#variable initialisation
w=500 #mass of liquid in N
g=9.806 #gravity in m/s*2
g1=3.5 #gravity in m/s*2
g2=18.0 #gravity in m/s*2
#Solution
m=w/g #mass
print "a)m=",round(m,2),"kg"
w1=m*g1 #weight
w2=m*g2
print "b)w1=",round(w1,2),"N"
print "w2=",round(w2,1),"N"
#variable initialisation
weight=400 # weight in N
gravity= 9.806 # gravity in m/s^2
F=800 # force in n
g=1.6 # gravity in m/s^2
#Solution
m=round(weight/gravity,3)
#mass is constant and does not change with location.By Newton's second law F=m*a"
a=F/m
#acceleration is independent of g.Hence both on the earth ,as well as on moon"
print "a=",round(a,3),"m/s^2"
#variable initialisation
RD=0.8 #relative density
rho_w=998 #density of water at 20 degree C in kg/m^3
g=9.81 #gravity
v=2.3 #viscosity in centistoke
#calculation
UnitWeight=((rho_w*RD)*g)/1000 #Unit Weight
print "(i) gamma=",round(UnitWeight,3),"kN/m^3"
v=2.3/(10**6) #viscosity in m^2/s
rho=round(rho_w*RD,1) #dymanic viscosity
mu=v*rho
print "(ii) mu="'%4.3E' % mu,"Pa.s."
from __future__ import division
import math
#variable initialisation
V=1.50 #Velocity in m/s
h=3 #Height in mm
mu=0.2 #Dynamic viscosity in Pa.s.
#Calculation
#Since the gap between the plates is very small,assume linear variation of velocity
#Let us take du/dy=a
a=V/(h/1000)
print "du/dy=",a,"(1/s)"
tow=mu*(a) #calculating shear stress on the bottom plate
print "tow=",int(tow),"N/m^2"
from sympy import Symbol,diff
#Variable Initialisation
mu=1.5 #coefficient of dynamic viscosity
#Calculation
y = Symbol('y')
u=4*y-y**2
du=diff(u,y,1) #differentiating u with respect to y
#tau=mu*(du) #Shear stress
y=0 #taking y as 0
tau_0=mu*(4-2*y)
print "tau_0=",tau_0,"Pa.s"
y=2 #taking y as 2
tau_2=mu*(4-2*y)
print "tau_2=",int(tau_2)
import math
#variable initialisation
N=240 #speed in rpm
D=90 #Diameter in mm
D1=95 #internal diameter in mm
mu=2 #dynamic viscosity in poise
L=50 #length in cm
#Calculation
mu=mu/10 #into Pa.s
omega=2*math.pi*N/60
V=omega*(D/1000)/2 #Circumferential velocity of the shaft
h=(D1-D)/2
#Assuming linear variation of velocity across the gap,
du=V #Velocity gradient
dr=h/1000
tau=mu*(du/dr) #Shear stress on the shaft
F_s=tau*(2*math.pi*(D/1000)/2)*(L/100) #shear force
T=F_s*((D/1000)/2)
P=(2*math.pi*N/60)*T #Power required
print "P=",round(P,1),"W"
import math
#variable initialisation
mu=6 #dynamic viscosity in poise
h=0.02 #radical clearence in mm
D=3 #diameter in cm
L=10 #length in cm
W_s=7.51 #effective weight of the sleeve in N
#Calculation
#tau=mu*(du/dr)
tau=(mu/(h/100)) #Shear stress
#shaer stress=submerged weight of the sleeve
#(2*math.pi*r*L)*tau=W_s
a=(2*math.pi*((D/100)/2)*(L/100)*tau)
V=W_s/a #velocity
print "V=",round(V*100,2),"cm/s"
import math
#variable initialisation
v=3.7E-4 #kinematic viscosity in m^2/s
rho=0.85 #relative density in kg/m^3
V=15 #velocity in cm
D=25 #diameter in cm
L=3.3 #length in m
D1=25.018 #internal diameter in cm
#Calculation
mu=v*rho*998
h=(D1-D)/(2E2)
tau=mu*((V/100)/h) #shear stress
A=math.pi*(D/100)*L #area
F_s=A*tau #Frictional resistance
print "F_s=",round(F_s/1000,3),"kN"
import math
#variable initialisation
D=10 #Diameter in cm
N=90 #rotations in rpm
D1=9.75 #diameter 2 in cm
