29 files changed, 518 insertions, 0 deletions
diff --git a/806/CH1/EX1.1/11.sce b/806/CH1/EX1.1/11.sce
new file mode 100644
index 000000000..00e0f8766
--- /dev/null
+++ b/806/CH1/EX1.1/11.sce
@@ -0,0 +1,14 @@
+clc
+pathname=get_absolute_file_path('11.sce')
+filename=pathname+filesep()+'11.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\11.txt')
+disp("a liquid has a viscosity of 0.005kg/m.s and density of 850kg/cm^3,calculate the kinematic viscosity in (a)SI units (b)USC units and (c)viscosity in USC units")
+disp("Solution:")
+v=u/p//kinematic viscosity
+V=v*(1/.3048)^2//kinematic viscosity in USC units
+U=u/47.9//viscosity in USC units
+disp("m^2/s",v,"Kinematic viscosity in SI units=")
+disp("ft^2/s",V,"Kinematic viscosity in USC units=")
+disp("slug/ft.s",U,"viscosity in USC units")
+diary(0)
+\ No newline at end of file
diff --git a/806/CH1/EX1.1/11.txt b/806/CH1/EX1.1/11.txt
new file mode 100644
index 000000000..978fca8ef
--- /dev/null
+++ b/806/CH1/EX1.1/11.txt
@@ -0,0 +1,22 @@
+ 
+ a liquid has a viscosity of 0.005kg/m.s and density of 850kg/cm^3,calculate the kinematic viscosity in (a)SI units (b)USC units and (c)viscosity in USC units   
+ 
+ Solution:   
+ 
+ Kinematic viscosity in SI units=   
+ 
+    0.0000059  
+ 
+ m^2/s   
+ 
+ Kinematic viscosity in USC units=   
+ 
+    0.0000633  
+ 
+ ft^2/s   
+ 
+ viscosity in USC units   
+ 
+    0.0001044  
+ 
+ slug/ft.s   
diff --git a/806/CH1/EX1.2/12.sce b/806/CH1/EX1.2/12.sce
new file mode 100644
index 000000000..310088f85
--- /dev/null
+++ b/806/CH1/EX1.2/12.sce
@@ -0,0 +1,12 @@
+clc
+pathname=get_absolute_file_path('12.sce')
+filename=pathname+filesep()+'12.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\12.txt')
+disp("A lubricated shaft rotates inside a concentric sleeve bearing at 1200rpm,the clearance is smaal with respect to the radius and hence a linear distribution velocity distribution may be assumed.What are the power requirements to rotate the shaft?R=2cm , L=6cm , dy=0.1mm , and u=0.2N.s/m^2.")
+disp("Solution:")
+t=u*du/dy*2*%pi/60*R*1000/100//shear stress in N/m^2
+T=t*2*%pi*R*L*R/1000000//torque in Nm
+P=T*du*2*%pi/60//Power in Watts
+disp("W",P,"Power required to rotate the shaft =")
+diary(0)
+\ No newline at end of file
diff --git a/806/CH1/EX1.2/12.txt b/806/CH1/EX1.2/12.txt
new file mode 100644
index 000000000..5a0ff5871
--- /dev/null
+++ b/806/CH1/EX1.2/12.txt
@@ -0,0 +1,11 @@
+ 
+ A lubricated shaft rotates inside a concentric sleeve bearing at 1200rpm,the clearance is smaal with respect to the radius and hence a linear distribution velocity  
+      distribution may be assumed.What are the power requirements to rotate the shaft?R=2cm , L=6cm , dy=0.1mm , and u=0.2N.s/m^2.                                    
