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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /3136 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '3136')
97 files changed, 5497 insertions, 0 deletions
diff --git a/3136/CH1/EX1.1/Ex1_1.sce b/3136/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..585705943 --- /dev/null +++ b/3136/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,19 @@ +clear all; clc;
+//This numerical is Ex 1_1E,page 9.
+Pso=20.5
+Psc=20.5*550//converting hp to fps system
+Qo=385
+Qc=385/449//converting gpm to ft^3/s
+E=0.83
+dp=E*Psc/(Qc*144)
+printf('The pressure rise is %g psi',dp)
+disp("After rounding off,pressure rise is 75.8 psi")
+dpr=75.8
+dHw=75.8*144/62.4//62.4 is accelaration due to gravity in fps system
+printf(' The head of water is %g ft of water',dHw)
+disp("After rounding off the value of head of water the answer is 175 ft of water.")
+dhwr=175//rounded off value of head of water
+sg=0.72//specific gravity of oil
+dHo=dhwr/sg
+printf(' The head of oil is %g ft of oil',dHo)
+disp("After rounding off the value of head of oil the answer is 243 ft of oil.")
diff --git a/3136/CH1/EX1.2/Ex1_2.sce b/3136/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..0de595936 --- /dev/null +++ b/3136/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,21 @@ +clear all; clc;
+//This numerical is Ex 1_1S,page 10.
+E=0.83//efficiency
+Ps=15300
+Q=87.4
+Qs=87.4/3600//flow rate in meter cube per sec
+rho=998
+g=9.81
+sg=0.72
+dp=E*Ps/Qs
+printf('\n The change in pressure (dp)is %g',dp)
+dpr=523000//rounded value of dp
+disp("The rounded off value of dp is 523kPa.")
+dHw=dpr/(rho*g)
+printf(' dHw is equal to %g m of water',dHw)
+disp("The rounded off value of dHw is 53.4 m of water.")
+dHwr=53.4//rounded off value of dHw
+disp("Thus we can determine head of oil.")
+dHoil=dHwr/sg
+printf(' dHoil is given by %g m of oil',dHoil)
+disp("The rounded off value of dHoil is 74.2 m of oil.")
diff --git a/3136/CH1/EX1.3/Ex1_3.sce b/3136/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..eb96f2420 --- /dev/null +++ b/3136/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,22 @@ +clear all; clc;
+//This numerical is Ex 1_2E,page 10.
+Q=12000
+A=3.5
+rho_a=0.0762
+E=0.85
+r=2.5//resistance of duct system
+V=Q/(60*A)
+printf('The air flow velocity at discharge is %0.2f ft/s',V)
+KE=(rho_a*(V^2))/(32.2*2)
+printf('\n The product is %0.2f lb/ft^2',KE)
+//PE=KE
+Hv=KE/62.4
+printf('\n The dynamic head is %0.3f ft',Hv)
+disp("The value of dynamic head in inches of water is 0.74.")
+Hvi=0.74//Head in inches
+Ht=r+Hvi
+printf('\n The total head is %0.2f inches of water',Ht)
+p_tot=Ht*62.4
+Ps=Q*p_tot/(60*12*E)
+printf('\n The shaft power is %0.1f ft-lb/s',Ps)
+disp("The shaft power is 7.2 hp.")
diff --git a/3136/CH1/EX1.4/Ex1_4.sce b/3136/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..c9eda978c --- /dev/null +++ b/3136/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,24 @@ +clear all; clc;
+//This numerical is Ex 1_2S,page 11.
+Q=340
+A=0.325
+V=Q/(60*A)
+printf('The air flow velocity at discharge is %0.1f m/s',V)
+rho_a=1.22
+Vr=17.4
+Hd=(rho_a*(Vr^2))/2
+printf('\n The dynamic pressure head is %0.1f Pa',Hd)
+Hdr=184.7//rounded off value of Hd
+rho_w=998//density of water=rhow
+g=9.81
+H=0.0635
+dp=rho_w*g*H//static pressure head
+printf('\n The static pressure head is %0.1f Pa',dp)
+dpr=621.7
+p_tot=Hdr+dpr
+printf('\n The total pressure head is %0.1f Pa',p_tot)
+p_tot=806.4
+E=0.85//efficiency
+Ps=Q*p_tot/(60*E)
+printf('\n The shaft power is %g W',Ps)
+disp("The shaft power is 5.376 kW.")
diff --git a/3136/CH1/EX1.5/Ex1_5.sce b/3136/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..a4b3190ef --- /dev/null +++ b/3136/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,19 @@ +clear all; clc;
+//This numerical is Ex 1_3E,page 11.
+H=295//net head in ft
+Q=148//water flow rate
+n=1800//rpm
+E=0.87//efficiency
+a=62.4//product of density and accelaration due to gravity
+omega=(n*2*%pi)/60
+dp=a*H
+printf('The pressure is %g lb/ft^2',dp)
+Ps=E*Q*dp
+printf('\n Output power is equal to %0.3f lb-ft/s',Ps)
+disp("The output output power can also be written as 2.37*10^6 lb-ft/s")
+disp("Output power in terms of horsepower is given by 4309hp.")
+Psr=2370000//rounded off value of Ps
+Torque=Psr/omega
+printf(' The output torque is %g lb-ft.',Torque)
+disp("The output torque can also be written as 12.57*10^3 lb-ft")
+
diff --git a/3136/CH1/EX1.6/Ex1_6.sce b/3136/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..449a5675f --- /dev/null +++ b/3136/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,18 @@ +clear all; clc;
+//This numerical is Ex 1_3S,page 12.
+H=90
+Q=4.2//water flow rate(in m^3/s)
+n=1800
+E=0.87//efficiency
+rho=998
+g=9.81
+omega=(n*2*%pi)/60
+dp=rho*g*H
+printf('The pressure is %g N/m^2',dp)
+Ps=E*Q*dp
+printf('\n Output power is equal to %0.3f N-m/s',Ps)
+disp("After rounding off the value of output power is 3220 kW.")
+Psr=3220000//rounded off value of Ps
+Torque=Psr/omega
+printf(' The output torque is %g N-m.',Torque)
+disp("After rounding off the output torque comes out to be 17.1*10^3 N-m.")
diff --git a/3136/CH10/EX10.1/Ex10_1.sce b/3136/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..c9dd3a476 --- /dev/null +++ b/3136/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,55 @@ +clear all; clc;
+//the values of omegas and energycoeeficient differ from the ones given in the book
+//the reasons for the same are mentioned in the code below
+H=85
+Q=16
+E=0.9//efficiency
+g=9.8
+rho=998
+
+P_o=E*rho*Q*g*H/1000
+printf("The estimated power (Po) is equal to %0.0f kW",P_o)
+
+
+disp("From figure 10.11,a Francis Turbine is selected. Then with the synchronous speed of 16 poles N is determined")
+N=120*60/16
+printf("N is equal to %grpm",N)
+
+
+N_s=(N*Q^0.5)/H^0.75
+printf("\nWe have value of Ns equal to %0.1f rpm(m^3/s)^0.5",N_s)
+
+
+Ksigma=2.11
+n=450/60
+g=9.8
+sigma=(Ksigma*n*Q^0.5)/((g*H)^0.75)
+printf("\n Value of sigma is equal to %0.2f",sigma)
+
+
+omega=(%pi*N)/30
+V=16
+omega_s=(omega*V^0.5)/((g*H)^0.75)//Answer given in the book is 1.33,this is because H has been wrongly substitued as 75. The correct substitution(H=85),gives the answer equal to 1.2157.
+thita=1.9
+K=1.054
+printf("\n Value of omegas is equal to %g",omega_s)
+
+
+disp("From figure 10.10 we have thita=1.9 for nq=Ns=64.3")
+disp("Since K*D*(g*H)^0.25/Q^0.5=thita, we can determine D.")
+
+D=(thita*(Q^0.5))/(((g*H)^0.25)*K)
+printf("\n Value of D is equal to %0.2f m ",D)
+
+
+disp("From figure 10.9 we have efficiency=0.95,which is close to the original estimation")
+D=1.34//value of D is approximately equal to 1.34
+k_phi=(%pi^2)/4
+k_psi=(%pi^2)/2
+flowcoeff=Q/(k_phi*n*(D^3))
+printf("The flow coefficient is equal to %0.3f",flowcoeff)
+
+
+energycoeff=(g*H)/(k_psi*(n^2)*(D^2))
+printf("\nThe energy coefficient is equal to %0.4f",energycoeff)//Answer given in the book is 1.47,this is because H has been wrongly substitued as 75. The correct substitution(H=85),gives the answer equal to 1.6713.
+
diff --git a/3136/CH10/EX10.2/Ex10_2.sce b/3136/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..f15622b8f --- /dev/null +++ b/3136/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,35 @@ +clear all; clc;
+
+disp("From psi=(g*H)/(omega^2*D^2) and N=30*omega/pi we get N=172.7/(psi^0.5)")
+disp("Also from phi=Q/(omega*D^3) and pi=Ps/(rho*omega^3*D^5) we get Q=0.353*phi*N and Ps=0.0087*N^3*pi")
+disp("Pick the points along 80% gate opening curve,read the values for phi,psi, and efficiency from figure 10.14")
+
+phi=[0.158 0.151 0.14 0.127 0.108 0.092 0.076 0.066];
+psi= [0.093 0.083 0.071 0.06 0.048 0.04 0.03 0.025];
+E= [55 56.5 58 62.5 69 71.5 67.5 60];//efficiency
+pai= [0.0078 0.0067 0.0058 0.0045 0.0034 0.0025 0.0015 0.001];
+
+N = zeros(1,length(phi));
+Ps = zeros(1,length(phi));
+Q = zeros(1,length(phi));
+
+for i = 1: length(phi)
+
+
+ N(i) = 172.7/sqrt(psi(i));
+ Ps(i) = 0.0087*N(i)^3*pai(i)*10^-3;
+ Q(i) = 0.353*phi(i)*N(i);
+end
+
+disp(" phi psi eff(%) pai N(rpm) Ps(mw) Q(m^3/s)")
+
+table = [phi' psi' E' pai' N' Ps' Q'];
+disp(table)
+
+plot(N,Ps,'o',N,Q,'d',N,E,'s')
+legend("Ps (mw)","Q (m^3/s)","Eff (%)",-1)
+xlabel("N (rpm)")
+ylabel("Ps (mW), Q (m^3/s) , eff (%)")
+set(gca(),"grid",[1 1])
+
+
diff --git a/3136/CH10/EX10.3/Ex10_3.sce b/3136/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..0d5bd474c --- /dev/null +++ b/3136/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,53 @@ +clear all; clc;
+D=3
+dn=0.08
+H=350
+En=0.82
+CVb=0.95
+Em=0.90
+Ev=0.96
+g=9.8
+
+V2=(2*En*g*H)^0.5
+printf(" The jet flow velocity is equal to %0.0f m/s",V2)
+
+Um=0.5*V2
+printf("\n Optimum wheel tangential velocity is Um is equal to %0.1f m/s",Um)
+
+N=(60*Um)/(%pi*D)
+printf("\n The rotating speed N is equal to %0.1f rpm",N)
+
+disp("Under the maximum utilization factor condition,we have beta3=90 degrees")
+disp(" Since delta Emax=(1+CVb)*U(V2-U),we get the equation delta Emax=1.95*(U^2)")
+delta_E_max=(1+CVb)*Um*(V2-Um)
+printf(" The value of deltaEmax is equal to %g N-m/kg",delta_E_max)
+
+An=(%pi/4)*(dn^2)
+Q=V2*An
+printf("\n The flow rate is %0.3f m^3/s",Q)
+
+m=998*Q
+Ps=Em*Ev*m*delta_E_max/1000
+printf("\n The total shaft power output is %0.1f kW",Ps)
+
+Ns=(N*(Q^0.5))/(H^0.75)
+printf("\n The specific speed can be calculated as %0.2f rpm((m^3/s)^0.5)/(m^0.75)",Ns)
+
+omega=%pi*N/30
+omega_s=omega*(Q^0.5)/((g*H)^0.75)
+printf("\n In non dimensional form , omegas is equal to %0.3f",omega_s)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH10/EX10.4/Ex10_4.sce b/3136/CH10/EX10.4/Ex10_4.sce new file mode 100755 index 000000000..b816f144e --- /dev/null +++ b/3136/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,82 @@ +clear all; clc;
+H=80
+Q=63
+Es=0.97
+N=400
+V3=25
+Dh3=2
+rh3=1/12
+g=32.2
+
+Ksigma=2.11
+n=N/60
+sigma=(Ksigma*n*(Q^0.5))/((g*H)^0.75)
+printf(" The value of sigma is equal to %0.2f ", sigma)
+
+disp("We have thita=2.4. Thita is also equal to Kt*D2*((g*H)^0.25)/(Q^0.5)")
+thita=2.4
+Kt=1.054
+D2=(thita*(Q^0.5))/(Kt*((g*H)^0.25))
+printf(" Thus the value of D2 is %0.1f ft",D2)
+
+D2r=2.5//rounded off D2
+U2=(D2r*N*%pi)/60
+printf("\n U2 is equal to %0.2f ft/s",U2)
+
+V2=(2*g*H*Es)^0.5
+printf("\n The inlet flow velocity V2 is equal to %0.2f ft/s",V2)
+
+disp("From the inlet velocity diagram for alpha2=20 degrees we have Vr2=V2*sinalpha2")
+alpha2=20
+Vr2=V2*(sin(alpha2*%pi/180))
+printf(" Vr2 is equal to %0.2f ft/s",Vr2)
+
+tanbeta2=Vr2/(V2*(cos(alpha2*%pi/180))-U2)
+printf("\n The value of tanbeta2 is %0.2f",tanbeta2)
+beta2=(atan(tanbeta2))*180/%pi
+printf("\n Thus value of beta2 is %0.1f degrees",beta2)
+
+disp("Selecting the incidence i=2.2 degrees we have betab2=62 degrees")
+
+disp("A2=Q/Vr2=(%pi*D2*b2) From this equation we can determine the value of b2.")
+A2=Q/Vr2
+printf(" A2 is equal to %0.2f ft^2",A2)
+b2=A2/(%pi*D2)
+printf("\n b2 is equal to %0.2f ft",b2)
+disp("Thus b2= 4 inches")
+
+disp("At the outlet with rh3=1 inch, setting gamma=15 degrees and V3=25")
+gamma1=15
+ A3=Q/V3
+printf(" The value of A3 is equal to %g ft^2",A3)
+ rt3=((A3*(cos(gamma1*%pi/180)))/%pi+(rh3^2))^0.5
+ printf("\n The value of rt3 is %0.2f ft",rt3)
+ disp("On converting the value of rt3 from feet to inches we get rt3=10.6inches")
+
+ rt3c=10.6//converted value of rt3
+ rh3c=1//converted value in inches
+ rm3=((((rt3c^2+ rh3c^2)/2)^0.5)/12)
+ printf(" The mean radius rm3 is equal to %0.3f ft",rm3)
+ Um3=26.3
+ tanbetam3=V3/Um3
+ printf("\n The value of tanbetam3 %0.2f",tanbetam3)
+ betam3=atan(tanbetam3)*180/%pi
+ printf("\n The value of betam3 whih is equal to betabm3 if no deviation is assumed is equal to %0.2f degrees",betam3)
+ disp("On rounding off we get the value og betam3=43.6 degrees")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH10/EX10.5/Ex10_5.sce b/3136/CH10/EX10.5/Ex10_5.sce new file mode 100755 index 000000000..9306fe7a1 --- /dev/null +++ b/3136/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,72 @@ +clear all; clc;
+disp("To use figure 10.21 we need the dimensional power specific speed. So the shaft power has to be estimated from figure 10.9 where the non dimensional omegas is needed.")
+
+Ve=5
+Hl=0.7
+
+H=80
+Q=63
+Es=0.97
+N=400
+V3=25
+Dh3=2
+rh3=1/12
+g=32.2
+
+omega=N*%pi/30
+omega_s=omega*(Q^0.5)/((g*H)^0.75)
+printf(" The value of omegas is %0.2f",omega_s)
+
+disp("We have efficiency=0.95")
+E=0.95
+rho=1.9378
+Ps=E*rho*g*Q*H/550//conversion factor =1/550
+printf(" The value of Ps %0.2f hp",Ps)
+
+Nsp=N*(Ps^0.5)/(80^1.25)
+printf("\nThe value of Nsp is equal to %0.2f rpm(hp^0.5)/ft^1.25",Nsp)
+
+disp("From figure 10.21, we obtain sigma approximately equal to 0.1 or NSPHavail/H is greater than or equal to 0.1")
+disp("NSPHavail =Ha-Z+Hl+Ve^2/(2*g)) and NSPH avail is greater than or equal to 8 ft.")
+disp("At T=70 degrees farenheit we have the value of Ha equal to")
+Ha=14.7*144/62.4
+printf(" %0.2f ft.",Ha)
+Hv=0.363*144/62.4
+printf("\n The value of Hv is equal to %0.2f ft",Hv)
+K=(Ve^2)/(2*g)
+NPSHavail=8
+printf("\n The value of (Ve^2)/(2*g) is equal to %0.2f ft",K)
+//In the book the value of Zmax is directly stated
+//I have used the given formualae and substiuted the values in it
+//let NPSHavail=8
+//then from the given formula we can find out the value of Zmax
+
+NPSH_avail=8
+H_vr=0.84//rounded off value
+Kr=0.39//rounded off value
+H_ar=33.9//rounded off value
+Z =H_ar-NPSH_avail+Hl+Kr-H_vr//Kr= rounded off value of (Ve^2)/(2*g)
+printf("\n The value of Zmax is equal to %0.1f ft",Z)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH11/EX11.1/Ex11_1.sce b/3136/CH11/EX11.1/Ex11_1.sce new file mode 100755 index 000000000..4ee16f625 --- /dev/null +++ b/3136/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,31 @@ +clear all; clc;
+
+Pe=1.5
+Eg=0.96//generator efficiency
+Em=0.94//transmission efficiency
+P=Pe/(Eg*Em)
+printf('\n The power is equal to %0.3f MW',P)
+disp("After converting to W the magnitude of power is equal to 1.662*10^6 W")
+
+Cp=0.47//from figure 11.10
+V=13
+rho=1.222
+disp(" Since P=Cp(0.5*rho*A*V^3),thus on substituting the values we get P=630.9A")
+A=P*10^6/(Cp*0.5*rho*V^3)// Since P=Cp*(0.5*rho*A*V^3)
+printf(' On substituing the value of P in P=630.9A we get A equal to %g',A)
+
+disp("After rounding off,the area is equal to 2634.7m^2")
+Ar=2634.7//rounded off A
+//A=R^2*pi
+R=sqrt(A/%pi)
+printf(' The Radius is equal to %g m',R)
+
+disp("After rounding off the,area is equal to 28.9m")
+Rr=28.9//rounded off
+D=2*Rr
+printf(' The Diameter is equal to %g m',D)
+
+omega=(V/R)*5.3// In the book diameter has been incorrectly substituted in place of radius(R). That is the reason why this particular answer doesn't match with the one given in the book.
+printf('\n Omega is equal to %0.2f rad/s',omega)
+N=(omega*30)/%pi//since N is proportional to omega and the answer for omega doesnt match with the answer given in the book(because of the aforementioned reason), the answer of N doesn't match either.
+printf('\n RPM is equal to %g rpm',N)
diff --git a/3136/CH11/EX11.2/Ex11_2.sce b/3136/CH11/EX11.2/Ex11_2.sce new file mode 100755 index 000000000..a4c97fbbd --- /dev/null +++ b/3136/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,64 @@ +clear all; clc;
+
+V= 40*(5280/3600)
+printf('V is equal to %0.2f ft/s',V)
+N=80
+omega=(N*%pi)/30
+printf('\n\nomega is equal to %0.2f rad/s',omega)
+rt=15
+rh=1
+Vt=(rt*omega)/V//tip velocity ratio
+printf('\n\nThe tip velocity ratio is equal to %0.2f ',Vt)
+
+Zb=12/Vt
+printf('\n \n Optimum number of blades is equal to %0.2f ',Zb)
+disp("On approximating,the optimum number of blades is equal to 5")
+
+rm=[(rt^2+rh^2)/2]^0.5
+printf('\nThe mean radius is equal to %0.2f ft',rm)
+
+Um=rm*omega
+printf('\n\nThe blade peripheral velocity at the mean radius is equal to %0.1f ft/s',Um)
+
+disp("Assuming V1=V")
+beta_1=(atan(Um/V))*180/%pi
+printf('\nThe relative flow angle at the inlet is equal to %0.1f degrees',beta_1)
+
+beta_2=65
+tanbetam=0.5*(tan(beta_1*%pi/180)+tan(beta_2*%pi/180))
+printf('\n\nThe value of tan of beta m is equal to %0.3f ',tanbetam)
+beta_m=(atan(tanbetam))*180/%pi
+printf('\n \n Mean relative flow angle (betam) is equal to %0.2f degrees',beta_m)
+
+Wm=V/cos(beta_m*%pi/180)
+printf('\n\nThe relative flow velocity (Wm) is equal to %0.1f ft/s',Wm)
+
+rho=0.0763
+gc=32.2
+c=1.2
+Cl=0.28
+Cd=0.015
+F_um=(rho*Wm^2*c*(Cl*cos(beta_m*%pi/180)-Cd*sin(beta_m*%pi/180))/(2*32.2))
+printf('\n\nThe tangential force (Fum), is equal to %0.2f lb/ft',F_um)
+
+delta_r=14//rt-rh=deltar
+Z_br=5//approximated value of Zb
+P_s=rm*F_um*delta_r*omega*Z_br/550
+printf('\n\nPs is approximately equal to %0.1f hp',P_s)
+
+A=%pi*rt^2
+A_r=707//rounding of A=706.9 to 707
+P_smax=((8/27)*(rho/gc)*707*58.67^3)/550
+printf('\n\nFrom the actuator theory,the maximum possible shaft power will be equal to %0.1f hp.',P_smax)
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH11/EX11.3/Ex11_3.sce b/3136/CH11/EX11.3/Ex11_3.sce new file mode 100755 index 000000000..aa2d3d66e --- /dev/null +++ b/3136/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,45 @@ +clear all; clc;
+
+V= 40//in mph
+V=58.9//in mph
+//more accurately
+V= 40*(5280/3600)
+a=0.27
+V1=V*(1-a)
+printf('V1 is equal to %0.1f ft/s',V1)
+
+N=60
+D=50
+U=(N*%pi/30)*(D/2)
+printf('\nU is equal to %0.1f ft/s',U)
+
+//from velocity triangle
+A=90+45
+printf('\nA is equal to %g degrees',A)
+
+//from cosine law
+W=(U^2+V1^2-2*U*V1*cos(A*%pi/180))^0.5
+printf('\nW is equal to %0.1f ft/s',W)
+
+//from sine law
+sinB=V1*(sin(A*%pi/180))/W
+printf('\nsinB is equal to %0.4f ft/s',sinB)
+
+B=asin(sinB)*180/%pi
+printf('\nB is equal to %0.1f degrees',B)
+
+setting_angle=85
+alpha=B-(90-setting_angle)
+printf('\nalpha is equal to %0.1f degrees',alpha)
+
+//from figure
+Cl=0.58
+Cd=0.027
+rho=0.0763
+gc=32.2
+c=1.2
+Wr=189.8//rounded off W
+Fu=rho*Wr^2*c*(Cl*sin(B*%pi/180)-Cd*cos(B*%pi/180))/(2*gc)
+printf('\nFu is equal to %0.3f lb/ft',Fu)
+
+disp("After rounding off the tangential force (Fu) is equal to 3.38 lb/ft")
diff --git a/3136/CH12/EX12.1/Ex12_1.sce b/3136/CH12/EX12.1/Ex12_1.sce new file mode 100755 index 000000000..57fa274ac --- /dev/null +++ b/3136/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,58 @@ +clear all; clc;
+
+disp("Assuming steady state flow,an adiabetic process and neligible difference between the inlet/outlet flow velocities.")
+T1=25+273
+p2=600
+p1=120
+k=1.4
+T_s2=T1*((p2/p1)^((k-1)/k))
+printf(" T_s2= %0.0f K",T_s2)
+
+disp("With constant Cp assumed, we have Eta_c=(Ts2-T1)/(T2-T1)")
+T_s2=472
+T1=298
+T2=503
+Eta_c=(T_s2-T1)/(T2-T1)
+printf(" Eta_c=%0.4f=84.8 percent",Eta_c)
+
+//let w_c/m=x
+Cp=1.004
+T1=298
+T2=503
+T1=298
+x=Cp*T1*((T2/T1)-1)
+printf("\n w_c/m=Cp*T1*((T2/T1)-1) %0.1f kJ/kg",x)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH12/EX12.2/Ex12_2.sce b/3136/CH12/EX12.2/Ex12_2.sce new file mode 100755 index 000000000..c4143b303 --- /dev/null +++ b/3136/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,71 @@ +clear all; clc;
+
+disp("The gas density at the inlet can be calculated as rho1=p1/(R*T1)")
+p1=14.7
+R=35.11
+T1=30+460
+rho_1=p1*144/(R*T1)
+printf(" rho1= %0.3f lbm/ft^3",rho_1)
+
+m=0.123*150/60
+printf("\n Hence we have mass flow rate m= %0.3f lbm/s",m)
+disp("Neglecting the mechanical and leakage and assuming ideal gas with constant Cp,we cam obtain T2")
+
+T1=490
+p2=40
+p1=14.7
+k=1.29
+Eta_c=0.87
+T2=T1*(1+[(p2/p1)^((k-1)/k)-1]/[Eta_c])
+printf(" T2=%0.0fR",T2)
+
+//let -w_c/m=x
+Cp=0.2007
+T2=632
+T1=490
+x=Cp*(T2-T1)
+printf("\n -w_c/m=%0.1f Btu/lbm",x)
+
+omega_c=0.307*28.5*778/550
+printf("\n -omega_c=%0.1f hp",omega_c)
+
+disp("For an isothermal process, we have -wc/m=p1*v1*ln(p2/p1)")
+p1=14.7
+v1=144/0.123
+p2=40
+x=p1*v1*log(p2/p1)
+printf(" Thus -wc/m= %0.2f lbf-ft/lbm",x)
+
+wc=0.307*17227/550
+printf("\n Hence -wc= %0.1f hp",wc)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH12/EX12.3/Ex12_3.sce b/3136/CH12/EX12.3/Ex12_3.sce new file mode 100755 index 000000000..5a29dfd02 --- /dev/null +++ b/3136/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,26 @@ +clear all; clc;
+
+disp("From the stam Mollier diagram,we have h1=1528 Btu/lbm,s1=1.82 Btu/(lbm_R at 250 psia and 1000 degrees Farenheit)")
+disp("Hence at p2=10 psia,we have hs2=1165 btu/lbm, where ss2=s1")
+disp("From Eta_t=(h1-h2)/(h1-hs2) we have")
+
+h2=1528-0.85*(1528-1165)
+printf("\n h2= %0.2f Btu/lbm",h2)
+
+m=1.2
+h1=1528
+h2=1219.5
+w_t=m*(h1-h2)
+printf("\n wt= %0.2f Btu/s",w_t)//Wrongly rounded off in the book. A more accurate answer is 370.2.
