.0f ',l,n)
diff --git a/1943/CH7/EX7.1/Ex7_1.sce b/1943/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..9b12b3f9e --- /dev/null +++ b/1943/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+p1=10//Initial pressure in bar
+T1=300+273//Initial temperature in K
+p2=2//Final pressure in bar
+m=1//Mass flow rate of steam in kg/s
+
+//Calculations
+px=(0.546*p1)//Critical pressure in bar
+ho=3052.2//Enthalpy in kJ/kg
+so=7.1229//Entropy in kJ/kg.K
+sx=so//Entropy in kJ/kg.K
+hx=2905.9//Enthalpy in kJ/kg
+vx=0.4125//Specific volume in m^3/kg
+Vx=(44.72*sqrt(ho-hx))//Critical velocity in m/s
+Ax=(vx/Vx)*10^4//Minimum area of the nozzle in sq.cm
+
+//Output
+printf('Minimum area of the nozzles is %3.3f sq.cm',Ax)
diff --git a/1943/CH7/EX7.10/Ex7_10.sce b/1943/CH7/EX7.10/Ex7_10.sce new file mode 100755 index 000000000..e45946eb3 --- /dev/null +++ b/1943/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,33 @@ + +clc
+clear
+//Input data
+a=20//Nozzle angle in degrees
+b2=30//Blade exit angle in degrees
+Vb=130//Mean blade speed in m/s
+V1=330//Velocity of steam in m/s
+f=0.8//Friction factor
+nn=0.85//Nozzle efficiency
+p1=20//Pressure in bar
+T1=250+273//Temperature in K
+p2=0.07//Pressure in bar
+rf=1.06//Reheat factor
+
+//Calculations
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Blade angle in degrees
+Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s
+Vr2=(f*Vr1)//Velocity in m/s
+dVw=(Vr1*cosd(b1))+(Vr2*cosd(b2))//Vecoity in m/s
+WD=(dVw*Vb)/1000//Workdone in kJ/kg
+nb1=((2*dVw*Vb)/V1^2)*100//Efficiency in percent
+nst=(nn*nb1)//Efficiency in percent
+nin=(nst*rf)*100//Efficiency in percent
+h1=2902.3//Enthalpy in kJ/kg
+s1=6.5466//Entropy in kJ/kg.K
+x2s=(s1-0.5582)/7.7198//Dryness fraction
+h2s=(163.16+x2s*2409.54)//Enthalpy in kJ/kg
+h12=(0.7041*(h1-h2s))//Change in enthalpy in kJ/kg
+n=ceil(h12/WD)//Number of stages
+
+//Output
+printf('(a) Work done in the stage per kg of steam is %3.2f kJ/kg \n Stage efficiency is %3.1f percent \n\n (b) Number of stages are %3.0f',WD,nst,n)
diff --git a/1943/CH7/EX7.11/Ex7_11.sce b/1943/CH7/EX7.11/Ex7_11.sce new file mode 100755 index 000000000..3570c3fdc --- /dev/null +++ b/1943/CH7/EX7.11/Ex7_11.sce @@ -0,0 +1,23 @@ + +clc
+clear
+//Input data
+d=800//Diameter in mm
+N=3000//Speed in rpm
+V1=300//Velocity in m/s
+a=20//Nozzle angle in degrees
+f=0.86//Frictional factor
+T=140//Axial thrust in N
+
+//Calculations
+Vb=((3.14*(d/1000)*N)/60)//Velocity in m/s
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Blade angle in degrees
+b2=b1//Blade angle in degrees
+Vr1=(V1*sind(a))/sind(b1)//Velocity in m/s
+Vr2=f*Vr1//Velocity in m/s
+w=(T/((Vr1*sind(b1))-(Vr2*sind(b2))))//Mass flow rate in kg/s
+dVw=(Vr2*cosd(b2))+(Vr1*cosd(b1))//Velocity in m/s
+P=(w*dVw*Vb*10^-3)//Power developed in kW
+
+//Output
+printf('Power deveoped in the blading is %3.2f kW',P)
diff --git a/1943/CH7/EX7.12/Ex7_12.sce b/1943/CH7/EX7.12/Ex7_12.sce new file mode 100755 index 000000000..688eca08f --- /dev/null +++ b/1943/CH7/EX7.12/Ex7_12.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+p1=15//Pressure in bar
+T1=300+273//Temperature in K
+p2=10//Pressure in bar
+nn=95//Nozzle efficiency in percent
+a=20//Nozzle angle in degrees
+x=5//The blade exit angle is 5 degrees less than the inlet angle
+f=0.9//Friction factor
+m=1350//Steam flow rate in kg/h
+
+//Calculations
+h1=3038.9//Enthalpy in kJ/kg
+s1=6.9224//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+t2s=250//Temperature in degree C
+h2s=2943.1//Enthalpy in kJ/kg
+V1=44.72*sqrt((nn/100)*(h1-h2s))//Velocity in m/s
+Vb=V1*(cosd(a)/2)//Velocity in m/s
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Blade angle in degrees
+b2=b1-x//Blade angle in degrees
+Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s
+Vr2=(f*Vr1)//Velocity in m/s
+dVw=(Vr1*cosd(b1))+(Vr2*cosd(b2))//Velocity in m/s
+dVa=(Vr1*sind(b1))-(Vr2*sind(b2))//Velocity in m/s
+Pa=(m/3600)*dVa//Axial thrust in N
+Pt=(m/3600)*dVw//Tangential thrust in N
+WD=(Pt*Vb*10^-3)//Diagram Power in kW
+dn=((WD*1000)/((1/2)*(m/3600)*V1^2))*100//Diagram efficiency in percent
+
+//Output
+printf('(a) Axial thrust is %3.2f N \n Tangential thrust is %3.2f N \n\n (b) Diagram Power is %3.3f kW \n\n (c) Diagram Efficiency is %3.1f percent',Pa,Pt,WD,dn)
diff --git a/1943/CH7/EX7.13/Ex7_13.sce b/1943/CH7/EX7.13/Ex7_13.sce new file mode 100755 index 000000000..a2881d8f5 --- /dev/null +++ b/1943/CH7/EX7.13/Ex7_13.sce @@ -0,0 +1,36 @@ + +clc
+clear
+//Input data
+V1=600//Velocity in m/s
+a=16//Nozzle angle in degrees
+Vb=120//Mean blade angle in degrees
+b2=18//Exit angle in degrees
+aa1=22//Exit angle in degrees
+b4=36//Exit angle in degrees
+m=5//Steam flow rate in kg/s
+f=0.85//Friction coefficient
+
+//Calculations
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Exit angle in degrees
+Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s
+Vr2=(f*Vr1)//Velocity in m/s
+a1=atand((Vr2*sind(b2))/((Vr2*cosd(b2))-Vb))//Angle in degrees
+V2=((Vr2*sind(b2))/sind(a1))//Velocity in m/s
+V3=(f*V2)//Velocity in m/s
+dVw1=(Vr1*cosd(b1))+(Vr2*cosd(b2))//Velocity in m/s
+dVa1=(V1*sind(a))-(V2*sind(a1))//Velocity in m/s
+b3=atand((V3*sind(aa1))/((V3*cosd(aa1))-Vb))//Angle in degrees
+Vr3=((V3*sind(aa1))/sind(b3))//Velocity in m/s
+Vr4=(f*Vr3)//velocity in m/s
+dVw2=(Vr3*cosd(b3))+(Vr4*cosd(b4))//Velocity in m/s
+dVa2=(V3*sind(aa1))-(Vr4*sind(b4))//Velocity in m/s
+udVw=(dVw1+dVw2)//Total velocity in m/s
+udVa=(dVa1+dVa2)//Total velocity in m/s
+Pt=(m*udVw*10^-3)//tangential thrust in kN
+Pa=(m*udVa*10^-3)//Axial thrust in kN
+WD=(Pt*Vb)//Power developed in kW
+nd=((2*udVw*Vb)/V1^2)*100//Diagram efficiency in percent
+
+//Output
+printf('(a) the tangential thrust is %3.3f kW \n (b) Axial thrust is %3.2f kN \n (c) Power developed is %3.2f kW \n (d) Diagram efficiency is %3.2f percent',Pt,Pa,WD,nd)
diff --git a/1943/CH7/EX7.14/Ex7_14.sce b/1943/CH7/EX7.14/Ex7_14.sce new file mode 100755 index 000000000..b6bb88921 --- /dev/null +++ b/1943/CH7/EX7.14/Ex7_14.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+a=17//Nozzle angle in degrees
+Vb=125//Blade velocity in m/s
+b2=22//Blade angle n degrees
+a1=26//Blade angle n degrees
+b4=30//Blade angle n degrees
+f=0.9//Friction factor
+a2=90//Axial angle in degrees
+
+//Calculations
+dVw=1040//Velocity in m/s from Velocity triangles Fig. E.7.14
+V1=575//Velocity in m/s from Velocity triangles Fig. E.7.14
+V4=75//Velocity of steam exiting stage in m/s from Velocity triangles Fig. E.7.14
+WD=(dVw*Vb)/1000//Diagram work in kJ/kg
+nd=((WD*1000)/((1/2)*V1^2))*100//Diagram efficiency in percent
+
+//Output
+printf('(a) Absolute velocity of steam leaving the stage is %3.0f m/s \n (b) the diagram work is %3.0f kJ/kg \n (c) the diagram efficiency is %3.2f percent',V4,WD,nd)