+ 
+ Solution:   
+ 
+ Power required to rotate the shaft =   
+ 
+    95.251282  
+ 
+ W   
diff --git a/806/CH1/EX1.3/13.sce b/806/CH1/EX1.3/13.sce
new file mode 100644
index 000000000..aab6d30bc
--- /dev/null
+++ b/806/CH1/EX1.3/13.sce
@@ -0,0 +1,10 @@
+clc
+pathname=get_absolute_file_path('13.sce')
+filename=pathname+filesep()+'13.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\13.txt')
+disp("a gas with molecular weight of 44 is at a pressure of 0.9MPa and a temperature of 20 degree celsius.Calculate its density")
+R=8312/M//gas constant(in m.N/kg.K)
+p=P/R/(273+T)*10^6//density in kg/m^3
+disp("kg/m^3",p,"Density p=")
+diary(0)
+\ No newline at end of file
diff --git a/806/CH1/EX1.3/13.txt b/806/CH1/EX1.3/13.txt
new file mode 100644
index 000000000..6a2280ad0
--- /dev/null
+++ b/806/CH1/EX1.3/13.txt
@@ -0,0 +1,8 @@
+ 
+ a gas with molecular weight of 44 is at a pressure of 0.9MPa and a temperature of 20 degree celsius.Calculate its density   
+ 
+ Density p=   
+ 
+    16.260056  
+ 
+ kg/m^3   
diff --git a/806/CH1/EX1.4/14.sce b/806/CH1/EX1.4/14.sce
new file mode 100644
index 000000000..791f2a478
--- /dev/null
+++ b/806/CH1/EX1.4/14.sce
@@ -0,0 +1,10 @@
+clc
+pathname=get_absolute_file_path('14.sce')
+filename=pathname+filesep()+'14.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\14.txt')
+disp("A liquid compressed in a cylinder has a volume of 1000 cm^3 at 1 MN/m^2 and volume of 995 cm^3 at 2MN/m^2.What is its bulk modulus of elasticity?")
+disp("Solution:")
+K=-(P2-P1)/(V2-V1)*V1//Modulus of elasticity
+disp("MPa",K,"Modulus of elasticity=")
+diary(0)
diff --git a/806/CH1/EX1.4/14.txt b/806/CH1/EX1.4/14.txt
new file mode 100644
index 000000000..856fdbd79
--- /dev/null
+++ b/806/CH1/EX1.4/14.txt
@@ -0,0 +1,14 @@
+ 
+ A liquid compressed in a cylinder has a 
+       volume of 1000 cm^3 at 1 MN/m^2 a 
+      nd volume of 995 cm^3 at 2MN/m^2.W 
+      hat is its bulk modulus of elastic 
+      ity?                               
+ 
+ Solution:   
+ 
+ Modulus of elasticity=   
+ 
+    200.  
+ 
+ MPa   
diff --git a/806/CH1/EX1.5/15.sce b/806/CH1/EX1.5/15.sce
new file mode 100644
index 000000000..d11cd408d
--- /dev/null
+++ b/806/CH1/EX1.5/15.sce
@@ -0,0 +1,12 @@
+clc
+pathname=get_absolute_file_path('15.sce')
+filename=pathname+filesep()+'15.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\15.txt')
+disp("With regard to Example 1.4 suppose the cylinder is a water quality sample bottle used to collect water samples at a predetermined depths.At deep Depths the sample Bottle has a smaller volume to collect 995 cm^3 due to compression.Suppose that analysis reveals that 15 mg of sediment are collected.What would be the difference in concerntration data measured shipboard where the pressure is atmospheric versus the in situ depths where the sample was collected?")
+disp("Solution:")
+C1=M/V1//Concerntration of shipboard
+C2=M/V2//concerntration of in situ depth
+disp("mg/cm^3",C1,"Concerntration of shipboard")
+disp("mg/cm^3",C2,"Concerntration of in situ depth")
+diary(0)
diff --git a/806/CH1/EX1.5/15.txt b/806/CH1/EX1.5/15.txt
new file mode 100644
index 000000000..c2d88f30b
--- /dev/null
+++ b/806/CH1/EX1.5/15.txt
@@ -0,0 +1,29 @@
+ 
+ With regard to Example 1.4 suppose the  
+      cylinder is a water quality sample 
+       bottle used to collect water samp 
+      les at a predetermined depths.At d 
+      eep Depths the sample Bottle has a 
+       smaller volume to collect 995 cm^ 
+      3 due to compression.Suppose that  
+      analysis reveals that 15 mg of sed 
+      iment are collected.What would be  
+      the difference in concerntration d 
+      ata measured shipboard where the p 
+      ressure is atmospheric versus the  
+      in situ depths where the sample wa 
+      s collected?                       