Torque=1.2 #torque in N.m
H=2.5 #height in cm
#Calculation
D=D/100 #into m
D1=D1/100
V=math.pi*D*N/60 #Tangential velocity
h=(D-D1)/2 #Radical clearence
tau=(V/h) #shear stress
area=2*math.pi*(D/2)*(H/100) #calculating area
F_s=tau*area #shear force
T=(F_s*(D/2)) #Torque
mu=1.2*100/T #Coefficient of viscosity
print "mu=",round(mu/100,3),"Pa.s"
#variable initialisation
sigma=0.073 #surface Tension at air-water interface in N/m
d=0.01 #diameter of air bubble in mm
#calculation
#air bubble has only one surface
R=(d/2)/1000
delta_p=(2*sigma)/R #Pressure difference in N/m^2
print "delta_p=",delta_p/1000,"kPa"
#variable initialisation
sigma=0.088 #surface Tension at soap-air interface in N/m
d=3 #diameter of air bubble in cm
#Solution
#soap bubble has two interfaces
R=0.03/2 #in m
delta_p=(4*sigma)/R #in N/m^2
print "delta_p=",round(delta_p,2),"N/m^2 above atmospheric pressure"
#variable initialisation
d1=6 #diameter in mm
d2=16 #diameter in mm
sigma=0.073 #surface tension of water in N/m
g=9.81 #gravity
rho=998 #density
#Solution
#assume angle of contact theta=0 degree
R1=0.006/2 #radius in m
R2=0.016/2 #raidus in m
r=(1/R1)-(1/R2)
h=(2*sigma*r)/(rho*g) #difference in water level
print "h=",round(h,4)*1000,"mm"
import math
#variable initialisation
d=3E-3 #diameter of tube
sigma=0.48 #surface tension in N/m
a=130 #angle of contact in degree C
rho=13600 #density of liquid in kg/m^3
g=9.81 #gravity
#solution
R=d/2
gamma=rho*g
e=math.radians(a) #converting degree into radians
c=math.cos(e) #cosine value
h=(2*sigma*c)/(gamma*R)
h1=round(h*1000,2)
print "Therefore,there is a capillary depression of",h1,"mm"
#variable initialisation
d=0.005 #diamter of grain in mm
sigma=0.073 #surface tension of air-water interface in N/m
g= 9.81 #gravity in m/s^2
de=998 #density
#solution
R=(d/2)/1000
ga=g*de
#by assuming theta = 0 degree
del_h=(2*sigma)/(ga*R) #height of water rise
print "Answer given in the book is wrong.It should be as,"
print "delta_h=",round(del_h,2),"m"
import math
#variable initialisation
M=28.97 #molecular weight of air
p=120E+3 #in abs
T=60 #Temperature in degree C
#solution
R=math.ceil(8312/28.97) #gas constant for air
T=T+273 #Temperature in K
rho=round(p/(R*T),3)
print "(i)density of air=",rho,"kg/m^3"
M1=44
R=int(math.ceil(8312/44)) # gas constant for co2
a=R*T
rho=p/a
print "(ii)density of co2=",round(rho,3),"kg/m^3"
#variable initialisation
p=100 #compression rate in abs
p1=80 #compression rate in abs
#solution
#In isothermal change K=p
KA=p #bulk modulus
print "KA=",KA,"kPa"
#In adiabatic change K=kp
k=1.4 #constant for gases
KB=int(k*p1) #bulk modulus
print "KB=",KB,"kPa"
print "KA<KB,gas A is more compressible than gas B,in the notified situation"
import math
#variable initialisation
k=1.43E+9 #bulk modulus of elasticity of kerosene in Pa
r=0.804 #relative density
#solution
rho=round(r*998,1) #density of kerosene in kg/m^3
C=math.ceil(math.sqrt(k/rho))
print "velocity of sound C=",int(C),"m/s"
import math
#variable initialisation
R=287 #gas constant
T=80 #temperature in degree C
#solution
T=273+80 #temperature in K
k=1.4 #for air
C=round(math.sqrt(k*R*T),1) #velocity of sound at 80 degree
print "C=",C,"m/s"