# Chapter 7 - Dimensional and Modal Analysis¶

## Ex 7.5 Page 312¶

In :
from __future__ import division
#input data
Nm=1000#Speed of the model in rpm
Hm=8#Head of the model in m
Pm=30#Power of the model in kW
Hp=25#Head of the prototype in m
DmDp=1/5#The scale of the model to original

#calculations
Np=((Hp/Hm)**(1/2))*(DmDp)*(Nm)#Speed of the prototype in rpm
Pp=(Pm)*((1/DmDp)**(5))*(Np/Nm)**(3)#Power developed by the prototype in kW
QpQm=((1/DmDp)**(3))*(Np/Nm)#Ratio of the flow rates of two pump(model and prototype)

#output
print '(1)Speed of prototype pump is %3.1f rpm\n(2)Power developed by the prototype pump is %3i kW\n(3)Ratio of the flow rates of two pumps is %3.4f'%(Np,Pp,QpQm)

(1)Speed of prototype pump is 353.6 rpm
(2)Power developed by the prototype pump is 4143 kW
(3)Ratio of the flow rates of two pumps is 44.1942


## Ex 7.6 Page 313¶

In :
#input data
Hp=85#Head of the prototype in m
Qp=(20000/3600)#Flow rate of the prototype in m**3/s
Np=1490#Speed of the prototype in rpm
Dp=1.2#Diameter of the prototype in m
dp=714#Density of the prototype fluid in kg/m**3
Pp=4#Power of the prototype in MW
Pm=500*10**-3#Power of the model in MW
Qm=0.5#Flow rate of the prototype in m**3/s
dm=1000#Density of the model fluid (water) in kg/m**3

#calculations
NpNm=(Qp/Qm)#Ratio of the speeds of the prototype and the model in terms of (Dm/Dp)**(3)
DmDp=1/(((NpNm)**(3))*(dp/dm)*(Pm/Pp))**(1/4)#The ratio of the diameters of model and the prototype or the scale ratio
NmNp=1/(NpNm*((DmDp)**(3)))#The speed ratio or the ratio of speeds of the model and the prototype
HmHp=((1/NmNp)**(2))*((1/DmDp)**(2))#The head ratio or the ratio of heads of the model and the prototype

#output
print '(1)The head ratio of the model is %3.1f\n(2)The speed ratio of the model is %3.1f\n(3)The scale ratio of the model is %3.1f'%(HmHp,NmNp,DmDp)
# Answer in the textbook is wrong

(1)The head ratio of the model is 1.0
(2)The speed ratio of the model is 3.3
(3)The scale ratio of the model is 0.3


## Ex 7.7 Page 315¶

In :
#input data
Np=400#The speed of the prototype in rpm
Qp=1.7#The discharge of the prototype in m**3/s
Hp=36.5#The head of the prototype in m
Pp=720#The power input of the prototype in kW
Hm=9#The head of the model in m
DmDp=1/6#The scale of model to prototype

#calculations
Nm=((Hm/Hp)**(1/2))*(1/DmDp)*Np#Speed of the model in rpm
Qm=((DmDp)**(3))*(Nm/Np)*(Qp)#Discharge of the model in m**3/s
Pm=((DmDp)**(5))*((Nm/Np)**(3))*Pp#Power required by the model in kW

#output
print '(a)Speed of the model is %3.2f rpm\n(b)Discharge of the model is %3.4f m**3/s\n(c)Power required by the model is %3.2f kW'%(Nm,Qm,Pm)

(a)Speed of the model is 1191.75 rpm
(b)Discharge of the model is 0.0234 m**3/s
(c)Power required by the model is 2.45 kW


## Ex 7.8 Page 316¶

In :
#input data
N1=1000#The running speed of the pump-1 in rpm
D1=0.3#The impeller diameter of pump-1 in m
Q1=0.02#The discharge of pump-1 in m**3/s
H1=15#The head developed by the pump-1 in m
N2=1000#The running speed of the pump-2 in rpm
Q2=0.01#The discharge of pump-2 in m**3/s

#calculations
D2=(((Q2/Q1)*(N1/N2))**(1/3))*(D1)#Impeller diameter of the pump-2 in m
H2=(((D2/D1)*(N2/N1))**(2))*(H1)#Head developed by the pump-2 in m

#output
print '(a)Impeller diameter of the pump-2 is %3.3f m\n(b)Head developed by the pump-2 is %3.2f m'%(D2,H2)

(a)Impeller diameter of the pump-2 is 0.238 m
(b)Head developed by the pump-2 is 9.45 m


## Ex 7.9 Page 316¶

In :
#input data
DmDp=1/10#The model ratio to prototype
Pm=1.84#Power developed by the model in kW
Hm=5#Head developed by the model in m
Nm=480#Speed of the model in rpm
Hp=40#Head developed by the prototype in m

#calculations
Np=((Hp/Hm)**(1/2))*(DmDp)*(Nm)#Speed of the prototype in rpm
Pp=((1/DmDp)**(5))*((Np/Nm)**(3))*Pm#Power developed by the prototype in kW
Nsp=((Np*((Pp)**(1/2)))/((Hp)**(5/4)))#Specific speed of the prototype
Nsm=((Nm*((Pm)**(1/2)))/((Hm)**(5/4)))#Specific speed of the prototype