+ 
+ Solution:   
+ 
+ Concerntration of shipboard   
+ 
+    0.015  
+ 
+ mg/cm^3   
+ 
+ Concerntration of in situ depth   
+ 
+    0.0150754  
+ 
+ mg/cm^3   
diff --git a/806/CH10/EX10.1/101.sce b/806/CH10/EX10.1/101.sce
new file mode 100644
index 000000000..a47b495c2
--- /dev/null
+++ b/806/CH10/EX10.1/101.sce
@@ -0,0 +1,22 @@
+clc
+pathname=get_absolute_file_path('101.sce')
+filename=pathname+filesep()+'101.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\101.txt')
+disp("A 75 mm diameter orific under a head of 4.88 m discharges 8900 N waterin 32.6 s.The trajectory was determined by measuring x=4.76 m for a drop of 1.22.determine the head loss per unit weight,the power loss,Cc,Cv,Cd")
+disp("Solution:")
+Vt=sqrt(2*g*H)//theoretical velocity
+t=sqrt(2*y/g)//actual time
+Va=x/t//actual velocity
+Cv=Va/Vt//coefficient of velocity
+Qa=F/p/T//actual discharge
+Cd=Qa/(%pi*d^2*Vt)*4//coefficient of discharge
+Cc=Cd/Cv//coefficient of contraction
+HL=H*(1-Cv^2)//Head loss
+PL=Qa*p*HL//power loss
+disp(Cd,"Cd=")
+disp(Cv,"Cv=")
+disp(Cc,"Cc=")
+disp("m.N/N",HL,"Head loss=")
+disp("W",PL,"Power loss=")
+diary(0)
+\ No newline at end of file
diff --git a/806/CH10/EX10.1/101.txt b/806/CH10/EX10.1/101.txt
new file mode 100644
index 000000000..b8bb70b76
--- /dev/null
+++ b/806/CH10/EX10.1/101.txt
@@ -0,0 +1,31 @@
+ 
+ A 75 mm diameter orific under a head of 4.88 m discharges 8900 N waterin 32.6 s.The trajectory was determined by measuring x=4.76 m for a drop of 1.22.determine t 
+      he head loss per unit weight,the power loss,Cc,Cv,Cd                                                                                                          
+ 
+ Solution:   
+ 
+ Cd=   
+ 
+    0.6437713  
+ 
+ Cv=   
+ 
+    0.9754098  
+ 
+ Cc=   
+ 
+    0.6600008  
+ 
+ Head loss=   
+ 
+    0.2370492  
+ 
+ m.N/N   
+ 
+ Power loss=   
+ 
+    64.715881  
+ 
+ W   
+ 
+    0.9754098  
diff --git a/806/CH4/EX4.5/45.sce b/806/CH4/EX4.5/45.sce
new file mode 100644
index 000000000..e76367382
--- /dev/null
+++ b/806/CH4/EX4.5/45.sce
@@ -0,0 +1,11 @@
+clc
+pathname=get_absolute_file_path('45.sce')
+filename=pathname+filesep()+'45.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\45.txt')
+disp("Water is flowing through an open channel at a depth of 2 m and a velocity of 3m/s.it then flows through a contracting chute into another channel where the depth is 1 m and the velocity is 10m/s.assuming frictionless flow,determine the difference in the elevation of the channel floors.")
+disp("Solution:")
+disp("The points 1 and 2 can be selected on the free surface and therefore p1=p2=0")
+disp("Therefore by bernoullis equation the difference in the elevation  of the channel is y=")
+y=(v2^2)/(2*p)+h2-h1-(v1^2)/(2*p)
+disp("m",y)
+\ No newline at end of file
diff --git a/806/CH4/EX4.5/45.txt b/806/CH4/EX4.5/45.txt
new file mode 100644
index 000000000..9717574b8
--- /dev/null
+++ b/806/CH4/EX4.5/45.txt
@@ -0,0 +1,15 @@
+ 
+ Water is flowing through an open channel at a depth of 2 m and a velocity of 3m/s.it then flows through a contracting chute into another channel where the depth i 
+      s 1 m and the velocity is 10m/s.assuming frictionless flow,determine the difference in the elevation of the channel floors.                                   
