# Chapter 23 : Testing of Engines¶

## Example 23.1 Page no : 340¶

In [1]:
import math

#Input data
n = 4.					#Four cylinder engine
N = 1200.					#Speed in r.p.m
BHP1 = 26.3					#Brake horse power in B.H.P
T = 11.3					#Average torque in kg
CV = 10000.					#Calorific value of the fuel in kcal/kg
m = 270.					#Flow rate in gm of petrol per B.H.P hour

#Calculations
BHP2 = (T*2*3.14*N)/4500					#Average B.H.P on 3 cylinders
IHP = BHP1-BHP2					#Average I.H.P of one cylinder
TIHP = (n*IHP)					#Total I.H.P
p = ((m/1000)*BHP1)/TIHP					#Petrol used in kg/I.H.P hr
nth = ((4500*60)/(427*p*CV))*100					#Indicated Thermal efficiency in percent

#Output
print 'Thermal efficiency is %3.1f percent'%(nth)

Thermal efficiency is 26.3 percent


## Example 23.2 Page no : 344¶

In [2]:
import math
n = 4					#Four cylinder engine
d = 0.1					#Diameter of piston in m
l = 0.15					#Stroke in m
RPM = 1600					#Speed in r.p.m
ap = (5.76*10**-4)					#Area of positive loop of indicator diagram in sq.m
an = (0.26*10**-4)					#Area of negative loop of indicator diagram in sq.m
L = 0.055					#Length of the indicator diagram in m
k = (3.5/10**-6)					#Spring constant in kg/m**2 per m

#Calculations
NA = (ap-an)					#Net area of the indicator diagram in sq.m
h = (NA/L)					#Average height of diagram in m
Pm = (h*k)					#Mean effective pressure in kg/m**2
IHP = (Pm*l*(3.14/4)*d**2*RPM*n)/4500					#Indicated Horse Power

#Output
print 'Indicated horse power of a four cylinder two stroke petrol engine is %3.1f'%(IHP)

Indicated horse power of a four cylinder two stroke petrol engine is 58.6


## Example 23.3 Page no : 348¶

In [3]:
import math

#Input data
n = 6					#Number of cylinders
d = 0.089					#Bore in m
l = 0.1016					#Stroke in m
vc = 3.183					#Compression ratio
rn = 55					#Relative efficiency in percent
m = 0.218					#Petrol consumption in kg/hp.hr
Pm = (8.4/10**-4)					#Indicated mean effective pressure in kg/m**2
N = 2500					#Speed in r.p.m

#Calculations
an = (1-(1/(vc-1)))*100					#Air standard efficiency in percent
nth = (rn*an)/100					#Thermal efficiency in percent
CV = ((4500*60)/(m*(nth/100)*427))					#Calorific value in kcal/kg
IHP = ((Pm*(3.14/4)*d**2*l*N*n)/(4500*2))					#Indicated horse power
p = (m*IHP)					#Petrol consumption in kg/hour

#Output
print '1) The calorific value of petrol is %i kcal/kg  \
\n2) Corresponding petrol consumption is %3.1f kg/hour'%(CV,p)

1) The calorific value of petrol is 9731 kcal/kg
2) Corresponding petrol consumption is 19.3 kg/hour


## Example 23.4 Page no : 349¶

In [5]:
import math

#Input data
n = 4.					#Number of cylinders
d = 0.2					#Bore in m
l = 0.3					#Stroke in m
N = 300.					#Speed in r.p.m
af = 5.					#Air to fuel ratio by volume. In textbook it is given as 4 which is wrong
nv = 78.					#Volumetric efficiency in percent
CV = 2200.					#Calorific value in kcal/cu.m at N.T.P
bth = 23.					#Brake thermal efficiency in percent

#Calculations
Vs = ((3.14/4)*d**2*l)					#Swept volume in cu.m
c = ((nv/100)*Vs)					#Total charge per stroke in cu.m
Vg = ((c/af)*N)					#Volume of gas used per min in cu.m at N.T.P
q = (CV*Vg)					#Heat supplied in kcal/min
BHP = ((bth/100)*q)/(4500./427)					#Brake horse power

#Output
print 'The volume of gas used per min is %3.3f cu.m at N.T.P  \
\nB.H.P of engine is %3.1f'%(Vg,BHP)