+
+w_t=370.3*778/550//converting units
+printf("\n On converting units, we have wt= %0.1f hp",w_t)//Since the basic value of w_t differs from that given in the book,so does the converted value.
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH12/EX12.4/Ex12_4.sce b/3136/CH12/EX12.4/Ex12_4.sce new file mode 100755 index 000000000..feeb6252c --- /dev/null +++ b/3136/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,33 @@ +clear all; clc;
+
+disp("From Ts2/T1=T3/Ts4=(p2/p1)^((k-1)/k),we have Ts2=901 R")
+disp("Ts4=1264.6R")
+disp("From Eta_c=(Ts2-T1)/(T2-T1) we have T2=980.2")
+disp("Hence w_c=Cp*(T2-T1)=149.4hp/(lbm/s)")
+disp("From Eta_t=(T3-T4)/(T3-Ts4),we have T4=1374.5R we hence wt=249.7hp/(lbm/s)")
+disp("w=w_t-w_c")
+w_t=249.7
+w_c=149.4
+w=w_t-w_c
+printf(" w=%0.1f hp/(lbm/s)",w)
+disp("q_in=Cp*(T3-T2)=383.5hp/(lbm/s)")
+w=100.2
+q_in=383.5
+Eta_th=w/q_in
+printf(" Eta_th=w/qin %0.2f =26 percent",Eta_th)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH12/EX12.5/Ex12_5.sce b/3136/CH12/EX12.5/Ex12_5.sce new file mode 100755 index 000000000..0868d5b57 --- /dev/null +++ b/3136/CH12/EX12.5/Ex12_5.sce @@ -0,0 +1,66 @@ +clear all; clc;
+
+disp("From Eta_hx=(T_dash-T_2)/(T_4-T_dash4),T_4dash=T_2")
+
+disp("we have T_2dash=T_2+Eta_hx*(T_4-T_2)")
+T_2=980.2
+Eta_hx=0.7
+T_4=1374.5
+T_2=980.2
+T_2dash=T_2+Eta_hx*(T_4-T_2)
+printf(" T_2dash=%0.1f R",T_2dash)
+
+Cp=0.339
+T_3=2110
+T_2dash=1256.2
+q_in=Cp*(T_3-T_2dash)
+printf("\n q_in %0.1f hp/(lbm/s)",q_in)
+
+Eta_th=99.8/289.4
+printf("\n Eta_th=%0.4f=34.6 percent",Eta_th)//in the book the value of Eta_th is rounded off to 34.6 and hence the same is done here.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH12/EX12.6/Ex12_6.sce b/3136/CH12/EX12.6/Ex12_6.sce new file mode 100755 index 000000000..e36fec961 --- /dev/null +++ b/3136/CH12/EX12.6/Ex12_6.sce @@ -0,0 +1,117 @@ +clear all; clc;
+
+disp("referring to figure A-14a,Va=650mph=953ft/s")
+Ta=510
+Va=953
+Cp=0.24*778*32.2
+T01=Ta+Va^2/(2*Cp)
+printf(" T01=%0.1f R",T01)
+
+disp("from Eta_i=(Ts01-Ta)/(T01-Ta),we have Ts01=Ta+Eta_i(T01-Ta)=577.9R")
+pa=10.5
+Ts01=577.9
+Ta=510
+//let x=k/(k-1)
+x=3.5
+p01=pa*(Ts01/Ta)^(x)
+printf(" p01=%0.1f psia",p01)
+
+p03=5*16.3
+printf("\n p03=p02=%0.1f psia",p03)
+
+T01=585.5
+Ts02=T01*(5^0.2857)
+printf("\n Ts02=%0.1f R",Ts02)
+
+disp("From Eta_c= (Ts02-T01)/(T02-T01),we have T)2=T01 +(Ts02-T01)/Eta_c=1012.7R")
+disp("For the turbojet w_c=w_t is assumed,we have T04=T03-(T02-T01)")
+T03=1810
+T02=1012.7
+T01=585.5
+T04=T03-(T02-T01)
+printf(" T04= %0.1f R",T04)
+
+disp("From Eta_t=(T03-T04)/(T03-Ts04)=1318.9R and p04=p03*(T_s04/T_03)^3.5=26.9 psia")
+
+disp("To check whether the flow is choked at the nozzle exit,we calculate te choking condition from Tc=2*T_04/(k+1)")
+T_04=1382.8
+k=1.4
+Tc=2*T_04/(k+1)
+printf(" Tc= %0.1f R",Tc)
+
+T04=1382.8
+Tc=1152.3
+Eta_n=0.92
+T_sc=T04-(T04-Tc)/Eta_n
+printf("\n T_sc= %0.1f R",T_sc)
+
+p04=26.9
+Tsc=1132.3
+T04=1382.8
+pc=p04*(Tsc/T04)^3.5
+printf("\n pc= %0.2f psia",pc)
+
+disp("Since pc>pa the flow is choked at the nozzle exit plane,and hence we have T5=Tc=1152.3R,ps=pc=13.36 psia")
+
+k=1.4
+R=53.33
+T5=1152.3
+Vj=(k*R*T5*32.2)^0.5//conversion factor=32.2
+printf(" Vj= %0.1f ft/s",Vj)
+
+p5=13.36
+rho5=p5*144/(R*T5)//144 = conversion factor
+printf("\n rho5 %0.4f lbn/ft^3",rho5)
+
+disp("m=rho*V*A")
+
+A5=60/(0.0313*1664.4)
+printf(" A5= %0.2f ft^2",A5)
+
+disp("F=m*(Vj-Va)+A5*(p5-pa)")
+F=60*(1664.4-953)/32.2+1.15*(13.36-10.5)*144
+printf(" F= %0.1f lbf",F)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH12/EX12.7/Ex12_7.sce b/3136/CH12/EX12.7/Ex12_7.sce new file mode 100755 index 000000000..b4f31b0e6 --- /dev/null +++ b/3136/CH12/EX12.7/Ex12_7.sce @@ -0,0 +1,46 @@ +clear all; clc;
+
+disp("Referring to figure A-12 and assuming pure saturated liquid at the condenser outlet,we have h9=hf=93 Btu/lbm from the steam table")
+disp("With p8=p9=2 psia and p6=100 psia,we have h6=h9+delta_p*144/(Eta_p*rho)")
+h9=93
+delta_p=100-2
+Eta_p=0.8
+rho=62.4*778
+h6=h9+delta_p*144/(Eta_p*rho)
+printf(" h6=%0.1f Btu/lbm",h6)
+
+disp("Assuming T6 to be approximately equal to T9=585 R with the definition of boiler efficiency Eta_b=(h7-h6)*m_w/[Cp*(T4-T5)*m_g]")
+h7=0.75*0.24*(1374.5-585)/0.10+93.4
+printf(" h7= %0.2f Btu/lbm",h7)
+
+disp("Assuming p6 to be approximately equal to p7=100 psia,we obtain s8s=s7=1.908 Btu/lbmR and h8s=1120 btu/lbm from the steam table")
+disp("From the definition of turbine efficiency,Eta_t=(h7-h8)/(h7-h8s) we can determine the value of h8")
+h8=1514.5-0.84*(1514.5-1120)
+printf(" h8=%0.0f Btu/lbm",h8)
+disp("Or w_t=h7-h8=331.4 Btu/lbm,w_p=0.4 Btu/lbm")
+
+disp("Hence the additional power output from the steam power plant per unit mass flow rate of gas through the gas turbine is delta_w=(w_t-w_p)*m_w*/m_g")
+delta_w=331*0.1
+printf(" delta_w= %0.1f Btu/lbm=hp/(lbm/s)",delta_w)
+
+Eta_th=(99.8+46.8)/(382.6)
+printf("\n The new thermal efficiency = %0.3f =38.3 percent",Eta_th)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH12/EX12.8/Ex12_8.sce b/3136/CH12/EX12.8/Ex12_8.sce new file mode 100755 index 000000000..33f45c338 --- /dev/null +++ b/3136/CH12/EX12.8/Ex12_8.sce @@ -0,0 +1,46 @@ +clear all; clc;
+
+disp("We have h4=104 Btu/lbm,h2=h1=53 Btu/lbm,p4=p1=20psia,s4=s-s3=0.226Btu/lbm-R and Hs3=122 Btu/lbm")
+h4=104
+hs3=122
+Eta_c=0.75
+h_3dash=h4+(hs3-h4)/Eta_c
+printf("h_3dash=%0.0f Btu/lbm",h_3dash)
+
+w_i=h_3dash-h4
+printf("\n The compressor work required per unit mass is w_i =%0.0f Btu/lbm",w_i)
+
+h1=53
+qi=h4-h1
+printf("\n The heat absorbed by the evaporator per unit mass =qi= %0.0f Btu/lbm",qi)
+
+beta1=qi/w_i
+printf("\n The coefficient of performance beta= %0.1f",beta1)
+
+m=10
+Pc=m*w_i
+printf("\n The total compressor power required Pc=%0.0f=240 Btu/s=340 hp",Pc)
+
+qi=51
+Qr=m*qi
+printf("\n The refrigeration capacity is Qr= %0.0f Btu/s=153 tons",Qr)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH2/EX2.1/Ex2_1.sce b/3136/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..f4e3dfded --- /dev/null +++ b/3136/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,24 @@ +clear all; clc;
+//This numerical is Ex 2_1,page 18.
+
+//These are examples given in the book,they are used to teach conversions from one unit to another
+
+rho=0.075//lbm/ft^3
+
+V=120
+
+RHO_c=rho/32.2//rho conerted to slug/ft^3
+
+disp("The initial value of rho was given in lb/ft^3. In order to convert it to slug/ft^3,we have used the conversion factor 1/32.2.")
+
+printf("\nThe initial value of rho was 0.075lb/ft^3 after coverting it is %0.5f slug/ft^3",RHO_c)
+
+p_d=RHO_c*V^2/2
+
+printf("\nHence the value of dynamic pressure is %0.2f (slug/ft^3)(ft/s)",p_d)
+
+p_dc=p_d/(12^2)//converting slug*s/ft^2 to lbf/in^2
+
+printf("\nThe final value of dynamic pressure is %0.3f lbf/in^2",p_dc)
+
+printf("\nHence we can say that the dynamic power is equal to %0.3f psi.",p_dc)
diff --git a/3136/CH2/EX2.10/Ex2_10.sce b/3136/CH2/EX2.10/Ex2_10.sce new file mode 100755 index 000000000..ad87c527e --- /dev/null +++ b/3136/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,24 @@ +clear all; clc;
+//This numerical is Ex 2_4E,page 31.
+
+disp("From phi equal to Qm/(Dm^3*Nm)=Qp/(Dp^3*Np) and psi=g*Hm/(Dm^2*Nm^2)=g*Hp/(Dp^2*Np^2), we have Qm=Qp*((Dm/Dp)^3)*(Nm/Np) and Hm=Hp*((Dm/Dp)^2)*(Nm/Np)^2)")
+
+Q_p=25
+//let x=Dm/Dp=1/5
+x=1/5
+N_m=3600
+N_p=1800
+H_p=160
+E=0.92
+//let y=rho*g=62.4
+y=62.4
+
+Q_m=Q_p*((x)^3)*(N_m/N_p)
+printf("\n The value of Q_m is equal to %g ft^3/s",Q_m)
+
+H_m=H_p*((x^2)*(N_m/N_p)^2)
+printf("\n The value of H_m is equal to %g ft",H_m)
+
+P_s=E*y*Q_m*H_m/550//1/550 is conversion factor
+printf("\n The value of P_s is equal to %0.02f hp",P_s)
+
diff --git a/3136/CH2/EX2.11/Ex2_11.sce b/3136/CH2/EX2.11/Ex2_11.sce new file mode 100755 index 000000000..3dfc2c6eb --- /dev/null +++ b/3136/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,25 @@ +clear all; clc;
+//This numerical is Ex 2_4S,page 31.
+disp("From phi equal to Qm/(Dm^3*Nm)=Qp/(Dp^3*Np) and psi=g*Hm/(Dm^2*Nm^2)=g*Hp/(Dp^2*Np^2), we have Qm=Qp*((Dm/Dp)^3)*(Nm/Np) and Hm=Hp*((Dm/Dp)^2)*(Nm/Np)^2)")
+
+Q_p=42
+//let x=Dm/Dp=1/5
+x=1/5
+N_m=3600
+N_p=1800
+H_p=50
+E=0.92
+rho=998
+g=9.81
+//let y=Q/60=0.011166
+y=0.011166
+
+Q_m=Q_p*((x)^3)*(N_m/N_p)
+printf("\n The value of Q_m is equal to %g cmm",Q_m)
+
+H_m=H_p*((x^2)*(N_m/N_p)^2)
+printf("\n The value of H_m is equal to %g m",H_m)
+
+Q_mr=0.67//rounded off
+P_s=E*rho*g*(y)*H_m
+printf("\n The value of P_s is equal to %0.1f W",P_s)
diff --git a/3136/CH2/EX2.2/Ex2_2.sce b/3136/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..4a5d9bbba --- /dev/null +++ b/3136/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,13 @@ +clear all; clc;
+//This numerical is Ex 2_2,page 18.
+
+// This numerical is used an example to teach conversion factors
+T0=600
+T1=550
+Cp=0.24
+halfVsquare=Cp*32.2*778*(T0-T1)
+printf("\n The value of half of V^2 is %g (Btu/slug*R)(lbf-ft/Btu)(R)",halfVsquare)
+printf("\n The value of half of V^2 ca also be written as %g lbf-ft/slug",halfVsquare)
+printf("\n The value of half of V^2 is also equal to %g (ft/s)^2",halfVsquare)
+V=sqrt(halfVsquare*2)
+printf("\n\n The value of V is equal to %0.1f ft/s",V)
diff --git a/3136/CH2/EX2.3/Ex2_3.sce b/3136/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..15e5c64e6 --- /dev/null +++ b/3136/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,12 @@ +clear all; clc;
+//This numerical is Ex 2_3,page 19.
+
+// This numerical is used an example to teach conversion factors
+
+rho=0.85*62.4
+p=50//in psi
+g=32.2
+disp("Since pressure is the product of density,gravitaional accelaration and head, we can convert pressure in psi to head in ft using suitable conversion factors.")
+H=p*144/( (rho/32.2)*32.2)
+printf("The value of head H is given by %0.1f lb/ft^2/((slug/ft^3)*(ft/s^2))",H)
+printf("\nThus the value of H is equal to %0.1f ft",H)
diff --git a/3136/CH2/EX2.4/Ex2_4.sce b/3136/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..c54097374 --- /dev/null +++ b/3136/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,12 @@ +clear all; clc;
+//This numerical is Ex 2_4,page 19.
+
+//converting dynamic pressure of air from 2_1 to wg
+
+pd=16.77//lbf/ft^2
+rho_w=62.4/32.2
+g=32.2
+disp("From pressure=density*gravitational accelaration *head, we can find out the value of head")
+h=pd/(rho_w*g)
+printf(" Hence the value of head H is %0.3f (lbf/ft^2)/(lbf/ft^3)",h)
+printf("\n The value of head H can also be given by %0.3f ft",h)
diff --git a/3136/CH2/EX2.5/Ex2_5.sce b/3136/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..88e4dcba8 --- /dev/null +++ b/3136/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,24 @@ +clear all; clc;
+//This numerical is Ex 2_1E,page 29.
+Q1=80
+N1=1000
+N2=1500
+delta_p1=150
+P_s1=8
+
+disp("From phi_2=phi_1 we have Q2/(D^3*N2)=Q1/(D^3*N1)")
+Q2=Q1*N2/N1
+printf(" The value of Q2 is equal to %g gpm \n",Q2)
+
+disp("From psi_1=psi_2 we have delta p 2/(rho*D^2*N2^2)=delta p 1/(rho*D^2*N1^2)")
+delta_p2=delta_p1*((N2/N1)^2)
+printf(" The value of delta_p2 is equal to %g psig \n",delta_p2)
+
+disp("From pi_2=pi_1 we have P_s2=P_s1*((N2/N1)^3)")
+P_s2=P_s1*(N2/N1)^3
+printf(" The value of P_s2 is equal to %g hp \n",P_s2)
+
+disp("The efficiencies are same at the corresponding points,so E1=E2")
+E1=Q1*delta_p1*0.00223*144/(550*P_s1)
+printf(" The value of E1=E2 is equal to %g \n",E1)
+disp("Thus the efficiency is equal to 87.57%")
diff --git a/3136/CH2/EX2.6/Ex2_6.sce b/3136/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..59ba4d5c2 --- /dev/null +++ b/3136/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,20 @@ +clear all; clc;
+//This numerical is Ex 2_1S,page 29.
+Q1=18.2
+N1=1000
+N2=1500
+delta_p1=10.3
+P_s1=6
+
+Q2=Q1*N2/N1
+printf("\n The value of Q2 is equal to %g m^3/h",Q2)
+
+delta_p2=delta_p1*((N2/N1)^2)
+printf("\n The value of delta_p2 is equal to %0.1f bars",delta_p2)
+
+P_s2=P_s1*(N2/N1)^3
+printf("\n The value of P_s2 is equal to %g kW",P_s2)
+
+E1=((Q1/3600)*delta_p1*10^2)/(P_s1)
+printf("\n The value of E1=E2 is equal to %g ",E1)
+disp("Thus the efficiency is equal to 86.8%")
diff --git a/3136/CH2/EX2.7/Ex2_7.sce b/3136/CH2/EX2.7/Ex2_7.sce new file mode 100755 index 000000000..2ce9dd9d8 --- /dev/null +++ b/3136/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,28 @@ +clear all; clc;
+//This numerical is Ex 2_2E,page 30.
+P_ho=30//hydraulic output power
+Q=5//flow rate at best efficiency point
+g=32.2
+rho=1.938
+Hp=320
+N=600
+
+delta_pm=P_ho*550/Q
+printf("Value of discharge head P_m %g lb/ft^2",delta_pm)
+
+Hm=delta_pm/(rho*g)
+printf("\n Value of H_m=%0.2f ft",Hm)
+disp("From the similarity law, H_p/H_m=((Np/Nm)^2)*((Dp/Dm)^2)")
+
+//let x=Hp/Hm
+x=Hp/Hm
+printf(" H_p/H_m =%0.2f",x)
+disp("Thus (N_p/N_m)*(D_p/D_m) is equal to 2.46")
+disp("Also the flow rate Q_p/Q_m=(N_p/N_m)*(D_p/D_m)^3")
+z=350/5//value of Qp/Qm
+printf(" Hence the value of Q_p/Q_m is equal to %g",z)
+disp("Thus D_p/D_m=5.33 ,N_p/N_m=0.461")
+//Let y=Np/Nm=0.461
+y=0.461
+N_m=N/y
+printf(" Thus N_m = %g rpm",N_m)
diff --git a/3136/CH2/EX2.8/Ex2_8.sce b/3136/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..40e54094b --- /dev/null +++ b/3136/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,31 @@ +clear all; clc;
+//This numerical is Ex 2_2S,page 30.
+//The value given in the book for N_m is 1315,but on calculating the true value is found out to be 1304.35
+
+P_ho=22.4//hydraulic output power
+Q=0.14//flow rate at best efficiency point
+g=9.8
+rho=998
+H_p=97.5
+N=600
+
+delta_pm=P_ho/Q
+printf("Value of discharge head P_m %g kPa",delta_pm)
+
+H_m=delta_pm*10^3/(rho*g)
+printf("\n Value of H_m=%g m",H_m)
+disp("From the similarity law, H_p/H_m=((N_p/N_m)^2)*((D_p/D_m)^2)")
+
+//let x=Hp/Hm
+H_mr=16.3//rounded off value
+x=H_p/H_mr
+printf(" H_p/H_m =%0.2f",x)
+disp("Thus (N_p/N_m)*(D_p/D_m) is equal to 2.45")
+disp("Also the flow rate Q_p/Q_m=(N_p/N_m)*(D_p/D_m)^3")
+z=9.9/0.14//value of Qp/Qm
+printf(" Thus the value of Q_p/Q_m is %0.1f",z)
+disp("Thus D_p/D_m=5.4 ,N_p/N_m=0.46")
+//Let y=Np/Nm=0.461
+y=0.46
+N_m=N/y//where N=600 and y=0.46
+printf(" Hence N_m = %g rpm",N_m)//value given in the book is 1315,but on calculating the true value is found out to be 1304.35
diff --git a/3136/CH2/EX2.9/Ex2_9.sce b/3136/CH2/EX2.9/Ex2_9.sce new file mode 100755 index 000000000..a851c38d1 --- /dev/null +++ b/3136/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,28 @@ +clear all; clc;
+//This numerical is Ex 2_3E,page 31.
+
+N=1500
+E=0.74
+Q=250*0.00223//0.00223 is conversion factor
+printf("Q is equal to %g",Q)
+H=18
+g=32.2
+Q_o=250//before converting Q
+
+disp("From the dimensional specific speed (N_s) and fig2.1 , we select a Francis type pump and efficiency is estimated to be equal to 74%")
+
+N_s=N*(Q_o^0.5)/(H^0.75)
+printf("\n N_s is equal to %0.0f rpm(gpm^0.5)/(ft^0.75)",N_s)
+disp("To find the approximate size,Figure 2.2 has to be used")
+
+omega=N*%pi/30
+printf("\n omega is equal to %0.0f",omega)
+
+omega_s=omega*(Q^0.5)/((g*H)^0.75)
+printf("\n omega_s is equal to %0.4f",omega_s)
+
+disp("From figure 2.2, it is obtained that delta_s=3.1")
+D=(3.1*(Q^0.5))/((g*H)^0.25)
+printf("\n Hence D is equal to %0.3f ft",D)
+disp("Hence D is equal to 5.7 in ")
+
diff --git a/3136/CH3/EX3.1/Ex3_1.sce b/3136/CH3/EX3.1/Ex3_1.sce new file mode 100755 index 000000000..e126cfe07 --- /dev/null +++ b/3136/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,46 @@ +clear all; clc;
+//This numerical is Ex 3_1E,page 43.
+N=800
+Q=1750//in gallon per minute
+r_1=3
+b_1=4
+r_2=9
+b_2=3
+
+omega=N*%pi/30
+printf('\n The angular velocity is %g rad/s',omega)
+omega_r=83.7
+disp("After rounding off the value of angular velocity is 83.7 rad/s")
+
+
+U_1=omega_r*r_1/12
+printf(' U_1=%g',U_1)
+disp("After rounding off the value of U_1 is 20.9 ft/s")
+U_1r=20.9//rounded value of U1
+
+
+U_2=omega_r*r_2/12
+printf(' U_2=%g',U_2)
+disp("After rounding off the value of U_2 is 62.7 ft/s")
+
+
+A_1=2*%pi*r_1*b_1/144
+printf(' A_1=%g ft^2',A_1)
+disp("After rounding off the value of A_1 is 0.523 ft^2")
+
+A_2=2*%pi*r_2*b_2/144
+printf(' A_2=%g ft^2',A_2)
+disp("After rounding off the value of A_2 is 1.18 ft^2")
+A_1r=0.523//rounded off
+A_2r=1.18//rounded off
+
+
+V_r1=(Q*0.00223)/(A_1r)
+printf(' The value of V_r1 is %g',V_r1)
+disp("The value of V_r1 after rounding off is 7.47 ft/s")
+
+V_r2=(Q*0.00223)/(A_2r)
+printf(' The value of V_r2 is %g',V_r2)
+disp("The value of V_r2 after rounding off is 3.27 ft/s")//actual value is 3.30,however the value given in the book is 3.27 ft/s
+
+
diff --git a/3136/CH3/EX3.10/Ex3_10.sce b/3136/CH3/EX3.10/Ex3_10.sce new file mode 100755 index 000000000..938224ec0 --- /dev/null +++ b/3136/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,22 @@ +clear all; clc;
+//This numerical is Ex 3_5S,page 53.
+
+//this numerical is based on numerical 3_4S
+//values found in the book for numerical 3_4S will be used to solve this numerical(3.5S)
+
+delta_Et=40.85
+U_1=6.9
+U_2=2.92
+V_m2=0.782
+V_2=0.782//since V2=Vm2
+
+V_r1=0.508
+W_u1=0.997
+W_1=(V_r1^2+W_u1^2)^0.5
+printf("\n W1 is equal to %0.2f ft/s",W_1)
+
+W_2=(U_2^2+V_m2^2)^0.5
+printf("\n W_2 is equal to %0.2f ft/s",W_2)
+
+Rt=0.5*[(U_1^2-U_2^2)+(W_2^2-W_1^2)]/(delta_Et)
+printf("\n Thus the value Rt is equal to %0.3f",Rt)
diff --git a/3136/CH3/EX3.11/Ex3_11.sce b/3136/CH3/EX3.11/Ex3_11.sce new file mode 100755 index 000000000..647960f5d --- /dev/null +++ b/3136/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,66 @@ +clear all; clc;
+//This numerical is Ex 3_6E,page 53.