+ 
+ Solution:   
+ 
+ The points 1 and 2 can be selected on the free surface and therefore p1=p2=0   
+ 
+ Therefore by bernoullis equation the difference in the elevation  of the channel is y=   
+ 
+    3.6381244  
+ 
+ m   
+ 
+-->exec('SCI/etc/scilab.quit','errcatch',-1);quit;
diff --git a/806/CH4/EX4.6/46.sce b/806/CH4/EX4.6/46.sce
new file mode 100644
index 000000000..fc58bb642
--- /dev/null
+++ b/806/CH4/EX4.6/46.sce
@@ -0,0 +1,16 @@
+clc
+pathname=get_absolute_file_path('46.sce')
+filename=pathname+filesep()+'46.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\46.txt')
+disp("A venturimeter,consisting of a converging portion followed by a throat of constant diameter and then gradually diverging portion,is used to detremine the flow of the pipe.The diameter at section 1 is 6 in. and that of section 2 is 4 in.,Find the discharge through the pipe when p1-p2=3psi and oil of sp grvty 0.9 is flowing")
+disp("Solution:")
+disp("From the continuity equation Q=A1v1=A2v2")
+disp("Therefore,v1=Q*16/pi , v2=Q*36/pi")
+disp("And moreover Z1=Z2")
+disp("Therefore discharge Q =")
+a1=%pi/16//sq.ft
+a2=%pi/36//sq.ft
+q=sqrt(a*144*(%pi^2)*2*32.185/((s*62.4)*(36^2-16^2)))
+disp("cfs",q)
+diary(0)
+\ No newline at end of file
diff --git a/806/CH4/EX4.6/46.txt b/806/CH4/EX4.6/46.txt
new file mode 100644
index 000000000..1933f4658
--- /dev/null
+++ b/806/CH4/EX4.6/46.txt
@@ -0,0 +1,18 @@
+ 
+ A venturimeter,consisting of a converging portion followed by a throat of constant diameter and then gradually diverging portion,is used to detremine the flow of  
+      the pipe.The diameter at section 1 is 6 in. and that of section 2 is 4 in.,Find the discharge through the pipe when p1-p2=3psi and oil of sp grvty 0.9 is flo 
+      wing                                                                                                                                                          
+ 
+ Solution:   
+ 
+ From the continuity equation Q=A1v1=A2v2   
+ 
+ Therefore,v1=Q*16/pi , v2=Q*36/pi   
+ 
+ And moreover Z1=Z2   
+ 
+ Therefore discharge Q =   
+ 
+    2.1677205  
+ 
+ cfs   
diff --git a/806/CH4/EX4.7/47.sce b/806/CH4/EX4.7/47.sce
new file mode 100644
index 000000000..a665e9b42
--- /dev/null
+++ b/806/CH4/EX4.7/47.sce
@@ -0,0 +1,19 @@
+clc
+pathname=get_absolute_file_path('47.sce')
+filename=pathname+filesep()+'47.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\47.txt')
+disp("The water supply reservoir as shown in given figure has an average depth of 20m, a surface area of 20km^2, ans an outlet whose centerline is 15m below the water surface.If the otflow diameter is 1 m, what is the outflow and its associated velocity?What would be the drwa down during one week and one day periods?")
