# Chapter 17 : Prime Movers¶

## Example 17.1 Page no 201¶

In :
from __future__ import division
print "Example 17.1 page no 201\n\n"
#fan are operating for transporting gas
#two fans fan(a)and fan(b)
D_a=46#diameter of blade of fan (a)
rpm_a=1575#operating speed of fan(a)
D_b=42#diameter of blade of fan(b)
rpm_b=1625#operating speed of fan(b)
h_p_a=47.5#power requirement of fan (a)
h_p_b=(rpm_b**3/rpm_a**3)*(D_b/D_a)**5*h_p_a#power requirement of fan(b)
print "\n power requirement h_p_b=%.2f bhp"%(h_p_b)#

Example 17.1 page no 201

power requirement h_p_b=33.10 bhp


## Example 17.2 Page no 201¶

In :
from __future__ import division
print "Example 17.2 page no 201\n\n"
rpm=1694#speed of fan
q=12200#flow rate of q_a
rpm_n=2100#new speed of fan
q_n=q*(rpm_n/rpm)#new flow rate
print "\nnew flow rate q_n=%.2f acfm"%(q_n)#
#applyingeq 17.5
P=5#pressure ,in
P_n=P*(rpm_n**2/rpm**2)#new pressure
print "\nnew pressureP_n=%.2f in H20"%(P_n)#
#required power is calculated using eq. 17.6
hp=9.25#power at 1694 speed
hp_n=hp*(rpm_n**3/rpm**3)#new power required
print "\n new powerhp_n=%.2f bhp"%(hp_n)#

Example 17.2 page no 201

new flow rate q_n=15123.97 acfm

new pressureP_n=7.68 in H20

new powerhp_n=17.62 bhp


## Example 17.3 Page no. 201¶

In :
from __future__ import division
print "\Example 17.3 page no 201\n\n"
# a gas stream in a process
P_l_m=4.4# minor pressure loss for duct work,valves etc,in
P_l_mz=6.4#major pressure loss due to pieces of equipment,in
P_drop=P_l_m+P_l_mz#total pressure drop
print "\n total pressure P_drop=%.2f in H20"%(P_drop)#
#applying eq 17.7
q=6500#flow rate ,acfm
neta=0.63#overall fan-motor effficiency
bhp=1.575e-4*q*P_drop/neta#brake horse power required
#1.575e-5 is aconversion factor for horse power
print "\n brake horse power bhp=%.2f bhp"%(bhp)#

\Example 17.3 page no 201

total pressure P_drop=10.80 in H20

brake horse power bhp=17.55 bhp


## Example 17.4 Page no 208¶

In :
from __future__ import division
print " Example 17.4 page no 208\n\n"
#a pump is in process
#given: parabolic pump pressure flow
#P=a-b*q**2  equation
#a and  b calculate from conditions
a=25
b=5
#then equation becomes P=25-5*q**2
#pressure at 1m**3/s  flow rate
q=1#flow rate,m**3/s
P=a-b*q**2#pressure
print "\n pressure P=%.2f kpa"%(P)#

 Example 17.4 page no 208

pressure P=20.00 kpa


## Example 17.6 Page no 214¶

In :
from __future__ import division
print "\n Example 17.6 page no. 214\n\n"
#the total head developed by a centrifugal pump is given by a equation
#hc=42-0.0047*q**2
#the pump is to be used in a water flow system in which the pump head in feet of water is given by eq.
#hp=12+0.0198*q**2
#for cal. flow rate hc=hp
q=35#from condition hc=hp,gpm
hc=42-0.0047*q**2#total head
print "\n total head hc=%.2f ft of water"%(hc)#
rho=62.40#density
q_c=0.078#flow rate in cfs unit
m_dot=rho*q_c#mass flow rate
print "\n m_dot mass flow rate =%.2f lb/s"%(m_dot)#
W_dot=m_dot*hc#fluid power requirement can be calculated
print "\n fluid power requirement W_dot=%.2f lbf.ft/s"%(W_dot)#
neta=.6#efficiency
W_dot_hp=.32#fluid power requirement in hp
bhp=W_dot_hp/neta#brake horse power
print "\n brake horse power bhp=%.2f bhp"%(bhp)#

 Example 17.6 page no. 214

total head hc=36.24 ft of water

m_dot mass flow rate =4.87 lb/s

fluid power requirement W_dot=176.40 lbf.ft/s

brake horse power bhp=0.53 bhp


## Example 17.8 Page no 216¶

In :
from __future__ import division
print " Example 17.8 page no 216\n\n"
#compressed air is to be employed in the nozzle
T1=520#temperature
P2=40#pressure
P1=14.7#atmosphric pressure
gamma=1.3#degree of freedom
R=1.987#gas constant
W_s=-(gamma*R*T1/(gamma-1))*((P2/P1)**((gamma-1)/gamma)-1)#compreesd energy requirement
print "\n energy requirement W_s=%.2f btu/lbmol of air"%(W_s)#
hp=W_s*(7.5/29)*778#power
print "\n power hp=%.2e ft .lbf/min"%(hp)#

 Example 17.8 page no 216

energy requirement W_s=-1163.53 btu/lbmol of air

power hp=-2.34e+05 ft .lbf/min