# Chapter 10 : Fluid flow in ducts¶

### Example 10.1 - Page No :405¶

In :
import math

# Variables
T = 30.;          			 #[degC] - temperature
d = 8.265*10**-4;  			 #[m] - diameter of the capillary viscometer
deltapbyL = -0.9364;  		 #[psi/ft] - pressure drop per unit length

# Calculations
deltapbyL = deltapbyL*(2.2631*10**4);  			 #[kg/m**2*sec**2] - pressure drop per unit length
Q = 28.36*(10**-6)*(1./60);
p = (0.88412-(0.92248*10**-3)*T)*10**3;  			 #[kg/m**3] - density
s = (math.pi*(d**2))/4.;
U = Q/s;
tauw = (d/4.)*(-deltapbyL);
shearrate = (8*U)/d;
mu = tauw/(shearrate);

# Results
print " The viscosity is  mu = %.3ef kg/m*sec = %.4f cP"%(mu,mu*10**3);
print " Finally, it is important to check the reynolds number to make sure the above equation applies"
Nre = (d*U*p)/(mu);
print " Nre = %d"%Nre
print " The flow is well within the laminar region and therefore the above equation applies";

 The viscosity is  mu = 5.135e-04f kg/m*sec = 0.5135 cP
Finally, it is important to check the reynolds number to make sure the above equation applies
Nre = 1214
The flow is well within the laminar region and therefore the above equation applies


### Example 10.2 - Page No :407¶

In :
# Variables
Nreold = 1214.;
Uold = 0.8810;
Nre = 13700.;
U = Uold*(Nre/Nreold);
Lbyd = 744.;
T = 30.;

# Calculations
# umsing the newton raphson method to calculate the value of f from the equation - 1/(f**(1/2)) = 4*math.log(Nre*(f**(1/2)))-0.4
f = 0.007119;
p = (0.88412-(0.92248*10**-3)*T)*10**3;  			 #[kg/m**3] - density
tauw = (1./2)*p*(U**2)*f;
deltap = tauw*(4.)*(Lbyd);
d = 0.03254/12;  			 #[ft]
L = Lbyd*d;

# Results
print " Pressure drop is  -deltap = %.3e N/m**2 = %.1f kpa = 130 psi"%(deltap,deltap*10**-3);
print " A pressure drop of 130 psi on a tube of length of %.3f ft is high and \
\nshows the impracticality of flows at high reynolds number in smaller tubes"%(L);

# Answer may vary because of rounding error.

 Pressure drop is  -deltap = 8.968e+05 N/m**2 = 896.8 kpa = 130 psi
A pressure drop of 130 psi on a tube of length of 2.017 ft is high and
shows the impracticality of flows at high reynolds number in smaller tubes


### Example 10.3 - Page No :414¶

In :
# Variables
u = 1./60;  			 #[m/sec] - velocity
p = 1000.;  			 #[kg/m**3] - density
mu = 1*10.**-3;  		 #[kg/m*sec] - vismath.cosity
d = 6*10.**-2;  		 #[m] - insid_e diameter of tube
L = 300.;  			     #[m] - length of the tube

# Calculations
Nre = (d*u*p)/(mu);
f = 16./Nre;
deltap = (4.*f)*(L/d)*((p*(u**2))/2.);

# Results
print "Nre = ",Nre,"therefore the flow is laminar"
print "f = " , f
print "Pressure drop -delta P = %.2f N/m**2  =  %.4f kPa  =  %.3e psi"%(deltap,deltap*10**-3,deltap*1.453*10**-4);

Nre =  1000.0 therefore the flow is laminar
f =  0.016
Pressure drop -delta P = 44.44 N/m**2  =  0.0444 kPa  =  6.458e-03 psi


### Example 10.4 - Page No :415¶

In :
from numpy import *

# Variables
# given
d = 6.*10**-2;  			 #[m] - insid_e diameter of tube
p = 1000.;        			 #[kg/m**3] - density
# for smooth pipe
Nre = array([10**4, 10**5]);
f = array([0.0076, 0.0045]);
mu = 10.**-3;     			 #[kg/m**2*s]
U = (Nre*mu)/(d*p);
L = 300.;  			 #[m] - length of the tube