+disp("Solution:")
+V2=sqrt(2*g*z1)//Velocity of liquid from the pipe
+Q=V2*%pi*d2^2/4//Discharge per sec
+disp("m^3/s",Q,"Discharge")
+v1=Q*24*3600//Discharge per day
+v2=Q*7*24*3600//Dicharge per week
+v=A*H//Original volume of liquid
+disp("Hence drop down in level for one day and one week are:")
+h1=v1/v*H//draw down of one day
+h2=v2/v*H//draw down of one week
+disp("m",h1,"draw down of one day")
+disp("m",h2,"draw down of one week")
+diary(0)
diff --git a/806/CH4/EX4.7/47.txt b/806/CH4/EX4.7/47.txt
new file mode 100644
index 000000000..4a8221360
--- /dev/null
+++ b/806/CH4/EX4.7/47.txt
@@ -0,0 +1,34 @@
+ 
+ The water supply reservoir as shown in  
+      given figure has an average depth  
+      of 20m, a surface area of 20km^2,  
+      ans an outlet whose centerline is  
+      15m below the water surface.If the 
+       otflow diameter is 1 m, what is t 
+      he outflow and its associated velo 
+      city?What would be the drwa down d 
+      uring one week and one day periods 
+      ?                                  
+ 
+ Solution:   
+ 
+ Discharge   
+ 
+    13.473642  
+ 
+ m^3/s   
+ 
+ Hence drop down in level for one day an 
+      d one week are:                    
+ 
+ draw down of one day   
+ 
+    0.0582061  
+ 
+ m   
+ 
+ draw down of one week   
+ 
+    0.4074429  
+ 
+ m   
diff --git a/806/CH7/EX7.1/71.sce b/806/CH7/EX7.1/71.sce
new file mode 100644
index 000000000..0ffc8a9a0
--- /dev/null
+++ b/806/CH7/EX7.1/71.sce
@@ -0,0 +1,15 @@
+clc
+pathname=get_absolute_file_path('71.sce')
+filename=pathname+filesep()+'71.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\71.txt')
+disp("A smooth flat plate 3 m wide and 30 m long is towed through still water at 20 degree celsius with a speed of 6m/s.Determine the drag on one side of the plate and the drag on the first 3 m of plate")
+disp("Solution:")
+R=U*l/v//Reynolds number
+Cd=0.455/(log10(R))^2.58//Constant
+D1=Cd*w*l*p*(U^2)/2//Drag force on whole plate
+D2=Cd*w*l1*p*U^2/2//Drag force on first 3 m
+disp("N",D1,"Drag force on whole plate")
+disp("N",D2,"Drag force on first 3m")
+
+diary(0)
diff --git a/806/CH7/EX7.1/71.txt b/806/CH7/EX7.1/71.txt
new file mode 100644
index 000000000..55756a872
--- /dev/null
+++ b/806/CH7/EX7.1/71.txt
@@ -0,0 +1,21 @@
+ 
+ A smooth flat plate 3 m wide and 30 m l 
+      ong is towed through still water a 
+      t 20 degree celsius with a speed o 
+      f 6m/s.Determine the drag on one s 
+      ide of the plate and the drag on t 
+      he first 3 m of plate              
+ 
+ Solution:   
+ 
+ Drag force on whole plate   
+ 
+    3176.7877  
+ 
+ N   
+ 
+ Drag force on first 3m   
+ 
+    317.67877  
+ 
+ N   
diff --git a/806/CH7/EX7.2/72.sce b/806/CH7/EX7.2/72.sce
new file mode 100644
index 000000000..2989457bb
--- /dev/null
+++ b/806/CH7/EX7.2/72.sce
@@ -0,0 +1,22 @@
+clc
+pathname=get_absolute_file_path('72.sce')
+filename=pathname+filesep()+'72.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\72.txt')
+disp("Dredging operations in a river yield large volumes of sediments whose smallest particle is measured to be coarse sand with a diameter of 4 microns and whose largest particle is coarse sand with a diameter of 1000 microns or 1mm.determine the settling velocity for each size class.Gw=9764 N/m^3;Ssand=2.65;Sclay=1.6;and the viscosity at 30 degree celsius is 0.8*10D-3 N.s/m^2")
+disp("Solution:")
+disp("(i)For clay particles:")
+disp("Assuming reynolds number less than 1")
