Chapter 4:Energy Equation and Its Applications

Example 4.1 Page No95

In [1]:
import math

#variable initialisation

d1=0.15                        #diameter of section A in m

Q=0.05                         #discharge in m^3/s

d2=0.30                        #diameter of section B in m

Z1=100                         #elevation at section A in m

Z2=107                         #elevation at section B in m

p1=30                          #pressure at section A  in kPa

H_L=2                          #energy loss in the pipe

g=9.81

#calculation

A1=(math.pi/4)*(d1*d1)        #calculating area for section A

V1=round(Q/A1,3)              #calculating volume 

A2=(math.pi/4)*(d2*d2)        #calculating area for section B

V2=round(Q/A2,4)              #calculating volume

ga=round(998*g/1000,2)        #unit length of water in kN/m^3

#By,Bernuolli equation

#when the flow is from A to B,taling the pressure as zero

p2=round((((p1/ga)+((V1*V1)/(2*g))+Z1)-(((V2*V2)/(2*g))+Z2+H_L))*ga,2) #calculating pressure

print "(i)p2=",p2,"kPa(guage)"

#when the flow is from B to A,taking pressure as zero

b=(p1/ga)+((V1*V1)/(2*g))+Z1

a=(((V2*V2/(2*g)))+Z2)-H_L

p2=round((b-a)*ga,2)                                                  #calculating pressure 

print "(ii)p2=",p2,"kPa(guage)"
(i)p2= -54.37 kPa(guage)
(ii)p2= -15.21 kPa(guage)

Example 4.3 Page No96

In [6]:
#variable initialisation

p1=0                                           #pressure in kPa

V1=0                                           #Velocity in m/s

p3=0                                           #pressure in kPa

Z3=20                                           #elevation at section 3 in m

H=4                                            #head in the tank in m

d=25                                           #diameter in mm

D=100                                          #Diameter of pipe in mm

g=9.81                                         #gravity 

RD_oil=0.8                                     #relative density in kg/m^3

#Calculation

#By continuity ctiterion,V2(math.pi/4)(D**2)=V3(math.pi/4)(d**2)=Q

#Applying Bernoullis equation to points 1 and 2 with centre line of pipe as datum and pressure as zero

#V2=(d/D)**2*V3

#Using this equation,(p1/ga)+(V1**2)/(2*g)+Z1)=(p3/ga)+((V3**2)/(2*g))+Z3+H_L

#H=V3**2/2g+20*V2**2/2g

#H_L=20*V2**2/(2*g)                            #given loss of pipe

V3=round(((2*g*H)/(Z3*((d/D)**4)+1))**(1/2),4)#Velocity in the pipe

Q=(math.pi/4)*((d/1000)**2)*(V3)*1000              #Discharge

print "Q=",round(Q,4),"L/s"

V2=round((d/D)**2*V3,4)

H_L=round(Z3*V2**2/(2*g),4)                   #Loss of head in the pipe

#Applying Bernoulli equation to points 1 and 2,

#(p1/ga)+(V1**2)/(2*g)+Z1)=(p3/ga)+((V3**2)/(2*g))+Z3+H_L

ga=round(RD_oil*998*(g/1000),3)

p2=round((H-(((V2**2)/(2*g))+p3+H_L)),4)       #Pressure at the base of the nozzle

print "p2=",round(p2*ga,3),"kPa"
Q= 4.1881 L/s
p2= 28.945 kPa

Example 4.4 Page No97

In [7]:
import math

#variable initialisation

d=0.30                         #diameter in m

d1=0.15                        #diameter in m

S=13.6                         #relative density of mercury 

g=9.81                         #gravity 

Q=0.120                        #discharge in L/s

#Calculation

#by continuity creiterion,

V1=round(Q/((math.pi/4)*(d*d)),4)

V2=round(Q/((math.pi/4)*(d1*d1)),4)

#Considering the elevation of section 1 as datum,

#by assumption ((p1-p2)/ga) is taken as a

#a=(S-1)*h

#By,Bernoulli equation for points 1 and 2

#a-0.8=((v2^2-V1^2)/2g)

h=((V2**2-V1**2)/(2*g))/(S-1)

print "h=",round(h*100,01),"cm"   #calculating magnitude

print "Deflection:The manometer limb  connected to section 1 will be having smaller column of mercury than the other limb"
h= 17.5 cm
Deflection:The manometer limb  connected to section 1 will be having smaller column of mercury than the other limb