# Calculations
deltap = zeros(2)
for i in range(2):
deltap[i] = (4*f[i])*(L/d)*((p*(U[i]**2))/2.);

# Results
print "for smooth pipe"
print "   Nre      f         -deltap";
print " %6.0f    %6.4f     %6.3f"%(Nre,f,deltap)
print " %6.0f    %6.4f     %6.3f"%(Nre,f,deltap)

# for commercial steel
Nre = array([10**4, 10**5]);
f = array([0.008 ,0.0053]);
U = (Nre*mu)/(d*p);
L = 300.;  			 #[m] - length of the tube
for i in range(2):
deltap[i] = (4*f[i])*(L/d)*((p*(U[i]**2))/2);

print "\nfor commercial steel pipe"
print "   Nre      f         -deltap";
print " %6.0f    %6.4f     %6.3f"%(Nre,f,deltap)
print " %6.0f    %6.4f     %6.3f"%(Nre,f,deltap)

# for cast iron pipe
Nre = array([10**4 ,10**5]);
f = array([0.009 ,0.0073]);
U = (Nre*mu)/(d*p);
L = 300.;  			 #[m] - length of the tube
for i in range(2):
deltap[i] = (4*f[i])*(L/d)*((p*(U[i]**2))/2);

print "\nfor cast iron pipe"
print "   Nre      f         -deltap";
print " %6.0f    %6.4f     %6.3f"%(Nre,f,deltap)
print " %6.0f    %6.4f     %6.3f"%(Nre,f,deltap)

for smooth pipe
Nre      f         -deltap
10000    0.0076     2111.111
100000    0.0045     125000.000

for commercial steel pipe
Nre      f         -deltap
10000    0.0080     2222.222
100000    0.0053     147222.222

for cast iron pipe
Nre      f         -deltap
10000    0.0090     2500.000
100000    0.0073     202777.778


### Example 10.5 - Page No :417¶

In :
# Variables
L = 300.;  			 #[m] - length of pipe
d = 0.06;  			 #[m] - insid_e diameter
deltap = 147.*10**3;  		 #[Pa] - pressure the pump can supply
ebyd = 0.000762;  			 # relative roughness
p = 1000.;  			     #[kg/m**3] - density
mu = 1.*10**-3;  			 #[kg/m*sec] - viscosity
tauw = (d*(deltap))/(4.*L);

# using the hit and trial method for estimation of flow velocity
# Calculations
# let
f = 0.005;
U = ((2*tauw)/(p*f))**(1./2);
Nre = (d*U*p)/mu;

# from the graph value of f at the above calculated reynolds no. and the given relative roughness(e/d)
f = 0.0054;
U = ((2*tauw)/(p*f))**(1./2);
Nre = (d*U*p)/mu;
# from the graph value of f at the above calculated reynolds no. and the given relative roughness(e/d)
f = 0.0053;
U = ((2*tauw)/(p*f))**(1./2);
Nre = (d*U*p)/mu;

# from the graph value of f at the above calculated reynolds no. and the given relative roughness(e/d)
f = 0.0053;
# At this point the value of f is deemed unchanged from the last iteration .Hence, the values obtained after the third iteration are the converged values

# Results
print " The maximum flow velocity is  U = %f m/sec"%(U);

# Answer may vary because of rounding error.

 The maximum flow velocity is  U = 1.665408 m/sec


### Example 10.6 - Page No :419¶

In :
# Variables
L = 300.;  			 #[m] - length of pipe
d = 0.06;  			 #[m] - insid_e diameter
deltap = 147.*10**3;  	 #[Pa] - pressure the pump can supply
ebyd = 0.000762;  		 # relative roughness
p = 1000.;  			 #[kg/m**3] - density

