# Ch-5 : Flow Through Nozzles and Diffusers¶

## Ex - 5.1 : Pg - 214¶

In [1]:
#initialisation of variables
c=300 #velocity in m/s
cp=1.005 #kj/kgk
g=1.4
t=478 #static temparature in k
p=15 #static pressure in bar
#CALCULATIONS
t0=t+((c)**2/(2*cp*1000))
x=(t0/t)**(g/(g-1))*p
#RESULTS
print 'stagnation temparature and stagnation pressure is %0.1f k and %0.2f bar'%(t0,x)
stagnation temparature and stagnation pressure is 522.8 k and 20.52 bar

## Ex - 5.2 : Pg - 215¶

In [2]:
#initialisation of variables
hg=2803.4 #kj/kg
c=300 #m/s
sg=6.1253 #kj/kgk
h2=2090.0 #kj/kg
#CALCULATIONS
h0=hg+((c)**2)/2000
c2=44.72*(h0-h2)**0.5
#RESULTS
print 'total enthalpy is %0.2f kj/kg'%(h0) #textbook answer is wrong
print '\nfinal stream is %0.2f m/s'%(c2) #textbook answer is wrong
total enthalpy is 2848.40 kj/kg

final stream is 1231.55 m/s

## Ex - 5.3 : Pg - 224¶

In [3]:
from __future__ import division
#initialisation of variables
R=0.2897 #kj/kgk
g=1.4
t1=313 #temparature in k
p1=20 #pressure in bar
p2=13 #pressure im bar
cp=1.0138 #kj/kgk
a=5*10**-4
#CALCULATIONS
rc=(2/(g+1))**(g/0.4)
t2=t1*(p2/p1)**((g-1)/g)
c2=44.72*(cp*(t1-t2))**(0.5)
rho=p2*100/(R*t2)
m=rho*c2*a
#RESULTS
print 'mass flow rate and velocity of air at exit are %0.3f kg/s and %0.2f kg/m*m*m'%(m,rho) #textbook answer slightly varies
mass flow rate and velocity of air at exit are 2.198 kg/s and 16.21 kg/m*m*m

## Ex - 5.4 : Pg - 225¶

In [4]:
#initialisation of variables
x=100 #x=h1-h* in kj/kg
m=120 #mass in kg
pi=(22/7)
y=501.5 #y=h1-h2 in kj/kg
v1=0.607 #volume
v2=6.477 #volume
#CALCULATIONS
c1=44.72*(x)**(0.5)
a1=m*v1/(c1*60)
d1=(4*a1/pi)**0.5
c2=44.72*(y)**(0.5)
a2=m*v2/(c2*60)
d2=(4*a2/pi)**0.5
#RESULTS
print 'area of cross section of throat and diameter of throat are %0.6f m*m and %0.4f m'%(a1,d1)
print '\narea of cross section at exit and diameter at exit are %0.5f m*m and %0.4f m'%(a2,d2)
area of cross section of throat and diameter of throat are 0.002715 m*m and 0.0588 m

area of cross section at exit and diameter at exit are 0.01294 m*m and 0.1283 m

## Ex - 5.5 : Pg - 226¶

In [5]:
#initialisation of variables
t1=593 #temparature in k
p2=1.05 #pressure in bar
p1=7 #pressure in bar
cp=1.005
p3=3.696 #pressure in bar
r=0.287 #kj/kgk
a=6.25*10**-4
g= 1.4 #ft/sec**2
R= 8.314
#CALCULATIONS
t2=t1*(p2/p1)**((g-1)/g)
c2=44.72*(cp*(t1-t2))**(0.5)
rho2=p2*100/(r*t2)
m2=rho2*c2*a
t3=t1*(p3/p1)**((g-1)/g)

c3=44.72*(cp*(t1-t3))**(0.5)

rho3=p3*100/(r*t3)
a3=m2/(rho3*c3)
#RESULTS
print 'exit velocity and mass flow rate are %0.f m/s and %0.3f kg/s'%(c2,m2)
print '\nthroat area is %0.4f m*m'%(a3)
exit velocity and mass flow rate are 706 m/s and 0.468 kg/s

throat area is 0.0004 m*m

## Ex - 5.6 : Pg - 229¶

In [6]:
#initialisation of variables
g=1.4 #gamma-const value
p1=4.5 #pressure in bar
p3=1.1 #pressure in bar
cp=1.005 #kj/kgk
rho4=0.5405 #density
rho3=0.9725 #density
t1=1023 #temparature in k
t2=852.16 #temparature in k
r=0.287 #cp-cv=const value
m=0.5 #mass
ieff=0.85 #isentropic efficiency
R= 8.314
#CALCULATIONS
p2=0.528*p1
t2=0.833*t1
c2=44.72*(cp*(t1-t2))**(0.5)
rho2=p2*100/(R*t2)
a2=m/(rho3*c2)
t3=t2*(p3/p2)**((g-1)/g)
t4=t2-(ieff*(t2-t3))
c3=44.72*(cp*(t1-t4))**(0.5)
rho3=p2*100/(R*t4)
a3=m/(rho4*c3)
#RESULTS
print 'throat area is %0.5f m*m'%(a2)
print '\nvelocity at exit,area at exit are %0.2f m/s and %0.6f m*m'%(c3,a3)
throat area is 0.00088 m*m