Example 4.7 Page No99

In [8]:
import math

#variable initialisation

p_A=4                    #vapour prssure in kPa(abs)

p1=95.48                 #atmospheric pressure in kPa

p2=95.48                 #atmospheric pressure in kPa

g=9.81                   #gravity

Z1=1.5                   #elevation at section 1 in m

Z_A=0                    #elevation at section A in m

ga=9.79                  #unit length of water in kN/m^3

V1=0                     #volume of section 1

#Calculation

#by applying Bernoullis equation to points 1 and A

#formula is,(p1/ga)+(V1**2)/(2*g)+Z1)=(pA/ga)+((V_A**2)/(2g))+Z_A

V_A=round(math.sqrt((((p1/ga)+(V1**2)/(2*g)+Z1)-(p_A/ga)+Z_A)*2*g),2)

V2=round(V_A/2,2)

print "V2=",V2,"m/s"

#by applying Bernoullis equation to points 1 and 2,with datum at point 2,

#(p1/ga)+(V1**2)/(2*g)+Z1)=(p2/ga)+((V_A**2)/(2g))+Z_A

c=(p1/ga)+((V1**2)/(2*g))+Z1

e=(p2/ga)+(V2**2)/(2*g)

L=e-c                    #Calculating length of pipe

print"L=",round(L,2),"m"
V2= 7.29 m/s
L= 1.21 m

Example 4.8 Page No99

In [9]:
import math

#variable initialisation

Q=0.015                           #discharge in m^3/s

d=0.05                            #diameter in m

V1=0                              #volume in m/s

Z1=0                              #elevation at section 1 in m

Z2=3.00                           #elevation at section 2 in m

H_L=1.5                           #Energy loss due to friction in m

p2=0 #pressure

ga=9.79                           #unit length of water in kN/m^3

g=9.81                            #gravity

#Calculation

V2=(Q*4)/(math.pi*(d**2))

#By applying Bernoullis equation to points 1 and 2,

#(p1/ga)+(V1**2)/(2*g)+Z1)=(p2/ga)+((V2**2)/(2g))+Z2

p1=(((p2/ga)+((V2**2)/(2*g))+Z2)+H_L-((V1**2)/(2*g)+Z1))*ga

print "p1=",round(p1,2),"kPa"     #Calculating air pressure
p1= 73.18 kPa

Example 4.16 Page No104

In [12]:
import math

from sympy import integrate,symbols,ln

#Variable Initialisation

x, y,z,c,r,r1,r2=symbols("x y z c r r1 r2")

b=0.4                                    #breadth in m

V0=2                                     #volume in m/s

r2=1.2                                   #radius in m

r1=0.8                                   #radius in m

g=9.81                                   #gravity

#Calculation

q=integrate(c/r, r)                      #Integrating v

s=q.subs(r,r2/r1)

C=(b*V0)/ln(r2/r1)                       #Finding C

V1=C/r1

print "V1=",round(V1,3),"m/s"            #Velocity at the inner wall

V2=C/r2

print "V2=",round(V2,3),"m/s"            #velocity at the outer wall

#Since the flow is irrotational,Bernoullis theorem can be applied across streamlines.

#(p1/ga)+(V1**2)/(2*g)+Z1)=(pb/ga)+((V2**2)/(2*g))+Z2

PD=(V1**2-V2**2)/(2*g)

print "The difference in pressure head between outer and inner wall is",round(PD,3),"m"
V1= 2.466 m/s
V2= 1.644 m/s
The difference in pressure head between outer and inner wall is 0.172 m

Example 4.19 Page No105

In [13]:
import math

#variable initialisation

Q=0.8                              #discharge in m^3/s

r=0.45                             #radius in m

Z1=40                              #elevation at section 1 in m

Z2=0                               #elevation at section 2 in m

P1=0                               #power at section 1 in m

V1=0                               #volume at section 1 in m

ga=9.79                            #unit length of water in kN/m^3

g=9.81                             #gravity

H_L=10                             #frictional loss

eta=0.85                           #turbine efficiency 

#Calculation

#(i)Water friction loss is neglected:

V2=round(Q/((math.pi/4)*(r**2)),2)

#Applying Bernoullis equation to points 1 and 2

#(p1/ga)+(V1**2)/(2*g)+Z1)=(pb/ga)+((V2**2)/(2*g))+Z2

H_T=round(((P1/ga)+(V1**2)/(2*g)+Z1)-(((V2**2)/(2*g))+Z2),2)     #Calculating heat extracted by the turbine

print"P=",round(ga*Q*H_T,1),"kW"                                 #power extracted by the turbine

print "(ii)when loses are included:"

#Applying Bernoullis equation to points 1 and 2

H_T=round(((P1/ga)+(V1**2)/(2*g)+Z1)-(((V2**2)/(2*g))+H_L+Z2),2)

Pn=ga*Q*H_T*eta                                                  #power output of the turbine

print "Pn=",round(Pn,1),"kW"
P= 303.2 kW
(ii)when loses are included:
Pn= 191.1 kW