# Calculations
mu = 1*10**-3;  			 #[kg/m*sec] - viscosity
Nvk = ((d*p)/mu)*((d*(deltap))/(2*L*p))**(1./2);

# From the fig  at given von karman no and relative roughness the value of f is-
f = 0.0055;
Nre = Nvk/(f**(1./2))
U = (Nre*mu)/(d*p);

# Results
print "von karman no. %.0f"%Nvk
print " Average velocity = %.2f m/sec"%(U);

von karman no. 7275
Average velocity = 1.63 m/sec


### Example 10.7 - Page No :422¶

In :
# Variables
L = 300.;  			 #[m] - length of pipe
d = 0.06;  			 #[m] - insid_e diameter
p = 1000.;  			 #[kg/m**3] - density
mu = 1.*10**-3;  			 #[kg/m*sec] - viscosity

# Calculations
Nre = array([10.**4, 10.**5]);
U = (Nre*mu)/(d*p);
N = (L/d)/45.;  			 # no of velocity heads

# Results
for i in range(2):
print "Nre = ",Nre[i]
print " -deltap  =  %.3f kPa  =  %.3f psi"%(deltap[i]*10**-3,deltap[i]*1.453*10**-4);

Nre =  10000.0
-deltap  =  1.543 kPa  =  0.224 psi
Nre =  100000.0
-deltap  =  154.321 kPa  =  22.423 psi


### Example 10.8 - Page No :439¶

In :
# Variables
mu = 6.72*10**-4;  		 #[lb/ft*sec] - viscosity
p = 62.4;  			     #[lb/ft**3] - density
S = 0.03322;  			 #[ft**2] - flow area
d = 0.206;  			 #[ft]
e = 1.5*10**-4;  		 # absolute roughness for steel pipe
ebyd = e/d;
Nre = 10.**5;

# friction factor as read from fig in book for the given reynolds no. and relative roughness is-
f = 0.0053;
U = (Nre*mu)/(p*d);
Q = U*S;
gc = 32.174;

# Calculations
# (a) equivalent length method
deltapbyL = f*(4/d)*(p*(U**2))*(1/(2*gc))*(6.93*10**-3);

# using L = Lpipe+Lfittings+Lloss;
Lfittings = 2342.1*d;
kc = 0.50;  			 #  due to contraction loss
ke = 1.;  			 # due to enlargement loss
Lloss = (kc+ke)*(1./(4*f))*d;
Lpipe = 137.;
L = Lpipe+Lfittings+Lloss;
deltap = deltapbyL*L;
patm = 14.696;  			 #[psi] - atmospheric pressure
p1 = patm+deltap;
print " a)The inlet pressure is p1 = %.1f psi"%(p1);

# (b) loss coefficient method
# using the equation deltap/p = -(Fpipe+Ffittings+Floss)
L = 137.;
kfittings = 52.39;
sigmaF = ((4.*f*(L/d))+kc+ke+kfittings)*((U**2)/(2*gc));
deltap = (p*sigmaF)/(144.);
p1 = patm+deltap;

# Results
print " b)The inlet pressure is  p1 = %.1f psi"%(p1);
print " Computation of the pressure drop by the loss coefficient method differs from the equivalent length \
\nmethod by less than 1 psi";

 a)The inlet pressure is p1 = 26.7 psi
b)The inlet pressure is  p1 = 27.2 psi
Computation of the pressure drop by the loss coefficient method differs from the equivalent length
method by less than 1 psi


### Example 10.9 - Page No :443¶

In :
# Variables
L1 = 50.;  			 #[m] - length of first pipe
L2 = 150.;  			 #[m] - length of second pipe
L3 = 100.;  			 #[m] - length of third pipe
d1 = 0.04;  			 #[m] - diameter of first pipe
d2 = 0.06;  			 #[m] - diameter of second pipe
d3 = 0.08;  			 #[m] - diameter of third pipe
deltap = -1.47*10**5;  			 #[kg/m*sec] - pressure drop
mu = 1*10.**-3;  			 #[kg/m*sec] - viscosity
p = 1000.;  			 #[kg/m**3] - density