+w1=d1^2*(S1-1)*p/(18*v)//Settling velocity of clay particles
+R1=d1*w1/v//reynolds number
+disp(R1,"R1=")
+disp("Since reynolds number is less than 1,hence")
+disp("m/s",w1,"Settling velocity for clay particles")
+disp("(ii)For sand particles:")
+disp("Assuming reynolds number greater than 1 and equal to 220 Cd=0.7")
+w2=sqrt(1.333*d2*p*(S2-1)/(Cd*r))//Settling velocity of clay particles
+R2=d2*w2*1e3/v//reynolds number
+disp(R2,"R2")
+disp("Since the reynolds number is greater than 1")
+disp("m/s",w2,"Hence the settling velovity =")
+diary(0)
diff --git a/806/CH7/EX7.2/72.txt b/806/CH7/EX7.2/72.txt
new file mode 100644
index 000000000..29bb09e63
--- /dev/null
+++ b/806/CH7/EX7.2/72.txt
@@ -0,0 +1,49 @@
+ 
+ Dredging operations in a river yield la 
+      rge volumes of sediments whose sma 
+      llest particle is measured to be c 
+      oarse sand with a diameter of 4 mi 
+      crons and whose largest particle i 
+      s coarse sand with a diameter of 1 
+      000 microns or 1mm.determine the s 
+      ettling velocity for each size cla 
+      ss.Gw=9764 N/m^3;Ssand=2.65;Sclay= 
+      1.6;and the viscosity at 30 degree 
+       celsius is 0.8*10D-3 N.s/m^2      
+ 
+ Solution:   
+ 
+ (i)For clay particles:   
+ 
+ Assuming reynolds number less than 1   
+ 
+ R1=   
+ 
+    3.255D-08  
+ 
+ Since reynolds number is less than 1,he 
+      nce                                
+ 
+ Settling velocity for clay particles   
+ 
+    0.0000065  
+ 
+ m/s   
+ 
+ (ii)For sand particles:   
+ 
+ Assuming reynolds number greater than 1 
+       and equal to 220 Cd=0.7           
+ 
+ R2   
+ 
+    219.41569  
+ 
+ Since the reynolds number is greater th 
+      an 1                               
+ 
+ Hence the settling velovity =   
+ 
+    0.1755325  
+ 
+ m/s   
diff --git a/806/CH8/EX8.2/82.sce b/806/CH8/EX8.2/82.sce
new file mode 100644
index 000000000..9b4886aa5
--- /dev/null
+++ b/806/CH8/EX8.2/82.sce
@@ -0,0 +1,15 @@
+clc
+pathname=get_absolute_file_path('82.sce')
+filename=pathname+filesep()+'82.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\82.txt')
+disp("A submarine moves through water at 30 ft/s.At a point A on the submarine 5 ft above the nose,the velocity of the submarine relative to the water is 50 ft/s.Determine the dynamic-pressure difference between this point and the nose,and determine the difference in total pressure between the two points.")
+disp("If the submarine is sationary and the water is movinf past it,The velocity at the nose is zero and the velocity at A is 50ft/s.By selecting the dynamic pressure at infinity as zero")
+E=q^2/2+g*hn
+p1=p*E//Pressure at nose
+p2=p*(E-q1^2/2)//Pressure at point A
+P1=p2-p1//Pressure difference at point A and nose
+P2=p*(g*hn-g*ha+(qn^2-q1^2)/2)
+disp("lb/ft^2",P1,"Hence dynamic pressure difference between point A and nose is=")
+disp("lb/ft^2",P2,"Hence total pressure difference between point A and nose is=")
+diary(0)
diff --git a/806/CH8/EX8.2/82.txt b/806/CH8/EX8.2/82.txt
new file mode 100644
index 000000000..e4947d3bd
--- /dev/null
+++ b/806/CH8/EX8.2/82.txt
@@ -0,0 +1,31 @@
+ 
+ A submarine moves through water at 30 f 
+      t/s.At a point A on the submarine  
+      5 ft above the nose,the velocity o 
+      f the submarine relative to the wa 
+      ter is 50 ft/s.Determine the dynam 
+      ic-pressure difference between thi 
+      s point and the nose,and determine 
+       the difference in total pressure  
+      between the two points.            