Example 4.17 Page No105

In [14]:
import math

#variable initialisation

Q=0.060                           #discharge in m^3/s

a=60                              #pumping water rate from tank in L/s

r1=0.15                           #radius in m

r2=0.10                           #radius in m

ga=9.79                           #unit length of water in kN/m^3

P=10                              #power delivered by the pump in kW

P_c=0                             #power delivered by section b in m

V_c=0                             #volume in m

Z_c=3                             #elevation at section c in m

Z_b=3                             #elevation at section b in m

H_L=1.2                           #heat delivered in m

g=9.81                            #gravity

#Calculation

V_a=round(Q/((math.pi/4)*(r1**2)),3)

V_b=round(Q/((math.pi/4)*(r2**2)),3)

#by applying power delivered formula,P=ga*Q*H_p

H_p=round(P/(Q*ga),2)                                                                      #calculating heat delivered by the pump

#By applying Bernoullis equation between C and A 

P_a=round((((P_c/ga)+((V_c**2)/(2*g))+Z_c)-((V_a**2)/(2*g)))*ga,4)

print "P_a=",round(P_a,3),"kPa"                                                            #Calculating pressure at A

#By applying Bernoullis equation between C and B with level at A as datum,

#(pc/ga)+(Vc**2)/(2*g)+Zc)=(pb/ga)+((Vb**2)/(2g))+Zb

P_b=round((((P_c/ga)+((V_c**2)/(2*g))+Z_c+H_p)-(((V_b**2)/(2*g))+Z_b+H_L))*ga-0.01,2)      #Calculating pressure at B

print "P_b=",P_b,"kPa"

print "when loses are considered:"

s=((P_c/ga)+((V_c**2)/(2*g))+Z_c+H_p)-(((V_b**2)/(2*g))+(Z_b+H_L)+(2*((V_b**2)/(2*g))))

print "P_b=",round((s*ga),1),"kPa"                                                         #Calculating pressure at B
P_a= 23.619 kPa
P_b= 125.75 kPa
when loses are considered:
P_b= 67.5 kPa

Example 4.20 Page No106

In [15]:
import math

#variable initialisation

Q=0.5                     #discharge in m^3/s

r=0.4                     #radius in m

r1=0.6                    #radius in m

P_a=30.0                  #pressure head at the upstream of turbine in m

P_b=-4                    #pressure head at the point B in draft tube in m

Z_a=2                     #elevation at section A in m

Z_b=0                     #elevation at section B in m

g=9.81                    #gravity

eta=0.90                  #turbine efficiency 

ga=9.79                   #unit length of water in kN/m^3

#Calculation

V_a=round(Q/((math.pi/4)*(r**2)),2)

V_b=round(Q/((math.pi/4)*(r1**2)),3)

#Applying Bernoullis equation to points A and B

#(P_a/ga)+(V_a**2)/(2*g)+Z_a)=(P_b/ga)+((V_b**2)/(2*g))+Z_b+H_t

H_T=round(((P_a)+(V_a**2)/(2*g)+Z_a)-((P_b)+((V_b**2)/(2*g))+Z_b)#calculating heat extracted
,2)

P=round(ga*Q*H_T*eta,2)                                          #Calculating power

print "Answer given in the book is wrong.It is,"

print "P=",P,"kW"
Answer given in the book is wrong.It is,
P= 161.46 kW

Example 4.23 Page No107

In [16]:
import math

#variable initialisation

Q=0.20                               #Discharge in m^3/s

r=0.25                               #radius at point 1 in m

r1=0.35                              #radius at point 2 in m

P1=120                               #pressure head at the upstream of turbine in m

ga=9.79                              #unit length of water in kN/m^3

alpha_1=1.1                          #kinetic energy correction factors for section 1

Z1=25                                #elevation at section 1 in m

alpha_2=1.5                          #kinetic energy correction factors for section 2

Z2=20                                #elevation at section 2 in m

g=9.81                               #gravity

#Calculation

V1=round(Q/((math.pi/4)*(r**2)),3)

V2=round((Q/((math.pi/4)*(r1**2))),3)

#Applying Bernoullis equation to points 1 and 2

#(P1/ga)+(V1**2)/(2*g)+Z1)=(P2/ga)+((V2**2)/(2*g))+Z2+H_L

#given H_L is,

H_L=round((1.2*(V1-V2)**2)/(2*g),3)          #calculating frictional loss

P2=round((((P1/ga)+alpha_1*((V1**2)/(2*g))+Z1)-(alpha_2*((V2**2)/(2*g))+Z2+H_L)),3)

print "P2=",round(P2*ga,2),"kPa"
P2= 172.45 kPa