# Calculation and Results
# for branch 1
S = (math.pi*(d1**2))/4;
Nvk = ((d1*p)/mu)*(-(d1*deltap)/(2*L1*p))**(1./2);
f = (1./(4*math.log10(Nvk)-0.4))**2;
U = (((-deltap)/p)*(d1/L1)*(2./4)*(1./f))**(1./2);
w1 = p*U*S;
print " For first branch w1 = %.2f kg/sec"%(w1);
# for branch 2
S = (math.pi*(d2**2))/4;
Nvk = ((d2*p)/mu)*(-(d2*deltap)/(2*L2*p))**(1./2);
f = (1./(4*math.log10(Nvk)-0.4))**2;
U = (((-deltap)/p)*(d2/L2)*(2./4)*(1./f))**(1./2);
w2 = p*U*S;
print " For second branch w2 = %.2f kg/sec"%(w2);
# for branch 3
S = (math.pi*(d3**2))/4;
Nvk = ((d3*p)/mu)*(-(d3*deltap)/(2*L3*p))**(1./2);
f = (1./(4*math.log10(Nvk)-0.4))**2;
U = (((-deltap)/p)*(d3/L3)*(2./4)*(1./f))**(1./2);
w3 = p*U*S;
print " For third branch w3 = %.2f kg/sec"%(w3);

# total flow rate w = w1+w2+w3
w = w1+w2+w3;
print " total flow rate is w = %.1f kg/sec"%(w);

 For first branch w1 = 4.74 kg/sec
For second branch w2 = 7.59 kg/sec
For third branch w3 = 20.42 kg/sec
total flow rate is w = 32.7 kg/sec


### Example 10.11 - Page No : 447¶

In :
# Variables
sp = 1.1;
p = sp*62.4;  			 #[lb/ft**3] - density
mu = 2*6.72*10**-4;  	 #[lb/ft*sec] - viscosity
Q = 400.;  			     #[gpm] - volumetric flow rate
e = 1.5*10**4;  		 #roughness of steel pipe
gc = 32.174;
kexit = 1.;
kentrance = 0.5;

# Calculations
# 4 in schedule pipe
d = 4.026/12;  			 #[ft]
U4 = Q/39.6;  			 #[ft/sec]
Lgv = 13.08;
Lglv = 114.1;
Le = 40.26;
Lpipe_4 = 22.;
Lfittings_4 = Lgv+Lglv+Le;
Lloss = 0;
L_4 = Lpipe_4+Lfittings_4+Lloss;
Nre_4 = (d*U4*p)/mu;
f = 0.00475;
Fpipe_4 = ((4*f*L_4)/d)*(U4**2)*(1/(2*gc));
Floss_4 = ((kentrance+0)*(U4**2))/(2*gc);

# 5 in schedule pipe
d = 5.047/12;
U5 = Q/62.3;
Lgv = 10.94;
Le = 75.71;
Lpipe_5 = 100.;
Lfittings_5 = Lgv+Le;
Lloss = 0.;
L_5 = Lpipe_5+Lfittings_5+Lloss;
Nre = (d*U5*p)/mu;
f = 0.00470;
Fpipe_5 = ((4*f*L_5)/d)*(U5**2)*(1./(2*gc));
Floss_5 = ((kexit+0)*(U5**2))/(2*gc);

# 6 in schedule pipe
d = 6.065/12;
U6 = Q/90.;
Lgv = 6.570;
Le = 30.36;
Lpipe_6 = 4.;
Lfittings_6 = Lgv+Le;
Lloss = 0.;
L_6 = Lpipe_6+Lfittings_6+Lloss;
Nre = (d*U6*p)/mu;
f = 0.00487;
Fpipe_6 = ((4*f*L_6)/d)*(U6**2)*(1./(2*gc));
kc = 0.50;
Floss_6 = kc*((U6**2)/(2*gc));
Ffittings = 0.;
deltap_6 = p*(Fpipe_6+Ffittings+Floss_6);