The volume of gas used per min is 0.441 cu.m at N.T.P
B.H.P of engine is 21.2


## Example 23.5 Page no : 353¶

In [7]:
import math

#Input data
d = 20.					#Bore in cm
l = 37.5					#Stroke in cm
r = 6.					#Compression ratio
IPm = 5.					#Indicated Mean effective pressure in kg/cm**2
ag = 6.					#Air to gas ratio
CV = 3070.					#Calorific value of gas in kcal/cu.m
T = 75.+273					#Temperature in K
p = 0.975					#Pressure in kg/cm**2
RPM = 240.					#Speed in r.p.m
g = 1.4					#Ratio of specific heats

#Calculations
Vs = (3.14/4)*d**2*l					#Stroke Volume in cu.m
Vg = (1/(r+1))*Vs					#Volume of gas in cylinder in cu.m per cycle
x = (Vg*(p/1.03)*(273/T))					#Volume at 'Vg' cu.m at 'p' kg/cm**2 and 'T' K are equivalent in cu.m
q = (CV*x)/10**6					#Heat added in kcal per cycle
IHP = (IPm*(Vs/100)*(RPM/2))/4500					#Indicated horse power
nth = ((IHP*4500)/(427*q*(RPM/2)))*100					#Thermal efficiency in percent
na = (1-(1/r**(g-1)))*100					#Air standard efficiency in percent
rn = (nth/na)*100					#Relative effeciency in percent

#Output
print 'The thermal efficiency is %3.1f percent  \
\nThe relative efficiency is %3.1f percent  \
\nIndicated horese power is %3.1f H.P'%(nth,rn,IHP)

The thermal efficiency is 36.0 percent
The relative efficiency is 70.3 percent
Indicated horese power is 15.7 H.P


## Example 23.6 Page no : 356¶

In [8]:
import math

#Input data
n = 4					#Number of cylinders
d = 6.25					#Diametre in cm
l = 9.5					#Stroke in cm
t = 678					#Torque in kg.m
N = 3000					#Speed in r.p.m
Vc = 60					#Clearance volume in c.c
be = 0.5					#Brake efficiency ratio based on the air standard cycle
CV = 10000					#Calorific value in kcal/kg
g = 1.4					#Ratio of specific heats

#Calculations
Vs = (3.14/4)*d**2*l					#Stroke volume in c.c per cylinder
r = ((Vs+Vc)/Vc)					#Compression ratio
na = (1-(1/r**(g-1)))					#Air standard efficiency
bth = (be*na)*100					#Brake thermal efficiency in percent
bhp = ((t/100)*2*3.14*N)/4500					#B.H.P in H.P
q = (bhp*(4500/427))/(bth/100)					#Heat supplied in kcal/min
F = (q*60)/CV					#Fuel consumption in kg/hour
P = (bhp*4500*2*100)/(n*Vs*N)					#pressure in kg/cm**2

#Output
print 'The fuel consumption is %3.2f kg/hour  \
\nThe brake mean effective pressure is %3.2f kg/cm**2'%(F,P)

The fuel consumption is 5.95 kg/hour
The brake mean effective pressure is 6.47 kg/cm**2


## Example 23.7 Page no : 360¶

In [11]:
import math

#Input data
n = 1.					#Number of cylinders
t = 30.					#Trail time in min
m = 5.6					#Oil consumption in l
CV = 9980.					#Calorific value of oil in kcal/kg
g = 0.8					#Specific gravity of oil
a = 8.35					#Average area of indicator diagram in sq.cm
l = 8.4					#Length of the indicator diagram in cm
is1 = 5.5					#Indicator spring scale
L = 147.5					#Brake load in kg
sp = 20.					#Spring balance reading in kg
d = 1.5					#Effective brake wheel diameter in m
N = 200.					#Speed in r.p.m
cyd = 30.					#Cylinder diameter in cm
l1 = 45.					#Stroke in cm
mw = 11.					#Jacket cooling water in kg/min
Tc = 35.+273					#Temperature rise of cooling water in K

#Calculations
mp = (a/l)*is1					#Mean effective pressure
ihp = ((mp*(l1/100)*(3.14/4)*cyd**2*(N/2))/4500)					#Indicated horse power in h.p
bhp = (L*3.14*d*N)/4500					#Brake horse power in h.p
nm = (bhp/ihp)*100					#Mechanical efficiency in percent
F = (m*(60/t)*g)					#Fuel consumption in kg/hour
Fc = (F/bhp)					#Specific fuel consumption in kg/B.H.P/hour
ith = ((ihp*(4500./427))/((F/60)*CV))*100					#Indicated thermal efficiency in percent