+disp("The velocity diagram is similar to that in figure 3.19")
+
+r_t=1.2
+r_h=0.7
+N=4800
+V_1=600
+
+r_m=[0.5*(r_t^2+r_h^2)]^0.5
+printf("\n r_m is equal to %0.3f ft",r_m)
+
+A=%pi*(r_t^2-r_h^2)
+printf("\n A is equal to %0.2f ft^2",A)
+
+r_m=0.982//rounding off rm
+U=(N*%pi*r_m)/30
+printf("\n U is equal to %0.1f ft/s",U)
+
+V_a1=V_1*cos(60*%pi/180)//angle is given as 60 degrees
+printf("\n V_a1 is equal to %0.2f ft/s",V_a1)
+
+V_u1=V_1*sin(60*%pi/180)//angle is given as 60 degrees
+printf("\n V_u1 is equal to %0.1f ft/s",V_u1)
+
+//let tan(betaf1)=(Vu1-U)/Va1=x
+x=(V_u1-U)/V_a1
+printf("\n tan(ßf1) is equal to %0.4f",x)
+beta_f1=(atan(x))*180/%pi
+printf("\n Hence ßf1= %0.2f degrees",beta_f1)
+
+disp("ßf1=ßb1. Also ßb2=ßf2=alpha_1=60 degrees")
+disp("From the velocity diagram we have V_u2=W_u1=V_u1-U")
+W_u1=V_u1-U
+printf("\n Thus Wu1 = %g ft/s",W_u1)
+
+//let l=Ps/m=U*(Vu1+Vu2)
+V_u2=W_u1//already stated above
+l=U*(V_u1+V_u2)
+printf("\n P_s/m=U*(V_u1+V_u2) is equal to %g (ft/s)^2",l)
+disp("Thus we can round it off to 29*10^5 (ft/s)^2")
+
+disp("Where m=rho*V_a1*A")
+rho=0.085/32.2
+m=rho*V_a1*A
+printf("\n m= %0.2f slug/s",m)
+
+disp("Thus we can determine P_s")
+l=2.69*10^5//rounded off value
+m=2.36//rounded off value
+P_s=(l*m)/550
+printf(" P_s =%0.0f hp",P_s)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH3/EX3.12/Ex3_12.sce b/3136/CH3/EX3.12/Ex3_12.sce new file mode 100755 index 000000000..c970457fd --- /dev/null +++ b/3136/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,71 @@ +clear all; clc;
+//This numerical is Ex 3_6S,page 54.
+
+disp("The velocity diagram is similar to that in figure 3.19")
+
+r_t=0.36
+r_h=0.21
+N=4800
+V_1=183
+
+r_m=[0.5*(r_t^2+r_h^2)]^0.5
+printf("\n rm is equal to %0.3f m",r_m)
+
+A=%pi*(r_t^2-r_h^2)
+printf("\n A is equal to %0.2f m^2",A)
+
+r_m=0.295//rounding off rm
+U=(N*%pi*r_m)/30
+printf("\n U is equal to %0.1f m/s",U)
+
+V_a1=V_1*cos(60*%pi/180)//angle is given as 60 degrees
+printf("\n V_a1 is equal to %0.2f m/s",V_a1)
+
+V_u1=V_1*sin(60*%pi/180)//angle is given as 60 degrees
+printf("\n V_u1 is equal to %0.1f m/s",V_u1)
+
+//let tan(betaf1)=(Vu1-U)/Va1=x
+x=(V_u1-U)/V_a1
+printf("\n tan(ß1) is equal to %g m/s^2",x)
+
+beta_f1=(atan(x))*180/%pi
+printf("\n We have ß_b1= ß_f1= %0.2f degrees",beta_f1)//value in book is 6.35 degrees Difference is obtained because actual value of x is substituted. Value substituted in the book is not 0.111 or 0.11147.
+
+disp("ß_f1=ß_b1. Also ß_b2=ß_f2=alpha_1=60 degrees")
+disp("From the velocity diagram we have V_u2=W_u1=V_u1-U")
+W_u1=V_u1-U
+printf("\n Thus W_u1 = %0.2f m/s",W_u1)
+
+//let l=Ps/m=U*(Vu1+Vu2)
+V_u2=W_u1//already stated above
+l=U*(V_u1+V_u2)
+printf("\n P_s/m=U*(V_u1+V_u2) is equal to %g (m/s)^2",l)
+disp("Thus we can round it off to 2.5*10^4 (m/s)^2")
+
+disp("Where m=rho*V_a1*A")
+rho=1.36
+V_a1=91.5//rounded off
+A=0.269//rounded off
+m=rho*V_a1*A
+printf("\n m= %0.1f kg/s",m)
+
+disp("Thus we can determine P_s")
+l=2.5*10^4//rounded off value
+m=33.5//rounded off value
+P_s=(l*m)
+printf(" P_s =%0.0f W",P_s)
+disp("Thus on rounding off P_s is equal to 837kW")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH3/EX3.13/Ex3_13.sce b/3136/CH3/EX3.13/Ex3_13.sce new file mode 100755 index 000000000..968e3e49a --- /dev/null +++ b/3136/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,60 @@ +clear all; clc;
+//This numerical is Ex 3_7E,page 54.
+
+//Velocity diagrams are not drawn. This is with Scilab team's permission.
+//The numerical part of the question has been solved.
+
+r_t=5.2
+r_h=3.5
+N=4500
+
+A=%pi*(r_t^2-r_h^2)
+printf(" A is equal to %0.2f in^2",A)
+disp("On converting to feet we get A= 0.322 ft^2")
+
+r_m=[0.5*(r_t^2+r_h^2)]^0.5
+printf("\n rm is equal to %0.2f in",r_m)
+disp(" On converting to feet we get rm =0.369ft")
+
+r_m=0.369//in feet
+U_m=(N*%pi*r_m)/30
+printf("\n U_m is equal to %0.0f ft/s",U_m)
+
+disp("From inlet velocity triangle we have V_1/sin(ß_1+pi/2)+U_m/sin(aplha_1-ß_1)")
+disp("Hence V_1=174*(sin(120)/sin(25))")
+
+V_1=174*(sin(120*%pi/180)/sin(25*%pi/180))
+printf(" V_1 = %0.2f ft/s",V_1)
+
+alpha_1=(55*%pi)/180//radians
+V_1=356.5//rounded off
+V_a=V_1*cos(alpha_1)
+printf("\n Thus V_a= %0.1f ft/s",V_a)
+
+rho=0.095
+V_a=204.5//rounded off
+A=0.322//rounded off
+m=rho*V_a*A
+printf("\n m= %0.3f lb/s",m)
+
+disp("From delta_E=U_m*V_u1 we have P_s=m*delta_E")
+m=6.25/32.2//in lbf
+U_m=174//rounded off
+V_u1=356.5*sin((55*%pi)/180)
+delta_E=U_m*V_u1
+P_s=m*delta_E
+printf(" P_s is equal to %0.1f ft-lbf/s",P_s)
+disp("On converting we get P_s = 17.9hp")
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH3/EX3.2/Ex3_2.sce b/3136/CH3/EX3.2/Ex3_2.sce new file mode 100755 index 000000000..69fc35fa3 --- /dev/null +++ b/3136/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,35 @@ +clear all; clc;
+//This numerical is Ex 3_1S,page 43.
+N=800
+Q=397//in meter cube per hour
+r_1=7.6
+b_1=10.2
+r_2=22.9
+b_2=7.6
+omega=N*%pi/30
+printf('The angular velocity is %g rad/s',omega)
+omegar=83.7
+disp("After rounding off the value of angular velocity is 83.7 rad/s")
+U_1=omega*r_1/100
+printf(' U_1=%g',U_1)
+disp("After rounding off the value of U_1 is 6.36 m/s")
+U_1r=6.36//rounded value of U1
+U_2=omega*r_2/100
+printf(' U_2=%g',U_2)
+disp("After rounding off the value of U_2 is 19.2 m/s")
+A_1=2*%pi*r_1*b_1
+printf(' A_1=%g cm^2',A_1)
+disp("After rounding off the value of A_1 is 487 cm^2")
+A_2=2*%pi*r_2*b_2
+printf(' A_2=%g cm^2',A_2)
+disp("After rounding off the value of A_2 is 1093.5 cm^2")
+A_1r=487//rounded off
+A_2r=1093.5//rounded off
+V_r1=(Q/3600)/(A_1r/10000)
+printf(' The value of V_r1 is %g',V_r1)
+disp("The value of V_r1 after rounding off is 2.26 m/s")
+V_r2=(Q/3600)/(A_2r/10000)
+printf(' The value of V_r2 is %g',V_r2)
+disp("The value of V_r2 after rounding off is 1.01 m/s")
+
+
diff --git a/3136/CH3/EX3.3/Ex3_3.sce b/3136/CH3/EX3.3/Ex3_3.sce new file mode 100755 index 000000000..3ccbe7b6e --- /dev/null +++ b/3136/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,39 @@ +clear all; clc;
+//This numerical is Ex 3_2E,page 44.
+
+V_r1=7.47
+U_1=20.9
+V_r2=3.27
+U_2=62.7
+
+//let x=tanbeta1
+x=V_r1/U_1
+printf("\n The value of ß_f1 is equal to %0.3f degrees",x)
+beta_f1=(atan(x))*180/%pi
+printf("\n Thus the value of ß_f1 is %0.2f degrees",beta_f1)
+
+V_1=V_r1
+W_1=(U_1^2+V_r1^2)^0.5
+printf("\n Thus the value of W_1 is %0.2f ft/s",W_1)
+
+beta_f2=beta_f1-10
+printf("\n Hence the value of beta_f2 is equal to %0.2f degrees",beta_f2)
+
+//rounding of value of betaf2 to be equal to 9.6
+beta_f2=9.6
+W_u2=V_r2/tan(beta_f2*%pi/180)
+printf("\n Hence the value of W_u2 is %0.1f ft/s",W_u2)
+
+V_u2=U_2-W_u2
+printf("\n Hence the value of V_u2 is equal to %0.1f ft/s",V_u2)
+
+
+//rounding off Wu2
+W_u2=19.3
+W_2=(W_u2^2+V_r2^2)^0.5
+printf("\n The value of W_u2 is equal to %0.3f ft/s",W_2)
+
+//rounding off Vu2
+V_u2=43.4
+V_2=(V_u2^2+V_r2^2)^0.5
+printf("\n Thus he value of V_2 is equal to %0.2f ft/s",V_2)
diff --git a/3136/CH3/EX3.4/Ex3_4.sce b/3136/CH3/EX3.4/Ex3_4.sce new file mode 100755 index 000000000..dae511432 --- /dev/null +++ b/3136/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,39 @@ +clear all; clc;
+//This numerical is Ex 3_2S,page 44.
+
+V_r1=2.26
+U_1=6.36
+V_r2=1.01
+U_2=19.2
+
+//let x=tan(beta_1)
+x=V_r1/U_1
+printf("\n The value of ß_f1 is equal to %0.3f degrees",x)
+beta_f1=(atan(x))*180/%pi
+printf("\n Thus the value of ß_f1 is %0.1f degrees",beta_f1)
+
+V_1=V_r1
+W_1=(U_1^2+V_r1^2)^0.5
+printf("\n Thus the value of W_1 is %0.2f m/s",W_1)
+
+beta_f2=beta_f1-10
+printf("\n Hence the value of ß_f2 is equal to %0.1f degrees",beta_f2)
+
+//rounding of value of betaf2 to be equal to 9.6
+beta_f2=9.6
+W_u2=V_r2/tan(beta_f2*%pi/180)
+printf("\n Hence the value of W_u2 is %0.2f m/s",W_u2)
+
+V_u2=U_2-W_u2
+printf("\n Hence the value of V_u2 is equal to %0.2f m/s",V_u2)
+
+
+//rounding off W_u2
+W_u2=5.97
+W_2=(W_u2^2+V_r2^2)^0.5
+printf("\n The value of W_2 is equal to %0.3f m/s",W_2)
+
+//rounding off V_u2
+V_u2=13.23
+V_2=(V_u2^2+V_r2^2)^0.5
+printf("\n Thus he value of V_2 is equal to %0.2f m/s",V_2)
diff --git a/3136/CH3/EX3.5/Ex3_5.sce b/3136/CH3/EX3.5/Ex3_5.sce new file mode 100755 index 000000000..6012432f4 --- /dev/null +++ b/3136/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,60 @@ +clear all; clc;
+//This numerical is Ex 3_3E,page 46.
+//the value given in the book for Um is 200.5,but on calculating the value comes out to be 200.3
+
+r_t=24
+r_h=10
+N=1250
+Q=53000
+
+r_m=(0.5*(r_t^2+r_h^2))^0.5
+printf("\n The mean radius rm is equal to %0.1f in",r_m)
+disp("Converting to feet we have r_m equal to 1.53 ft")
+
+
+//rm in feet equals 1.53
+r_m=1.53
+U_m=((N*%pi)/30)*r_m
+printf(" Um = %0.1f ft/s",U_m)//the value given in the book for Um is 200.5,but on calculating the value comes out to be 200.3
+
+A=(%pi*(r_t^2-r_h^2))/144
+printf("\n The value of A is %0.1f ft^2",A)
+
+//rounding off A to be 10.4
+A=10.4
+V_a2=Q/(A*60)
+V_a1=V_a2
+printf("\n V_a1=V_a2 %0.1f ft/s",V_a2)
+
+beta_f1=(atan(U_m/V_a1))*180/%pi
+beta_b1=beta_f1
+printf("\n beta_b1=beta_f1 %0.1f degrees",beta_b1)
+
+
+rho_w=62.4
+rho_a=0.0762
+g=32.2
+H_w=2/12
+disp("We know that g*Ha=(rho_w/rho_a)*g*H_w=U_mV_u2=Um*(U_m-V_a2*tan(beta_f2))")
+disp("Hence we can find out the value of tan(ß_f2).Thus we can determine value of ß_f2.")
+//Let n=U_m*(U_m-V_a2*tan(beta_f2)) and m=(rho_w/rho_a)*g*H_w
+m=(rho_w/rho_a)*g*H_w
+//therefore tan_betaf2=(U_m-m/U_m)/V_a2
+tanbeta_f2=(U_m-m/U_m)/V_a2
+beta_f2=(atan(tanbeta_f2))*180/%pi
+printf("\n Thus the value of ß_f2 is equal to %0.1f degrees",beta_f2)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH3/EX3.6/Ex3_6.sce b/3136/CH3/EX3.6/Ex3_6.sce new file mode 100755 index 000000000..8a0a33cf9 --- /dev/null +++ b/3136/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,55 @@ +clear all; clc;
+//This numerical is Ex 3_3S,page 46.
+
+r_t=0.6
+r_h=0.25
+N=1250
+Q=1500
+
+r_m=(0.5*(r_t^2+r_h^2))^0.5
+printf("\n The mean radius rm is equal to %0.2f in. ",r_m)
+
+U_m=((N*%pi)/30)*r_m
+printf(" On converting,U_m = %0.1f m/s",U_m)
+
+A=(%pi*(r_t^2-r_h^2))
+printf("\n The value of A is %0.4f m^2",A)
+
+
+V_a2=Q/(A*60)
+V_a1=V_a2
+printf("\n V_a1=V_a2 %0.1f m/s",V_a2)
+
+beta_f1=(atan(U_m/V_a1))*180/%pi
+beta_b1=beta_f1
+printf("\n beta_b1=beta_f1 %0.1f degrees",beta_b1)
+
+
+rho_w=998
+rho_a=1.22
+g=9.8
+H_w=5/100
+U_m=60.2//rounding off Um
+disp("We know that g*H_a=(rho_w/rho_a)*g*H_w=U_mV_u2=U_m*(U_m-V_a2*tan(beta_f2))")
+disp("Hence we can find out the value of tanbet_f2")
+//Let n=U_m*(U_m-V_a2*tan(beta_f2)) and m=(rho_w/rho_a)*g*H_w
+m=(rho_w/rho_a)*g*H_w
+//therefore tanbeta_f2=(U_m-m/U_m)/V_a2
+tanbeta_f2=(U_m-m/U_m)/V_a2
+beta_f2=(atan(tanbeta_f2))*180/%pi
+printf("\n Thus the value of beta_f2 is equal to %0.1f degrees",beta_f2)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH3/EX3.7/Ex3_7.sce b/3136/CH3/EX3.7/Ex3_7.sce new file mode 100755 index 000000000..e6e971300 --- /dev/null +++ b/3136/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,80 @@ +clear all; clc;
+//This numerical is Ex 3_4E,page 49.
+
+//the value of deltaE2 slightly differs from the value given in the book. However the calculated value is correct for te given substitutions
+
+N=120
+r_1=1.8
+b_1=0.3
+
+omega=(N*%pi)/30
+printf("\n Omega is equal to %0.3f rad/s",omega)
+
+U_1=r_1*omega
+printf("\n U_1 is equal to %0.1f ft/s",U_1)
+
+r_m2=(0.5*(1^2+0.4^2))^0.5
+//rounding off the value of rm2
+r_m2=0.761
+
+U_2=r_m2*omega
+printf("\n U_2 is equal to %0.2f ft/s",U_2)
+
+A_1=2*%pi*r_1*b_1
+printf("\n A_1 is equal to %0.2f ft^2",A_1)
+
+r_t2=1.0
+r_h2=0.4
+b_2=0.5
+A_2=%pi*(r_t2+r_h2)*b_2
+printf("\n A_2 is equal to %0.2f ft^2",A_2)
+
+disp("Assume swirl free flow at discharge that is V_u2=0 and ß_f1=ß_b1, ß_f2=ß_b2.")
+
+V_m2=U_2*tan(15*(%pi/180))
+printf("\n So V_m2=U_2tan15 is equal to %0.2f ft/s",V_m2)
+disp("Thus now we can determine Q")
+
+V_m2=2.56//rounding off
+Q=V_m2*A_2
+disp("Q=V_m2*A_2=V_r1*A_1")
+printf("\n Thus Q is equal to %0.2f ft^3/s",Q)
+
+disp("Since the value of Q,A_1 is known,we can determine the value of V_r1")
+V_r1=Q/A_1
+printf(" The value of V_r1 is equal to %0.2f ft/s",V_r1)
+
+beta_f1=27
+W_u1=V_r1/tan((beta_f1*%pi)/180)
+printf("\n W_u1 is equal to %0.2f ft/s",W_u1)
+
+V_u1=U_1-W_u1
+printf("\n The value of V_u1 is %0.2f ft/s",V_u1)
+
+U_1=22.6//rounding off
+V_u1=19.3//rounding off
+delta_Et=U_1*V_u1
+printf("\n delta_Et is equal to %0.1f ft^2/s^2",delta_Et)//the value given in the book is 437.1,but the actual value is as calculated for the given values of U1 and Vu1
+
+m=(62.4/32.2)*5.63
+P_s=m*delta_Et
+printf("\n Thus P_s is equal to %0.1f ft*lbf/s",P_s)//since value of deltaEt differs from the one given in the book,so does the value of Ps
+disp("Converting P_s to hp we get 8.65hp")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH3/EX3.8/Ex3_8.sce b/3136/CH3/EX3.8/Ex3_8.sce new file mode 100755 index 000000000..1e1e88a77 --- /dev/null +++ b/3136/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,80 @@ +clear all; clc;
+//This numerical is Ex 3_4S,page 50.
+//the value of deltaE2 slightly differs from the value given in the book. However the calculated value is correct for te given substitutions
+
+N=120
+r_1=54.8
+b_1=9.1
+
+omega=(N*%pi)/30
+printf("\n Omega is equal to %0.3f rad/s",omega)
+
+U_1=r_1*omega/100
+printf("\n U_1 is equal to %0.1f m/s",U_1)
+
+r_m2=(0.5*(30.5^2+ 12.2^2))^0.5
+
+U_2=r_m2*omega/100
+printf("\n U_2 is equal to %0.2f m/s",U_2)
+
+A_1=2*%pi*r_1*b_1/10000
+printf("\n A1 is equal to %0.3f m^2",A_1)
+
+r_t2=30.5
+r_h2=12.2
+b_2=15.2
+A_2=%pi*(r_t2+r_h2)*b_2/10000
+printf("\n A_2 is equal to %0.3f m^2",A_2)
+
+disp("Assume swirl free flow at discharge that is V_u2=0 and ß_f1=ß_b1, ß_f2=ß_b2.")
+
+V_m2=U_2*tan(15*(%pi/180))
+printf("\n So V_m2=U_2tan15 is equal to %0.3f m/s",V_m2)
+disp("Thus now we can determine Q")
+
+Q=V_m2*A_2
+disp("Q=V_m2*A_2=V_r1*A_1")
+printf("\n Thus Q is equal to %0.3f m^3/s",Q)
+
+disp("Since the value of Q,A_1 is known,we can determine the value of V_r1")
+Q=0.159
+V_r1=Q/A_1
+printf(" The value of V_r1 is equal to %0.4f m/s",V_r1)//actual values are taken,hence a 0.0005 difference in answer is observed
+
+beta_f1=27//ßf1
+V_r1=0.508//rounding off Vr1
+W_u1=V_r1/tan((beta_f1*%pi)/180)
+printf("\n W_u1 is equal to %0.4f m/s",W_u1)
+
+W_u1=0.997//rounding off
+U_1=6.9//rounding off
+V_u1=U_1-W_u1
+printf("\n The value of V_u1 is %0.2f m/s",V_u1)//0.02 difference obtained because of substituting the values as they ahve been found out
+
+
+V_u1=5.92//as substituted in the book
+delta_Et=U_1*V_u1
+printf("\n deltaEt is equal to %0.2f m^2/s^2",delta_Et)//
+
+m=998*0.782*0.204
+P_s=m*delta_Et
+printf("\n Thus P_s is equal to %0.0f W",P_s)
+disp("Converting P_s to kW we get 6.503kW")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH3/EX3.9/Ex3_9.sce b/3136/CH3/EX3.9/Ex3_9.sce new file mode 100755 index 000000000..d2288d59e --- /dev/null +++ b/3136/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,26 @@ +clear all; clc;
+//This numerical is Ex 3_5E,page 52.
+
+//this numerical is based on numerical 3.4E
+//values found in the book for numerical 3.4E will be used to solve this numerical(3.5E)
+
+delta_Et=437.1
+U_1=22.6
+U_2=9.56
+V_m2=2.56
+V_2=2.56
+
+V_r1=1.66
+W_u1=3.26
+W_1=(V_r1^2+W_u1^2)^0.5
+printf("\n W_1 is equal to %0.2f ft/s",W_1)
+
+W_2=(U_2^2+V_m2^2)^0.5
+printf("\n W_2 is equal to %0.2f ft/s",W_2)
+
+V_u1=19.3
+V_1=(V_r1^2+V_u1^2)^0.5
+printf("\n V_1 is equal to %0.2f ft/s",V_1)
+
+Rt=0.5*[(U_1^2-U_2^2)+(W_2^2-W_1^2)]/(delta_Et)
+printf("\n Thus the value Rt is equal to %0.3f",Rt)
diff --git a/3136/CH4/EX4.1/Ex4_1.sce b/3136/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..58dcf1392 --- /dev/null +++ b/3136/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,40 @@ +clear all; clc;
+
+N=1750
+r_2=0.15
+U_2=N*%pi*r_2/30
+printf("With U_2= %0.1f m/s",U_2)
+
+b_2=0.012
+
+A_2=2*%pi*r_2*b_2
+printf(" and A_2= %0.4f m^2",A_2)
+
+disp("Equations 4.3 and 4.4 can be rewritten.")
+disp("H_a=µ_s*(U_2/g)*(U_2-Q/(A_2*tanß_b2))-kQ^2...(4.3)")
+disp("δ_Ha/δ_Q=(-µ_s*(U_2/(g*A_2*tanß_b2)))-2*k_Q...(4.4)")
+disp("Thus the equations become")
+
+disp("(27.5/9.8)*(27.5-(4.5/(1000*0.0113*tan(60))))*µs-k*(4.5/1000)^2=60")
+disp("(27.5/(9.8*0.0113*tan(60)))*µs+(2*(4.5/1000))*k=2.5*1000")
+
+disp("from equation 4.3 µ_s=(60+(2.025*10^-5)*k)/76.25")
+disp("from equation 4.4 µ_s=(25-(9*10^-5)*k)/1.434")
+disp("Solving for µs and k we get µ_s=0.852 and k=2.642*10^5")
+
+H_a=60
+Q=4.5/1000
+k=2.642*10^5
+ETA_h=H_a/(H_a+k*(Q^2))
+printf(" Thus ETA_h= %0.4f",ETA_h)
+
+P_s=3.2*10^3
+rho=1000
+g=9.8
+ETA_oa=(rho*Q*g*H_a)/P_s
+printf("\n ETA_oa= %0.3f",ETA_oa)//answer given in the book is0.825,however the correct answer is0.827.
+
+disp("Eta_m=0.946")
+
+
+
diff --git a/3136/CH4/EX4.10/Ex4_10.sce b/3136/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..784abaee4 --- /dev/null +++ b/3136/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,41 @@ +clear all; clc;
+//Value of s differs. The one given in the book is incorrect. On calculation the value is equal to the one provided here.
+
+disp("The dynamic pressure at the inlet is calculated from pd1=rho*V1^2/2")
+
+Q=220*0.00223//0.00223 is conversion factor
+D=5
+A=%pi*(D^2)/4
+V_1=(Q*144)/(A)
+printf(" V1=Q/A= %0.2f ft/s",V_1)
+
+rho=62.4
+V_1=3.60
+p_d1=(rho*(V_1^2))/(2*144*32.2)
+printf(" \npd1= %0.3f psi",p_d1)
+
+disp("The vapor pressure of water at 80 degrees Farenheit is obtained from steam tables as pv=0.507psia.")
+disp("The required NPSH is designated at the condition such that discharge pressure drops for 3% from the non cavitation condition.")
+
+dpdash=0.97*70
+printf(" Hence from the above test data,the correspondin dpdash= %0.2f psi",dpdash)
+
+p_1dash=7.2+(7.5-7.2)*(67.9-67.5)/0.5
+printf("\n p1dash can also be obtained %0.2f psia",p_1dash)
+
+disp("Also NPSPR=p1dash+pd1-pv")
+p_1dash=7.44//rounded off
+p_d1=0.087//rounded off
+p_v=0.507
+NPSPR=p_1dash+p_d1-p_v
+printf("\n Hence NPSPR = %0.2f psia",NPSPR)
+
+NPSPR=7.02//rounded off
+NPSPHR=2.307*NPSPR
+printf("\n NPSPHR= %0.2f ft",NPSPHR)
+
+N=3500
+Q=220
+NPSHR=16.2
+S=(N*(Q^0.5))/((NPSHR)^0.75)
+printf("\n S= %0.0f rpm*(gpm^0.5)/(ft^0.75))",S)//S= 6419 but on calculation it is found out that S=6429. Hence the value given here is correct.
diff --git a/3136/CH4/EX4.11/Ex4_11.sce b/3136/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 000000000..5a14b827f --- /dev/null +++ b/3136/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,35 @@ +clear all; clc;
+
+disp("With H=k*(Q^2)+3,the constant k can be obtained.")