+ 
+ If the submarine is sationary and the w 
+      ater is movinf past it,The velocit 
+      y at the nose is zero and the velo 
+      city at A is 50ft/s.By selecting t 
+      he dynamic pressure at infinity as 
+       zero                              
+ 
+ Hence dynamic pressure difference betwe 
+      en point A and nose is=            
+ 
+  - 2418.75  
+ 
+ lb/ft^2   
+ 
+ Hence total pressure difference between 
+       point A and nose is=              
+ 
+  - 2730.0334  
+ 
+ lb/ft^2   
diff --git a/806/CH8/EX8.3/83.sce b/806/CH8/EX8.3/83.sce
new file mode 100644
index 000000000..90666a4f9
--- /dev/null
+++ b/806/CH8/EX8.3/83.sce
@@ -0,0 +1,13 @@
+clc
+pathname=get_absolute_file_path('83.sce')
+filename=pathname+filesep()+'83.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\83.txt')
+disp("A source with strength 0.2m^3/s and a vortex with strength 1 m^2/s are located at the origin.Determine the equations for velocity potential and stream function.What are the velocity components at x=1 m,y=0.5 m?")
+disp("Solution:")
+r=sqrt(x^2+y^2)
+Vr=1/(10*%pi*r)
+Vt=1/(2*%pi*r)
+disp("m/s",Vr,"Vr=")
+disp("m/s",Vt,"Vt=")
+diary(0)
diff --git a/806/CH8/EX8.3/83.txt b/806/CH8/EX8.3/83.txt
new file mode 100644
index 000000000..44a79db57
--- /dev/null
+++ b/806/CH8/EX8.3/83.txt
@@ -0,0 +1,22 @@
+ 
+ A source with strength 0.2m^3/s and a v 
+      ortex with strength 1 m^2/s are lo 
+      cated at the origin.Determine the  
+      equations for velocity potential a 
+      nd stream function.What are the ve 
+      locity components at x=1 m,y=0.5 m 
+      ?                                  
+ 
+ Solution:   
+ 
+ Vr=   
+ 
+    0.0284705  
+ 
+ m/s   
+ 
+ Vt=   
+ 
+    0.1423525  
+ 
+ m/s   
diff --git a/806/DEPENDENCIES/101.sci b/806/DEPENDENCIES/101.sci
new file mode 100644
index 000000000..3b45cd5d1
--- /dev/null
+++ b/806/DEPENDENCIES/101.sci
@@ -0,0 +1,7 @@
+x=4.76//m
+y=1.22//m
+d=0.075//m(Diameter of orifice)
+H=4.88//m(Head)
+F=8900//N
+p=9810//N/m^3(Weight density of water)
+T=32.6//s
+\ No newline at end of file
diff --git a/806/DEPENDENCIES/1025.sci b/806/DEPENDENCIES/1025.sci
new file mode 100644
index 000000000..6c1e24f33
--- /dev/null
+++ b/806/DEPENDENCIES/1025.sci
@@ -0,0 +1,7 @@
+Cd=0.74//coefficient of discharge
+h1=2//m
+p=15//kPa(Air pressure)
+S=0.92//Specific gravity of oil
+d=0.07//m(Diameter of opening)
+g=9.81//m/s^2(Acceleration due to gravity)
+P=9.810//N/m^3(Weight density of water)
+\ No newline at end of file
diff --git a/806/DEPENDENCIES/1026.sci b/806/DEPENDENCIES/1026.sci
new file mode 100644
index 000000000..51531ef72
--- /dev/null
+++ b/806/DEPENDENCIES/1026.sci
@@ -0,0 +1,8 @@
+Cd=0.74//coefficient of discharge
+h1=2//m
+p=15//kPa(Air pressure)
+S=0.92//Specific gravity of oil
+d=0.07//m(Diameter of opening)
+g=9.81//m/s^2(Acceleration due to gravity)
+P=9.810//N/m^3(Weight density of water)
+Cv=0.96//coefficient of velocity
+\ No newline at end of file