diff --git a/1943/CH7/EX7.15/Ex7_15.sce b/1943/CH7/EX7.15/Ex7_15.sce new file mode 100755 index 000000000..84fdbc19a --- /dev/null +++ b/1943/CH7/EX7.15/Ex7_15.sce @@ -0,0 +1,45 @@ + +clc
+clear
+//Input data
+p1=35//Pressure in bar
+T1=350+273//Temperature in K
+p2=0.07//Pressure in bar
+x=1/4//Fraction of drop in isentropic enthalpy
+a=20//Nozzle angle in degrees
+nn=88//Nozzle efficiency in percent
+y=0.2//Velocity ratio
+b2=30//Exit blade angle in degrees
+b4=30//Exit blade angle in degrees
+f=0.9//Friction coefficienct
+in=75//Internal efficiency of the turbine in percent
+
+//Calculations
+h1=3106.4//Enthalpy in kJ/kg
+s1=6.6643//Entropy in kJ/kg.K
+x2s=(s1-0.5582)/7.7198//dryness fraction
+h2s=(163.16+x2s*2409.54)//Enthalpy in kJ/kg
+dh=(h1-h2s)//Change in enthalpy in kJ/kg
+h13s=x*dh//Change in enthalpy in kJ/kg
+h13=(nn/100)*h13s//Change in enthalpy in kJ/kg
+V1=(44.72*sqrt(h13))//Velocity in m/s
+Vb=(y*V1)//Velocity in m/s
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Angle in degrees
+Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s
+Vr2=(f*Vr1)//Velocity in m/s
+dVw1=(Vr1*cosd(b1))+(Vr2*cosd(b2))//Velocity in m/s
+V2=sqrt((Vr2*sind(b2))^2+((Vr2*cosd(b2))-Vb)^2)//Velocity in m/s
+V3=f*V2//Velocity in m/s
+b3=atand((V3*sind(b2))/((V3*cosd(b2))-Vb))//Angle in degrees
+Vr3=((V3*sind(b2))/sind(b3))//Velocity in m/s
+Vr4=f*Vr3//Velocity in m/s
+dVw2=(Vr3*cosd(b3))+(Vr4*cosd(b4))//Velocity in m/s
+dVw=(dVw1+dVw2)//Velocity in m/s
+nb1=((2*dVw*Vb)/V1^2)*100//Efficiency in percent
+ns=(nn*nb1)/100//Efficiency in percent
+ht=(in/100)*dh//Total change in enthalpy in kJ/kg
+hc=(ns/100)*h13s//Total change in enthalpy in kJ/kg
+pp=(hc/ht)*100//Percentage of enthalpy
+
+//Output
+printf('Efficiency of first stage is %3.2f percent \n Percentage of the total power developed by the turbine is %3.2f percent',ns,pp)
diff --git a/1943/CH7/EX7.16/Ex7_16.sce b/1943/CH7/EX7.16/Ex7_16.sce new file mode 100755 index 000000000..80c82af05 --- /dev/null +++ b/1943/CH7/EX7.16/Ex7_16.sce @@ -0,0 +1,26 @@ + +clc
+clear
+//Input data
+R=50//Percentage of reaction
+b1=35//Angle in degrees
+q=b1//Angle in degrees
+b2=20//Angle in degrees
+a=b2//Angle in degrees
+N=1500//Speed in rpm
+d=0.67//Mean diameter in m
+p=1.5//Pressure in bar
+x=0.96//Dryness fraction
+w=3.6//Flow rate in kg/s
+
+//Calculations
+Vb=(3.14*d*N)/60//Velocity in m/s
+V1=(Vb*(sind(180-b1)/sind(b1-b2)))//Veocity in m/s
+Vr1=(Vb*(sind(b2)/sind(b1-b2)))//Velocity in m/s
+dVw=(V1*cosd(a))+(Vr1*cosd(q))//Velocity in m/s
+v1=(0.001052+x*1.15937)//Specific volume in m^3/kg
+hb=((w*v1)/(3.14*d*V1*sind(a)))*1000//Required height in mm
+P=(w*dVw*Vb)/1000//Power developed in kW
+
+//Output
+printf('(a) the required height of blading is %3.1f mm \n (b) the power developed by the ring is %3.3f kW',hb,P)
diff --git a/1943/CH7/EX7.17/Ex7_17.sce b/1943/CH7/EX7.17/Ex7_17.sce new file mode 100755 index 000000000..7a248a456 --- /dev/null +++ b/1943/CH7/EX7.17/Ex7_17.sce @@ -0,0 +1,25 @@ + +clc
+clear
+//Input data
+N=400//Speed in rpm
+P=5//Power in MW
+m=6//Flow rate in kg/kWh
+b2=20//Blade angle in degrees
+a=b2//Angle in degrees
+x=1.35//Velocity ratio
+p=1.2//Pressure in bar
+x1=0.95//Steam quality
+Dh=12//Ratio of Dm and hb
+
+//Calculations
+w=(m*P*1000)/3600//Mass flow rate in kg/s
+v1=(0.0010468+x1*1.454)//Specific volume in m^3/kg
+hb=((w*v1)/(Dh*3.14*x*((Dh*N)/60)*3.14*sind(a)))^(1/3)*1000//Blade height in mm
+Vb=((3.14*Dh*(hb/1000)*N)/60)//velocity in m/s
+V1=(x*Vb)//Velocity in m/s
+dVw=((2*V1*cosd(a))-Vb)//velocity in m/s
+WD=(w*dVw*Vb*10^-3)//Diagram power in kW
+
+//Output
+printf('(a)Blade height is %3.0f mm \n (b) the diagram power is %3.2f kW',hb,WD)
diff --git a/1943/CH7/EX7.18/Ex7_18.sce b/1943/CH7/EX7.18/Ex7_18.sce new file mode 100755 index 000000000..2250a3224 --- /dev/null +++ b/1943/CH7/EX7.18/Ex7_18.sce @@ -0,0 +1,28 @@ + +clc
+clear
+//Input data
+N=3000//Speed in rpm
+Vb=100//Mean blade speed in m/s
+x=0.56//Velocity ratio
+a=20//Blade angle in degrees
+b2=a//Blade angle in degrees
+v=0.65//Specific volume in m^3/kg
+h=25//Mean height in mm
+n=5//Number of pairs of blades
+
+//Calculations
+V1=(Vb/x)//Velocity in m/s
+Vr2=V1//Velocity in m/s
+Dm=(Vb*60)/(3.14*N)//Diameter in m
+w=((3.14*Dm*h*V1*sind(a))/v)/1000//Mass flow rate in kg/s
+ws=(w*3600)//Mass flow rate in kg/hr
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Blade angle in degrees
+Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s
+dhmb=(1/2)*(Vr2^2-Vr1^2)/1000//Change in enthalpy in kJ/kg
+dsta=(2*dhmb)//Change in enthalpy of stage in kJ/kg
+dsta5=(n*dsta)//Total Change in enthalpy of stage in kJ/kg
+Dp=(w*dsta5)//Diagram power in kW
+
+//Output
+printf('Mass flow rate of steam is %3.3f kg/s \n Useful enthalpy drop is %3.2f kJ/kg \n The diagram power is %3.1f kW',w,dsta5,Dp)
diff --git a/1943/CH7/EX7.19/Ex7_19.sce b/1943/CH7/EX7.19/Ex7_19.sce new file mode 100755 index 000000000..398435d6b --- /dev/null +++ b/1943/CH7/EX7.19/Ex7_19.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+P=8//Power in MW
+N=5000//Speed in rpm
+p=40//pressure in bar
+T=500//Temperature in degree C
+p2=0.1//Pressure in bar
+in=0.85//Internal efficiency of turbine
+nm=0.96//Mechanical efficiency
+nn=0.92//Nozzle efficiency
+a=15//Nozzle angle in degrees
+Vb=300//Blade velocity in m/s
+
+//Calculations
+V1=(2*Vb)/cosd(a)//Velocity in m/s
+dh=((V1/44.72)^2/nn)//Change in enthalpy in kJ/kg
+h1=3445.3//Enthalpy in kJ/kg
+s1=7.0901//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+x2=(s2-0.6493)/7.5009//Dryness fraction
+h2s=(191.83+x2*2392.8)//Enthalpy in kJ/kg
+h12s=(h1-h2s)//Change in enthalpy in kJ/kg
+n=(h12s/dh)//Number of stages
+w=((P*1000)/(in*nm))/h12s//Mass flow rate in kg/s
+h13=(nn*dh)//Change in enthalpy in kJ/kg
+h3=h1-h13//Enthalpy in kJ/kg
+v3=0.17//Specific volume in m^3/kg
+A=(w*v3)/V1//Area in m^2
+hm=(A/(((Vb*60)/N)*sind(a)))*1000//Height in mm
+
+//Output
+printf('(a) the number of impulse stages are%3.0f \n (b) the nozzle height is %3.1f mm',n,hm)
diff --git a/1943/CH7/EX7.2/Ex7_2.sce b/1943/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..45bb2b82c --- /dev/null +++ b/1943/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,35 @@ + +clc
+clear
+//Input data
+p1=10//Initial pressure in bar
+T1=300+273//Initial temperature in K
+p2=1//Final pressure in bar
+x=0.15//Friction loss of the isentropic enthalpy drop
+ms=1//Steam flow rate in kg/s
+d=25//Exit diameter of the nozzles in mm
+
+//Calculations
+px=(0.546*p1)//Critical pressure in bar
+h1=3052.2//Enthalpy in kJ/kg
+s1=7.1276//Entropy in kJ/kg
+s2s=s1//Entropy in kJ/kg
+h2s=2916.2//Enthalpy in kJ/kg
+Vx=(44.72*sqrt(h1-h2s))//Critical velocity in m/s
+h3s=2605//Enthalpy in kJ/kg