+
+H=10*144/62.4
+printf("We have H= %0.2f ft and Q=25gpm",H)
+
+Q_1=25
+k=(H-3)/(Q_1^2)
+printf("\nThus k= %0.4f ft/(gpm^2)",k)
+
+disp("Thus the H/Q relationship can be tabulated and plotted as follows")
+
+Q=[0 5 10 15 20 25 30];
+Hdash = zeros(1,length(Q));
+
+for i = 1: length(Q)
+
+
+ Hdash(i)=(0.0321*((Q(i))^2)+3);
+end
+
+table = [Q' Hdash'];
+disp(" Q(gpm) H(ft)")
+disp(table)
+
+plot(Q,Hdash)
+
+xlabel("Q (gpm)")
+ylabel("H(ft)")
+
+disp("(b) Rewrite H=0.0321*(Q^2)+3 into Q=5.581*((H-3)^0.5)")
+disp("From Qc=2Q and Hc=H,we have Qc=11.16*(Hc-3)^0.5 or Qc^2=124.6*(Hc-3)")
+disp("Or Hc=0.00802*Qc^2+3")
+disp("It is the expression of the combined system")
diff --git a/3136/CH4/EX4.12/Ex4_12.sce b/3136/CH4/EX4.12/Ex4_12.sce new file mode 100755 index 000000000..ec8fd40db --- /dev/null +++ b/3136/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,28 @@ +clear all; clc;
+
+disp("The single pump performance is given as follows")
+Q=[0 70 130 172 186 190 195 200];
+H=[1600 1500 1275 920 785 540 435 285];
+table=[Q' H'];
+disp(" Q(gpm) H(ft)")
+disp(table)
+
+disp("a) Two connected in series")
+Q_1=[0 70 130 172 186 190 195 200];
+H_1=[3200 3000 2550 1840 1570 1080 870 570];
+table1=[Q_1' H_1']
+disp(" Q_1(gpm) H_1(ft)")
+disp(table1)
+
+disp("b) Two connected in parallel")
+Q_2=[0 140 260 344 372 380 390 400];
+H_2=[1600 1500 1275 920 785 540 435 285];
+table2=[Q_2' H_2']
+disp(" Q2(gpm) H2(ft)")
+disp(table2)
+
+plot(Q,H,'o',Q_1,H_1,'d',Q_2,H_2,'s')
+legend("H(ft))","H1(ft)","H2(ft)",-1)
+xlabel("Q (gpm)")
+ylabel("H(ft)")
+set(gca(),"grid",[1 1])
diff --git a/3136/CH4/EX4.13/Ex4_13.sce b/3136/CH4/EX4.13/Ex4_13.sce new file mode 100755 index 000000000..9089db9eb --- /dev/null +++ b/3136/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,48 @@ +clear all; clc;
+
+disp("The specific speeds for all three options can be calculated")
+
+N_sa=3600*(500^0.5)/(350^0.75)
+printf("Nsa= %0.0f rpm*(gpm^0.5)/(ft^0.75)",N_sa)
+
+N_sb=4320*(500^0.5)/(350^0.75)
+printf("\nNsb= %0.0f ",N_sb)
+
+N_sc=3600*((500^0.5)/(175^0.75))
+printf("\nNsc= %0.0f",N_sc)
+
+disp("From figures 4.20 and 4.36, we oobtain Effa=0.75, Dsa=1.7,Effb=0.78,Dsb=1.4,Effc=0.80,Dsc=1.31")
+
+D_sa=1.7
+Q=500
+H=350
+D_a=D_sa*(Q^0.5)/(H^0.25)
+printf("\nDa=Dsa*(Q^0.5)/(H^0.25)= %0.1f in",D_a)
+
+D_sb=1.4
+D_b=D_sb*(Q^0.5)/(H^0.25)
+printf("\nDb= %0.1f in",D_b)
+
+D_sc=1.1
+D_c=D_sc*(Q^0.5)/(H^0.25)
+printf("\nDc= %0.1f in",D_c)//for Dsc=1.1 the value determined is correct. The value of Dc given in the book is incorrect.
+
+Q=500*0.00223//conversion from gpm to cubic ft/s
+//let x=rho*g
+x=62.4
+P_h=Q*x*H
+printf("\nThe output hydraulic power is equal to %0.0f ft-lbf/s = 44.3hp=33kW",P_h)
+
+P_a=33/(0.75*0.9)
+printf("\nThe power required for option a= %0.1f kW",P_a)
+
+P_b=33/(0.78*0.95*0.9)
+printf("\nThe power required for option b= %0.1f kW",P_b)
+
+P_c=33/(0.80*0.9)
+printf("\nThe power required for option c= %0.1f kW",P_c)
+
+
+
+
+
diff --git a/3136/CH4/EX4.2/Ex4_2.sce b/3136/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..3560463b6 --- /dev/null +++ b/3136/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,58 @@ +clear all; clc;
+
+disp("The relevant formulae are given as : ")
+disp("Eta_1 =((62.4*0.00223*Q_1*H_1/(550*P_s1)))*100")
+disp("Q_2=Q_1*N_2/N_1")
+disp("H_2=H_1*(N_2/N_1)^2")
+disp("P_s2=P_s1*((N_2/N_1)^3)")
+disp("H and P_s at Q=0 are obtained with extrapolition")
+
+Q_1= [0 285 435 540 785 920 1275];
+H_1= [205 200 195 190 186 172 130];
+P_s1=[28 31 36 42 44 49 58];
+
+Eta_1 = zeros(1,length(Q_1));
+Q_2=zeros(1,length(Q_1));
+H_2=zeros(1,length(Q_1));
+P_s2=zeros(1,length(Q_1));
+Eta_2=zeros(1,length(Q_1));
+
+for i = 1: length(Q_1)
+
+
+ Eta_1(i) =((62.4*0.00223*Q_1(i)*H_1(i)/(550*P_s1(i))))*100;//multiplied by 100 to get answer in percentage
+ Q_2(i) = Q_1(i)*1850/1350;//Since Q2=Q1*N2/N1 and N2=1850 and N1=1350
+ H_2(i) = H_1(i)*((1850/1350)^2);//Since H2=H1*((N2/N1)^2)
+ P_s2(i)=P_s1(i)*((1850/1350)^3)
+ Eta_2(i)=Eta_1(i)
+end
+
+table = [Q_1'H_1'P_s1'Eta_1'Q_2'H_2'P_s2'Eta_2'];
+disp(" Q_1(gpm) H_1(ft) P_s1(hp) Eta_1(%) Q_2(gpm) H_2(ft) P_s2(hp) Eta_2(%)")
+disp(table)
+
+figure()
+plot(Q_1,H_1,'o',Q_1,P_s1,'d',Q_1,Eta_1,'s')
+legend("H_1(ft)","P_s1(hp)","Eta_1(%)",-1)
+xlabel("Q_1(gpm)")
+ylabel("H_1(ft), P_s1(hp) , Eta_1(%)")
+set(gca(),"grid",[1 1])
+
+figure(1)
+plot(Q_2,H_2,'o',Q_2,P_s2,'d',Q_2,Eta_2,'s')
+legend("H_2(ft)","Ps_2(hp)","Eta_2(%)",-1)
+xlabel("Q_2(gpm)")
+ylabel("H_2(ft), Ps_2(hp) , Eta_2(%)")
+set(gca(),"grid",[1 1])
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH4/EX4.3/Ex4_3.sce b/3136/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..d8aa50843 --- /dev/null +++ b/3136/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,30 @@ +clear all; clc;
+
+disp("The expression Ps=rho*Q*gH/Eta is used to calculate the shaft power for al finite flow rate conditions.For the shutoff condition,since Eta is zero,Ps has to be extrapolated. Also 0.1H is plotted instead of H in the same chart since its order of magnitude is higher than others.")
+
+disp("The efficiency and NPSHR are read from the constant contours with interpolation and Ps is calculated with the above equation.")
+
+Q = [40 80 120 160 200 220];
+OneTenthH = [27 26.5 25.5 24 21.5 20];
+eff = [30 42.5 52 56.7 57 54.5];
+NPSHR = [5 5.5 6.1 7.3 11.5 16];
+
+Ps = zeros(1,length(Q));
+
+for i = 1: length(Q)
+
+
+ Ps(i)=(62.4*0.00223*Q(i)*OneTenthH(i)*10)/((eff(i)/100)*550);//550 is conversion factor
+
+end
+
+disp(" Q(gpm) 1/10H(ft) eff(%) NPSHR(ft) Ps(hp)")
+
+table=[Q' OneTenthH' eff' NPSHR' Ps' ];
+disp(table)
+
+plot(Q,OneTenthH,'o',Q,NPSHR,'d',Q,Ps,'s',Q,eff,'*')
+legend("OneTenthH(ft)","NPSHR(ft))","Ps(hp)","eff(%)",-1)
+xlabel("Q(gpm)")
+ylabel("OneTenthH(ft), NPSHR(ft) , Ps(hp) eff(%)")
+set(gca(),"grid",[1 1])
diff --git a/3136/CH4/EX4.4/Ex4_4.sce b/3136/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..c50d44993 --- /dev/null +++ b/3136/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,17 @@ +clear all; clc;
+
+disp("From figure 4.7a, H=235ft, Q=170 gpm at b.e.p. of 57.5%. ")
+disp("The corresponding NPSHR is 8.7ft. From the steam table we have pv=3.73 psia at T=150 degrees Farenheit")
+disp("Also NPSPA=pt-pf+rho*g*Z-pv is greater than or equal to NPSPR ")
+disp("Thus we can determine the value of Z")
+
+NPSHR=8.7
+p_t=20
+p_f=1.5*((170/50)^2)
+p_v=3.73
+//Let x=rho*g=62.4
+x=62.4
+
+Z=NPSHR-[p_t-p_f-p_v]*(144/(x))//144 is conversion factor
+
+printf(" Thus the value of Zis equal to %0.1f ft",Z)
diff --git a/3136/CH4/EX4.5/Ex4_5.sce b/3136/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..9fb2112d5 --- /dev/null +++ b/3136/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,25 @@ +clear all; clc;
+
+disp("The b.e.p. of all four trims are read as follows")
+Dia=[9.25 8 7 6];
+Q=[220 170 155 130];
+H=[330 235 170 120];
+Ps=[29 17 12.5 8];
+eff=[62.5 57.5 53.5 49.5];
+table=[Dia' Q' H' Ps' eff']
+disp("Dia(in) Q(gpm) H(ft) Ps(hp) eff(%)")
+disp(table)
+
+disp("From the similarity laws Q is proportional to D, H proportional to D^2 and Ps proportional to D^3")
+
+disp("The performance should be")
+
+Di=[9.25 8 7 6];
+Q=[220 190 166 143 ];
+H=[330 246 189 139];
+P_s=[29 18.7 12.5 7.9];
+eff=[62.5 62.5 62.5 62.5];
+
+disp("Dia(in) Q(gpm) H(ft) P_s(hp) eff(%)")
+table=[Dia' Q' H' P_s' eff']
+disp(table)
diff --git a/3136/CH4/EX4.6/Ex4_6.sce b/3136/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 000000000..7ce7c8c0c --- /dev/null +++ b/3136/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,17 @@ +clear all; clc;
+
+disp("From figure 4.7a,we have NPSHR=8.7ft=2.65m at Q=170gpm for the b.e.p. condition of the 8in impeller.")
+
+disp("From figure 4.18 we have deltaNPSH=0.6ft=0.18m and pv=69kPa at T=90 degrees celsius for water. ")
+disp("The total NPSPA is equal to pt-rho*g*Z-pf-pv and the total NPSPR is equal to rho*g*(NPSHR-deltaNPSH)")
+disp("To avoid cavitation, at the pump inlet ,it is required to have NPSPA>NPSPR")
+disp("Hence it is required (pt-rho*g*Z-pf-pv)>rho*g*(NPSHR-deltaNPSH)")
+p_f=10
+p_v=69
+rho=998
+g=9.8
+Z=2
+NPSHR=2.65
+deltaNPSH=0.18
+p_t=p_f+p_v+rho*g*(Z+NPSHR-deltaNPSH)/1000//1000 is conversion factor
+printf("\n Thus pt is equal to %0.1f kPA",p_t)
diff --git a/3136/CH4/EX4.7/Ex4_7.sce b/3136/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..46b0f23cc --- /dev/null +++ b/3136/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,154 @@ +clear all; clc;
+
+disp("(a) Start wit calculation of specific speed")
+
+N=2400
+Q=1200
+H=450
+N_s=(N*Q^0.5)/(H^0.75)
+printf("Ns= %0.0f rpm*gpm^0.5/ft^0.75",N_s)
+
+disp("From figure 4.20 we obtain overall efficiency is equal to 0.76")
+Q=1200
+delta_p=0.00223*62.4*450
+ETA_oa=0.76
+P_s=(Q*delta_p)/(ETA_oa)
+printf("Since Ps=(Q*delta_p)/(ETA_oa),P_s= %0.0f ft-lbf/s",P_s)
+
+disp("To determine the inlet configuration:")
+N=2400
+omega=N*%pi/30
+printf("Omega = %0.0frad/s",omega)
+tau=P_s/omega
+printf("\nτ= %0.0f lbf-ft",tau)
+
+disp("Hence the shaft diameter Dsh can be determined")
+sigma_s=10000
+D_sh=(16*tau/(144*%pi*sigma_s))^(1/3)
+printf("D_sh= %0.4f ft =1.34 inches",D_sh)
+
+disp("A pure radial inet confuguratin isbselected that is Dt1=Dh1=D1(see figure 4.19).The impeller eyee area and leading edge diameters can be calculated from the following equations.")
+
+disp("U_1=0.5*omega*D_1")
+disp("Q=A_1*V_1=A_e*V_e")
+disp("V_1=1.1*V_e")
+disp("ß_f1=taninverse(V1/U1)")
+
+disp("Where A_1=ε_1*pi*D_1*b_1")
+disp("Where A_e=pi*D_e^2/4")
+disp("With φ_1=Q/(A_e*U_e)=Q/(pi*omega*re^3)=27")
+disp("Thus we obtain re=2.8 in and selecting r1=3.2 inches the final results are obtained as Ve=Q/(pi*D_e^2/4)=15.6ft")
+disp("V_1=17.2ft/s , A_1=22.4in^2, b2=1.24in with ε_1=0.9 assumed. Also U_1=66.9ft/s,ß_f1=14.4 degrees,ß_b1=17 degrees and W_1=69.1 ft/s")
+
+disp("(c) The impeller discharge is determined ")
+omega=251
+Q=2.676
+//elt x=g*H=14490
+x=14490
+omega_s=omega*(Q^0.5)/((x)^0.75)
+printf("\nomegas is %0.3f",omega_s)
+
+delta_s=9
+D_2=delta_s*(Q^0.5)/((x)^0.25)
+printf("\nD2= %0.4f ft = 16.1in",D_2)
+
+D_2=1.342//rounded off
+omega=251
+U_2=D_2*omega/2
+printf("\n U_2= %0.2f",U_2)//in the book the answer has been rounded off to 168.5
+
+disp("Selecting ß_b2=27.5 degrees")
+Zb=6.5*((1+0.4)/(1-0.4))*sin(((17+27.5)/2)*%pi/180)
+printf("\nZb= %0.2f is approximately equal to 6",Zb)
+MUs=1-%pi*sin((27.5/6)*%pi/180)
+printf("\nThus µs= %0.3f",MUs)//The answer given in the book is 0.76,but the more accurate answer is 0.749
+
+ETA_h=1-0.8/((1200)^0.25)
+printf("\n ETA_h= %0.3f",ETA_h)
+
+disp("(3) Substituting these values into the Euler equation ,we obtain values of Vu2 and Vr2")
+ETA_h=0.864
+U_2=168.5
+V_u2=x/(ETA_h*U_2)
+printf("\n Vu2= %0.2f ft/s",V_u2)//value given in the book is equal to 99.6,however the more accurate value is equal to 99.53
+
+disp("Vu2=µ*(U2-Vr2/tanßb2)")
+disp("Hence we can find out the value of Vr2=19.4 ft/s")
+V_r2=19.4
+V_u2=99.6//rounded off value taken in the book
+U_2=168.5
+W_2=[V_r2^2+(U_2-V_u2)^2]^0.5
+printf("W2= %0.2f ft/s",W_2)
+
+disp("Hence W2/W1 >1 , which is not appropriate. A different D2 has to be selected,say D2=15in.")
+disp("So we have U_2=159.6 ft/s,V_u2=106.9 ft/s , V_rt=8.5ft/s, and W_2=50.7 ft/s")
+disp("Hence W_2/W_1=0.74, which is acceptable")
+
+disp("The outlet dimensions are calculated as A2=Q/Vr2")
+Q=1200
+V_r2=8.5
+A_2=(Q*0.00223)/V_r2//0.00223 is conversion factor to convert gallons per minute to cubic feet per second
+printf("\nThus A_2= %0.3f ft^2",A_2)
+disp("On converting, we get A2=45.3 in^2")
+disp("A_2=ε_2*pi*D_2*b_2, where ε_2=1-(Zb*t)/(pi*D_2*sin(ß_b2))")
+Zb=6
+t=0.25
+D_2=15
+beta_b2=27.5*%pi/180//converting to radians
+epsilon_2=1-(Zb*t)/(%pi*D_2*sin(beta_b2))
+printf("Thus ε2 = %0.2f",epsilon_2)
+A_2=45.3//in inches
+b_2=A_2/(epsilon_2*%pi*D_2)
+printf("\nThus we can determine b_2 which is equal to %0.0f in",b_2)
+
+epsilon_1=1-(6*0.25)/(%pi*6.4*sin(17*%pi/180))
+printf("\nε_1= %0.2f",epsilon_1)
+disp("b_1 has to be adjusted")
+b_1=1.3*0.9/0.74
+printf("\nThus b_1 is equal to %0.2f in",b_1)
+
+disp("The blade profile is constructed by assuming a linear blade angle distribution")
+disp("ßb=ßb1+(ßb2-ßb1)*(r-r1)/(r2-r1))=17+2.44(r-3.2)")
+disp("where ßb is in degrees and r is in inches.")
+disp("Also deltatheta can be expressed as (180/pi)*deltar/(r*tanßb)")
+disp("If we select deltar=0.43 in,then ßb,deltatheta,theta,dL and L can be calculated and tabulated as follows")
+
+r=[3.2 3.63 4.06 4.49 4.92 5.35 5.78 6.21 6.64 7.07 7.5];
+beta_1=[17 18.05 19.1 20.15 21.2 22.25 23.3 24.35 25.4 26.45 27.5];
+dthita=[25.188 20.831 17.528 14.957 12.913 11.259 9.8993 8.768 7.8157 7.006 6.3116 ];
+thita=[0 25.188 46.018 63.546 78.502 91.415 102.67 112.57 121.34 129.16 136.16];
+dL=[1.4709 1.388 1.3143 1.2484 1.1892 1.1358 1.0873 1.0431 1.0026 0.9655 0.9314];
+L=[0 1.471 2.859 4.173 5.422 6.611 7.747 8.834 9.877 10.88 11.85];
+r_2=[3.2 3.63 4.06 4.49 4.92 5.35 5.78 6.21 6.64 7.07 7.5];
+beta_2=[17 17.6 18.2 18.8 19.4 20 20.6 21.2 21.8 22.4 23];
+dthita_2=[25.19 21.4 18.46 16.12 14.22 12.65 11.34 10.23 9.279 8.456 7.74];
+thita_2=[0 25.2 46.6 65 81.2 95.4 108 119 130 139 147];
+dL_2=[1.471 1.422 1.377 1.334 1.295 1.257 1.222 1.189 1.158 1.129 1.101];
+L_2=[0 1.471 2.893 4.27 5.604 6.899 8.157 9.379 10.57 11.73 12.85];
+
+table=[r' beta_1' dthita' thita' dL' L' r_2' beta_2' dthita_2' thita_2' thita_2' dL_2' L_2']
+disp("The table is in the given order:beta,dthita,thita,dL,L.r,beta,dthita,thita,thita,dL,L")
+disp(table)
+
+disp("The table shows that with ßb2=27.5 degrees the blade lemgth L=11.85 in.")
+sigma_s=6*11.85/(%pi*15)
+printf(" Hence the solidity is sigma_s= %0.2f which is too low.",sigma_s)
+
+sigma_s=9*12.85/(%pi*15)
+printf("\n Hence the new ßb2=23 degrees and Zb=9 are selected so that we hae final value of sigma_s=%0.2f.",sigma_s)
+
+disp("The final values of the impeller outlet dimensions are revised:")
+disp("A2=0.194 ft^2=27.9 in^2, b2=0.67 in , with mu_s=0.86, epsilon_2=0.88 , Vr2=13.8 ft/s , W2=51.8 ft/s")
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH4/EX4.8/Ex4_8.sce b/3136/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..79c77575b --- /dev/null +++ b/3136/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,40 @@ +clear all; clc;
+
+disp("From example 4.7 we have Q=1200gpm,H=450ft,N=2400rpm")
+disp("Furthermore Ns=851 rpm*(gpm^0.5)/(ft^0.75),D2=15in,b2=0.67in,U2=156.9 ft/s.")
+disp("Also,ßb2=23 degrees,µs=0.86,V2=13.8ft/s and Vu2=106.9ft/s")
+disp("Hence we have the angular momentum L=r2*Vu2 ")
+L=15*106.9/(2*12)
+printf(" Thus L= %0.2f dt/s^2",L)
+
+disp("Fom figure 4.24 we have K3=0.46 and Ns=851.")
+disp("Hence the volute exit velocity Vt=K3*(2*g*H)^0.5")
+V_3=0.46*sqrt(2*32.2*450)
+printf("\n Thus Vt= %0.2f ft/s",V_3)
+
+disp("Also A_t=Q/V_t ")
+A_t=1200*0.00223/78.3
+printf(" At= %0.3f ft^2 which is equal to 4.92 in^2",A_t)
+disp("Also from Figure 4.24 we have (D3-D2)/D2=0.08")
+
+r_3=7.5*(1+0.08)
+printf("\n r3= %0.2f",r_3)
+disp("If the trapezoid with an included angle of 30 degrees is selected for bassic geometry of the volute cross section,the dimensions from each section can be calculated from Athita*Vthita=(Q*thita)/(2*pi)")
+
+disp("r_c*V_thita=CL")
+disp("A_thita=0.5*h*(a+b)")
+disp("(a-b)/(2*h)=tan(15)")
+disp("x_c=h*(2a+b)/(3*(a+b))")
+disp("rc=r3+x_c")
+
+disp("where b=1.82 b2=1.206 in. is selected")
+disp("The equations are to be solved with iterations and performed with a spreadsheet.")
+disp("The initial velocity is selected as V_thita=V_t")
+disp("Then A_thita is calculated from 1,a and h from(3) and (4), and x_c and r_c from (5) and (6)")
+disp("Then the improved V_thita is obtained from (2) and the process is repeated until the discrepancies of the repeated values are small enough.")
+disp("In the final design the trapezoid is modified by roundidng off the corners and the area may be increassed by 5 to 10 percent to accomodate the blockage due to the boundary layer and sme secondary flows.")//final answers have not been provided in the book
+
+
+
+
+
diff --git a/3136/CH4/EX4.9/Ex4_9.sce b/3136/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..5119c6279 --- /dev/null +++ b/3136/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,37 @@ +clear all; clc;
+//Values of tenPs0 and eff slightly vary from those given in the book. On calculating it was found out that the values given here are more accurate than those given in the textbook
+
+disp("Use the motor calibration curves to convert input electrical power to shaft power for the corresponding rotating speed. The result is provided below.")
+
+Q1=[40 80 120 140 160 180 200 220];
+dp1=[77 76 70 66 64 62 55 48];
+Ps1=[5.2 5.6 6 6.5 7.2 7.6 7.75 7.8];
+N1=[1775 1772 1765 1760 1755 1753 1750 1747];
+table=[Q1' dp1' Ps1' N1'];
+disp(" Q1 dp1(psi) Ps1(hp) N1(rpm)")
+disp(table)
+
+disp("Convert each operating condition of flow rate,pressure rise and shaft power from the test rotating speed to the rated speed of 1750rpm.")