# 3/4 in 18 gauge tube
d = 0.652112/12;
L_3by4 = 15.;
U_3by4 = (Q*0.962)/100.;
Floss_3by4 = 100.*(kexit+kentrance)*((U_3by4**2.)/2.);
Nre = d*U_3by4*p*(1./mu);
f = 0.08*((Nre)**(-1./4))+0.012*((d)**(1./2));
deltap_3by4 = ((4*f*p*L_3by4)/d)*((U_3by4**2)/(2*gc));
Fpipe_3by4 = 100.*((4.*f*L_3by4)/d)*((U_3by4**2.)/(2.*gc));
deltap_spraysystem = 25.; 			 #[psi]
Fspraysystem = (deltap_spraysystem/p)*(144.);
delta_p = (p*(kexit+kentrance))*((U_3by4**2.)/(2.*gc))
Fpipe = Fpipe_4+Fpipe_5+Fpipe_6;
Floss = Floss_4+Floss_5+Floss_6+Floss_3by4;
ws = 0. + (((15.**2)-0)/(2*gc))+38.9+382.5;
w = (Q*p)/(7.48);
Ws = (ws*w)/(33000.);
efficiency = 0.6;
Ws_actual = Ws/efficiency

# Results
print " The power supplied to the pump is  %.1f hp"%(Ws_actual);

 The power supplied to the pump is  78.8 hp


### Example 10.12 - Page No :454¶

In :
# Variables
kexit = 1.;
kentrance = 0.5;
Q = 400.;  			 #[gpm] - volumetric flow rate
gc = 32.174;

# for 4 inch pipe
d = 4.026;  		 #[inch]
L = 22.;  			 #[ft]
Lbyd = (L*12)/(d);

# Calculation and Results
# adding the contributions due to fittings
Lbyd = Lbyd+3*13+340+4*30;
N = Lbyd/45.;
N = N+kentrance+0;
U4 = Q/39.6;  			 #[ft/sec]
Fpipe_4 = (N*(U4**2))/(2*gc);
print " F4 in.pipes  =  %.2f ft*lbf/lbm"%(Fpipe_4);

# for 5 inch pipe
L = 100.;  			     #[ft]
d = 5.047;  			 #[inch]
Lbyd = (L*12.)/(d);

# valves contributes 26 diameters and six elbows contribute 30 diameters ecah;therefore
Lbyd = Lbyd+26+6*30;
N = Lbyd/45.;  			 # no. of velocity heads
N = N+kexit+kentrance;
U5 = Q/62.3;
Fpipe_5 = (N*(U5**2))/(2*gc);
print " F5 in.pipes  =  %.2f ft*lbf/lbm"%(Fpipe_5);

# for 6 inch pipe
d = 6.065;  		 #[inch]
L = 5.;  			 #[ft]
Lbyd = (L*12.)/(d);

# adding the contributions due to fittings
Lbyd = Lbyd+1*13+2*30;
N = Lbyd/45;
N = N+0+kentrance;
U6 = Q/90.;
Fpipe_6 = (N*(U6**2))/(2*gc);
print " F6 in.pipes  =  %.3f ft*lbf/lbm"%(Fpipe_6);
F_largepipes = Fpipe_4+Fpipe_5+Fpipe_6;
print " Flarge pipes  =  %.2f ft*lbf/lbm"%(F_largepipes);

 F4 in.pipes  =  20.69 ft*lbf/lbm
F5 in.pipes  =  7.28 ft*lbf/lbm
F6 in.pipes  =  0.719 ft*lbf/lbm
Flarge pipes  =  28.68 ft*lbf/lbm


### Example 10.14 - Page No :459¶

In :
# Variables
l = 0.09238;
rh = 0.1624*l;
L = 300.;
de = 4.*rh;
p = 1000.;  			 #[kg/m**3]
mu = 10.**-3;  			 #[kg/m*sec]
Uavg = 1.667;