+V1=(44.72*sqrt((h1-h2s)+(0.85*(h2s-h3s))))//Velocity in m/s
+s3s=s1//Entropy in kJ/kg
+x3s=(s3s-1.3025)/6.0579//Dryness fraction
+h3s=(417.46+(x3s*2258.01))//Enthalpy in kJ/kg
+h2s3=((1-x)*(h2s-h3s))//Enthalpy in kJ/kg
+h3=h2s-h2s3//Enthalpy in kJ/kg
+x3=(h3-417.46)/2258.01//Dryness fraction
+v3=(0.001043+(x3*1.694))//Specific volume in m^3/kg
+v2s=0.416//Specific volume in m^3/kg
+vx=v2s//Specific volume in m^3/kg
+Ax=(ms/Vx)*vx*10^4//Minimum area in cm^2
+A1=(ms*v3)/V1*10^4//Area in cm^2
+n=(A1*4)/(3.14*(d/10)^2)//Number of nozzles
+
+//Output
+printf('Minimum area of the nozzles is %3.2f cm^2 \n the number of nozzles are %3.0f',Ax,n)
diff --git a/1943/CH7/EX7.20/Ex7_20.sce b/1943/CH7/EX7.20/Ex7_20.sce new file mode 100755 index 000000000..a6662702b --- /dev/null +++ b/1943/CH7/EX7.20/Ex7_20.sce @@ -0,0 +1,24 @@ + +clc
+clear
+//Input data
+p=1.5//Pressure in bar
+x1=0.9//Dryness fraction
+m=7//Steam flow rate in kg/s
+N=3000//Turbine speed in rpm
+x=0.7//Velocity ratio
+y=0.75//Velocity ratio
+a=20//Exit angle in degrees
+b2=a//Angle in degrees
+hx=1/10//Fraction of height
+
+//Calculations
+v=0.001052+x1*1.15937//Specific volume in m^3/kg
+Dm=((m*v*60)/(3.14*hx*y*3.14*N))^(1/3)//Diameter in m
+hb=Dm*1000*hx//Height in mm
+Vb=(3.14*Dm*N)/60//Velocity in m/s
+dVw=((2*x*Vb*cosd(a)/sind(a))-Vb)//Velocity in m/s
+P=(m*dVw*Vb)/1000//Power developed in kW
+
+//Output
+printf('Height of the moving blades at exit is %3.1f mm \n Power developed in the blade row is %3.2f kW',hb,P)
diff --git a/1943/CH7/EX7.21/Ex7_21.sce b/1943/CH7/EX7.21/Ex7_21.sce new file mode 100755 index 000000000..f4c2ae929 --- /dev/null +++ b/1943/CH7/EX7.21/Ex7_21.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+p=40//Pressure in bar
+T=500//Temperature in degree C
+p1=0.1//Pressure in bar
+a=16//Nozzle angle in degrees
+N=3000//Speed in rpm
+
+//Calculations
+h1=3445.3//Enthalpy in kJ/kg
+s1=7.0901//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+x2s=(s2-0.6493)/7.5009//Dryness fraction
+h2s=(191.83+x2s*2392.8)//Enthalpy in kJ/kg
+V1=44.72*sqrt(h1-h2s)//Velocity in m/s
+Vb=V1*(cosd(a)/2)//Velocity in m/s
+Dm=(Vb*60)/(3.14*N)//Diameter in m
+V2=44.72*sqrt((h1-h2s)/2)//Velocity in m/s
+Vb2=V2*cosd(a)//Velocity in m/s
+Dm2=(Vb2*60)/(3.14*N)//Diameter in m
+V3=44.72*sqrt((h1-h2s)/4)//Velocity in m/s
+Vb3=V3*(cosd(a)/2)//Velocity in m/s
+Dm3=(Vb3*60)/(3.14*N)//Diameter in m
+V4=44.72*sqrt(h1-h2s)//Velocity in m/s
+Vb4=V4*(cosd(a)/4)//Velocity in m/s
+Dm4=(Vb4*60)/(3.14*N)//Diameter in m
+V5=44.72*sqrt((h1-h2s)/(2*4))//Velocity in m/s
+Vb5=V5*cosd(a)//Velocity in m/s
+Dm5=(Vb5*60)/(3.14*N)//Diameter in m
+
+//Output
+printf('The mean diameter of the wheel if the turbine were of \n (a) single impulse stage is %3.2f m \n (b) single 50 percent reaction stage is %3.1f m \n (c) four pressure (or Rateau) stages is %3.2f m \n (d) one two-row Curtis stage is %3.3f m \n (e) four stage 50 percent reaction stages is %3.2f m',Dm,Dm2,Dm3,Dm4,Dm5)
diff --git a/1943/CH7/EX7.22/Ex7_22.sce b/1943/CH7/EX7.22/Ex7_22.sce new file mode 100755 index 000000000..710a899f3 --- /dev/null +++ b/1943/CH7/EX7.22/Ex7_22.sce @@ -0,0 +1,32 @@ + +clc
+clear
+//Input data
+p=150//Pressure in bar
+T=600//Temperature in degree C
+Vb=300//Velocity in m/s
+nn=95//Nozzle efficiency in percent
+a=15//Nozzle angle in degrees
+a1=25//Angle in degrees
+
+//Calculations
+h1=3582.3//Enthalpy in kJ/Kg
+s1=6.6776//Entropy in kJ/kg.K
+s2=s1//Entropy in kJ/kg.K
+x2s=(s2-0.6493)/7.5009//Dryness fraction
+h2s=(191.83+x2s*2392.8)//Enthalpy in kJ/kg
+h12s=(h1-h2s)//Difference in enthalpy in kJ/kg
+V1=(Vb*2)/cosd(a)//Velocity in m/s
+dhs=(V1/44.72)^2/(nn/100)//Change in enthalpy in kJ/kg
+n1=ceil(h12s/dhs)//Number of stages
+V2=(Vb/cosd(a1))//Velocity in m/s
+dhs2=(V2/44.72)^2/(nn/(2*100))//Change in enthalpy in kJ/kg
+n2=h12s/dhs2//Number of stages
+V3=(Vb*4)/cosd(a)//Velocity in m/s
+dhs3=(V3/44.72)^2/(nn/100)//Change in enthalpy in kJ/kg
+dhhs3=(h12s-dhs3)//Change in enthalpy in kJ/kg
+n3=dhhs3/dhs//Number of stages
+n4=dhhs3/dhs2//Number of stages
+
+//Output
+printf('Number of stages required in \n (a) all simple impulse stages are %3.0f \n (b) all 50 percent reaction stages are %3.0f \n (c) a 2-row Cutris stage follwed by simple impulse stages are %3.0f \n (d) a 2-row Cutris stage followed by 50 percent reaction stages are %3.0f',n1,n2,n3,n4)
diff --git a/1943/CH7/EX7.23/Ex7_23.sce b/1943/CH7/EX7.23/Ex7_23.sce new file mode 100755 index 000000000..46ea919c8 --- /dev/null +++ b/1943/CH7/EX7.23/Ex7_23.sce @@ -0,0 +1,28 @@ + +clc
+clear
+//Input data
+p1=20//Pressure in bar
+T=400//Temperature in degree C
+p2=0.1//Pressure in bar
+n=4//Number of stages
+ns=75//Stage efficiency in percent
+
+//Calculations
+h16s=(3250-2282)//Change in enthalpy in kJ/kg
+h12s=(h16s/n)//Change in enthalpy in kJ/kg
+p=[8,2.6,0.6]//pressures in bar from Mollier chart
+h12=(ns/100)*h12s//Change in enthalpy in kJ/kg
+h23s=(3060-2800)//Change in enthalpy in kJ/kg
+h23=(ns/100)*h23s//Change in enthalpy in kJ/kg
+h34s=(2870-2605)//Change in enthalpy in kJ/kg
+h34=(ns/100)*h34s//Change in enthalpy in kJ/kg
+h45s=(2680-2410)//Change in enthalpy in kJ/kg
+h45=(ns/100)*h45s//Change in enthalpy in kJ/kg
+h5=2470//Enthalpy in kJ/kg
+rf=(h12s+h23s+h34s+h45s)/h16s//Reheat factor
+nth=((h12+h23+h34+h45)/h16s)*100//Internal efficiency in percent
+nin=(ns*rf)//Internal efficiency in percent
+
+//Output
+printf('The interstage pressures are %i bar, %3.1f bar, %3.1f bar \n The reheat factor is %3.3f \n The turbine internal efficiency is %3.1f percent',p(1),p(2),p(3),rf,nin)
diff --git a/1943/CH7/EX7.25/Ex7_25.sce b/1943/CH7/EX7.25/Ex7_25.sce new file mode 100755 index 000000000..85100d8c1 --- /dev/null +++ b/1943/CH7/EX7.25/Ex7_25.sce @@ -0,0 +1,29 @@ + +clc
+clear
+//Input data
+n=20//Number of stages
+P=12//Power in MW
+p=15//Pressure in bar
+T=350//Temperature in degree C
+p1=0.14//Pressure in bar
+ns=75//Stage efficiency in percent
+rf=1.04//Reheat factor
+p2=1//Pressure in bar
+a=20//Angle in degrees
+v=0.7//Velocity ratio
+h=1/12//Blade height in terms of mean blade diameter
+
+//Calculations
+nint=(ns/100)*rf//Internal efficiency
+dhs=855//Enthalpy in kJ/kg
+dha=ceil(nint*dhs)//Actual enthalpy change in kJ/kg
+w=(P*1000)/dha//Mass flow rate in kg/s
+Vb=(sqrt((dha/n)/((((2/v)*cosd(a))-1)*10^-3)))//Velocity in m/s//
+vg=1.694//Specific volume in m^3/kg
+Dm=sqrt((w*vg)/(3.14*h*(Vb/v)*sind(a)))//Diameter in m
+N=((Vb*60)/(3.14*Dm))//Speed in rpm
+
+//Output
+printf('(a) the flow rate of steam required is %3.2f kg/s \n (b) the mean blade diameter is %3.3f m \n (c) the speed of the rotor is %3.0f rpm',w,Dm,N)
+//In textbook, Vb is given wrong as 141.4 m/s instead of 140.6 m/s. Hence the answers are different.