+Q0 = zeros(1,length(Q1));
+dp0 = zeros(1,length(Q1));
+tenPs0= zeros(1,length(Q1));
+eff=zeros(1,length(Q1));
+for i = 1: length(Q1)
+
+
+ Q0(i)=Q1(i)*(1750/N1(i));
+ dp0(i)=dp1(i)*(1750/N1(i))*(1750/N1(i));
+ tenPs0(i)=10*Ps1(i)*(1750/N1(i))*(1750/N1(i))*(1750/N1(i));//values slightly differ from those given in the book,however on calculation the values given here are more accurate than those given in the book
+ eff(i)=100*(Q0(i)*dp1(i)*144)/(449*550*tenPs0(i)/10);//since eff depends on tenPs0,hence they differ too
+end
+
+table2=[Q0' dp0' tenPs0' eff'];
+disp(" Q0(gpm) dp0(psi) tenps(hp) eff(%)")
+disp(table2)
+
+plot(Q0,dp0,'o',Q0,tenPs0,'d',Q0,eff,'s')
+legend("dp(psi) ","10Ps(hp)","Eff (%)",-1)
+xlabel("Q(gpm))")
+ylabel("dp(psi, 10Ps(hp),Eff(%))")
+set(gca(),"grid",[1 1])
+
diff --git a/3136/CH5/EX5.1/Ex5_1.sce b/3136/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..a98d296f7 --- /dev/null +++ b/3136/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,98 @@ +clear all; clc;
+
+disp("From figure 5.3a, we have Cl=0.36, Cd=0.017 and alpha=1.8 degrees at maximum L/D")
+
+N=3500
+C_l=0.36
+alpha=1.8
+C_d=0.017
+r_t=13.25
+r_h=3.25
+Zb=7
+Q=11560//cfm
+gamma_1=70
+alpha=1.8
+
+A=%pi*(r_t^2-r_h^2)
+printf("\n A is equal to %0.1f in^2",A)
+disp("On converting it in terms of feet we have A= 3.6 ft^2")
+
+r_m=[0.5*(r_t^2+r_h^2)]^0.5
+printf("\n rm is equal to %0.2f in",r_m)
+disp("On converting it in terms of feet we have rm=0.804 ft ")
+
+r_m=0.804//in feet
+U_m=(N*%pi*r_m)/30
+printf("\n Um is equal to %0.1f ft/s",U_m)
+
+r_m=9.65//in inches
+s=(2*%pi*r_m)/Zb
+printf("\n s= %0.2f in",s)
+
+disp("Va=V1=Q/A")
+
+A=3.6//in square feet
+V_1=Q/(A*60)//divided by 60 to get answer in terms of ft/s
+printf("\n V1=Va= %0.2f ft/s",V_1)
+
+U_m=294.7//rounded off
+V_1=53.5//rounded off
+beta_1=(atan(U_m/V_1))*180/%pi
+printf("\n ß1= %0.1f degrees",beta_1)
+
+beta_m=gamma_1+alpha
+printf("\n ßm= %0.1f degrees",beta_m)
+
+disp("From tanßm=(tanß1+tanß2)/2 ,ß2=30.1 degrees; Then Wmcosßm=Va")
+V_a=53.5//in ft/s
+W_m=V_a/(cos(beta_m*%pi/180))
+printf("\n So Wm= %0.1f ft/s",W_m)
+
+W_m=171.3//rounded off
+rho=0.0762/32.2
+cm=3.5
+s=8.66//rounded off
+disp("hence we have deltapr=((rho*Wm^2)/2)*(cm/s)*(Cl*sin(betam*%pi/180)-Cd*cos(betam*%pi/180))")
+delta_pr=((rho*W_m^2)/2)*(cm/s)*(C_l*sin(beta_m*%pi/180)-C_d*cos(beta_m*%pi/180))
+printf("\n deltapr= %0.4f lb/ft^2 ",delta_pr)
+disp("On rounding off we get deltapr= 4.73 lbf/ft^2")
+disp("Thus deltapr=0.0328psia=0.91 in.wg")
+
+disp("Across the stator, from the velocity diagram, we have Wu2=Va*tanß2 and V2=(Va^2+(Um-Wu2)^2)^0.5")
+
+beta_2=30.1*%pi/180
+V_a=53.5//rounded off
+W_u2=V_a*tan(beta_2)
+printf(" Thus Wu2 is equal to %0.0f ft/s",W_u2)
+
+U_m=294.7//rounded off
+V_2=(V_a^2+(U_m-W_u2)^2)^0.5
+printf("\n V2= %0.0f ft/s",V_2)
+
+disp("So assuming V_3=V_a")
+
+V_a=53.5//rounded off
+V_2=269//rounded off
+eta_s=0.85//efficiency
+rho=0.0762/32.2
+delta_ps=(((eta_s*rho)/2)*(V_2^2-V_a^2))/144//144 is conversion factor
+disp("delta_ps=(((etas*rho)/2)*(V2^2-Va^2))")
+printf("\n deltaps is equal to %0.3f psia",delta_ps)
+disp("On converting the unit deltaps = 1.12 ft which is equal to 13.44 inches of water")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH5/EX5.2/Ex5_2.sce b/3136/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..e2af6a966 --- /dev/null +++ b/3136/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,163 @@ +clear all; clc;
+
+disp("a)")
+disp("Convert: 1) Q=5 m^3/s=10595 cfm")
+disp("2) rhoa=0.0761lbm/ft^3")
+
+disp("3) SP=deltap/(rhow*g)")
+delta_ps=500
+rho_w=1000
+g=9.8
+SP=delta_p/(rho_w*g)
+printf("Hence SP = %0.3f m",SP)
+disp("Thus SP= 2.01 in.wg.")
+
+disp("b)")
+disp("Calculating the specific speed:Ns")
+N_s=1500*((10575)^0.5)/(2.01)^0.75
+printf("The value of Ns is equal to %0.2f rpm*(cfm^0.5)/(in of water^0.75)",N_s)
+
+N=1500
+omega=N*%pi/30
+printf("\nOmega = %0.0f rad/s",omega)
+
+omega_s=157*(5^0.5)/((500/1.22)^0.75)
+printf("\nSo omegas = %0.2f ",omega_s)
+
+disp("From figure 5.10b,we select v=Dh/Dt=0.5 and the blade number Zb=6*v/(1-v)")
+v=0.5
+Zb=6*v/(1-v)
+printf("Hence Zb= %0.2f",Zb)
+
+disp("From figure 2.2,the specific diameter obtained as deltas is approximately equal to 1.5")
+D_t=1.5*(5^0.5)/((500/1.22)^0.25)
+printf("\nHence Dt = %0.3f m",D_t)
+
+v=0.5
+D_t=0.74//rounded off
+D_h=v*D_t
+printf("\nDh= %0.2f m",D_h)
+
+A=%pi*(D_t^2)*(1-v^2)/4
+printf("\nAlso A = %0.4f m^2",A)
+
+D_m=((D_t^2+D_h^2)/2)^0.5
+printf("\nDm = %0.4f m",D_m)
+
+A=0.322//rounded off
+Q=5
+V_a=Q/A
+printf("\nVa= %0.1f m/s",V_a)
+
+U_m=omega*D_m/2
+printf("\nUm = %0.2f",U_m)
+
+V_a=15.5//rounded off
+PHI_m=V_a/U_m
+printf("\nPHIm = %0.3f",PHI_m)
+
+disp("Now from figure 5.10c we can obtain Phim*(s/c)=0.65")
+//let s/c=x
+PHI_m=0.337//rounded off
+x=0.65/PHI_m
+printf("\nThus (s/c)= %0.2f",x)
+
+disp("Here s= pi*Dm/Zb")
+s= %pi*D_m/Zb
+printf("\n Thus s= %0.3f m",s)
+
+disp("Assuming V1=V3=Va=15.5 m/s the total head can be calculated from g*Ht=deltaps/rho+Va^2/2")
+//let y=g*Ht=deltaps/rho+(Va^2)/2
+rho_a=1.22
+y=delta_ps/rho_a+(V_a^2)/2
+printf("\ng*Ht = %0.0f (m/s)^2",y)
+disp("or TP=2.59 in.wg")
+
+disp("c)")
+ETA_h=0.77
+V_u2=(y)/(ETA_h*U_m)//Since y=(g*Ht)
+printf("\nVu2 is equal to %0.0f m/s",V_u2)
+
+beta_1=(atan(U_m/V_a))*180/%pi
+printf("\nß1= %0.2f degrees",beta_1)
+
+beta_2=(atan(((U_m-V_u2)/V_a)))*180/%pi
+printf("\n ß2= %0.2f degrees",beta_2)
+
+//let m=tanßm=0.5*(tanß1+tanß2)
+beta_1=71.3//rounded off
+beta_2=63.4//rounded off
+m=0.5*(tan(beta_1*%pi/180)+tan(beta_2*%pi/180))
+printf("\ntanßm=0.5*(tanß1+tanß2) =%0.3f",m)
+beta_m=(atan(m))*180/%pi
+printf("\n ßm = %0.0f",beta_m)
+
+disp("We know that Cl=2*(s/c)*(tanß1-tanß2)*cosßm")
+x=1.93//rounded off
+beta_1=71.3
+beta_2=63.4
+beta_m=68
+//Let a=tanß1
+//Let b=tanß2
+//Letc=cosßm
+a=tan(beta_1*%pi/180)
+b=tan(beta_2*%pi/180)
+c=cos(beta_m*%pi/180)
+a=2.95//rounded off
+b=2.0//rounded off
+c=0.374//rounded off
+Cl=2*x*(a-b)*c//Since x=(s/c)
+ printf("\nCl= %0.2f ",Cl)
+
+ disp("This is the cascade coefficient required.To use the isolated airfoil data,we obtain K=1.2 from figure 5.9 with gamma=60 degrees. Hence we can determine Cli")
+
+ C_li=1.37/1.2
+ printf("\nCli= %0.2f",C_li)
+
+disp("d)")
+disp("If NACA 4312 airfoil selection is selected,at Alpha=12 degrees,Cli=1.14 and Cl/Cd=L/D=12")
+disp("Substituting the above mentioned data in Rr=Wmu/Um=phi*tan(ßm)")
+phi=0.337
+d=tan(beta_m*%pi/180)
+R_r=phi*d
+printf("\n Thus Rr= %0.3f",R_r)
+disp("deltar is approximately=deltas is approximately=0.08")
+Eta_h=0.337*(((0.834-(0.337*0.08))/(0.337+(0.08*0.834)))+((1-0.834-(0.337*0.08))/(0.337+(0.08*(1-0.834)))))
+printf("\nETAh= %0.2f",Eta_h)
+disp("Etah=0.80 is approximately equal to 0.77")
+disp("Also gamma=betam-alpha=68-12=56 degrees")
+disp("c=s/1.93")
+c=s/1.93
+printf("\nThus c = %0.2f m",c)
+
+disp("e)")
+disp("Double check the data obtained with those given in Figure 5.1.It is shown that Etas=0.74,Ds=0.35=c.")
+SP=2.01
+CFM=10595
+Dt=0.35*(CFM^0.5)/(SP^0.25)
+printf("\nHence the value of Dt= %0.1f in",D_t)
+disp("On converting,Dt=0.77m")
+disp("It is close to what we have. However, some alternative design maybe performed with the selection of a little higher hub-tip ratio v and other availabe airfoil sections")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH5/EX5.3/Ex5_3.sce b/3136/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..886dc35c4 --- /dev/null +++ b/3136/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,23 @@ +clear all; clc;
+
+disp("From equation 5.11 with Vu=constant,we have (Vu^2/r)dr+(Va)dVa=0 ")
+disp("Or Vu^2lnr+Va^2/2=C")
+disp("Or Vu^2*ln(r/rh)=(Vah^2-Va^2)/2")
+disp("Or Va^2=Vah^2-2*Vu^2*ln(r/rh)")
+
+V_ah=70
+//let m=Vah^2
+m=V_ah^2
+printf("\n Vah^2 = %0.0f",m)
+
+V_u=500
+//let n=Vu^2
+n=V_u^2
+printf("\n Vu^2= %0.0f",n)
+
+//let x=Vu^2*2
+x=n*2//Since n=Vu^2
+printf("\n 2*Vu^2= %0.0f",x)
+
+disp("Thus Va=(4900-5000*ln(0.5r))^0.5 m/s")
+disp("and at the tip,Vat=27.1m/s")
diff --git a/3136/CH5/EX5.4/Ex5_4.sce b/3136/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..27624fdc1 --- /dev/null +++ b/3136/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,52 @@ +clear all; clc;
+
+disp("Assume uniform axial flow at the inlet")
+disp("(a) At the hub")
+disp("V_a=15.5m/s")
+
+omega=157
+D_h=0.37
+U_h=0.5*omega*D_h
+printf(" Uh= %0.2f m/s",U_h)
+
+disp("Vu2=(Vmu2*Dm)/Dh")
+V_mu2=15
+D_m=0.585
+V_u2=(V_mu2*D_m)/D_h
+printf(" Hence Vu2= %0.1f m/s",V_u2)
+
+V_a=15.5
+disp("tan(ß1)=(Uh/Va). Hence we can determine value of ß1")
+//let x=Uh/Va
+x=U_h/V_a
+beta_1=(atan(x))*180/%pi
+printf(" ß1= %0.1f degrees",beta_1)
+
+disp("tan(ß2)=(Uh-Vu2)/Va")
+//Let y=(Uh-Vu2)/Va
+y=(U_h-V_u2)/V_a
+beta_2=(atan(y))*180/%pi
+printf(" ß2= %0.2f degrees \n",beta_2)
+
+disp("(b) At the tip")
+
+U_t=0.5*(157*0.74)
+printf("\n Ut= %0.2f m/s",U_t)
+
+V_u2=(15*0.585)/0.74
+printf("\n Vu2= %0.2f m/s",V_u2)
+
+//let p=atan(58.09/15.5)
+p=(atan(58.09/15.5))*180/%pi
+printf("\n ß1 = %0.0f degrees",p)
+
+//let q=atan((58.09-11.86)/15.5)
+q=atan((58.09-11.86)/15.5)*180/%pi
+printf("\n ß2 = %0.3f degrees",q)
+disp("On rounding off ß2=71.4")
+
+
+
+
+
+
diff --git a/3136/CH5/EX5.5/Ex5_5.sce b/3136/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..0733379de --- /dev/null +++ b/3136/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,87 @@ +clear all; clc;
+
+disp("Since the pressure changes are small compared with the barometric pressure,constant densities are assumed that is rho3 =rho2=rhoa where rhoa=pa/(RTa)")
+p_a=14.6
+T_a=535
+R=53.3
+rho_a=(p_a*144)/(R*T_a)//144 is the conversion factor
+printf("rhoa= %0.4f lbm/ft^3",rho_a)
+
+A2=5*6.5
+printf("\nA2 equals %g in^2",A2)
+disp("On converting A2=0.225ft^2")
+
+A3=%pi*6.065^2/4
+printf("\nA3 equals %0.1f in^2",A3)
+disp("On converting A3=0.2007ft^2")
+
+disp("From (rho3*V3^2)/2=rhow*g*pv3,V3 can be calculated")
+rho_w=62.4
+g=32.2
+rho_3=0.0737
+j=sqrt(2*rho_w*g/(rho_3*12))
+printf("V3=%0.1f * pv3^0.5 ft/s",j)
+
+disp("The inlet flow rate can be calculated as Q=Q3=V3*A3=0.2007*60*V3=12.04*V3")
+disp("Also the dynamic pressure can be calculated as pv2=pv3*((A3/A2)^2)*(rho3/rho2)=(0.2/0.225)^2*pv3=0.79pv3")
+
+disp("The total pressure pt2=ps2+pv2")
+disp("To correct these data to a fixed speed of 3500rpm,the fan laws can be used as Qdash=Q*3500/N,pt2dash=pt2*((3500/N)^2) and Hdash=H*((3500/N)^3).")
+disp("The total efficiency can be calculated as ETAt=(rhow*g*pt2dash*Qdash)/(Hdash)")
+disp("On simplifying ETAt=Qdash*pt2dash/(6346*Hdash)")
+
+p_v3=[2.4 2.1 1.9 1.5 1.2 0.8 0.4];
+p_s2=[2.5 4.2 6.0 6.8 7.6 8.5 9.5];
+N=[3450 3520 3500 3420 3430 3500 3520];
+H=[1.1 1.25 1.49 1.55 1.67 1.72 1.81];
+
+V_3=zeros(1,length(p_v3));
+Q=zeros(1,length(p_v3));
+p_v2=zeros(1,length(p_v3));
+p_t2=zeros(1,length(p_v3));
+Qdash=zeros(1,length(p_v3));
+tenpt2dash=zeros(1,length(p_v3));
+tenHdash=zeros(1,length(p_v3));
+eta_t=zeros(1,length(p_v3));
+
+for i = 1: length(p_v3)
+
+
+ V_3(i) = 67.4*sqrt(p_v3(i));
+ Q(i) = 12.04*V_3(i);
+ p_v2(i) =0.79*p_v3(i);
+ p_t2(i)= p_s2(i)+p_v2(i);
+ Qdash(i)= Q(i)*(3500/(N(i)));
+ tenpt2dash(i)= 10*p_t2(i)*((3500/N(i))^2);
+ tenHdash(i)= 10*H(i)*((3500/(N(i)))^3);
+ eta_t(i)= ((Qdash(i)*(tenpt2dash(i))/10)/(6346*(tenHdash(i))/10))*100;
+end
+
+disp("The table is in the order given in the book,that is pv3, ps2, N, H, V3, Q, pv2, pt2, Qdash, tenpt2dash, tenHdash and etat.")
+table=[p_v3' p_s2' N' H' V_3' Q' p_v2' p_t2' Qdash' tenpt2dash' tenHdash' eta_t' ];
+disp(table)
+
+plot(Q,tenpt2dash,'o',Q,tenHdash,'d',Q,eta_t,'s')
+legend("tenpt2dash (inches of water)","tenHdash (hp)","eta_t (%)",-1)
+xlabel("Q(cfm)")
+ylabel("tenpt2dash (inches of water), tenHdash (hp) , eta_t (%)")
+set(gca(),"grid",[1 1])
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH5/EX5.6/Ex5_6.sce b/3136/CH5/EX5.6/Ex5_6.sce new file mode 100755 index 000000000..e555d8faa --- /dev/null +++ b/3136/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,16 @@ +clear all; clc;
+
+disp("Since dBt=10*[((log10^dBb)/10)+(10^dBf/10)] we can determine dBf")
+
+//let x=10^(dBt/10)
+x=10^7
+
+//let y=10^(dBb/10)
+y=10^6.5
+
+//let z=x-y
+z=x-y
+
+disp("dBf=10*(log(10^(dBt/10)+10^(dBb/10)))")
+dBf=10*[log10(z)]
+printf(" Thus dBf= %0.2f",dBf)
diff --git a/3136/CH5/EX5.7/Ex5_7.sce b/3136/CH5/EX5.7/Ex5_7.sce new file mode 100755 index 000000000..699b6d047 --- /dev/null +++ b/3136/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,32 @@ +clear all; clc;
+
+disp("The average flow velocity V is determined as V=Q/A")
+Q=7000
+A=%pi*(1.5^2)/4
+V=Q/A
+printf(" Thus V = %0.2f fpm",V)
+disp("Thus V is approximately equal to 4000fpm")
+
+disp("So we have 100% effective discharge duct length Le=[2.5+(4000-2500)/1000]*D")
+D=1.5
+Le=[2.5+(4000-2500)/1000]*D
+printf("\n Since D=1.5 ft, Le is equal to %0.2f ft",Le)
+
+//let x=L1/Le
+L1=0.75
+x=L1/Le
+printf("\n Since L1=0.75,L1/Le=%0.3f",x)
+
+disp("So the line V in figure 5.18 can be used to obtain SEF1=0.255in.wg")
+disp("For the two elbows we have L2/D and C2")
+//let y=L2/D
+L2=3
+y=L2/D
+printf("\n The value of L2/D is equal to %0.2f",y)
+disp("From figure 5.19a,C2=4.2")
+
+disp("Hence we can obtain SEF2=4.2*(rho*V^2/2)")
+rho=0.075
+V=4000
+SEF2=4.2*((rho/32.2)*(V/60)^2/2)*(12/62.4)//constants used are conversion factors
+printf("\n SEF2 is equal to %0.2f in.wg",SEF2)
diff --git a/3136/CH5/EX5.8/Ex5_8.sce b/3136/CH5/EX5.8/Ex5_8.sce new file mode 100755 index 000000000..54af5278d --- /dev/null +++ b/3136/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,58 @@ +clear all; clc;
+
+disp("We have the duct resistance deltap=0.25*((7000/5000)^2)")
+delta_p=0.25*((7000/5000)^2)
+printf(" deltap %0.2f in.wg at 7000cfm",delta_p)
+
+disp("The density in the building can be calculated from rho=p/R*T")
+p=12.5
+R=53.3
+T=85+460
+rho=p*144/(R*T)//144 is convesrion factor
+printf("\n rho= %0.5f lbm/ft^3",rho)
+
+disp("So the equivalent SPe of the required fan at STP can be found out")
+SPe=0.49*0.075/0.0619
+printf("\n SPe = %0.4f in.wg.",SPe)
+
+disp("Form figure 5.24a,at SP=0.5,Q1=6894,N1=347,BHP1=0.76 and Q2=7660, N2=360, BHP2=0.88. Hence for Q=7000cfm we can determine Ndash")
+Ndash=347+(360-347)*((7000-6894)/(7660-6894))
+printf("\n Ndash is equal to %0.2f rpm",Ndash)
+
+BHPdash=0.76+(0.88-0.76)*0.138
+printf("\n BHPdash= %0.4f",BHPdash)
+
+disp("At Sp=5/8=0.625 in.wg.,Q1=6894,N1=375,BHP1=0.92 and Q2=7660,N2=387,BHP2=1.05,so for Q=7000cfm we can determine Ndbldash and BHPdbldassh")
+
+Ndbldash=375+((387-375)*0.138)
+printf("\n Ndbldash %0.2f rpm",Ndbldash)
+
+BHPdbldash=0.92+(1.05-0.92)*0.138
+printf("\n BHPdbldash %0.4f hp",BHPdbldash)
+
+disp("Again interpolating for SPe=0.593,we have values of N and BHP as mentioned below ")
+N=348.8+(376.6-348.8)*((0.593-0.5)/(0.625-0.5))
+printf("\n N= %0.0frpm",N)
+BHP=0.776+(0.938-0.776)*0.744
+printf("\n BHP= %0.4f hp",BHP)
+
+disp("Correcting for the reduced air density,the actual BHP is determined")
+BHP=0.896*0.0619/0.075
+printf("\n Actual BHP is equal to %0.4f",BHP)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH5/EX5.9/Ex5_9.sce b/3136/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 000000000..844d5b875 --- /dev/null +++ b/3136/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,9 @@ +clear all; clc;
+
+disp("1.Convert the system specification to Qdash=10000cfm with the correesponding total pressure loss")
+delta_p_ldash=0.5*((10000/5000)^2)
+printf("\n delta(pl)dash= %0.0f in.wg.",delta_p_ldash)
+
+disp("2. Draw a system line(which is a straight line with slope of 2 on a log-log chart)passing this point and intercept at 700rpm")
+
+disp("Read the operating condition as Q=11000cfm,deltap=2.51 in.wg. and Ps=5.6hp")
diff --git a/3136/CH6/EX6.1/Ex6_1.sce b/3136/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..7a146d0fa --- /dev/null +++ b/3136/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,70 @@ +clear all; clc;
+
+disp("Pick the data rows for the flow rates of 10241 and 13965 cfm and list them in the first four columns of the table below.")
+disp("Convert the flow rate Q,static pressure SP and brake horsepower BHP values for various rpm into 300rpm based on the fan laws.")
+disp("That is Q1=Q*(300/N), SP1=SP((300/N)^2),BHP1=BHP*((300/N)^3)")
+disp("Calculate the static efficiency using ETAs=(Q*SP)/(6346*BHP)")
+disp("The results are plotted into the chart")
+
+
+
+
+
+Q=[10241 10241 10241 10241 10241 10241 10241 13965 13965 13965 13965 13965 13965 13965 13965 ];
+SP= [0.25 0.375 0.5 0.625 0.75 1 1.25 0.25 0.375 0.5 0.625 0.75 1 1.25 1.5];
+N= [300 321 341 363 385 427 466 380 396 413 428 444 473 506 537];
+BHP= [0.82 1.03 1.23 1.45 1.68 2.19 2.75 1.58 1.86 2.16 2.43 2.72 3.3 3.89 4.51];
+
+Q1= zeros(1,length(Q));
+SP1 = zeros(1,length(Q));
+BHP1 = zeros(1,length(Q));
+ETA=zeros(1,length(Q));
+
+for i = 1: length(Q)
+
+ Q1(i)=Q(i)*(300/N(i))
+ SP1(i)=SP(i)*((300/N(i))^2)
+ BHP1(i)=BHP(i)*((300/N(i))^3)
+ ETA(i)=(Q(i)*SP(i))/(6346*BHP(i))
+
+end
+
+//disp(" phi psi eff(%) pai N(rpm) Ps(mw) Q(m^3/s)")
+
+table = [Q' SP' N' BHP' Q1' SP1' BHP1' ETA'];
+disp(" Q SP N BHP Q1 SP1 BHP1 ETA")
+disp(" (cfm) (in.wg) rpm (cfm) (in.wg)")
+disp(table)
+
+plot(Q1,SP1,'o',Q1,BHP1,'d',Q1,ETA,'s')
+legend("SP1(in.wg)","BHP1(hp)","ETA",-1)
+xlabel("Q(cfm)")
+ylabel("SP1(in.wg), BHP1(hp) , ETA")
+set(gca(),"grid",[1 1])
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH6/EX6.10/Ex6_10.sce b/3136/CH6/EX6.10/Ex6_10.sce new file mode 100755 index 000000000..be54d5c45 --- /dev/null +++ b/3136/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,71 @@ +clear all; clc;
+
+disp("From table 6.1 at 1.25 SP,the rotating speeds for Q1=11172cfm and Q2=12103cfm are N1=474rpm and N2=483 rpm respectively")
+Ns=474+[(483-474)*(12000-11172)]/[12103-11172]
+printf(" Hence the rotating speed for the selected fan is determined by inetrpolation %0.0f rpm\n\n",Ns)
+
+disp("Select a few data points around 482 rpm from table 6.1 as:")
+Q=[14896 12103 11172 11172 10241 7448];
+N=[490 448 436 474 466 360];
+SP=[1.0 1.0 1.0 1.25 1.25 0.75];
+BHP=[3.66 2.67 2.40 2.97 2.75 1.2];
+
+disp(" Q(cfm) N(rpm) SP(in.wg) BHP(hp)")
+table=[Q' N' SP' BHP']
+disp(table)
+
+disp("Convert them into conditiond of 482 rpm according to the similarity laws,resulting in")
+Q1=[14653 13021 12350 11360 10593 9972];
+SP1=[0.967 1.16 1.22 1.29 1.34 1.34];
+BHP1=[3.5 3.20 3.24 3.12 3.04 2.88];
+table1=[Q1' SP1' BHP1']
+disp(" Q(cfm) SP(in.wg) BHP(hp)")
+disp(table1)
+
+disp("The system curve can be calculated from the following table")
+
+Q2=[10000 11000 12000 13000 14000];
+H2=[0.87 1.05 1.25 1.47 1.70];
+
+sqrQ2 = zeros(1,length(Q2));
+
+for i = 1: length(Q2)
+
+
+ sqrQ2(i) = [Q2(i)]^2;
+
+end
+
+
+table2=[Q2' H2']
+disp(" Q(cfm) H(in.wg)")
+disp(table2)
+
+disp("The system curve can be calculated from H versus Q^2. It is plotted as shown.")
+//The system curve has not been provided in the book for this numerical. However they have mentioned that the parameters for the curve are H and Q^2,and as such has been plotted here.