#Output
print 'a) I.H.P is %3.1f  \
\nb) B.H.P is %3.1f  \
\nc) Mechanical efficiency is %3.1f percent  \
\nd) Specific fuel consumption is %3.2f kg/B.H.P/hour  \
\ne) Indicated thermal efficiency is %3.1f percent'%(ihp,bhp,nm,Fc,ith)

a) I.H.P is 38.6
b) B.H.P is 30.9
c) Mechanical efficiency is 79.9 percent
d) Specific fuel consumption is 0.29 kg/B.H.P/hour
e) Indicated thermal efficiency is 27.3 percent


## Example 23.8 Page no : 361¶

In [12]:
import math

#Input data
d = 15.					#Diameter in cm. In textbook it is given wrong as 39
l = 45.					#Stroke in cm
f = 9.5					#Total fuel used in litres
CV = 10500.					#Calorific value in kcal/kg
n = 12624.					#Total no. of revolutions
imep = 7.24					#Gross i.m.e.p in kg/cm**2
pimep = 0.34					#Pumping i.m.e.p in kg/cm**2
L = 150.					#Net load on brake in kg
db = 1.78					#Diameter of the brake wheel drum in m
dr = 4.					#Diameter of rope in cm
cw = 545.					#Cooling water circulated in litres
Tc = 45.					#Cooling water temperature rise in degree C
g = 0.8					#Specific gravity of oil

#Calculations
ihp = ((imep-pimep)*(l/100)*3.14*d**2*n)/(4500*60)					#I.H.P in h.p
q = (f*g*CV)/60					#Heat supplied in kcal/min
bhp = (L*3.14*(db+(dr/100))*n)/(4500*60)					#B.H.P in h.p
qbhp = (bhp*4500)/427					#Heat equivalent of B.H.P in kcal/min
qw = (cw*Tc)/60					#Heat lost to jacket cooling water in kcal/min
dq = (q-(qbhp+qw))					#Heat unaccounted in kcal/min

#Output
print 'Heat supplied is %3.0f kcal/min  \
\nHeat equivalent of B.H.P is %3.0f kcal/min  \
\nHeat lost to jacket cooling water is %3.0f kcal/min  \
\nHeat unaccounted is %3.0f kcal/min'%(q,qbhp,qw,dq)

Heat supplied is 1330 kcal/min
Heat equivalent of B.H.P is 422 kcal/min
Heat lost to jacket cooling water is 409 kcal/min
Heat unaccounted is 499 kcal/min


## Example 23.9 Page no : 364¶

In [14]:
import math

#Input data
d = 27.					#Diameter in cm
l = 45.					#Stroke in cm
db = 1.62					#Effective diameter of the brake in m
t = (38.*60+30)					#Test duration in sec
CV = 4650.					#Calorific value in kcal/m**3 at N.T.P
n = 8080.					#Total no. of revolutions
en = 3230.					#Total number of explosions
p = 5.75					#Mean effective pressure in kg/cm**2
V = 7.7					#Gas used in m**3
T = 15.+273					#Atmospheric temperature in K
pg = 135.					#pressure of gas in mm of water above atmospheric pressure
hb = 750.					#Height of barometer in mm of Hg
L = 92.					#Net load on brake in kg
w = 183.					#Weigh of jacket cooling water in kg
Tc = 47.					#Cooling water temperature rise in degree C

#Calculations
ihp = (p*(l/100)*(3.14/4)*d**2*en)/(4500*(t/60))					#I.H.P in h.p
bhp = (L*3.14*db*n)/(4500*(t/60))					#B.H.P in h.p
pa = (hb+(pg/13))					#Pressure of gas supplied in mm of Hg
Vg = (V*(273/T)*(pa/760))					#Volume of gas used at N.T.P in m**3
q = (Vg*CV)/(t/60)					#Heat supplied per minute in kcal
qbhp = (bhp*4500)/427					#Heat equivalent of B.H.P in kcal/min
qc = (w/(t/60))*Tc					#Heat lost to jacket cooling water in kcal/min
qra = (q-(qbhp+qc))					#Heat lost to exhaust, etc in kcal/min