#output
print '(a)Power developed by the prototype is %3i kW\n(b)Speed of the prototype is %3.2f rpm\n(c)Specific speed of the prototype is %3.1f\n(d)Specific speed of the model is %3.1f\n Thus the specific speed of the model is equal to the prototype and thus it is verified'%(Pp,Np,Nsp,Nsm)

(a)Power developed by the prototype is 4163 kW
(b)Speed of the prototype is 135.76 rpm
(c)Specific speed of the prototype is 87.1
(d)Specific speed of the model is 87.1
Thus the specific speed of the model is equal to the prototype and thus it is verified


## Ex 7.10 Page 317¶

In :
from __future__ import division
#input data
DmDp=1/10#The model ratio to prototype
Hm=5#The head developed by the model in m
Hp=8.5#The head developed by the prototype in m
Pp=8000*10**3#The power developed by the prototype in W
Np=120#The speed of running of the prototype in rpm
d=1000#density of the water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2
n0=0.85#Overall efficiency of the prototype

#calculations
Nm=((Hm/Hp)**(1/2))*(1/DmDp)*(Np)#Speed of the mpdel in rpm
Qp=Pp/(d*g*n0*Hp)#Discharge from the prototype in m**3/s
Qm=((DmDp)**(3))*(Nm/Np)*(Qp)#Discharge from the model in m**3/s
Pm=((DmDp)**(5))*((Nm/Np)**(3))*(Pp)*10**-3#Power of the model in kW

#output
print '(a)Speed of the model is %3.1f rpm\n(b)Discharge from the model is %3.3f m**3/s\n(c)Power of the model is %3.1f kW'%(Nm,Qm,Pm)

(a)Speed of the model is 920.4 rpm
(b)Discharge from the model is 0.866 m**3/s
(c)Power of the model is 36.1 kW


## Ex 7.11 Page 318¶

In :
#input data
P1=6600#Initial power developed by the turbine in kW
N1=100#Initial speed of the turbine in rpm
H1=30#Initial head of the turbine in m
H2=18#Final head of the turbine in m

#calculations
N2=N1*((H2/H1)**(1/2))#The final speed of the turbine in rpm
P2=P1*((H2/H1)**(3/2))#The final power developed by the turbine in kW

#output
print '(1)The final speed of the turbine is %3.2f rpm\n(2)The final power developed by the turbine is %3i kW'%(N2,P2)

(1)The final speed of the turbine is 77.46 rpm
(2)The final power developed by the turbine is 3067 kW


## Ex 7.12 Page 319¶

In :
#input data
H1=25#The initial head on the turbine in m
N1=200#The initial speed of the turbine in rpm
Q1=9#The initial discharge of the turbine in m**3/s
n0=0.9#Overall efficiency of the turbine
H2=20#The final head on the turbine in m
d=1000#density of the water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2

#calculations
N2=N1*((H2/H1)**(1/2))#The final speed of the turbine in rpm
Q2=Q1*((H2/H1)**(1/2))#The final discharge of the  turbine in m**3/s
P1=n0*d*g*Q1*H1*10**-3#Power produced by the turbine initially in kW
P2=P1*((H2/H1)**(3/2))#Power produced by the turbine finally in kW

#output
print '(a)The final speed of the turbine is %3.2f rpm\n(b)The final discharge of the  turbine is %3.2f m**3/s\n(c)Power produced by the turbine initially is %3.3f kW\n(d)Power produced by the turbine finally is %3.2f kW'%(N2,Q2,P1,P2)

(a)The final speed of the turbine is 178.89 rpm
(b)The final discharge of the  turbine is 8.05 m**3/s
(c)Power produced by the turbine initially is 1986.525 kW
(d)Power produced by the turbine finally is 1421.44 kW


## Ex 7.13 Page 320¶

In :
#input data
P1=5000*10**3#The initial power produced in W
H1=250#The initial head produced in m
N1=210#The initial speed of turbine in rpm
n0=0.85#Overall efficiency of the turbine
H2=160#The final head produced in m
d=1000#density of the water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2

#calculations
Nu=N1/((H1)**(1/2))#The unit speed of the turbine
Pu=P1/((H1)**(3/2))*10**-3#The unit power of the turbine
Q1=P1/(d*g*n0*H1)#The initial discharge of the turbine in m**3/s
Qu=Q1/((H1)**(1/2))#The unit discharge of the turbine
Q2=Qu*((H2)**(1/2))#The final discharge of the turbine in  m**3/s
N2=Nu*((H2)**(1/2))#The final speed of the turbine in rpm
P2=Pu*((H2)**(3/2))#The final power of the turbine in kW
Ns=(N2*((P2)**(1/2)))/((H2)**(5/4))#The specific speed of the turbine

#output
print '(a)The unit speed of the turbine is %3.2f\n(b)The unit power of the turbine is %3.3f\n(c)The unit discharge of the turbine is %3.3f\n(d)The final discharge of the turbine is %3.2f m**3/s\n(e)The final speed of the turbine is %3.2f rpm\n(f)The final power of the turbine is %3.1f kW\n(g)The specific speed of the turbine is %3.2f'%(Nu,Pu,Qu,Q2,N2,P2,Ns)

(a)The unit speed of the turbine is 13.28
(b)The unit power of the turbine is 1.265
(c)The unit discharge of the turbine is 0.152
(d)The final discharge of the turbine is 1.92 m**3/s
(e)The final speed of the turbine is 168.00 rpm
(f)The final power of the turbine is 2560.0 kW
(g)The specific speed of the turbine is 14.94