+plot(sqrQ2,H2)
+xlabel("Q^2 ")
+ylabel("H")
+set(gca(),"grid",[1 1])
+xtitle("System curve: H versus Q squared")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH6/EX6.2/Ex6_2.sce b/3136/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..a073072e3 --- /dev/null +++ b/3136/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,53 @@ +clear all; clc;
+
+disp("Assume as given Na=2400 rpm")
+disp("T1a=68 degrees Farenheit=528R,p1=14.7 psia,p2a=8.5psig=23.2 psia")
+disp("Qa=1800cfm,Eta=0.70 Nb=3600rpm and T1b=50 to 95 degrees Farenheit")
+
+Q_a=1800
+N_b=3600
+N_a=2400
+Q_b=Q_a*(N_b/N_a)
+printf("\n Qb= %0.0f cfm",Q_b)
+
+disp("((p2/p1)_b)^((k-1)/k)-1=(T1a/T1b)*[((p2/p1)_a)-1]*((Nb/Na)^2)")
+disp("On simplifying ((p2/p1)_b)=0.313*(T1a/T1b)")
+disp("We obtain p2b=14.7*[1+0.313*(T1a/T1b)]^3.5=36.6 to 39.3 psia")
+disp("Also Psb=rho_1b*Q_b*H_i/Eta={(p_1b/(R*T_1b))*Qb*Cp*T_1b*[(p2/p1)^((k-1)/k)-1]}/Eta")
+disp("Psb=[(3.5*14.7*144*2700/(550*60))*0.313*(T1a/T1b)]/0.7")
+disp("Psb=271*(T1a/T1b)=257.8 to 280.6hp")
+
+disp("From rho_0*Q_dash=rho_1*Q,we have the flow rate measured at the standard condition,Q_dashb=(rho_1b/rho_0)*Qb=(T0/T1b)*Qb=2795 to 2568 cfm")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH6/EX6.3/Ex6_3.sce b/3136/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..022d923de --- /dev/null +++ b/3136/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,84 @@ +clear all; clc;
+//Answer of H_ib given in the book is 5830,however it is incorrect. The correct answer is found out to be 5837.4883
+
+disp("rho_a=p/(R*T)")
+rho_a=(13.7*0.491*144)/(53.33*(-10+460))
+printf(" rho_a= %0.5f lbm/ft^3",rho_a)
+
+disp("Qb=Q*(Nb/Na)")
+Q=(180/0.0403)*(12000/25000)
+printf(" q= %0.0f cfm",Q)
+
+disp("Hb=Ha*((Nb/Na)^2)")
+disp("we obtain")
+disp("((p2/p1)^((k-1)/k)_b)=[((p2/p1)^((k-1)/k)_a)-1]*((Nb/Na)^2)*(T1a/T1b)+1")
+//let x=((p2/p1)^((k-1)/k)_b)
+p2=34.5
+p1=13.7
+//let y=((k-1)/k)
+y=0.2857
+Nb=12000
+Na=25000
+T_1a=450
+T_1b=535
+x=[(p2/p1)^(y)-1]*((Nb/Na)^2)*(T_1a/T_1b)+1
+printf(" ((p2/p1)^((k-1)/k)_b)=%0.4f\n",x)
+
+disp("So p2b=((1.0585)^3.5)*(p1b)")
+p_1b=30
+p_2b=((1.0585)^3.5)*(p_1b)
+printf(" p2b= %0.1f in.Hg",p_2b)
+
+disp("H_ib=[(k*R*T_1b)/(k-1)]*[((p2/p1)^((k-1)/k)_b)-1]")
+k=1.4
+R=53.33
+T_1b=535
+p2=36.6
+p1=30
+y=0.2857
+H_ib=[(k*R*T_1b)/(k-1)]*[((p2/p1)^y)-1]
+printf(" H_ib=%0.4f ft-lbf/lbm",H_ib)//Answer of H_ib given in the book is 5830,however it is incorrect. The correct answer is found out to be 5837.4883 on calculating
+
+disp("rho_b*Qb=mb")
+mb=(30*0.491*144)/(53.33*535)*2144
+printf(" mb= %0.1f lbm/min",mb)
+
+disp("Thus Psb=mb*Hib/Eta")
+P_sb=159.4*5830/(60*550*0.70)
+printf(" Psb= %0.1f hp",P_sb)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH6/EX6.4/Ex6_4.sce b/3136/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..6adac7dc4 --- /dev/null +++ b/3136/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,21 @@ +clear all; clc;
+
+Q_c=1000
+N_a=3550
+N_c=4000
+Q_a=Q_c*(N_a/N_c)
+printf("\n Qa=Qc*(Na/Nc)=%0.1f cfm",Q_a)
+
+disp("H=Cp*T1*[(p2/p1)^((k-1)/k)-1]")
+disp("Ha/Hc=((Na/Nc)^2)")
+disp("Where ka=1.40,Cpa=0.24 Btu/lbm*degrees Farenheit,,T1a=530R,((ka-1)/ka)=0.2857 and kc=1.30")
+disp("Cpc=0.20Btu/(lbm*degrees Farenheit),T1c=550R,((kc-1)/kc)=0.2307, we have:")
+disp("(Cpa*T1a)*[[(p2/p1)^((k-1)/k)-1]]_a/{(Cpc*T1c)*[[(p2/p1)^((k-1)/k)-1]]_c}=((3550/4000)^2)")
+//let x=(Cpa*T1a)*[[(p2/p1)^((k-1)/k)-1]]_a/{(Cpc*T1c)*[[(p2/p1)^((k-1)/k)-1]]_c}
+x=((3550/4000)^2)
+printf(" Thus the value is equal to %0.3f",x)
+
+disp("Or")
+disp("(0.24*530*[(p2/p1)^(0.2857-1]_a)/(0.20*550*[(19.7/17.7)^(0.2307)-1])=0.787")
+disp("Hence ((p2/p1)^0.2857)_a=1+(0.787*0.025)/1.156=1.017")
+disp("p2a=15.6 psia")
diff --git a/3136/CH6/EX6.5/Ex6_5.sce b/3136/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..3e8c17b5d --- /dev/null +++ b/3136/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,17 @@ +clear all; clc;
+//the answer of p02b given in the book is 94.1,however the right answer is 92.05 which is rounded off to 92.1 here
+
+p_02b=90+0.5*0.00237*((500)^2)/144
+printf("\n p02b= %0.1f psia",p_02b)
+
+//let x=T02/T01
+x=(480+460)/(60+460)
+printf("\n T02/T01= %0.4f which is also =(94.1/14.7)^((0.2857)/(Eta_p)) ",x)//answer given in the book is 94.1,which is substituted
+disp("From (0.2857/Eta_p)*ln(94.1/14.7)=ln(1.807) we can obtain the polytropic efficiency")
+Eta_p=(log(94.1/14.7)/log(1.807))*0.2857
+printf("Eta_p= %0.4f",Eta_p)
+
+disp("0.85=[(4.5)^(0.2857)-1]/[(4.5)^(0.2857/Eta_p)-1")
+disp("Thus 0.5638/((4.5^(0.2857/Eta_p))-1)")
+disp("We have polytropic efficiency for compressor A,Eta_pa=0.878.")
+disp("Hence,compressor bB is moe efficient")
diff --git a/3136/CH6/EX6.6/Ex6_6.sce b/3136/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..7810a61f7 --- /dev/null +++ b/3136/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,80 @@ +clear all; clc;
+
+disp("The specifications are N=1800 rpm=188.5 rad/s,Q=14000cfm=233.3 cfs,and sp=5 in.wg")
+N=1800
+Q=233.3
+sp=5
+Ns=N*(Q^0.5)/(sp^0.75)
+disp("Ns=%0.0f rpm*(cfm^0.5)/(in water^0.75)")
+
+disp("From Figure 5.1,selecting the air foil blade,we have Etas=0.80 and Ds=d2*(sp^0.25)/Q^0.5=0.33")
+
+Q=14000
+sp=5
+Eta_s=0.8
+P_s=Q*sp/(Eta_s*6356)
+printf(" Ps=%0.2f hp",P_s)//incorrectly rounded off in the book
+
+d2=0.33*(14000^0.5)/(5^0.25)
+printf("\n d2=%0.0f in",d2)
+
+tau=(13.7*550)/(1800*%pi/30)
+printf("\n τ=%0.0f lbf-ft",tau)
+
+tau=40
+s_s=10000
+ds=[(16*tau)/(%pi*s_s)]^(1/3)
+printf("\n The shaft diameter is found out to be %0.2f in",ds)
+disp("Where s_s is the assumed maximum allowable shear stress of the shaft material.")
+disp("Hence d_d is set as 0.75in and d_h=1.0in or rh=0.5in. Using the maximum W1 criteria,we ave ßf1=35.2 degrees and r1=9.88 inches which is rounded off to 10 inches.")
+
+disp("Also we have b1=5 inches and ßb1=ßf1+i=37 degrees")
+omega=188.5
+r2=13/12
+U2=omega*r2
+printf(" To determine outlet dimensions we have U2= %0.1f ft/s",U2)
+
+disp("For a straight blade we have ßdashb2=52 degrees, so try ßb2=45 degrees")
+disp("Zb=26 and mu_s=0.794")
+disp("Since the static pressure is specified but the total pressure is needed in the Euler equation,the discharge flow is needed.")
+disp("Setting b2=b1=5 incheswe can determine the values of A2 and A3")
+
+r2=13
+b2=5
+A2=2*%pi*r2*b2/144
+printf(" A2=%0.3f ft^2",A2)
+
+A2=2.83
+A3=1.5*A2
+printf("\n The fan discharge area A3 is set as A3=1.5*A2=%0.2f ft^2",A3)
+disp("Hence we have V3=Q/A3=54.8 ft/s")
+
+sp=62.4*5/12
+rho=0.00237
+V3=54.8
+tp=sp+(0.5)*rho*(V3^2)
+printf(" tp=%0.2f lbf/ft^2",tp)
+
+disp("Also Eta_t=Eta_s*(tp/sp)=Eta_imp*Eta_v*Eta_m and assuming Eta_v*Eta_m=0.95")
+Eta_imp=((0.8*29.5)/26)/(0.95)
+printf(" So we hae Eta_imp= %0.4f",Eta_imp)
+
+disp("So from the Euler equation, we have (29.5/0.00237)=154.8*(204.2-(82.4/tanßb2))")
+disp("ß_b2=33.6 degrees")
+disp("Since Zb and mu_s are related to ßb2,we have to repeat the calculations")
+disp("The new alues Zb=20 and mu_s=0.781 and ß_b2=33.9 degrees are obtained")
+disp("Hence the final values are 20 and ß_b2=33.9 degrees")
+
+disp("From equation(6.12) we have(dΘ/dr)=(1/(2*tanßf1)*[2*(r/r1)*(1/r1)]")
+disp("Hence tanßdash_b2=[dr/(r*dΘ)_r2]=tanßf1*((r1/r2)^2)")
+tan_beta_dashb2=tan((35.2*%pi/180)*[(10/13)^2])
+printf(" Thus the value of tanßdash_b2= %0.3f",tan_beta_dashb2)//Answer f=given in the book is 0.417,however the more correct answer is 0.380
+betab2dash=((atan(tan_beta_dashb2))*180/%pi)
+printf("\n ß_dash_b2= %0.1f degrees",betab2dash)//Since value of tanß'b2 is different,ß'b2 is different
+disp("This value is smaller than ßb2 which is expected from 6.13c")
+
+
+
+
+
+
diff --git a/3136/CH6/EX6.7/Ex6_7.sce b/3136/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 000000000..fbb6862be --- /dev/null +++ b/3136/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,193 @@ +clear all; clc;
+
+disp("Inlet Configuration selection: Mr1,t=0.75 and ßf1=25 degrees")
+M_r1t=0.75
+beta_f1=25*%pi/180
+M_1t=M_r1t*sin(beta_f1)
+printf(" M_1t=%0.3f",M_1t)
+
+disp("T1=To1/(1=((k-1))*(M_lt^2)/2")
+T1=530/(1+0.2*(0.317^2))
+printf("\n Thus the value of T1= %0.1fR",T1)
+
+a1=(1.4*53.33*32.2*519.6)^0.5
+printf("\n a1= %0.2f ft/s",a1)//answer provided here is more accurate
+
+V_1t=0.317*1117.6
+printf("\n V1=%0.1f ft/s",V_1t)
+
+W_1t=0.75*1117.6
+printf("\n W_1t= %0.1f ft/s",W_1t)
+
+U_1t=[(838.2^2)-(354.3^2)]^0.5
+printf("\n U_1t= %0.1f ft/s",U_1t)
+
+omega=1623
+U_1t=759.6
+r_1t=U_1t/omega
+printf("\n So we have r_1t= %0.3f ft=5.6in",r_1t)
+
+//let x=k/k-1
+x=3.5
+po1=14.7
+T_1=519.6
+T_o1=530
+p1=po1*[(T_1/T_o1)^x]
+printf("\n So we have p1= %0.1f psia",p1)
+
+p1=13.7
+R=53.33
+T1=519.6
+rho1=(p1*144)/(R*T_1)
+printf("\n rho1= %0.4f lbm/ft^3",rho1)
+
+m=17
+rho1=0.0713
+V1=354.3
+A1=m/(rho1*V1)
+printf("\n A1= %0.3f ft^2=96.9 in ^2,assuming uniform inlet flow.\n\n",A1)
+
+disp("From pi*[(r_1t^2)-(r_1h^2)]=A1")
+disp("We have r_1h={[(r_1t^2)-A1]/pi}^0.5=0.72 in")
+disp("U_1h=97.0 ft/s")
+
+disp("ß_f1,h=taninverse(V_1h/U_1h)")
+V_1h=354.3
+U_1h=97
+beta_f1h=(atan(V_1h/U_1h))*180/%pi
+printf(" ßf1= %0.1f degrees",beta_f1h)
+
+disp("(B) Outlet Configuration From")
+
+Cp=0.24
+To1=530
+//let y=(k-1)/k=0.2857
+y=0.2857
+//let m=po2/po1
+m=2.5
+H_ad=Cp*To1*778*[(m)^(y)-1]
+printf(" H_ad=H_ad=Cp*To1*778[(po2/po1)^((k-1)/k)-1] %0.0f ft-lbf/lbm ",H_ad)
+
+m=17
+rho1=0.0713
+Q1=m/rho1
+printf("\n Q1=%0.1f ft^3/s",Q1)
+
+N=15500
+H_ad=29614
+Q1=238.4
+Ns=N*(Q1^0.5)/((H_ad)^0.75)
+printf("\n We obtain specific speed %0.0f rpm*{(ft^3/s)^0.5}(ft-lbf/lbm)^0.75 ",Ns)
+
+disp("Fom figure 6.7 we have estimated Eta_c=0.87 and Ds=D2*(H_ad)/(Q1^0.5)=1.50")
+D2=1.5*(238.4^0.5)/(29614^0.25)
+printf("\n D2=%0.4f ft=21.2 inches or r2=10.6 in. and U2=r2*omega=1433.7 ft/s",D2)
+
+disp("Referring to figure 6.16c,we have T_sso3=T_o1*(po3/po1)^((k-1)/k)")
+//((k-1)/k)=x
+x=0.2857
+T_o1=530
+//let l=po3/po1
+l=2.5
+T_sso3=T_o1*(l^x)
+printf(" T_sso3= %0.1f R ",T_sso3)
+disp("From Eta_c=(T_sso3-(T_o1))/(To3-To1) we have To2=To3")
+//let l=To2/To3
+l=(1/0.87)*(688.6-530)+530
+printf(" To2/To3 %0.1fR",l)
+disp("Also from Etam=U2*Vu2/[Cp*(T_o2-T_o1)]")
+
+Vu2=0.95*0.24*778*32.2*(712.3-530)/(1433.7)
+printf(" With the estimated Eta_m=0.95,Vu2=%0.1f ft/s",Vu2)
+
+disp("Flow coefficient φ=Vm2/U2=0.30 ")
+Vm2=0.3*1433.7
+printf(" Vm2=%0.1f ft/s",Vm2)
+
+disp("W2=[Vm2^2+(U2-Vu2)^2]^0.5=827.9 ft/s")
+disp("V2=[Vu2^2+Vm2^2]^0.5=844.1 ft/s")
+
+W1t=838.2
+W2=827.9
+Df=W1t/W2
+printf(" Hence we have diffusion factor Df=%0.3f. The value is less than 1.9 which is okay.",Df)
+
+disp("The impeller efiiciency can be estiamted form the losses fraction X=(1-Eta_imp)/(1-Eta_c) is approximately =0.6 ")
+Eta_imp=1-0.6*(1-0.87)
+printf("\n Eta_imp %0.3f",Eta_imp)
+
+disp("Hence from Eta_imp=(T_so2-T_o1)/(T_o2-T_o1),we have T_so2=T_o1+Eta_imp*(To2-T_o1)=698R and po2=po1*(T_so2/T_o1)^(k/(k-1))")
+po1=14.7
+T_so2=698.1
+T_o1=530
+//Let x=(k/(k-1))
+x=3.5
+po2=po1*(T_so2/T_o1)^x
+printf(" po2=%0.2f psia ",po2)
+
+disp("Then from the energy equation we have T2=T_o2-V2^2/(2*Cp)=653.0R")
+disp("Hence p2=p02*(T2/To2)^(k/(k-1))=28.4psia and rho2=p2/(R*T2)=0.117lbm/ft^3")
+
+disp("Selecting Zb=16 and using the Stanitz fromula for the slip coefficient we have Vdash_u2=Vu2+0.63*pi*U2/Zb")
+Vu2=726.3
+U2=1433.7
+Zb=16
+ Vdash_u2=Vu2+0.63*%pi*U2/Zb
+printf(" Vdash_u2= %0.1f ft/s",Vdash_u2)
+
+disp("tanß_b2=Vm2/(U2-Vdash_u2)")
+Vm2=430.1
+U2=1433.7
+Vdash_u2=903.6
+tan_beta_b2=Vm2/(U2-Vdash_u2)
+printf(" tanß_b2=%0.2f ",tan_beta_b2)
+
+betab2=(atan(tan_beta_b2))*180/%pi
+printf("\n ß_b2= %0.0f degrees",betab2)
+
+disp("With blade thickness t=0.15,contraction factor is determined")
+Zb=16
+t=0.15
+betab2=39*%pi/180
+D2=21.3
+epsilon2=1-[(Zb*t)/(sin(betab2))]/(%pi*D2)
+printf(" Thus epsilon2= %0.2f",epsilon2)
+
+disp("Hence from the mass equation we can determine b2")
+b2=17/(0.117*430.1*%pi*1.765*0.94)
+printf(" b2=%0.4f=0.7 8inch",b2)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH6/EX6.8/Ex6_8.sce b/3136/CH6/EX6.8/Ex6_8.sce new file mode 100755 index 000000000..3258ac015 --- /dev/null +++ b/3136/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,65 @@ +clear all; clc;
+
+disp("Using figure 6.7 the specific speed Ns=N(V1^0.5)/(H_ad^0.75) should be calculated based on the inlet volumetric flow rate V1 and the adiabetic head per stage H_ad")
+p1=14.7
+R=53.3
+T1=530
+rho1=p1*144/(R*T1)//144 is conversion factor.the actual formula is rho1=p1/(R*T1)
+printf(" From rho1=p1/(R*T1) =%0.3f lbm/ft^3, we have V1",rho1)
+
+V1=100/0.075
+printf("\n Thus V1= %0.0f ft^3/s.",V1)
+
+disp("Also from equation (A.3), we have total adiabetic head ΣH_ad=Cp*T1*[(pe/pi)^((k-1)/k)-1]")
+Cp=0.24
+T1=530
+pe=50
+pi=14.7
+//Let y= (k-1)/k
+y=0.2857
+H_ad=Cp*T1*778*[{(pe/pi)^(y)}-1]//778 is conversion factor
+printf(" Thus ΣH_ad= %0.0f lbf-ft/lbm \n",H_ad)
+
+disp("The specific speeds for different number of stages are calculated as follows ")
+
+Ns=4800*(1333^0.5)/(41436^0.75)
+printf(" For n=1 , Ns=4800*(1333^0.5)/(41436^0.75)=%0.0f ",Ns)
+
+Ns=4800*(1333^0.5)/(13812^0.75)
+printf("\n For n=3 , Ns=4800*(1333^0.5)/(13812^0.75)=%0.2f ",Ns)
+
+Ns=4800*(1333^0.5)/(10359^0.75)
+printf("\n For n=4 , Ns=4800*(1333^0.5)/(10359^0.75)=%0.0f ",Ns)
+
+Ns=4800*(1333^0.5)/(6906^0.75)
+printf("\n For n=6 , Ns=4800*(1333^0.5)/(6906^0.75)=%0.0f ",Ns)
+
+Ns=4800*(1333^0.5)/(5179^0.75)
+printf("\n For n=8 , Ns=4800*(1333^0.5)/(5179^0.75)=%0.0f ",Ns)
+
+disp("Reading the curves in figure 6.7,the best efficient point is around Ns=280 with Eta=0.90")
+disp("ds=D*(H_ad^0.25)/(V1^0.5)=0.7")
+disp("Hence 8 stage mixed flow impellers are selected.")
+
+D=0.7*(1333^0.5)/((41436/8)^0.25)
+printf(" The impeller diameter is calculated to be = %0.1f ft",D)
+
+m=100
+H_ad=41436
+Eta=0.9
+Ps=m*H_ad/(Eta*550)//converting units
+printf("\n The total required shaft power Ps=m*H_ad/(Eta)=%0.0f hp",Ps)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH6/EX6.9/Ex6_9.sce b/3136/CH6/EX6.9/Ex6_9.sce new file mode 100755 index 000000000..7e8d593bd --- /dev/null +++ b/3136/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,93 @@ +clear all; clc;
+
+disp("The inlet specific volume is calculated from the ideal gas equation")
+R=53.3
+T1=580
+p1=65
+nu1=(R*T1)/(144*p1)
+printf(" nu1=%0.3f ft^3/lbm=",nu1)
+
+disp("Q1=m*v1=4594 cfm")
+
+R=53.3
+T1=580
+//let y=(n/(n-1))
+y=2.625
+p2=250
+p1=65
+H_oa=R*T1*(y)*[(p2/p1)^(1/y)-1]
+printf(" The overall adiabetic head is calculated as H_o/a= %0.0f ft-lbf/lbm",H_oa)
+
+y=(0.75*1.4)/(1.4-1)
+printf("\n Where n/(n-1)=(Eta_p*k)/(k-1)=%0.3f",y)
+
+disp("From figure 6.7 , a centrifugal compressor with speed N=10000rpm is appropriate for the present application")
+
+disp("To use figure 6.7,the specific speed can be calculated from Ns=N*(V1^0.5)/(H_ad^0.75)")
+V1=4954/60
+printf(" Where V1= %0.1f cfs",V1)
+disp("H_ad=(H_o/a)/Eta_s is the head for each stage .")
+disp("Selecting the number of stages to be 2,4, and 6,the head for each stage and specific speed can be calculated,then the total-to=total adiabetic efficiencies can be read ")
+disp("The required shaft horse power can be calculated with the volumetric and mechanical efficiencies assumed to be 0.98 and 0.95 respectively")
+disp("That is Ps=(m*(H_o/a)/(33000*Eta_ad*Eta_v*Eta_m)")
+
+H_oa=54417
+V1=82.6
+Eta_v=0.98
+Eta_m=0.95
+N=10000
+
+StageNo=[2 4 6];
+Eta_ad=[0.72 0.83 0.87];
+
+
+H_ad= zeros(1,length(StageNo));
+Ns = zeros(1,length(StageNo));
+Ps = zeros(1,length(StageNo));
+
+for i = 1: length(StageNo)
+
+
+ H_ad(i) =H_oa/(StageNo(i));
+ Ns(i) =N*(V1^0.5)/(H_ad(i)^0.75);
+ Ps(i) = (m*H_oa)/(33000*Eta_ad(i)*Eta_v*Eta_m);
+end
+
+disp("StageNo H_ad Ns Eta_ad Ps")
+disp(" (ft-lbf/lbm) (hp)")
+table = [StageNo' H_ad' Ns' Eta_ad' Ps'];
+disp(table)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH7/EX7.1/Ex7_1.sce b/3136/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..b32e0581b --- /dev/null +++ b/3136/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,60 @@ +clear all; clc;
+
+disp("From (poe/poi)=[(Toi+ns*deltaTo)]^(k*eff*p/(k-1))")
+disp("From the above mentioned equation we can find out the value of deltaTo")
+delta_To=((8^(1/(3.5*0.87)))-1)*530/7
+printf("deltaTo= %0.1f degrees Farenheit",delta_To)
+
+Q=450
+r_m1=9
+b=3
+V_a1=(Q*144)/(2*%pi*r_m1*b)
+printf("\n Thus Va1 = %0.0f fps",V_a1)
+
+N=12000
+U=(N*%pi*9)/(30*12)
+printf("\n N= %0.1f fps",U)
+
+disp("deltaho=U*Va*(tanß1-tanß2) and R=(Φ/2)*(tanß1+tanß2")
+//let y= tanß1-tanß2
+y=[(0.24*778*74.2*32.2)/(942.5*382)]
+printf("\n Thus tanß1-tanß2= %0.2f",y)
+
+//let x=tanß1+tanß2
+x=(0.5*2*942.5/382)
+printf("\n Thus tanß1+tanß2= %0.3f",x)
+
+disp("Hence we get tanß1=1.853")
+tanbeta1=1.853
+beta_1=(atan(tanbeta1))*180/%pi
+printf("\n The value of ß1= %0.1f degrees",beta_1)
+disp("Also tanß2=0.613")
+tanbeta2=0.613
+beta_2=(atan(tanbeta2))*180/%pi
+printf("\n The value of ß2=α1= %0.1f degrees",beta_2)
+
+disp("Ps=m*Cp*ns*deltaTo/etam")
+P_s=(0.075*450*0.24*778*7*74.2)/(550*0.95)
+printf("\n The total power required is Ps= %0.0f hp",P_s)
+
+disp("The adiabetic efficiency is given as ((poe/poi)^(((k-1)/k))-1/((Toe/Toi)-1)")
+ETA_ad=0.811/0.979
+printf("\n Thus adiabetic efficiency is %0.4f",ETA_ad)//answer given in the book is 0.827,but this is more accurate
+
+disp("ETAc=ETAm*ETAad")
+ETA_m=0.95
+ETA_ad=0.827
+ETA_c=ETA_m*ETA_ad
+printf(" Hence ETAc= %0.3f ",ETA_c)
+disp("Thus ETAc=78.6%")
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH7/EX7.2/Ex7_2.sce b/3136/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..9cb018fef --- /dev/null +++ b/3136/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,89 @@ +clear all; clc;
+
+r_h=15
+r_t=24
+r_m=((r_h^2+r_t^2)/2)^0.5
+printf("rm= %0.0f in",r_m)
+
+N=6000
+r_m=20
+U_m=N*%pi*r_m/(12*30)
+printf("\n U_m= %0.2f ft/s",U_m)
+
+disp("We have psia=lamda*psi=lambda*phi*(tanßm1-tanßm2)")
+//let x=tanßm1-tanßm2
+x=(0.24*778*32.2*35)/(0.92*1047.2*450)
+printf("\n Hence we can find out tanßm1-tanßm2= %0.3f",x)
+
+disp("From equation 7.2B for Rm=0.5 we have αm1=ßm2")
+disp("We get values of tanαm1+tanßm1=2.325 and tanαm2+tanßm2=2.325")
+
+disp("Hence we have tanßm1-tanαm1=0.485")
+
+tanalpham1=0.92
+alpham_1=((atan(tanalpham1)))*180/%pi
+printf("\n Thus αm1= %0.1f degrees",alpham_1)
+
+tanalpham2=1.405
+alpham_2=((atan(tanalpham2)))*180/%pi
+printf("\n Thus αm2= %0.2f degrees",alpham_2)
+
+disp("To determine the flow angles at the hub and tip,we use the free vortex condition of Vur=const, or rhtanαh=rttanαt=rmtanαm.Hence the flow angles can be determined.")