diff --git a/1943/CH7/EX7.26/Ex7_26.sce b/1943/CH7/EX7.26/Ex7_26.sce new file mode 100755 index 000000000..e0e41290b --- /dev/null +++ b/1943/CH7/EX7.26/Ex7_26.sce @@ -0,0 +1,32 @@ + +
+clc
+clear
+//Input data
+V1=600//Velocity in m/s
+Vb=120//Mean blade velocity in m/s
+a=16//Nozzle angle in degrees
+b=[18,21,35]//Exit angles in degrees
+m=5//Steam flow rate in kg/s
+h=25//Nozzle height in mm
+v=0.375//Specific volume in m^3/kg
+p=25//Pitch in mm
+t=0.5//Thickness in mm
+kb=0.9//Constant
+
+//Calculations
+l=((m*v)/(sind(a)*V1*(h/1000)*kb))//Length of the nozzle arc in m //Length of the nozzle arc is calculated wrong as 0.454m instead of 0.5 m
+b1= atand((V1*sind(a))/((V1*cosd(a))-Vb))//Angle in degrees
+Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s
+Vr2=kb*Vr1//Velocity in m/s
+V2=sqrt(Vr2^2+Vb^2-2*Vr2*Vb*cosd(b(1)))//Velocity in m/s
+V3=291//Velocity in m/s
+b3=atand((V3*sind(b(2)))/((V3*cosd(b(2)))-Vb))//Angle in degrees
+Vr3=((V3*sind(b(2)))/sind(b3))//Velocity in m/s
+Vr4=(Vr3*kb)//Velocity in m/s
+hb1=((m*v*(h/1000))/(l*((p/1000)*sind(b(1))-(t/1000))*Vr2))*1000//Height in mm
+hn=((m*v*(h/1000))/(l*((p/1000)*sind(b(2))-(t/1000))*V3))*1000//Height in mm
+hb2=((m*v*(h/1000))/(l*((h/1000)*sind(b(3))-(t/1000))*Vr4))*1000//Height in mm
+
+//Output
+printf('Blade heights at the exit of each row: \n First row of moving blades is %3.1f mm \n Fixed row of guide blades is %3.1f mm \n Second row of moving blades is %3.1f mm',hb1,hn,hb2)
diff --git a/1943/CH7/EX7.27/Ex7_27.sce b/1943/CH7/EX7.27/Ex7_27.sce new file mode 100755 index 000000000..4b83a1369 --- /dev/null +++ b/1943/CH7/EX7.27/Ex7_27.sce @@ -0,0 +1,76 @@ + +clc
+clear
+//Input data
+P=200//Power in MW
+p=180//Pressure in bar
+T=550//Temperature in degree C
+P1=600//Power in MW
+p1=300//Pressure in bar
+T1=580//Temperature in degree C
+nt=90//Turbine efficiency in percent
+
+//Calculations
+h1=3430//Enthalpy in kJ/kg
+h2s=3040//Enthalpy in kJ/kg
+h12s=(h1-h2s)//Enthalpy in kJ/kg
+h12=(nt/100)*h12s//Enthalpy in kJ/kg
+h2=3070//Enthalpy in kJ/kg
+v2=0.06//Specific volume in m^3/kg
+h4=3560//Enthalpy in kJ/kg
+h3s=2000//Enthalpy in kJ/kg
+h13s=(h1-h3s)//Enthalpy in kJ/kg
+h13=(nt/100)*h13s//Enthalpy in kJ/kg
+w=(P*10^3)/h13//Mass flow rate in kg/s
+Vbm=350//Velocity in m/s
+N=3000//Speed in rpm
+a=25//Angle in degrees
+Dm=(Vbm*60)/(3.14*N)//Diameter in m
+hD=0.3//Assuming (hb/Dm)max
+hb=(hD*Dm)//Height in m
+Ab=(3.14*Dm*hb*0.9*sind(a))//Flow area in m^2
+V1=(Vbm/cosd(a))//Velocity in m/s
+Vo=(Ab*V1)//Volume flow rate in m^3/s
+v=(Vo/w)//Specific volume in m^3/kg
+h5s=2456//Enthalpy in kJ/kg
+p5=0.36//Pressure in bar
+h45s=(h4-h5s)//Enthalpy in kJ/kg
+h45=(nt/100)*h45s//Enthalpy in kJ/kg
+h5=h4-h45//Enthalpy in kJ/kg
+x5=0.952//Dryness fraction
+h56s=(h5-2340)//Enthalpy in kJ/kg
+h56=(nt/100)*h56s//Enthalpy in kJ/kg
+h6=h5-h56//Enthalpy in kJ/kg
+v6=18//Specific volume in m^3/kg
+mm=(Vo/v6)//Maximum mass flow that one stage can accommodate in kg/s
+np=(w/mm)//Number of parallel exhausts
+rp=(p1/4)//Reheat pressure in bar
+xh1=3410//Enthalpy in kJ/kg
+xh2s=3015//Enthalpy in kJ/kg
+xh12s=xh1-xh2s//Enthalpy in kJ/kg
+xh12=(nt/100)*xh12s//Enthalpy in kJ/kg
+xv2=0.035//Specific volume in m^3/kg
+xh4=3060//Enthalpy in kJ/kg
+xh3s=1960//Enthalpy in kJ/kg
+xh13s=xh1-xh3s//Enthalpy in kJ/kg
+xh3=(xh1-xh13s)//Enthalpy in kJ/kg
+xw=(P1*10^3)/xh13s//Mass flow rate in kg/s
+xvm=(Vo/xw)//Maximum specific volume in m^3/kg
+Vf=(xw*xv2)//Volume flow rate in m^3/s
+xh5s=2300//Enthalpy in kJ/kg
+xh45s=xh4-xh5s//Enthalpy in kJ/kg
+xh45=(nt/100)*xh45s//Enthalpy in kJ/kg
+xh5=xh4-xh45s//Enthalpy in kJ/kg
+xv5=1.25//Specific volume in m^3/kg
+xx5=0.86//Dryness fraction
+xh6s=2050//Enthalpy in kJ/kg
+xh56s=xh5-xh6s//Enthalpy in kJ/kg
+xh56=(nt/100)*xh56s//Enthalpy in kJ/kg
+xh6=(xh5-xh56)//Enthalpy in kJ/kg
+xv6=12//Specific volume in m^3/kg
+xx6=0.792//Dryness fraction
+xmm=(Vo/xv6)//Maximum mass flow in kJ/kg
+xnp=ceil(xw/xmm)//Number of parallel exhausts
+
+//Output
+printf('Number of parallel exhausts in : \n (a)condition a are %i \n (b)condition b are %i',np,xnp)
diff --git a/1943/CH7/EX7.28/Ex7_28.sce b/1943/CH7/EX7.28/Ex7_28.sce new file mode 100755 index 000000000..ade35ba92 --- /dev/null +++ b/1943/CH7/EX7.28/Ex7_28.sce @@ -0,0 +1,32 @@ + +clc
+clear
+//Input data
+P=100//Power in MW
+T=550//temperature in degree C
+p=0.1//Pressure in bar
+m=500000//Mass flow rate in kg/h at rated load
+mo=25000//Mass flow rate in kg/h at zero load
+x=[1/4,1/2,3/4,1]//Fraction of load
+
+//Calculations
+b=(m-mo)/(P*10^3)//Steam rate in kg/kWh
+y1=(x(1)*(P*10^3))//For one-fourth load
+s1=(mo/y1)+b//Steam rate in kg/kWh
+y2=(x(2)*(P*10^3))//For one-fourth load
+s2=(mo/y2)+b//Steam rate in kg/kWh
+y3=(x(3)*(P*10^3))//For one-fourth load
+s3=(mo/y3)+b//Steam rate in kg/kWh