velocity at exit,area at exit are 794.29 m/s and 0.001165 m*m

## Ex - 5.7 : Pg - 231¶

In [7]:
#initialisation of variables
p1=5 #pressure in bars
h1=2709 #kj/kg
h2=2649.5 #kj/kg
v2=0.6059 #volume flowrate in m*m*m/kg
m=2 #mass  in kg
v3=6.5098 #volume flowrate in m*m*m/kg
h1=2714.0 #kj/kg
h2=2649.5 #kj/kg
h3=2247.4 #kj/kg
eff=0.9 #efficiency
#CALCULATIONS
p2=0.578*p1
c2=44.72*(h1-h2)**(0.5)
a2=m*v2/c2
x=eff*(h1-h3) #x=h1-h3
c3=44.72*(x)**(0.5)
a3=m*v3/c3
#RESULTS
print 'velocity and area at throat are %0.1f m/s and %0.6f m*m'%(c2,a2)
print '\nvelocity and area at exit are %0.2f m/s and %0.5f m*m'%(c3,a3)
velocity and area at throat are 359.2 m/s and 0.003374 m*m

velocity and area at exit are 916.42 m/s and 0.01421 m*m

## Ex - 5.8 : Pg - 233¶

In [8]:
#initialisation of variables
t1=323 #temp in k
c1=300 #velocity in m/s
c2=100 #velocity in m/s
cp=1.005 #kj/kgk
p1=10 #pressure in bar
p3=14 # pressure in bar
g= 1.4#ft/sec**2
#CALCULATIONS
t2=t1+(c1**2-c2**2)/(2*cp*1e3)
p2=p1*(t2/t1)**(g/(g-1))
t2s=t1*(p3/p1)**((g-1)/g)
h3=cp*t2s
x=(0.5*((c1)**2-(c2)**2))/1000 #x=h2-h1
h1=cp*t1
eff=(h3-h1)/(x)
#RESULTS
print 'diffuser efficiency is %0.4f '%(eff)
diffuser efficiency is 0.8189

## Ex - 5.9 : Pg - 232¶

In [9]:
#initialisation of variables
t1=323 #temparature in k
t2=362.8 #temparature in k
c1=300 #velocity in m/s
c2=100 #velocity in m/s
cp=1.005 #kj/kgk
p1=10 #pressure in bar
p3=14 # pressure in bar
g=1.4
#CALCULATIONS
tx=t1+((c1)**2/(2*cp*1000))
po1=p1*(tx/t1)**(g/(g-1))
po2=p3*(tx/t2)**(g/(g-1))
tpr=po2/po1
rrr=(po2-p1)/(po1-p1)
#RESULTS
print 'total pressure ratio and ram recovery ratio are %0.3f  and %0.4f '%(tpr,rrr)
total pressure ratio and ram recovery ratio are 0.932  and 0.8143

## Ex - 5.10 : Pg - 235¶

In [10]:
#initialisation of variables
h1=2724.7 #kj/kg under 3 bar pressure
s1=6.991 #kj/kgk under 3 bar pressure
sf2=1.530 #kj/kgk
sfg2=5.597 #kj/kgk
hf2=504.7 #kj/kg
hfg2=2201.6 #kj/kg
vg2=0.8854
a2=3*10**-4 #area in m*m
v1=0.6056 #m*m*m/kg
p1=3 #bar
p2=2 #bar
n=1.3
t1=406.54 #temparature in k
ps=0.917 #bar
v2=0.8273 #m*m*m/kg
#CALCULATIONS
x2=(s1-sf2)/(sfg2)
h2=hf2+(x2*hfg2)
v2=x2*vg2
c2=44.72*(h1-h2)**(0.5)
m1=a2*c2/v2
v2=v1*(p1/p2)**(1/n)
c3=((-2*n/n-1)*p1*v1*((p2/p1)**((n-1)/n)-1))**0.5*543.53
m2=a2*c3/v2
t2=t1*(p2/p1)**((n-1)/n)
de=2/ps
#RESULTS
print 'mass flow rate is %0.4f '%(m2)
print '\ndegree of super saturation is %0.2f '%(de)
mass flow rate is 0.1375

degree of super saturation is 2.18