+
+tanalphah1=0.92*20/15
+printf("\n tanαh1= %0.3f",tanalphah1)
+alpha_h1=((atan(tanalphah1)))*180/%pi
+printf("\nThus αh1= %0.1f degrees",alpha_h1)
+
+tanalphat1=0.92*20/24
+printf("\n\n tanαt1= %0.3f",tanalphat1)
+alpha_t1=((atan(tanalphat1)))*180/%pi
+printf("\n Thus αt1= %0.1f degrees",alpha_t1)
+
+tanalphah2=1.405*20/15
+printf("\n\n tanαh2= %0.3f",tanalphah2)
+alpha_h2=((atan(tanalphah2)))*180/%pi
+printf("\n Thus αh1= %0.1f degrees",alpha_h2)
+
+tanalphat2=1.405*20/24
+printf("\n\n tanαt2= %0.3f",tanalphat2)
+alpha_t2=((atan(tanalphat2)))*180/%pi
+printf("\n Thus αh1= %0.1f degrees",alpha_t2)
+
+disp("The degree of reaction at the hub and tip can be determined.")
+Rh=1-((1-0.5)/((15/20)^2))
+Rt=1-((1-0.5)/((24/20)^2))
+printf("\n Rh= %0.2f",Rh)
+printf("\n Rt= %0.2f",Rt)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH7/EX7.3/Ex7_3.sce b/3136/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..4f70408d7 --- /dev/null +++ b/3136/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,44 @@ +clear all; clc;
+
+disp("From the previous numerical wee have ßm1=αm2=54.5 degrees, tanßm1=1.405")
+disp("ßm2=αm1=42.6 degrees. Thus tanßm2=0.92")
+
+disp("tanßm1-tanßm2=1.55(1+1.5*(s/c)),thus we can determine s/c")
+//let x= s/c
+x=[1.55/(1.405-0.92)-1]/1.5
+printf("Thus (s/c)= %0.2f",x)
+
+disp("Also b/c=3,we have c=(rt-rh)/3")
+rt=24
+rh=15
+c=(rt-rh)/3
+printf("\n c= %0.0f in",c)
+
+s=1.47*c
+printf("\n Hence we determine s to be equal to %0.1f in",s)
+
+rm=20
+Zb=2*%pi*rm/s
+printf("\n Zb= %0.0f",Zb)
+
+disp("The blade angles can be estimated from ßb1=ßm1-i and ßb2=ßm2-δ")
+disp("Where i=3 degrees and δ=m*Θ*((s/c)^0.5)")
+
+//let n=a/c
+n=0.5
+disp("m=0.23*((2*(a/c))^2)+0.1*(ßm2/50)")
+disp("Θ=ßm1-ßm2")
+thita=11.9
+m=0.23+(0.1*42.6/50)//for circular blade
+printf("\n m= %0.3f",m)
+m=0.315
+x=1.47
+delta=m*thita*(x^0.5)
+printf("\n δ= %0.1f degrees",delta)
+
+beta_b1=54.5-3
+printf("\n ßb1= %0.1f degrees",beta_b1)
+
+beta_b2=42.6-4.5
+printf("\n ßb2= %0.1f degrees",beta_b2)
+
diff --git a/3136/CH7/EX7.4/Ex7_4.sce b/3136/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..bd7b1f79d --- /dev/null +++ b/3136/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,70 @@ +clear all; clc;
+
+disp("We have psia=lamda*psi=lambda*phi*(tanßm1-tanßm2) and R=0.5*φ(tanß1+tanß2) we can obtain the values of tanß1 and tanß2")
+//let x=tanßm1-tanßm2
+x=0.35/(0.92*0.5)
+printf("\n tanßm1-tanßm2= %0.3f",x)
+
+//let y=tanß1+tanß2
+y=2*0.5/0.5
+printf("\nThus tanß1+tanß2= %0.3f",y)
+
+disp("Hence tanß1=1.38")
+tanbeta1=1.38
+beta_1=((atan(tanbeta1)))*180/%pi
+printf("\nThus the value of ß1 is equal to %0.1f degrees",beta_1)
+
+disp("tanß2=0.619")
+tanbeta2=0.619
+beta_2=((atan(tanbeta2)))*180/%pi
+printf("\nThus the value of ß1 is equal to %0.1f degrees",beta_2)
+
+disp("For each stage we have psi=Cp*deltaTos/((Um)^2)")
+delta_T_os=0.35*(920^2)/6012
+printf("\nHence deltaTos= %0.1f R",delta_T_os)
+
+Cp=0.24*778*32.2
+printf("\nWhere Cp = %0.0f ft-lbf/slug",Cp)
+
+disp("For overall compressor form equation 47.4 we have (Poe/poi)=[(1+deltaToe/Toi)]^(k*eff*p/(k-1))")
+delta_T_oe=530*[(4.5^(0.2857/0.9))-1]
+printf("\nThus deltaToe= %0.0f R",delta_T_oe)
+
+disp("The number of stages can be calculated as ns=deltaToe/deltaTos")
+delta_T_oe=324
+delta_T_os=49.3
+ns=delta_T_oe/delta_T_os
+printf("\nThus ns= %0.2f",ns)
+disp("ns is approximately equal to 7")
+
+disp("Hence the actual values are: ")
+delta_T_oe=7*49.3
+printf("deltaToe= %0.1f R",delta_T_oe)
+
+//let f=poe/po1
+f=(1+(345.1/530))^(0.9*3.5)
+printf("\n poe/po1= %0.2f",f)
+
+disp("The adiabetic efficiency is given as ((poe/poi)^(((k-1)/k))-1/((Toe/Toi)-1)")
+//let k-1/k=d
+d=0.2857
+f=4.85
+
+ETA_ad=((f^(d))-1)/((345.1/530))
+printf("\nETAad= %0.4f",ETA_ad)
+//let r=ETAd*100
+r=ETA_ad*100
+printf("\n Thus ETAad= %0.2f percent",r)
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH7/EX7.5/Ex7_5.sce b/3136/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 000000000..82fe6b821 --- /dev/null +++ b/3136/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,253 @@ +clear all; clc;
+
+disp("Assume constant axial flow velocity Va=500 ft/s , no prewhirl and hence V1=Va for the first stage and frre vortex conditions for all stages.")
+
+disp("1. First Stage")
+
+T1=530-((500^2)/(2*6012))
+printf("T1= %0.1f R",T1)
+
+p1=14.7*((509.2/530)^3.5)
+printf("\n p1= %0.2f psia",p1)
+
+rho_1=(12.78*144)/(53.33*509.2)
+printf("\n rho_1= %0.5f lm/ft^3",rho_1)
+
+disp("With a selected nu=rh/rt=0.45, A=m/(rho_1*Va)")
+m=65
+rho_1=0.0677
+Va=500
+A=m/(rho_1*Va)
+printf("\n A= %0.2f ft^2 = 276.5in^2",A)
+
+disp("A=pi*(r_t^2)*(1-nu^2)")
+A=276.5
+nu=0.45
+r_t=sqrt(A/(%pi*(1-(nu^2))))
+printf("\n r_t= %0.1f in",r_t)
+
+disp("rm=7.6in")
+
+disp("Ut=rt*omega")
+N=15000
+omega=N*%pi/30
+rt=10.5/12//in feet
+Ut=rt*omega
+printf("\n Ut %0.1f ft/s",Ut)
+
+rh=4.7/12//in feet
+Uh=rh*omega
+printf("\n Uh= %0.1f ft/s",Uh)
+
+rm=7.6/12//in feet
+Um=rm*omega
+printf("\n Um= %0.1f ft/s",Um)//answer given in the book is 996.6,however 994.8 is more accurate
+
+disp("Without inlet whirl flow,we have Wt=(Ut^2+Va^2)^0.5")
+Ut=1374.4
+Va=500
+Wt=sqrt(Ut^2+Va^2)
+printf("\n Thus Wt = %0.0ft/s",Wt)
+
+Uh=615.2
+Va=500
+Wh=sqrt(Uh^2+Va^2)
+printf("\n Thus Wh = %0.1ft/s",Wh)
+
+Um=996.6//This is the answer substituted in the book,although as mentioned earlier 994.8 is more accurate.
+Va=500
+Wm=sqrt(Um^2+Va^2)
+printf("\n Thus Wm = %0.1ft/s",Wm)
+
+tanbeta1m=Um/Va
+printf("\n tanß1m= %0.2f",tanbeta1m)
+
+beta1m=((atan(tanbeta1m)))*180/%pi
+printf("\n ß1m= %0.1f degrees",beta1m)
+
+disp("With a1=(k*R*T1)^0.5=1309ft/s,the relative Mach numbers can be determined.")
+Wt=1463
+a1=1309
+M_rt=Wt/a1
+printf("\n M_rt= %0.2f",M_rt)
+
+Wh=792.8
+M_rh=Wh/a1
+printf("\nM_rh= %0.3f",M_rh)
+
+Wm=1115.0
+M_rm=Wm/a1
+printf("\n M_rm= %0.2f",M_rm)
+
+disp("Hence the relative flow at the leading edge is transonic,and a supersonic blade might be needed.")
+disp("From the previous sum,to achieve a pressure ratio of 4.5,we have deltaTos=324/7=46.3R.")
+disp("So deltaTos=45R is selected at the first stage.")
+disp("From Cp*deltaTos=lambda*Um*Va*(tanß1m-tanß2m),with lambda=0.95,we obtain tanß2m=1.42. Thus ß2m=54.8 degrees")
+
+tanbeta1=1.99
+tanbeta2=1.42
+Va=500
+Um=996.6
+R=(Va/(2*Um))*(tanbeta1+tanbeta2)
+printf("\n R=(Va/(2*Um))*(tanbeta1+tanbeta2)= %0.3f",R)
+
+disp("We have nu=(1-R)^0.5")
+nu=sqrt(1-R)
+printf("\n nu= %0.2f",nu)
+
+disp("This is less than the selected value and is acceptable.")
+beta1=63.3
+beta2=54.8
+deHaller=cos(beta1*%pi/180)/cos(beta2*%pi/180)
+printf("\n Also checking the deHaller number W2/W1=cosß1/cosß2=%0.2f >0.70,it is also acceptable",deHaller)
+
+disp("Before proceeding to the next stages,the following parameters have to be specified. The summation of deltaTos should add upto 324degrees.So the following arramgement is chosen")
+Stage_number=[1 2 3 4 5 6 7 ];
+deltaTos_R=[45 46 47 47 47 47 45];
+lambda=[0.95 0.93 0.90 0.89 0.88 0.86 0.85];
+R=[0.855 0.8 0.7 0.63 0.60 0.55 0.5];
+table=[Stage_number' deltaTos_R' lambda' R']
+disp("The table from left to right has values of Stagenumber deltaTos lamdbda and R")
+disp(table)
+
+disp("2.Second Stage")
+
+To1=530+45
+printf("To1= %0.0f R",To1)
+
+po1=14.7*((1+(0.9*45/530))^3.5)
+printf("\n po1= %0.0f psia",po1)
+
+disp("With the specified values of deltaTos,lamda and R,equations 7.1 and 7.2A can be solved for tanß1 and tanß2.")
+disp("Hence ß1=62.1 degrees and ß2=52.4 degrees,where W2/W1=0.76>0.70")
+beta1=62.1*%pi/180//converting to radians
+V1=[(Va^2)+(Um-(Va*tan(beta1)))^2]^0.5
+printf("Also we have V1=[(Va^2)+(Um-(Va*tan(beta1)))^2]^0.5= %0.2f ft/s",V1)//answer in the book is 502.6,however the value found out here is more accurate
+
+Cp=6012
+V1=502.6
+T1=To1-((V1^2)/(2*Cp))
+printf("\nT1= %0.0f R",T1)
+
+po1=19
+T1=554
+To1=575
+p1=po1*((T1/To1)^3.5)
+printf("\np1= %0.1f psia",p1)
+
+disp("rho1= p1/(R*T1)= 0.0813 lb_m/ft^3")
+
+
+
+disp("A=m/(rho1*Va)=230")
+disp("A is also = 2*pi*r_m*b")
+disp("Where b=rt-rh and rm is the same as that for the first stage. Hence we have rt=10 in,rh=5.2 in and nu=0.52,which is greater than nu_min(0.447) ")
+
+disp("3.Third to Seventh stage")
+disp("The calculation procedure for these stages will be similar to the second stage. The calculations have not been repeated in the book,and hence the same is done here.")
+disp("The results are as follows")
+
+Stagenumber=[1 2 3 4 5 6 7];
+To1=[530 575 621 668 715 762 809];
+po1=[14.7 19 24.3 30.5 37.9 46.3 55.9];
+rt=[10.5 10 9.6 9.3 9.1 8.9 8.8];
+rh=[4.7 5.2 5.6 5.9 6.1 6.3 6.5];
+rm=[7.6 7.6 7.6 7.6 7.6 7.6 7.6 ];
+beta1=[63.3 62.1 59.7 57.5 56.5 54.7 52.4];
+beta2=[54.9 52.3 47.2 43.2 41.4 38 34.8]
+table2=[Stagenumber' To1' po1' rt' rh' rm' beta1' beta2']
+disp("Stageno. To1 po1 rt rh rm beta1 beta2")
+disp(table2)
+
+disp("The final discharge stagnation pressure can also be checked with the flow across the last stage such that poe=po3=po1(1+ETAs*(deltaTos/To1))^(k/(k-1))")
+po1=55.9*(1+(0.9*45/809))^3.5
+printf("po1= %0.1f psia",po1)
+//let n=poe/poi
+n=66.3/14.7
+printf("\n This checks with poe/poi = %0.2f",n)
+
+
+
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diff --git a/3136/CH7/EX7.6/Ex7_6.sce b/3136/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..a457b6371 --- /dev/null +++ b/3136/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,18 @@ +clear all; clc;
+
+disp("We have psi=lambda*[1-φ*(tanα1+tanß2)] and R=0.5+[(φ/2)*(tanß2-tanα1)]")
+disp("We have tanα1+tanß2=(1-psi/lamda)=1.01")
+disp("tanα1-tanß2=(1-2R)/φ=-0.545,or psi=lambda*(1-1.01*φ) and R=0.5+0.272*φ Hence φdash=0.495,assuming lamda to be constant we can determine the values of psi_dash and R_dash")
+
+lambda=0.9
+//Let (tanalpha1+tanbeta2)=x
+x=1.01
+phi_dash=0.495
+psi_dash=lambda*[1-(phi_dash*(x))]
+printf("\n psi_dash= %0.2f",psi_dash)
+
+//Let= 0.5*(tanbeta2-tanalpha1=y
+y=0.272
+R_dash=0.5+[(phi_dash*0.272)]
+printf("\n R_dash= %0.3f",R_dash)
+
diff --git a/3136/CH8/EX8.1/Ex8_1.sce b/3136/CH8/EX8.1/Ex8_1.sce new file mode 100755 index 000000000..e89251287 --- /dev/null +++ b/3136/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,29 @@ +clear all; clc;
+
+disp("From ETAad=(1-((p0e/p0i)^(ETAs*(k-1)/k)))/(1-((p0e/p0i)^(k-1)/k)))")
+ETA_ad=(1-((14.7/200)^(0.85/3.5)))/(1-(14.7/200)^(0.2857))
+printf("\n ETAad= %0.3f",ETA_ad)
+
+disp("T_soe/T_0i=((p0e/p0i)^((k-1)/k))")
+//let T_soe/T_0i=w
+w=(14.7/200)^0.2857
+printf("\n T_soe/T_0i= %0.4f",w)
+
+T_soe=0.4743*(1800+460)
+printf("\n T_soe= %0.0f R",T_soe)
+
+disp("ETAad=(T_0i-T_0e)/(T_0i-T_soe)=0.893")
+T_0i=T_soe/w//since T_soe/T_0i=w
+T_0e=T_0i-(ETA_ad*(T_0i-T_soe))
+printf(" Thus T_0e= %0.0f R=739 degrees Farenheit",T_0e)
+
+disp("Also for impulse turbine, we have ß2=ß3.Hence from delta_H0=Um*(V_u2-V_u3)=Um*[Um+Va*tanß2-(Um-Va*tan(ß3))]=2*Um*Va*tanß")
+delta_h0=2*750*400*tan(50*%pi/180)
+printf("\n delta_h0= %g is approximately equal to 7.15*10^5 ((ft/s)^2)",delta_h0)
+//Let 0.24*778*32.2*delta_T0=u
+u=7.15*10^5
+delta_T0=u/(0.24*778*32.2)
+printf("\n\n Or 0.24*778*32.2*delta_T0=7.15*10^5,we have temperature rise per stage = %0.0f degrees Farenheit",delta_T0)
+
+n_s=(1800-739)/119
+printf("\n Hence the number of stages n_s= %0.2f is approximately equal to 9",n_s)
diff --git a/3136/CH8/EX8.2/Ex8_2.sce b/3136/CH8/EX8.2/Ex8_2.sce new file mode 100755 index 000000000..c41a9fb4f --- /dev/null +++ b/3136/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,132 @@ +clear all; clc;
+
+disp("Referring to figure 8.3,we have T_s2=T_01*((p2/p01)^((k-1)/k))")
+T_01=2860
+p2=180
+p01=250
+k=1.4//k=(Cp/Cv)
+T_s2=T_01*((p2/p01)^((k-1)/k))
+printf(" T_s2= %0.0f R",T_s2)
+
+disp(" From ε_s=(h2-hs2)/((V_2^2)/2),we have Cp*(T2-Ts2=ε_s*(V_2^2)/2.Combining with Cp*(T_02-T_2)=((V_2^2)/2),where T02=T01, we have Cp*(T_02-Ts2)=(1+εs)*((V_2^2)/2)")
+disp("V2=[2*Cp*(T_02-T_s2)/(1+εs)]^0.5")
+T_02=T_01
+Cp=0.24*778*32.2
+epsilon_s=0.07
+T_s2=2604
+V2=[2*Cp*(T_02-T_s2)/(1+epsilon_s)]^0.5
+printf(" V2= %0.0f ft/s",V2)
+
+V2=1696
+alpha2=65*%pi/180//converting to radians
+V_u2=V2*sin(alpha2)
+printf("\n Hence we have V_u2=V2*sin(α2) = %0.0f ft/s",V_u2)
+
+Va=V2*cos(alpha2)
+printf("\n Va=V2*cos(alpha2)= %0.1f ft/s",Va)
+
+disp("T2=T_02-(V_2^2)/(2*Cp)")
+T_02=2860
+V2=1696
+Cp=0.24*778*32.2
+T2=T_02-(V2^2)/(2*Cp)
+printf(" Hence we have T2= %0.0f R ",T2)
+
+disp("Since V1=Va,we have T1=T_01-((V1^2)/(2*Cp))")
+V1=716.8
+T_01=2860//2860R
+T1=T_01-((V1^2)/(2*Cp))
+printf("T1= %0.0f R",T1)
+
+disp("From delta_E=Cp*delta_T0s=U*(V_u2+V_u3)=U*V_u2, we have U=Ps/(m*V_u2)")
+Ps=375*550*32.2//converting unit of Ps
+m=3
+V_u2=1537
+U=Ps/(m*V_u2)
+printf("U=%0.0f ft/s",U)
+
+disp("φ=Va/U")
+Va=716.8
+U=1440
+phi=Va/U
+printf("φ=%0.3f",phi)
+
+tanbeta3=U/Va
+printf("\n tanß3= %0.2f",tanbeta3)
+
+beta3=((atan(tanbeta3))*180/%pi)
+printf("\n ß3= %0.1f degrees",beta3)
+
+alpha3=0
+phi=0.498
+alpha2=65*%pi/180
+R=1+((phi/2)*(tan(alpha3)-tan(alpha2)))
+printf("\n R= %0.3f",R)
+
+disp("Also from the velocity diagram in figure 8.4,we have tanß2=tanα2-(1/φ)=0.136,so ß2=7.8 degrees")
+disp("Similarly we have W3=Va/cosß3=1606 ft/s and W2=Va/cosß2=723.5 ft/s")
+disp("Across the rotor we have h2+(W2^2)/2=h3+(W3^2)/2. Hence T3=T2+(W2^2)-(W3^2)/(2*Cp)=2450R")
+disp("We have Ts3=T3-εr*(W3^2)/(2*Cp)=2424R")
+disp("Also p3=p2*(Ts3/T2)^(k/(k-1))=136.9 psia")
+ETAs=(1+(0.12*(1606^2)+0.07*(1696^2)*(2450/2621))/(2*0.24*778*32.2*(2817-2450)))^-1
+printf(" From equation 8.2 we have ETAs= %0.4f",ETAs)
+
+//Let j=0.498/2
+j=0.498/2
+//Let k=0.12*[(sec(63.5*%pi/180))^2]
+k=0.12*[(sec(63.5*%pi/180))^2]
+//Let l=0.07*(2450/2621)*[(sec(65*%pi/180))^2]
+l=0.07*(2450/2621)*[(sec(65*%pi/180))^2]
+//let m=tan(63.5*%pi/180)+tan(7.8*%pi/180)
+m=tan(63.5*%pi/180)+tan(7.8*%pi/180)
+
+ETAs=[1+((j*(k+l))/m)]^-1
+printf("\n From equation 8.3 we have ETAs= %0.4f",ETAs)
+
+disp("Also ETAs can be calculated from ETAs=(T_01-T_03)/(T_01-T_ss03)")
+disp("We have T_03=T3+V3^2/(2*Cp)")
+disp("p03=p3*(T_03/T3)^(k/(k-1))")
+disp("T_ss03=T_01*(p_03/(p_01)*((k-1)/k))")
+ETAs=(2860-2493)/(2860-2450)
+printf(" Hence we have ETAs= %0.3f",ETAs)
+
+
+
+
+
+
+
+
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diff --git a/3136/CH8/EX8.3/Ex8_3.sce b/3136/CH8/EX8.3/Ex8_3.sce new file mode 100755 index 000000000..cec110aeb --- /dev/null +++ b/3136/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,106 @@ +clear all; clc;
+
+disp("Use the velocity diagram shown in figure 8.2 or 8.4")
+disp("We have Vatanß2=Vatanα2-Um")
+disp("Or tanß2=tanα2-Um/Va")
+
+tanbeta2=tan(75*%pi/180)-1200/500
+printf(" Thus tanß2= %0.2f",tanbeta2)
+beta2=((atan(tanbeta2))*180/%pi)
+printf("\n Thus ß2= %0.2f",beta2)
+
+disp("Also Vatanß3=Vatanα3+Um")
+disp("tanß3=tanα3+Um/Va")
+tanbeta3=tan(10*%pi/180)+1200/500
+printf(" tanß3= %0.2f",tanbeta3)
+beta3=((atan(tanbeta3))*180/%pi)
+printf("\n ß3= %0.2f",beta3)
+
+disp("From Cp*deltaT0s=deltaE=Um(Vu2+Vu3)=UmVa(tanα2+tanα3)")
+deltaE=1200*500*(tan(75*%pi/180)+tan(10*%pi/180))
+printf(" Thus Cp*deltaT0s=deltaE= %g=2.34*10^6 ((ft/s)^2)",deltaE)
+
+deltaT0s=(2.34*(10^6))/(0.24*778*32.2)
+disp("deltaT0s=(2.34*(10^6))/(0.24*778*32.2)")
+printf(" Thus deltaT0s= %0.2f R which is rounded off to 890R",deltaT0s)
+
+disp("Hence neglecting leakage and mechanical losses , we have shaft power output Ps=mCpdeltaT0s")
+Ps=50*2.34*(10^6)/(32.2*550)
+printf("\n Hence we have Ps= %0.2f hp wich is rounded off to 6607hp",Ps)
+
+disp("The degree of reaction at the mean radius can be determined from equation 8.5A")
+R=(500/(2*1200))*(tan(68.8*%pi/180)-tan(53.1*%pi/180))
+printf("\n Thus R = %0.3f",R)
+
+disp("To determine the radii the flow area A2 can be determined from m=rho2*A2*Va. The density rho2 can be determined from p2 and T2 which can be caalculated as follows.")