+y4=(x(4)*(P*10^3))//For one-fourth load
+s4=(mo/y4)+b//Steam rate in kg/kWh
+h1=3511//Enthalpy in kJ/kg
+xs1=6.8142//Entropy in kJ/kg.K
+xs2=xs1//Entropy in kJ/kg.K
+x2s=(xs2-0.6493)/7.5009//Dryness fraction
+h2s=191.83+x2s*2392.8//Enthalpy in kJ/kg
+nR=((h1-h2s)/(h1-191.83))*100//Rankine efficiency in percent
+nac=((P*10^3*3600)/(m*(h1-191.83)))*100//Actual efficiency in percent
+nTG=((P*10^3*3600)/(m*(h1-h2s)))*100//Turbogenerator efficiency in percent
+
+//Output
+printf('(a) Steam rate at: \n One-fourth load is %3.2f kg/kWh \n Half load is %3.2f kg/kWh \n Three-fourth load is %3.2f kg/kWh \n Full load is %3.1f kg/kWh \n\n (b) Rankine cycle efficiency is %3.1f percent \n (c) Actual efficiency at full load is %3.1f percent \n (d) The turbogenerator efficiency at full load is %3.1f percent',s1,s2,s3,s4,nR,nac,nTG)
diff --git a/1943/CH7/EX7.3/Ex7_3.sce b/1943/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..0dbdd63d9 --- /dev/null +++ b/1943/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,28 @@ + +clc
+clear
+//Input data
+p1=7.8//Pressure in bar
+t1=180+273//Temperature in K
+p2=1.03//pressure in bar
+m=3.6//flow rate of air in kg/s
+g=1.4//Ratio of specific heats
+R=287//Characteristic gas constant in J/kg.K
+cp=1.005//Specific heat in kJ/kg.K
+
+//Calculations
+pxpo=(2/(g+1))^(g/(g-1))//Ratio of pressure
+px=pxpo*p1//Critical pressure in bar
+txto=(2/(g+1))//Ratio of temperatures
+tx=t1*txto//Critical temperature in K
+vx=(R*tx)/(px*10^5)//Critical specific volume in m^3/kg
+Vx=sqrt(g*R*tx)//Critical velocity in m/s
+Ax=((m*vx)/Vx)*10^6//Critical area in mm^2
+tot1=(p1/p2)^((g-1)/g)//Ratio of temperatures
+t1i=t1/tot1//Temperature in K
+v1=(R*t1i)/(p2*10^5)//Specific volume in m^3/kg
+V1=44.72*sqrt(cp*(t1-t1i))//Velocity in m/s
+A1=((m*v1)/V1)*10^6//Area in mm^2
+
+//Output
+printf('Area of throat is %3.0f mm^2 \n Exit area is %i mm^2',Ax,A1)
diff --git a/1943/CH7/EX7.4/Ex7_4.sce b/1943/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..cb663182b --- /dev/null +++ b/1943/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,32 @@ + +clc
+clear
+//Input data
+p1=3.8//pressure in bar
+T1=450+273//Tempereture in K
+p2=1//pressure in bar
+m=16//Flow rate in kg/s
+Cd=0.98//coefficient of discharge
+nv=0.93//nozzile effeciency
+cp=1.11//Specific heat in kJ/kg.K
+g=1.333//Ratio of specific heats
+
+//Calculations
+pxpo=(2/(g+1))^(g/(g-1))//Pressure ratio
+px=pxpo*p1//Critical pressure in bar
+TxTo=2/(g+1)//Temperature ratio
+Tx=T1*TxTo//Critical temperature in K
+Vx=44.72*sqrt(cp*(T1-Tx))//critical velocity in m/s
+R=(cp*(g-1)*1000)/g//Characteristic gas constant in J/kg.K
+vx=(R*Tx)/(px*10^5)//Critical specific volume in m^3/kg
+ws=(m/Cd)//Mass flow rate in kg/s
+Ax=(ws*vx)/Vx//Critical area in m^2
+T1sTo=(p2/p1)^((g-1)/g)//Temperature ratio
+T1s=T1*T1sTo//Temperature in K
+T1i=(T1-(nv*(T1-T1s)))//Temperature in K
+v1=(R*T1i)/(p2*10^5)//Specific volume in m^3/kg
+V1=44.72*sqrt(cp*(T1-T1i))//Velocity in m/s
+A1=(ws*v1)/V1//Area in m^2
+
+//Output
+printf('Throat raea is %3.4f m^2 \n Exit arae is %3.4f m^2',Ax,A1)
diff --git a/1943/CH7/EX7.5/Ex7_5.sce b/1943/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 000000000..3eff3bc22 --- /dev/null +++ b/1943/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,43 @@ + +clc
+clear
+//Input data
+p1=20//pressure in bar
+T1=300+273//Tempereture in K
+p2=3//pressure in bar
+m=0.3//Flow rate in kg/s
+n=1.3//Adiabatic constant
+Cd=0.98//Coefficient of discharge
+Cv=0.92//Coefficient of velocity
+
+//Calculations
+vo=0.1255//Specific volume in m^3/kg
+px=(0.546*p1)//Critical pressure in bar
+vx=(p1/px)^(1/n)*vo//Critical specific volume in m^3/kg
+Vx=sqrt(n*px*10^5*vx)//Critical velocity in m/s
+Ax=((m*vx)/Vx)*10^6//Critical area in m^2
+v1vo=(p1/p2)^(1/n)//Ratio of specific volumes
+v1=(vo*v1vo)//Specific volume in m^3/kg
+V1=sqrt(2*((n/(n-1))*10^5*((p1*vo)-(p2*v1))))//Velocity in m/s
+A1=((m*v1)/V1)*10^6//Area in mm^2
+ho=3050//Enthalpy in kJ/kg
+hx=2920//Enthalpy in kJ/kg
+h1s=2650//Enthalpy in kJ/kg
+ws=(m/Cd)//Flow rate in kg/s
+Vsx=44.72*sqrt(ho-hx)//Velocity in m/s
+V1s=44.72*sqrt(ho-h1s)//Velocity in m/s
+Vo1=(V1s*Cv)//Velocity in m/s
+hoh1=(V1/44.72)^2//Change in enthalpy in kJ/kg
+h1=ho-hoh1//Enthalpy in kJ/kg
+x1=(h1-561.47)/2163.8//Dryness fraction
+vo1=(0.001073+(x1*0.6047))//Specific volume in m^3/kg
+Ao1=((ws*vo1)/Vo1)*10^6//Exit nozzle area in mm^2
+Vox=(Vsx*Cv)//Velocity in m/s
+hohx=(Vox/44.72)^2//Change in enthalpy in kJ/kg
+hox=(ho-hohx)//Enthalpy in kJ/kg
+vox=0.22//Specific volume in m^3/kg
+Aox=((ws*vox)/Vox)*10^6//Critical area in m^2
+
+//Output
+printf('(a) Area of throat is %3.1f mm^2 \n Exit area is %3.1f mm^2 \n\n (b) Area of throat is %3.1f mm^2 \n Exit area is %3.1f mm^2',Ax,A1,Aox,Ao1)
+//In textbook, Ao1 is given wrong.