# Calculations
Nre = (de*Uavg*p)/mu;
f = 0.0053;
deltap = ((4.*f*L)/de)*(p*(Uavg**2)*(1./2));

# Results
print " Pressure drop -deltap  =  %.3e kg/m*s  =  %.3e N/m**2  =  %.1f kPa"%(deltap,deltap,deltap*10**-3);

 Pressure drop -deltap  =  1.473e+05 kg/m*s  =  1.473e+05 N/m**2  =  147.3 kPa


### Example 10.15 - Page No :466¶

In :
# Variables
Q = 400.;         			 #[gpm]
p = 1.1*62.4;  	    		 #[lbm/ft**3]
mu = 2.*(6.72*10**-4);  	 #[lb/ft*sec]
e = 1.5*10**4;

# Calculations
# 4 inch schedule pipe
d = 0.3355;
S = (math.pi*(d**2))/4;
U4 = Q/39.6;
ebyd = e/d;
w = 3671./60;
pm = 13.45*62.4;
g = 32.1;
gc = 32.174;
deltaz = 2.5;
deltap = (g/gc)*(pm-p)*(deltaz);
betaa = ((1.)/(1.+((2*p*gc)*(deltap))*(((0.61*S)/w)**2)))**(1./4);
d2 = betaa*d;
Nre2 = (4*w)/(math.pi*d2*mu);
a = (1./30)*4.026;
b = (1./4)*(2.013-1.21);
c = (1./8)*(2.42);
if a<b :
if a<c :
opt = a;
else:
opt = c;
else:
if b<c:
opt = b;
else:
opt = c;

# Results
print " The pertinent orifice details are  orifice diameter  =  %.3f in  corner taps, \
\n square edge orifice plate not over %.3f in thick"%(d2*12,opt);

 The pertinent orifice details are  orifice diameter  =  2.425 in  corner taps,
square edge orifice plate not over 0.134 in thick


### Example 10.16 - Page No :470¶

In :
# Variables
Q = 400.;     	    		        #[gpm]
p = 1.1*62.4;     			        #[lbm/ft**3]
mu = 2*(6.72*10**-4);  			    #[lb/ft*sec]
e = 1.5*10**4;

# Calculations
# 4 inch schedule pipe
d = 0.3355;
S = (math.pi*(d**2))/4;
U4 = Q/39.6;
ebyd = e/d;
w = 3671./60;
pm = 13.45*62.4;
g = 32.1;
gc = 32.174;
Nre = (d*U4*p)/mu;
if Nre>10**4:
c = 0.98;

deltaz = 2.5;
deltap = (g/gc)*(pm-p)*(deltaz);
betaa = ((1.)/(1+((2*p*gc)*(deltap))*(((c*S)/w)**2)))**(1./4);
d2 = betaa*d;

# Results
print " The pertinentr details of the venturi design are Throat diameter  =  %.2f inch \
\n Approach angle   =  25 Divergence angle  =  7"%(d2*12);

 The pertinentr details of the venturi design are Throat diameter  =  1.95 inch
Approach angle   =  25 Divergence angle  =  7


### Example 10.17 - Page No :477¶

In :
# Variables
Uzmax = 3.455;  			 #[ft/sec]
m = 32;
a1 = -0.3527;
a2 = -0.6473;
rbyro = 0.880;

# Calculations
UzbyUzmax = 1+a1*(rbyro**2)+a2*(rbyro**(2*m));
Uz = Uzmax*(UzbyUzmax);
Uzavg = (4./9)*Uzmax+(5./18)*(Uz+Uz);

# Results
print " the average velocity is  Uzavg  =  %.2f ft/sec  \
\n Thus, in this Example  there is an  inherent error of 5.5 percent, even before any experimental errors are introduced"%(Uzavg);

 the average velocity is  Uzavg  =  2.93 ft/sec
Thus, in this Example  there is an  inherent error of 5.5 percent, even before any experimental errors are introduced