+disp("From Cp*T02=CpT2+V2^2/2 and V2= VA/cosα2")
+V2=500/(cos(75*%pi/180))
+printf("\n V2= %0.0f ft/s",V2)
+
+T2=2000-(1932^2)/(2*0.24*778*32.2)
+printf("\n Thus we have T2= %0.0f degrees Farenheit = 2150R",T2)
+
+disp("From the definition of loss coefficient εe we have Ts2=T2-εsV2^2/(2*Cp)")
+Ts2=1690-(0.08*(1932^2))/(2*.24*778*32.2)
+printf("\n Ts2= %0.2f degrees Farenheit which is equal to 2125.2R",Ts2)
+
+//Let x= P2/p01
+x=(2125.2/2460)^(1.4/0.4)
+printf("\n and P2/p01= %0.2f",x)
+
+
+P2=200*0.60
+printf("\n P2= %0.0f psia",P2)
+
+disp("Hence the density can be calculated as rho2=p2/(R*T2)")
+
+rho2=120*144/(53.3*2150)
+printf("\n Thus ro2=%0.3f lbm/ft^3",rho2)
+
+A2=50/(0.151*500)
+printf("\n A2=m/(rho2*Va)=%0.3f ft^2",A2)
+
+//let y=rt^2-rh^2
+y=0.662/%pi
+printf("\n rt^2-rh^2=A2/pi=%0.3f",y)
+
+rm=30*1200/(8000*%pi)
+printf(" and rm= %0.2f ft",rm)
+
+disp("rt^2+rh^2=2*rm^2=4.09ft^2")
+disp("and 1.466ft,rh=1.393ft")
+rt=1.466
+rh=1.393
+b=rt-rh
+printf(" b= %0.3f ft=0.88in",b)
+
+
+
+
+
+
+
+
+
+
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+
diff --git a/3136/CH8/EX8.4/Ex8_4.sce b/3136/CH8/EX8.4/Ex8_4.sce new file mode 100755 index 000000000..3dc0d8667 --- /dev/null +++ b/3136/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,47 @@ +clear all; clc;
+
+disp("For N/(T01^1/2)=40,the following data of p02/p01,τ/p01 and Eta can be obtained from figure 8.9a")
+//let x=p02/p01
+x=[0.70 0.75 0.8];
+//let y=τ/p01
+y=[8.7 5.3 2.2];
+Eta=[0.81 0.64 0.41];
+//let z=Ps/(p01*((T01)^1/2))
+z=[0.066 0.040 0.017];
+//let i=m*((T01)^1/2)/p01
+i=[2.48 2.34 2.0];
+table=[x' y'Eta' z' i'];
+disp(" The columns of the table are in the order p02/p01 τ/p01 Eta Ps/(p01*((T01)^1/2)) and m*((T01)^1/2)/p01")
+disp(table)
+
+disp("The power and mass flow rate have to be obtained with the following manipulations. ")
+disp("Frpm Ps=τ*omega, wee obtain:")
+disp("Ps/(p01*((T01)^0.5))=τ*N*pi/(30*550*p01*((T01)^0.5))")
+disp("Also from Ps/m=Eta*Cp*T01[1-(p02/p01)^((k-1)/k)] we obtain")
+disp("m*(T01^0.5)/p01={[Ps/(p01*T01^0.5)]/(Eta*Cp)}*[1-(p02/p01)^((k-1)/k)]^-1")
+disp("Where (k-1)/k=0.40/1.4")
+
+//Let j=(k-1)/k=0.40/1.4
+j=0.40/1.4
+printf("Thus=(k-1)/k %0.4f",j)
+
+disp("And Cp= 0.24*Btu/(1bm*R)")
+Cp=0.24*778/(550)
+printf("Thus Cp= %0.4f hp-s/(lbm*R)",Cp)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH8/EX8.5/Ex8_5.sce b/3136/CH8/EX8.5/Ex8_5.sce new file mode 100755 index 000000000..a4b811d58 --- /dev/null +++ b/3136/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,49 @@ +clear all; clc;
+
+disp("From Um=rm*omega")
+disp("We have rm=30*Um/(N*pi)")
+rm=30*850/(8000*%pi)
+printf(" Thus rm= %0.2f ft=12.1 in",rm)
+
+disp("rm=(rt^2+rh^2/2)^0.5 or rt^2+rh^2=293.8")
+disp("Combined with b=rt-rh=4 in,we have rt^2-4rt-138.9=0,thus rt=13.95 in")
+
+disp("and rh=9.95 in")
+
+disp("To find the number of stages required the exhaust air temperature can be estimated as T0e=T0i*(p0e/p0i)^((k-1)/k)")
+T0e=1400*(14.7/60)^0.2857
+printf(" Thus T0e= %0.1f R",T0e)
+
+disp("The maximum energy available per unit mass of air is delta_Hs=Cp*(T0i-T0e)")
+delta_Hs=0.24*(1400-936.7)
+printf(" delta_Hs= %0.1f Btu/lbm",delta_Hs)
+
+disp("The maximum energy transfer per stage with an impulse turbine is deltaEi=2*Um^2")
+delta_Ei=2*(850^2)/(32.2*778)
+printf(" delta_Ei= %0.2f Btu/lbm",delta_Ei)
+
+disp("Hence the required number of stages is ETAsi=delta_Hs/delta_Ei")
+ETAsi=111.2/57.68
+printf(" ETAsi= %0.2f which is approximately equal to 2",ETAsi)
+
+disp("With the reaction turbine stages,it will be delta_Er=Um^2")
+delta_Er=850^2/(32.2*778)
+printf(" delta_Er= %0.2f Btu/lbm",delta_Er)
+disp("And ETAse=3.85 is approximatly equal to 4")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH8/EX8.6/Ex8_6.sce b/3136/CH8/EX8.6/Ex8_6.sce new file mode 100755 index 000000000..5c558d424 --- /dev/null +++ b/3136/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,128 @@ +clear all; clc;
+
+disp("From figure 8.14c we have Pl=620hp at N=18400rpm. Pick a point on the curve of N/(T01^0.5)=18400/(530^0.5)=800")
+disp("In figure 8.14a,say p02/p01=5")
+disp("So we have")
+
+disp("m(T01^0.5)/p01 ETAc p02(psia) m(lbm/s) p03(psia) p03/p04")
+disp(" 5.7 0.85 73.5 3.64 71.5 4.86")
+
+disp("where p02=5*14.7")
+ p02=5*14.7
+ printf(" Thus p02= %0.2f",p02)
+
+disp("m=5.7*14.7/(530^0.5)")
+ m=5.7*14.7/(530^0.5)
+ printf(" m= %0.2f",m)
+
+disp("73.5-2")
+ p03=73.5-2
+ printf(" m= %0.2f psia",p03)
+
+//Let i=p03/p04
+i=71.5/14.7
+printf("\n p03/p04= %0.2f",i)
+
+disp("Then from figure 8.14b,with p03/p04 and m3=m2 we have")
+disp("m(T01^0.5)/p01 T03(R) N/(T03^0.5) ETAt")
+disp("2.56 2528 366 0.87")
+
+disp("where T03=(2.56*71.5/3.64)^2")
+T03=(2.56*71.5/3.64)^2
+printf(" T03= %0.2f",T03)
+
+disp("N/T03=18400/(2528^0.5)")
+//let k=N/T03
+k=18400/(sqrt(2528))
+printf(" Thus T03= %0.2f",k)
+
+disp("So from equations (8.1),(7.4) and(8.11) we have:")
+disp("delta_T034=ETAt*T03*[1-(p04/p03)^((k-1)/k)]")
+delta_T034=0.87*2528*[1-(4.86)^(-0.248)]
+printf(" delta_T034 = %0.0f R",delta_T034)
+
+disp("delta_T012=(T01/ETAc)*[(p02/p01)^((k-1)/k)-1]")
+delta_T012=(530/0.85)*[(5^0.2857)-1]
+printf(" delta_T012= %0.0fR",delta_T012)
+
+P0=3.64*(0.28*713-0.24*(364/0.95))
+printf(" \n and P0=3.64*(0.28*713-0.24*(364/0.95))= %0.0f Btu/s=554hp, which is less than Pl",P0)
+
+disp("So we pick another point on the same curve , say p02/p01=5.2, and repeat the calculations ")
+
+disp("m(T01^0.5)/p01 ETAc p02(psia) m(lbm/s) p03(psia) p03/p04 m(T01^0.5)/p01 T03(R) N/(T03^0.5) ETAt")
+disp("5.6 0.88 76.4 3.57 74.4 5.06 2.55 2824 346 0.85")
+
+delta_T034=0.85*2824*[1-(5.06^(-0.248))]
+printf("\n The new delta_T034= %0.2fR",delta_T034)//the book has rounded off the value to 794R,the value calculated in this code is more accurate
+
+delta_T012=(530/0.88)*[(5.2^0.2857)-1]
+printf("\n delta_T012= %0.0fR",delta_T012)
+
+P0=3.57*(0.28*794-0.24*362/0.95)
+printf("\n Net output power P0= %0.0f Btu/s=660hp, which is much greater than Pl \n\n",P0)
+
+disp("Pick another point say p02/p01=5.15")
+
+disp("m(T01^0.5)/p01 ETAc p02(psia) m(lbm/s) p03(psia) p03/p04 m(T01^0.5)/p01 T03(R) N/(T03^0.5) ETAt")
+disp("5.65 0.87 75.7 3.61 73.7 5.01 2.55 2710 353 0.86")
+
+delta_T034=0.86*2710*[1-(5.01^(-0.248))]
+printf("\n\n From new values delta_T034= %0.0f R",delta_T034)
+
+delta_T012=(530/0.87)*[(5.15^(0.2857))-1]
+printf("\n and delta_T012= %0.0f R",delta_T012)
+
+P0=3.61*(0.28*768-0.24*(364/0.95))
+printf("also P0= %0.1f Btu/s =628hp",P0)
+
+disp("P0 is close to Pl")
+disp("So the running point is around p02/p01=5.15, m(T01^0.5)/p01=5.65")
+disp("and N/(T01^0.5)=800 on the compressor characteristics")
+disp("It is not too close to the surge line and hence is safe.")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
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+
+
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+
diff --git a/3136/CH8/EX8.7/Ex8_7.sce b/3136/CH8/EX8.7/Ex8_7.sce new file mode 100755 index 000000000..63280be96 --- /dev/null +++ b/3136/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,166 @@ +clear all; clc;
+
+disp("Velocity diagrams at the rotor inlet and outlet are given.")
+
+disp("Velocities at the rotor inlet can be calculated .")
+
+r2=5/12
+N=15000
+U2=r2*N*%pi/30
+printf(" U2=r2*N*pi/30 = %0.1f ft/s",U2)
+
+alpha2=85*%pi/180//converting to radians
+V2=U2/sin(alpha2)
+printf("\n V2= %0.0f ft/s",V2)
+
+Vr2=U2/(tan(alpha2))
+printf("\n Vr2=W2= %0.1f ft/s",Vr2)
+
+disp("Hence from Cp(T02-T2)=V2^2/2, where T02=T01, we have")
+T02=2000
+V2=657
+Cp=0.24*778*32.2
+T2=T02-(V2^2)/(2*Cp)
+printf("\n T2=T02-(V2^2)/(2*Cp) =%0.1fR",T2)
+
+disp("From εn=(T2-T2_dash)/((V2^2)/(2*Cp)), we have")
+T2=1964.1
+epsilon_n=0.08
+V2=657
+T2_dash=T2-epsilon_n*V2^2/(2*Cp)
+printf(" T2_dash=T2-epsilon_n*V2^2/(2*Cp)= %0.1fR",T2_dash)
+
+p01=50
+T2_dash=1961.2
+T01=2000
+//let i=(k/(k-1))
+i=3.5
+p2=p01*(T2_dash/T01)^i
+printf("\n p2= %0.1f psia",p2)
+
+p2=46.7
+R=53.3
+T2=1964.1
+rho2=p2*144/(R*T2)//conversion factor=144
+printf("\n rho2= %0.3f lb,/ft^3",rho2)
+
+rho2=0.064
+Vr2=57.3
+A2=(2*%pi*5*2/144)
+m=rho2*Vr2*A2
+printf("\n So the mass flow rate m=rho2*Vr2*A2= %0.2f lbm/s",m)
+
+disp("Assuming whirl-free flow at the rotor outlet under the design condition, we have")
+U2=654.5
+delta_E=(U2)^2
+printf("\n delta_E=U2*Vu2=U2^2 %g ((ft/s)^2)=428370/(32.2*550)=24.2hp/(lbm/s)",delta_E)
+
+m=1.60
+delta_E=24.2//after converting to new units
+Ps=m*delta_E
+printf("\n Ps= %0.1f hp",Ps)
+
+rm3=2.06/12
+Um3=rm3*N*%pi/30
+printf("\n Um3= %0.2f ft/s",Um3)
+
+beta3=30*%pi/180//converting to radians
+V3=Um3/(tan(beta3))
+printf("\n V3 = %0.0f ft/s",V3)
+
+W3=Um3/sin(beta3)
+printf("\n W3 = %0.2f ft/s",W3)//the value has been rounded off to 539.2 in the book,however the value found here is more accurate
+
+disp("The turbine efficiency can be determined from equations 8.12 and 8.13. Without detailed calculations wthe result is given as ETAt=0.691")
+
+disp("The exhaust pressure/temperature can be determined from te following calculations with the help of figure 8.21")
+delta_E=428370/(32.2*778)
+printf(" From Cp(T01-T03)=delta_E= %0.1fBtu/lbm",delta_E)
+
+disp("(T01-T03)/(T01-T3dash)=Etat=0.691")
+
+T03=2000-(17.1/0.24)
+printf(" T03= %0.0f R",T03)
+
+T01=2000
+T03=1929
+ETAt=0.691
+T3_dash=T01-(T01-T03)/ETAt
+printf("\n T3_dash=T01-(T01-T03)/ETAt %0.0fR =",T3_dash)
+
+//let i=k/(k-1)
+i=3.5
+p01=50
+T3_dash=1897
+T01=2000
+p3=p01*(T3_dash/T01)^i
+printf("\n p3=p01*(T3_dash/T01)^i= %0.1f psia",p3)
+
+T2=1964.4
+p3=41.6
+p2=46.7
+// Let l=(k-1)/k
+l=0.2857
+T3_dbldash=T2*(p3/p2)^(l)
+printf("\n T3_dbldash=T2*(p3/p2)^(1/i)=%0.2fR",T3_dbldash)//answer given in the book is 1900.3 R,however the value tabulated here is more accurate
+
+T3_dbldash=1900.3
+epsilon_r=0.45
+W3=539.2
+Cp=0.24*32.2*778
+T3=T3_dbldash+epsilon_r*(W3^2)/(2*Cp)
+printf("\n T3= %0.1f R",T3)
+
+
+
+
+
+
+
+
+
+
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+
diff --git a/3136/CH9/EX9.1/Ex9_1.sce b/3136/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..bf850de14 --- /dev/null +++ b/3136/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,191 @@ +clear all; clc;
+
+rm=8/12
+N=7500
+U=rm*N*%pi/30
+printf("The peripheral velocity is calculated as U= %0.1f ft/s\n\n",U)
+
+disp("From equation 9.1 we have U/V1=sinα1/4")
+//let x= U/V1
+alpha1=70*%pi/180
+x=(sin(alpha1))/4
+printf("\n Thus U/V1= %0.4f",x)
+
+V1=U/x
+printf("\n Thus V1= %0.1f ft/s",V1)
+
+disp("From velocity diagram at station1 we have V1sinα1-W1sinß1=U and V1cosα1=W1cosß1 or W1sinß1")
+//let y= W1sinß1
+V1=2228.8
+U=523.6
+y=V1*sin(alpha1)-U
+printf("\n Hence W1sinß1= %0.1f ft/s",y)
+
+//Let z=W1cosß1
+z=V1*cos(alpha1)
+printf("\n Thus W1cosß1= %0.1f ft/s",z)
+
+disp("Hence tanß1=2.06")
+tanbeta1=2.06
+beta1=(atan(tanbeta1))*180/%pi
+printf("\n Thus beta1= %0.1f degrees and W1=1746 ft/s",beta1)
+
+disp("At station 2 we have W2sinß2-V2sinα2=U and V2cosα2=W2cosß2,with W2=W1=1746ft/s and ß1=ß2=64.1 degrees")
+//Let l=V2sinα2
+l=1746*sin(64.1*%pi/180)-523.6
+printf(" Thus V2sinα2=%0.0f ft/s",l)
+
+//m=V2cosα2
+m=1746*cos(64.1*%pi/180)
+printf("\n V2cosα2 %0.2f ft/s",m)
+
+disp("Hence tanα2=1.373")
+tanalpha2=1.373
+alpha2=((atan(tanalpha2)*180/%pi))
+printf(" Hence α2= %0.2f degrees",alpha2)
+
+disp("Hence V2=1295.2 ft/s")
+
+disp("At station 3 we have V3sinα3-W3sinα3=U=523.6ft/s")
+disp("Also W3cosß3=V3cosα3")
+//let n=V3cosα3
+V3=1295.2
+alpha3=53.9*%pi/180
+n=V3*cos(alpha3)
+printf(" Thus W3cosß3= %0.1f ft/s",n)
+
+disp("Hence tanß3=0.685")
+tanbeta3=0.685
+beta3=((atan(tanbeta3))*180/%pi)
+printf(" Hence ß3= %0.1f degrees",beta3)
+
+disp("Thus W3=925.1 ft/s")
+
+disp("Also W4=W3=925.1ft/s")
+disp("ß4=ß3=34.4 degrees")
+disp("And V4=VaV1cosß4")
+beta4=34.4*%pi/180
+//let y=Va*V1
+y=925.0848
+V4=y*cos(beta4)
+printf(" Thus V4= %0.1f ft/s",V4)
+disp("α4=0 degrees")
+
+disp("From these velocities,the energy transfers of the rotors can be calculated")
+
+U=523.6
+V1=2228.8
+alpha1=70*%pi/180
+V2=1295.2
+alpha2=53.9*%pi/180
+delta_E1=U*(V1*sin(alpha1)+V2*sin(alpha2))
+printf(" Thus delta_E1= %0.1f ((ft/s)^2)",delta_E1)
+
+delta_E1=1.643*(10^6)/(32.2*778)//converting units from (ft/s)^2 to Btu/lb
+printf("\n On converting to Btu/lb we have delta_E1=%0.1f Btu/lb",delta_E1)
+
+V3=1295.2
+alpha3=alpha2
+delta_E2=U*(V3*sin(alpha3))
+printf("\n delta_E2=%0.1f ((ft/s)^2)",delta_E2)
+delta_E2=0.546*(10^6)/(32.2*778)
+printf("\n On converting to Btu/lb we have delta_E2=%0.1f Btu/lb",delta_E2)
+
+delta_Ec=65.6+21.8
+printf("\n Hence the total energy transfer is delta_Ec= %0.1f Btu/lb",delta_Ec)
+disp("To compare with that calculated with equation9.3,we have delta_Ec=8*U^2")
+delta_Ec=8*(U^2)
+printf(" delta_Ec= %0.2f ((ft/s)^2)",delta_Ec)
+delta_Ec=2.19*10^6/(32.2*778)//converting units
+printf("\n On converting we have delta_Ec= %0.2f Btu/lb",delta_Ec)//answer given in the book is 87.5,however 87.42 is more accurate
+
+disp("The difference is due to round off error.")
+disp("The static enthalpies and pressure at stations 1,2,3 and 4 are same for the ideal case and can be calculated from h1=h01-((V1)^2)/2 ")
+disp("Where h01=h0i=1405Btu/lb from the Mollier diagram for p0i=3000 psia,T01=950 degrees Farenheit")
+//let l=(V1^2)/2
+V1=2228.8
+ l=(V1^2)/(2*32.2*778)
+printf(" Thus (V1^2)/2 = %0.0f Btu/lb",l)
+
+disp("Hence we have h1=1306 Btu/lb and p1=1400psia")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
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+
+
diff --git a/3136/CH9/EX9.2/Ex9_2.sce b/3136/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..41f25d76f --- /dev/null +++ b/3136/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,56 @@ +clear all; clc;
+
+disp("To use Figure 9.8,with Qf=ΣU^2/delta_Hs=2*g_c*lambda^2*R_H")
+disp("The value of R_H can be estimated with equation 8.4.")
+disp("Using k=1.3 for steam and suusming ETA_p=0.90 we have ETAad=[1-(p_e/p_i)^(ETAp*(k-1)/k)]/[1-(p_e/p_i)^(k-1)/k]=0.931")
+
+ETA_ad=0.931
+ETA_p=0.90
+R_H=ETA_ad/ETA_p
+printf(" R_H=ETA_ad/ETA_p= %0.3f",R_H)
+
+disp("For impulse stages,the optimal efficiencies occur at lambda=U/V2=sinα2/2=0.47 with alpha2=70 degrees")
+QF=2*25052*(0.47^2)*1.035
+printf(" So Qf can be calculated as %0.0f",QF)
+
+disp("From figure 9.8, the efficiency can be estimated as ETA=83%")
+
+disp("From the Mollier diagram in figure A1 we have hi=1525 Btu/lbm,hse=1150 Btu/lbm,with s_i=s_es=1.8Btu/lb-R")
+delta_Hs=1525-1150
+printf(" Hence delta_Hs=%0.0f Btu/lbm",delta_Hs)
+
+summation_sqr(U)=11455*375
+printf("\n So we have ΣU^2=%0.0f ((ft/s)^2)",summation_sqr(U))
+
+disp("With 10 identical stages,we have U^2=429562")
+sqr(U)=429562
+U=sqrt(sqr(U))
+printf(" Thus U= %0.0f ft/s",U)
+
+omega=3600*%pi/30
+D=2*U/omega
+printf("\n The turbine diameter D= %0.3f ft",D)//The answer has been incorrectly rounded off to 3.47 in the book. A more accurate answer is provided here.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
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+
+
diff --git a/3136/CH9/EX9.3/Ex9_3.sce b/3136/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..7edaab739 --- /dev/null +++ b/3136/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,40 @@ +clear all; clc;
+
+disp("The tangential velocity at the rotor mean radius is Um")
+rm=1.5
+N=3600
+Um=rm*N*%pi/30
+printf("Um=rm*N*pi/30= %0.1f",Um)
+
+disp("From the velocity diagram in figure 8.11 for the impulse stages we have delta_h0")
+disp("delta_h0=delta_Et=UmVu2-UmVu3/g_c=UmVu2/g_c=2((Um)^2)/gc")
+Um=565.5
+gc=32.2
+delta_h0=2*((Um)^2)/gc
+printf("\n delta_h0= %0.0f lbf-ft/lbm=25.5 Btu/lbm",delta_h0)
+
+disp("From the Mollier diagram in appendix A, we have hoi=1565 Btu/lbm")
+disp("For the stages with constant mean radi,we have hoi-hoe=n_s*delta_hoe or hoe=h0i-n_s*delta_hoe")
+h_oe=1565-(12*25.5)
+printf(" hoe= %0.0f Btu/lbm.",h_oe)
+
+disp("Also from ETA_ad=(hoi-hoe)/(hoi-hsoe), we have hsoe=hoi-(hoi-hoe)/ETA_ad")
+h_soe=1565-306/0.85
+printf(" hsoe= %0.0f Btu/lbm",h_soe)
+
+disp("From s_soe=Soi=1.69 Btu/(lbm-R),we have poe=50 psia and Toe=450 degrees Farenheit")
+disp("Also from given efficiencies we have Ps=ETAv*ETAm*m*n_s*delta_ho")
+Ps=0.92*0.90*7*12*19863/550
+printf(" Ps=%0.0f hp",Ps)
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH9/EX9.4/Ex9_4.sce b/3136/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..123dd97b5 --- /dev/null +++ b/3136/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,45 @@ +clear all; clc;
+
+disp("Based on the law of Willian we have m_t=0.5+CP_s")
+C=(7-0.5)/2512
+printf(" Where C= %0.4f lbm/(hp-s)\n",C)
+
+disp("So we have SR=mt/Ps=0.5/Ps+(2.6*10^-3)")
+disp("And HR=Q_h*SR")
+
+
+disp("OR at full load,")
+SR_1=(0.5/2512)+(2.6*10^-3)
+printf("\n SR_1= %0.4f lbm/(hp-s)=10.1 lbm/(hp-h)",SR_1)
+
+HR_1=1750*10.1
+printf("\n HR_1=%0.0f Btu/(hp-h)",HR_1)
+
+
+disp("at 50% load,")
+SR_2=(0.5/1256)+(2.6*10^-3)
+printf("\n SR_2= %0.4f lbm/(hp-s)=10.8 lbm/(hp-h)",SR_2)
+
+HR_2=1750*10.8
+printf("\n HR_2=%0.0f Btu/(hp-h)",HR_2)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH9/EX9.5/Ex9_5.sce b/3136/CH9/EX9.5/Ex9_5.sce new file mode 100755 index 000000000..ad0a39750 --- /dev/null +++ b/3136/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,34 @@ +clear all; clc;
+
+disp("To calculate the thermal efficiency,the units have to be consisitent. With hp=0.707 Btu/s=2545Btu/h, we have ETAth=Ps+Qe/Qin")
+ETA_th=(2512*0.707+1259*7)/(7*1750)
+printf(" Thus ETA_h= %0.3f",ETA_th)
+
+ETA_th=2512*0.707/(7*1750)
+printf("\n For the simple shaft power system,we have ETA_th= %0.3f",ETA_th)
+
+ETA_th=2545/17675
+printf("\n From the heat rate,ETA_th=2545/HR %0.3f",ETA_th)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3136/CH9/EX9.6/Ex9_6.sce b/3136/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..ba0bbfd86 --- /dev/null +++ b/3136/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,142 @@ +clear all; clc;
+
+disp("The enthalpies at various points have t be determined first. For the steam turbine cycle,fro the Mollier diagram or steam tables,we have h1=hf1=83.6")
+
+h2=83.6+(0.0185*(1000-1.5)*144)/(778*0.80)
+printf(" h2=h1+nu*deltap/ETA_p =%0.1f Btu/lbm",h2)
+
+disp("h3=1447 Btu/lbm")
+
+disp("s_s4=s3=1.61 Btu/(lbm-R)")
+
+disp("hs4=925 Btu/lbm")
+
+disp("Hence from ETAst=(h3-h4)/(h3-hs4) we have h4=h3-(h3-hs4)*ETAst")
+h3=1447
+hs4=925
+ETA_st=0.85
+h4=h3-(h3-hs4)*ETA_st
+printf(" h4= %0.0f Btu/lbm",h4)
+
+h4=1003
+h2=87.9
+h1=83.6
+ETA_ths=[(h3-h4)-(h2-h1)]/(h3-h2)
+printf("\n The thermal efficiency of the steam turbine cycle is then obtained as ETA_th,s=(Wst-Wp)/(Qin,s)=%0.4f ",ETA_ths)//it has been rounded off to 32.3 in the book
+
+disp("For the gas turbine cycle, an ideal gas with constant Cp is assumed for the working gas. With Cp=0.24 Btu/(lbm-R) and k=1.4 we have T6")
+T5=540
+//Let n=p6/p5 and m= (k-1)/k
+n=15
+m=0.2857
+ETAc=0.82
+T6=T5+[T5*({(n)^(m)}-1)]/ETAc
+printf("\n T6= %0.0fR=849 degrees Farenheit",T6)
+
+T7=2560
+//let b=(1/15)^0.2857
+b=0.461
+ETA_gt=0.85
+T8=T7-T7*[1-b]*ETA_gt
+printf("\n T8=%0.0f R= 928 degrees Farenheit",T8)
+
+disp("Which should be greater than T3")
+T7=2560
+T8=1388
+T6=1309
+T5=540
+ETA_thg=[(T7-T8)-(T6-T5)]/(T7-T6)
+printf("\n The thermal efficiency of the gas turbine is obtained as %0.3f=32.2 percent",ETA_thg)
+
+disp("From the energy balance equation across the HRSG,we have m_g*Cp*(T8-T9)")
+disp("ms/mg=[Cp*(T8-T9)]/(h3-h2)")
+//let x=ms/mg
+Cp=0.24
+T8=928
+T9=450
+h3=1447
+h2=87.9
+x=[Cp*(T8-T9)]/(h3-h2)
+printf("\n Thus ms/mg=%0.3f",x)
+
+disp("Hence the thermal efficiency of the combined cycle is obtained as ETA_th,c=[(Wgt-Wc)+(ms/mg)*(Wst-Wp)]/[Cp*(T7-T6)]")
+ETA_thc=[0.24*(1172-769)+0.084*(439.7)]/(0.24*1251)
+printf("\n ETA_th,c= %0.3f=44.5 percent",ETA_thc)
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