diff --git a/1943/CH7/EX7.6/Ex7_6.sce b/1943/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..c553f880e --- /dev/null +++ b/1943/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+p1=5//Pressure of steam in bar
+V=100//Velocity in m/s
+p2=1.5//Exit pressure in bar
+At=1280//Throat area in mm^2
+Ae=1600//Exit area in mm^2
+rp=0.58//Critical pressure ratio
+
+//Calculations
+ho=2749//Enthalpy in kJ/kg
+so=6.822//Entropy in kJ/kg.K
+px=(rp*p1)//Critical pressure in bar
+sx=so//Entropy in kJ/kg.K
+xx=(sx-1.660)/5.344//Dryness fraction
+hx=(556+(xx*2168))//Enthalpy in kJ/kg
+Vx=sqrt(((ho+((V^2*10^-3))/2)-hx)*(2/10^-3))//Velocity in m/s
+vx=(xx*0.6253)//Specific volume in m^3/kg
+w=(At*10^-6*Vx)/vx//Mass flow rate in kg/s
+s1s=sx//Entropy in kJ/kg.K
+x1s=(so-1.434)/5.789//Dryness fraction
+h1s=(467+x1s*2226)//ENthalpy in kJ/kg
+z=((Vx^2*10^-3)/2)-hx//z value
+//By iteratio scheme
+x1=0.932//Dryness fraction
+v1=1.080//Specific volume in m^3/kg
+h1=2542//Enthalpy in kJ/kg
+V1=652.2//Velocity in m/s
+nn=((hx-h1)/(hx-h1s))//Nozzle efficiency
+
+//Output
+printf('Mass flow rate is %3.3f kg/s \n Nozzle efficiency is %3.3f',w,nn)
diff --git a/1943/CH7/EX7.7/Ex7_7.sce b/1943/CH7/EX7.7/Ex7_7.sce new file mode 100755 index 000000000..6f219377c --- /dev/null +++ b/1943/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,32 @@ + +clc
+clear
+//Input data
+p1=5//Pressure in bar
+T1=200+273//Temperature in K
+p2=2//Pressure in bar
+m=0.3//Mass flow rate in kg/s
+n=1.3//Adiabatic index
+
+//Calculations
+vo=0.4249//Specific volume in m^3/kg
+ho=2855.4//Enthalpy in kJ/kg
+so=7.0592//Entropy in kJ/kg.K
+x1=0.972//Dryness fraction
+h1=(504.7+x1*2201.9)//Enthalpy in kJ/kg
+v1=x1*0.8857//Specific volume in m^3/kg
+V1=44.72*sqrt(ho-h1)//Velocity in m/s
+A1=((m*v1)/V1)*10^6//Area in mm^2
+rp=(p1/p2)^(1/n)//Specific volume ratio
+vR=(vo*rp)//Specific volume in m^3/kg
+VR=sqrt(2*((n/(n-1))*(p1*vo-p2*vR)*10^5))//Velocity in m/s
+AR=((m*vR)/VR)*10^6//Area in mm^2
+TR=T1/(p1/p2)^((n-1)/n)//Temperature in K
+tR=(TR-273)//Temperature in degree C
+ts=120.23//Saturation temperature at pressure p1 in degree C
+ds=ts-tR//Degree of subcooling in degree C
+ps=1.4327//Saturation pressure at tR in bar
+dsu=(p2/ps)//Degree of supersaturation
+
+//Output
+printf('(a) Exit area when the flow is in equilibrium throughout is %3.0f mm^2 \n (b) Exit area when the flow is supersaturated is %3.1f mm^2 \n (i) The degree of supercooling is %3.2f degree C \n (ii) The degree of supersaturation is %3.3f',A1,AR,ds,dsu)
diff --git a/1943/CH7/EX7.8/Ex7_8.sce b/1943/CH7/EX7.8/Ex7_8.sce new file mode 100755 index 000000000..bf6c2ce35 --- /dev/null +++ b/1943/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,34 @@ + +clc
+clear
+//Input data
+p1=5//Pressure in bar
+T1=200//Temperature in degree C
+p2=2//Pressure in bar
+m=0.3//Mass flow rate in kg/s
+n=1.3//Adiabatic index
+nn=0.92//Nozzle efficiency
+cp=1.925//mean specific heat in kJ/kg.K
+x=[2.308,1943]//pv*10^3 = 2.308(h-1943)
+
+//Calculations
+vo=0.4249//Specific volume in m^3/kg
+ho=2855.4//Enthalpy in kJ/kg
+so=7.0592//Entropy in kJ/kg.K
+x1=0.972//Dryness fraction
+h1=(504.7+x1*2201.9)//Enthalpy in kJ/kg
+v1=x1*0.8857//Specific volume in m^3/kg
+V1=44.72*sqrt(ho-h1)//Velocity in m/s
+h=ho-h1//Change in enthalpy in kJ/kg
+hoq=nn*h//Change in enthalpy in kJ/kg
+VQ=44.72*sqrt(hoq)//Velocity in m/s
+toq=(hoq/cp)//Temperature difference in degree C
+tQ=(T1-toq)//Temperature in degree C
+TQ=tQ+273//Temperature in K
+vQ=((p1*100*vo)/(T1+273))*(TQ/T1)//Specific volume in m^3/kg
+A1=((m*vQ)/VQ)*10^6//Area in mm^2
+vQ=(x(1)*(ho-hoq-x(2)))/(10^3*p2)//Specific volume in m^3/kg
+A11=((m*vQ)/VQ)*10^6//Area in mm^2
+
+//Output
+printf('Exit area is %3.1f mm^2 which upon checking is %3.0f mm^2',A1,A11)
diff --git a/1943/CH7/EX7.9/Ex7_9.sce b/1943/CH7/EX7.9/Ex7_9.sce new file mode 100755 index 000000000..a0f05fba1 --- /dev/null +++ b/1943/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,26 @@ + +clc
+clear
+//Input data
+V1=1000//Speed in m/s
+Vb=400//Peripheral velocity in m/s
+a=20//Nozzle angle in degree
+m=0.75//Mass flow in kg/s
+f=80//Percentage reduction of relative velocity
+
+//Calculations
+b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Blade angle in degree
+V=342//Velocity from E7.9 in m/s
+Vr1=V/sind(b1)//Velocity in m/s
+dVw=(2*Vr1*cosd(b1))//Velocity in m/s
+Pt=(m*dVw)//Tangential thrust in N
+WD=(Pt*Vb)/1000//Diagram power in kW
+nD=(WD/(0.5*m*V1^2*10^-3))*100//Diagram efficiency in percent
+Pa=0//Axial thrust in N
+Vr2=(f/100)*Vr1//Velocity in m/s
+Pa2=m*sind(b1)*(Vr1-Vr2)//Axial thrust in N
+WD2=(m*(Vr1+Vr2)*cosd(b1)*Vb)/1000//Diagram power in kW
+nD2=(WD2/(0.5*m*V1^2*10^-3))*100//Diagram efficiency in percent
+
+//Output
+printf('Blade Angle is %3.2f degrees \n\n Neglecting the friction effects \n Tangential force is %3.2f N \n Axial thrust is %i N \n Diagram efficiency is %3.1f percent \n\n Considering the friction effects \n Axial thrust is %3.1f N \n Diagram Power is %3.2f kW \n Diagram efficiency is %3.2f percent',b1,Pt,Pa,nD,Pa2,WD2,nD2)
diff --git a/1943/CH8/EX8.1/Ex8_1.sce b/1943/CH8/EX8.1/Ex8_1.sce new file mode 100755 index 000000000..8ea897e93 --- /dev/null +++ b/1943/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,43 @@ + +clc
+clear
+//Input data
+ws=250;//The amount of steam received by the surface condenser in t/h
+tsat=40;//The saturated temperature in degree centigrade
+m=12;//The amount of moisture present in the steam in percentage
+tc1=32;//The inlet temperature of cooling water in degree centigrade
+tc2=38;//The outlet temperature of cooling water in degree centigrade
+p=0.078;//The pressure inside the condenser in bar
+V=1.8;//velocity of circulating water in m/s
+do=0.0254;//The outer diameter of the condenser tubes in m
+T=0.00125;//The thickness of the condenser tubes in m
+pi=3.141;//Mathematical constant of pi
+U=2.6;//The overall heat transfer coefficient in kW/m^2K
+Cpc=4.187;//The specific heat of water in kJ/kgK
+R=0.287;//Real gas constant in kJ/kgK
+P=1000;//The density of water in kg/m^3
+
+//Calculations
+x2=0.88;//The quality of the steam
+hfg=2407;//The enthalpy of evaporation at 40 degreeC in kJ/kg
+h=x2*hfg;//The change in enthalpy in kJ/kg
+di=do-(2*T);//The inner diameter of the condenser tubes in m
+wc=[[(ws*1000)/3600]*h]/(Cpc*(tc2-tc1));//Mass flow rate of water in kg/s
+psat=0.07375;//The saturated pressure at 40 degree centigrade in bar
+pair=(p-psat)*100;//The pressure of air in kPa
+vf=0.001008;//Specific volume of saturated liquid in m^3/kg
+vfg=19.544;//Specific volume of vapour in m^3/kg
+v2=vf+(x2*vfg);//Specific volume at 40 degree centigrade in m^3/kg
+wair=[pair*[(ws*1000)/3600]*v2]/(R*(tsat+273));//Mass flow rate of air in kg/s
+t1m=(8-2)/log(8/2);//The mean temperature in degree centigrade
+Ao=[[(ws*1000)/3600]*h]/(U*t1m);//The area of the tubes in m^2
+n=(wc*(4/pi)*(1/di^2)*(1/(P*V)));//The number of tubes
+l=Ao/(pi*do*n);//Yhe length of tubes in m
+
+//Output
+printf('(a) The rate of flow of cooling water = %3.1f kg/s \n (b) The rate of air leakage into the condenser shell = %3.3f kg/s \n (c) The length of tubes = %3.2f m \n (d) The number of tubes = %3.0f ',wc,wair,l,n)
+
+
+
+
+
diff --git a/1943/CH8/EX8.2/Ex8_2.sce b/1943/CH8/EX8.2/Ex8_2.sce new file mode 100755 index 000000000..8bde8e1b2 --- /dev/null +++ b/1943/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,42 @@ + +clc
+clear
+//Input data
+ws=20;//The amount of dry saturated steam received by a surface condencer in t/h
+tsh=40;//The temperature of dry saturated steam in degree centigrade
+wa=(0.35/1000);//The mass flow rate of air per 1000 kg of steam in kg
+tc=38;//The temperature at which condensate leaves the temperature in degree centigrade
+tm=10;//The temperature at which makeup water is supplied in degree centigrade
+tc1=32;//The inlet temperature of cooling water in degree centigrade
+tc2=38;//The outlet temperature of cooling water in degree centigrade
+tas=27;//The temperature of air along with steam in degree centigrade
+psat=0.07384;//The saturated pressure at 40 degree C in bar
+vg=19.52;//The specific volume at 40 degree C in m^3/kg
+R=0.287;//Real gas constant in kJ/kgK
+Cpc=4.187;//The specific heat of water in kJ/kgK
+Cp=1.005;//The specific heat of air in kJ/kgK
+
+//Calculations
+pair=[(wa*R*(tsh+273)*1000)/vg]*10^-5;//The pressure of air in bar
+psat1=0.06624;//The saturated pressure at 38 degree C in bar
+vg1=21.63;//The specific volume at 38 degree C in m^3/kg
+pair1=psat-psat1;//The pressure of air in bar
+wa1=(ws*1000)*wa;//Mass of air removed per hour in kg/h
+V1=((wa1*R*(273+tc2)*1000))/(pair1*10^5);//Volume of air remove per hour
+ws1=V1/vg1;//The mass of steam accompanying air in kg/h
+psat2=0.03564;//The saturated pressure at 27 degree C in bar
+vg2=38.81;//The specific volume at 27 degree C in m^3/kg
+pair2=psat-psat2;//The pressure of air in bar
+V2=(wa1*R*1000*(tas+273))/(pair2*10^5);//Volume of air removed per hr in m^3/hr
+ws2=V2/vg2;//The mass of steam accompanying air in kg/h
+ws3=ws1-ws2;//Saving mass of steam by using seperate extraction in kg/hr
+Q3=[ws3*Cpc*(tc-tm)]/3600;//Saving in heat supply in the boiler in kW
+V=[(V1-V2)/V1]*100;//Percentage reduction in air ejector load in %
+hc=159.3;//Enthalpy at 38 degree C in kJ/kg
+hs1=2574.3;//Enthalpy at 40 degree C in kJ/kg
+hs2=2550.3;//Enthalpy at 27 degree C in kJ/kg
+Q2=[[(ws*1000)*(hs1-hc)]-[(wa1*(Cp*(tsh-tas)))]-(ws3*hs2)]/3600;//The amount of heat in kW
+wc=Q2/(Cpc*(tc2-tc1));//The mass flow rate of water in kg/s
+
+//Output
+printf('(a) The rate of saving of condensate and the rate of saving in the heat supply in the boiler due to seperate air extraction pump = %3.3f kW \n (b) The percentage reduction in air ejector load due to this seperate air extraction method = %3.1f percent \n (c) The rate of cooling water flow = %3.0f kg/s ',Q3,V,wc)
diff --git a/1943/CH8/EX8.3/Ex8_3.sce b/1943/CH8/EX8.3/Ex8_3.sce new file mode 100755 index 000000000..7b9673c65 --- /dev/null +++ b/1943/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,32 @@ + +clc
+clear
+//Input data
+tw3=30;//The inlet temperature of water in degree centigrade
+wc=1.15;//Mass flow rate of cooling water in kg per kg air
+tdb1=20;//The dry bulb temperature of air in degree centigrade
+R1=60;//Relative humidity of air while entering in percentage
+tdb2=28;//The dry bulb temperature while leaving in degree centigrade
+R2=90;//Relative humidity of air while leaving in percentage
+tm=20;//The temperature of makeup water in degree centigrade
+Cpc=4.187;//The specific heat of water in kJ/kgK
+G=1;//Mass flow rate of dry air in kg/s
+
+//Calculations
+twb1=15.2;// from psychrometric chart The wet bulb temperature while entering in degree centigrade
+twb2=26.7;// from psychrometric chart The wet bulb temperature while leaving in degree centigrade
+h1=43;//The enthalpy from chart for dry air in kJ/kg dry air
+h2=83.5;//The enthalpy from chart in kJ/kg dry air
+W1=0.0088;//Humidity in kg water vapour/kg dry air
+W2=0.0213;//Humidity in kg water vapour/kg dry air
+hw3=125.8;//Enthalpy of water entering the tower in kJ/kg
+hw=84;//Enthalpy of makeup water in kJ/kg
+hwc=[(G/wc)*[(h2-h1)-(W2-W1)*hw]];//The change in enthalpy of water in kJ/kg
+tw4=tw3-(hwc/Cpc);//The exit temperature of water in degree centigrade
+ta=tw4-twb1;//The approach temperature in degree centigrade
+tr=tw3-tw4;//The range temperature in degree centigrade
+x=G*(W2-W1);//Fraction of water evaporated in kg/kg dry air
+
+//Output
+printf(' (a) The temperature of water leaving the tower = %3.1f degree centigrade \n (b) The fraction of water evaporated = %3.4f kg/kg dry air \n (c) The approach of the cooling tower = %3.1f degree centigrade \n The Range of the cooling tower = %3.1f degree centigrade ',tw4,x,ta,tr)
+
diff --git a/1943/CH8/EX8.4/Ex8_4.sce b/1943/CH8/EX8.4/Ex8_4.sce new file mode 100755 index 000000000..c2c123b64 --- /dev/null +++ b/1943/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,42 @@ + +clc
+clear
+//Input data
+tw3=45;//The temperature of warm water in degree centigrade
+wc1=6;//The cooling water inflow in kg/s
+V=10;//volume flow of ID fan in m^3/s
+Ws=4.90;//Heat absorbed by air in kW
+ti=20;//The temperature of air entering the tower in degree centigrade
+R=60;//The relative humidity in percentage
+to=26;//The temperature of air leaving the tower in degree centigrade
+p=1.013;//The constant pressure throughout the tower in bar
+r=0.287;//Real gas constant in kJ/kgK
+Cpc=4.187;//The specific heat of water in kJ/kgK
+Cp=1.005;//The specific heat of air in kJ/kgK
+
+//Calculations
+ps=0.0234;//The pressure at 20 degreec in bar
+ps1=(R/100)*ps;//The pressure of water vapour in bar
+pa1=p-ps1;//The pressure of air in bar
+G1=(pa1)/(r*10^-3*(ti+273));//The mass flow rate of dry air in kg/s
+w1=(ps1*10^5*V)/(0.4619*10^3*(ti+273));//Mass flow rate of vopour in kg/s
+W1=w1/G1;//Moisture flow in kg vap/kg dry air
+ps2=0.0336;//The pressure at 26 degree C at exit in bar
+pw2=0.0336;//The pressure of water vapour at 26 degree C at exit in bar
+W2=(0.622)*(pw2/(1-pw2));//oisture flow in kg vap/kg dry air
+G2=G1;//The mass flow rate of dry air in kg/s
+w2=W2*G2;//Moisture flow at exit in kg/s
+wm=w2-w1;//Makeup water required in kg/s
+wc2=wc1-wm;//The cooling water outflow in kg/s
+hw3=Cpc*tw3;//The enthalpy of warm water in kJ/kg
+hg=2538.1;//The enthalpy of gas at 20 degree C in kJ/kg
+tsat=12;//The saturation temperature in degree centigrade
+pw1=0.01404;//The pressure at 12 degree C in bar
+hw1=hg+(1.88*(ti-tsat));//The enthalpy of warm water in kJ/kg
+hw2=2548.4;//The enthalpy of evaporation at 26 degree C in kJ/kg
+hw4=[[G1*[(Cp*(to-ti))+W2*hw2-W1*hw1]-Ws]-(wc1*hw3)]/-wc2;//The enthapy of warm water at point 4 in kJ/kg
+E=hw4/Cpc;//Exit water temperature in degree centigrade
+
+//Output
+printf('(a) The final temperature of the water = %3.2f degree centigrade \n (b)The amount of makeup water required per second = %3.4f kg/s ',E,wm)
+
diff --git a/1943/CH9/EX9.1/Ex9_1.sce b/1943/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..4727df1f1 --- /dev/null +++ b/1943/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,22 @@ + +clc
+clear
+//Input data
+mp=1.007277;//Atomic Mass of proton in amu
+mn=1.008665;//Atomic Mass of neutron in amu
+me=0.00055;//Atomic Mass of electron in amu
+mo=15.99491;//Atomic Mass of oxygen in amu
+np=8;//Number of protons in oxygen
+ne=8;//Number of electrons in oxygen
+nn=8;//Number of neutrons in oxygen
+a=931;//One amu in MeV
+No=16;//Number of nucleons in oxygen
+
+//Calculations
+m=(np*mp)+(ne*me)+(nn*mn)-mo;//The mass defect in amu
+B=m*a;//Binding energy in MeV
+Bn=B/No;//Binding energy per nucleon
+
+//Output
+printf('The mass defect = %3.5f amu \n The binding energy per nucleon = %3.2f MeV ',m,Bn)
+
diff --git a/1943/CH9/EX9.10/Ex9_10.sce b/1943/CH9/EX9.10/Ex9_10.sce new file mode 100755 index 000000000..364c436aa --- /dev/null +++ b/1943/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,17 @@ + +clc
+clear
+//Input data
+f=3.5;//Mass fraction of U-235 in the fuel in percentage
+G=180;//Energy per fission in Mev
+F=10^13;//The neutron flux in neutrons/cm^2s
+sf=577;//Fission cross section of U-235 in barns
+M=1.602*10^-13;//One MeV in J
+
+//Calculations
+N=2.372*(f/100)*10^22;//The fuel density for a uranium oxide fuel in nuclei/cm^3
+q=G*N*sf*10^-24*F;//The rate of energy release in MeV/cm^3s
+qg=q*M;//The rate of energy release in W/cm^3
+
+//Output
+printf('The specific energy release rate for a light water uranium reactor = %3.2f W/cm^3',qg)
diff --git a/1943/CH9/EX9.11/Ex9_11.sce b/1943/CH9/EX9.11/Ex9_11.sce new file mode 100755 index 000000000..cbe32c26e --- /dev/null +++ b/1943/CH9/EX9.11/Ex9_11.sce @@ -0,0 +1,16 @@ + +clc
+clear
+//Input data
+P=1;//The operating power of a reactor in W
+K=1.0015;//The effective multiplication factor of Reactor becomes suppercritical
+t=0.0001;//The average neutron life in s
+t1=1.0001;//Neutron life time in s
+
+//Calculations
+d=(K-1)/K;//The reactivity
+Z=(d*P)/t;//The number of neutrons
+n=exp(Z)/10^6;//Neutron density * 10^6
+
+//Output
+printf('The reactor power level at the end of 1s is %3.3f MW',n)
diff --git a/1943/CH9/EX9.2/Ex9_2.sce b/1943/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..6b0ba6856 --- /dev/null +++ b/1943/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,16 @@ + +clc
+clear
+//Input data
+amr=226.095;//Atomic mass of radium in amu
+AC=6.023*10^23;//Avogadro constant in molecules/g.mol
+h=1620;//Half life of radium in years
+
+//Calculations
+D=(0.6931/(h*365*24*3600));//The decay constant in 1/s
+Na=AC/amr;//Number of atoms per gram of radium
+Ao=D*Na;//Initial activity in dis/s
+
+//Output
+disp(D,"The decay constant (in s^-1) = ");
+disp(Ao,"The initial activity of 1 g of radium 226 in dis/s) = ");
diff --git a/1943/CH9/EX9.3/Ex9_3.sce b/1943/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..2b55faea5 --- /dev/null +++ b/1943/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,18 @@ + +clc
+clear
+//Input data
+F=190;//Each fission of U-235 yeilds in MeV
+a=85;//Assuming the Neutrons absorbed by U-235 cause fission in percentage
+b=15;//Non fission capture to produce an isotope U-236 in percentage
+Q=3000;//The amount of thermal power produced in MW
+
+//Calculations
+E=F*1.60*10^-13;//Each fission yields a useful energy in J
+N=1/E;//Number of fissions required
+B=[(10^6)*(N*86400)]/(a/100);//One day operation of a reactor the number of U-235 nuclei burned is in absorptions per day
+M=(B*235)/(6.023*10^23);//Mass of U-235 consumed to produce one MW power in g/day
+M1=M*3;//Mass of U-235 consumed to produce 3000 MW power in g/day
+
+//Output
+printf('The fuel consumed of U-235 per day = %3.1f g/day ',M1)
diff --git a/1943/CH9/EX9.4/Ex9_4.sce b/1943/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..a3f5378c2 --- /dev/null +++ b/1943/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,14 @@ + +clc
+clear
+//Input data
+sa1=10;//Cross section of nucleus in barns
+N=2200;//Neutrons in m/s
+En1=0.1;//Kinetic energy of neutrons increases in eV
+En2=0.02525;//Kinetic energy of neutron in eV
+
+//Calculations
+sa2=sa1/[(En1/En2)^0.5];//The cross section of neutrons in barns
+
+//Output
+printf('The cross section of neutrons = %3.2f barns ',sa2)
diff --git a/1943/CH9/EX9.5/Ex9_5.sce b/1943/CH9/EX9.5/Ex9_5.sce new file mode 100755 index 000000000..9e9172569 --- /dev/null +++ b/1943/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,17 @@ + +clc
+clear
+//Input data
+U1=99.285;//Uranium consists of U-238 in percentage
+U2=0.715;//Uranium consists of U-235 in Percentage
+E=0.025;//The energy of neutrons in eV
+sc=2.72;//Capture cross section for U-238 in barns
+sf=0;//fission cross section for U-238 in barns
+sc1=101;//Capture cross section for U-235 in barns
+sf1=579;//fission cross section for U-235 in barns
+
+//Calculations
+sa=(U1/100)*(sc+sf)+(U2/100)*(sc1+sf1);//The microscopic absorption cross section of natural uranium in barns
+
+//Output
+printf("The microscopic absorption cross section of natural uranium = %3.1f barns ',sa)
diff --git a/1943/CH9/EX9.6/Ex9_6.sce b/1943/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..783299b00 --- /dev/null +++ b/1943/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,14 @@ + +clc
+clear
+//Input data
+p=1;//The density of water in g/cm^3
+sch=0.332;//The microscopic capture cross section of hydrogen in barn
+sco=0.0002;//The microscopic capture cross section of oxygen in barn
+
+//Calculations
+N=(6.023*10^23)*p/18;//Number of molecules of water per cm^3
+scw=(2*N*sch*10^-24)+(N*sco*10^-24);//The microscopic capture cross section of water in cm^-1
+
+//output
+printf('The microscopic capture cross section of water = %3.4f cm^-1 ',scw)
diff --git a/1943/CH9/EX9.7/Ex9_7.sce b/1943/CH9/EX9.7/Ex9_7.sce new file mode 100755 index 000000000..16c4c4219 --- /dev/null +++ b/1943/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,21 @@ + +clc
+clear
+//Input data
+m=230;//The amount of boron piece in g
+mw=10;//The molecular weight of boron
+R=9.57*10^13;//Reaction rate in cm^-3s^-1
+d=2.3;//Density of boron in g/cm^3
+sa=755;//Absorbption cross section in barns
+ss=4;//Elastic scattering cross section in barns
+
+//Calculations
+st=sa+ss;//The total cross section in barns
+N=(d*6.023*10^23)/mw;//The number density of neutrons in cm^-3
+S=N*st*10^-24;//Number density of neutrons for total in cm^-1
+F=R/S;//Neutron flux in cm^-2s^-1
+L=1/S;//Average distance a neutron travels before it is absorbed in cm
+
+//Output
+disp(F,"The thermal neutron flux (in cm^-2s^-1) = ");
+disp(L,"The average distance that a neutron travels before it is absorbed (in cm) = ");
diff --git a/1943/CH9/EX9.8/Ex9_8.sce b/1943/CH9/EX9.8/Ex9_8.sce new file mode 100755 index 000000000..da86ffc3a --- /dev/null +++ b/1943/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,14 @@ + +clc
+clear
+//Input data
+Eni=4.8;//The energy of the newly born electron in MeV
+Enf=0.025;//The energy of the electron after slow down in eV
+A=12;//The mass number of the graphite (carbon)
+
+//Calculations
+L=1-[[(A-1)^2/(2*A)]*log((A+1)/(A-1))];//The logarithmic energy decrement
+n=(log(Eni*10^6/Enf))/L;//The number of collisions required to slowdown the neutron
+
+//Output
+printf('The logarithmic energy decrement representing the neutron energy loss per elastic collision = %3.3f \n The number of collisions necessary = %3.0f ',L,n)
diff --git a/1943/CH9/EX9.9/Ex9_9.sce b/1943/CH9/EX9.9/Ex9_9.sce new file mode 100755 index 000000000..82ffc481b --- /dev/null +++ b/1943/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,24 @@ + +clc
+clear
+//Input data
+f=100;//The reactor is fuelled of natural uranium in tonnes
+A=238.05;//The atomic mass of natural uranium
+F=10^13;//The average thermal neutron flux in neutrons/cm^2s
+A1=235.04;//The atomic mass of U-235
+sf=579;//The fission cross section of U-235 in barns
+sc=101;//The capture cross section of U-235 in barns
+E=200;//The energy released per fission in MeV
+P=0.715;//U-235 in natural uranium in percentage
+N=2200;//The average thermal neutron in m/s
+
+//Calculations
+n=[(f*1000)*6.023*10^26*(P/100)]/A;//The number of U-235 atoms in the reactor in atoms
+R=(sf*10^-24)*F*n;//The rate of fission in the reactor in fissions/s
+T=R*E*1.602*10^-19;//Thermal power of the reactor in MW
+Rr=T/f;//Rating the reactor MW/tonne
+Rc=[[(R*A1*60*60*24)]/(6.023*10^26)];//The rate of consumption of U-235 by fission in kg/day
+Rcc=Rc*1000;//The rate of consumption of U-235 by fission in g/day
+
+//Output
+printf('(a) The rating of the reactor = %3.2f MW/tonne \n (b)The rate of consumption of U-235 per day = %3.3f kg/day (or) %3.0f g/day ',Rr,Rc,Rcc)
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