#Output
print 'Heat supplied is %3.1f kcal/min  \
\nHeat equivalent of B.H.P is %3.0f kcal/min  \
\nHeat lost to jacket cooling water is %3.1f kcal/min  \
\nHeat lost to exhaust radiation etc. is %3.1f kcal/min'%(q,qbhp,qc,qra)

Heat supplied is 882.0 kcal/min
Heat equivalent of B.H.P is 230 kcal/min
Heat lost to jacket cooling water is 223.4 kcal/min
Heat lost to exhaust radiation etc. is 428.6 kcal/min


## Example 23.10 Page no : 366¶

In [15]:
import math

#Input data
d = 25.					#Bore in cm
l = 50.					#Stroke in cm
N = 240.					#Speed in r.p.m
n = 100.					#Number of times fires per minute
qc = 0.3					#Quantity of coal gas used in cu.m per minute
h = 100.					#Head in mm of water
bp = 1.03					#Barometric pressure in kg/cm**2
T = 15.+273					#Temperature in K
ma = 2.82					#Mass of air used in kg per minute
R = 29.45					#Characteristic gas constant in kg.m/kg.K

#Calculations
gp = (bp+(100/13.6)*(bp/76))					#Gas pressure in kg/cm**2
Vc = (qc*(gp/bp)*(273/T))					#Volume of coal gas at N.T.P in cu.m per minute
Vce = (Vc/n)					#Volume of coal gas per explosion in cu.m at N.T.P
va = (ma*R*273)/(bp*10**4)					#Volume of air taken in at N.T.P in cu.m per min
V = ((va-(((N/2)-n)*Vce))/(N/2))					#Volume in cu.m
tV = (V+Vce)					#Total volume of charge in cu.m at N.T.P
Vs = ((3.14/4)*(d**2*l)*10**-6)					#Swept volume in cu.m
nv = (tV/Vs)*100					#Volumetric efficiency in percent

#Output
print 'a) the charge of air per working cycle as measured at N.T.P is %3.5f cu.m  \
\nb) the volumetric efficiency is %3.1f percent'%(tV,nv)

a) the charge of air per working cycle as measured at N.T.P is 0.02094 cu.m
b) the volumetric efficiency is 85.4 percent


## Example 23.11 Page no : 368¶

In [16]:
import math

#Input data
d = 18.					#Diameter in cm
l = 24.					#Stroke in cm
t = 30.					#Duration of trail in min
r = 9000.					#Total number of revolutins
e = 4445.					#Total number of explosions
mep = 5.85					#Mean effective pressure in kg/cm**2
Nl = 40.					#Net load on brake wheel in kg
ed = 1.					#Effective diameter of brake wheel in meter
tg = 2.3					#Total gas used at N.T.P in m**3
CV = 4600.					#Calorific value of gas in kcal/m**3 at N.T.P
ta = 36.					#Total air used in m**3
pa = 720.					#Pressure of air in mm of Hg
Ta = 18.+273					#Temperature of air in K
da = 1.293					#Density of air at N.T.P in kg/m**3
Te = 350.+273					#Temperature of exhaust gases in K
Tr = 18.+273					#Room temperature in K
Cp = 0.24					#Specific heat of exhaust gases in kJ/kg.K
twc = 81.5					#Total weight of cylinder jacket cooling water in kg
dT = 33.					#Rise in temperature of jacket cooling water in degree C
R = 29.45					#Characteristic gas constant in kg.m/kg.degree C

#Calculations
ihp = (mep*(l/100)*(3.14/4)*d**2*(e/t))/4500					#Indicated horse power in h.p
bhp = (Nl*3.14*r*ed)/(4500*t)					#Brake horse power in h.p
qs = (tg/t)*CV					#Heat supplied at N.T.P in kcal
qbhp = (bhp*4500)/427					#Heat equivalent of B.H.P in kcal/min
ql = (twc/t)*dT					#Heat lost to cylinder jacket cooling water in kcal/min
VA = (ta*(273/Ta)*(pa/760))					#Volume of air used at N.T.P in m**3
WA = (VA*da)/t					#Weight of air used per min in kg
WG = (1.03*tg*10**4)/(R*273)					#Weight of gas in kg
Wg = (WG/t)					#Weight of gas per minute in kg
We = (WA+Wg)					#Total weight of exhaust gases in kg
qle = (We*(Te-Tr)*Cp)					#Heat lost of exhaust gases in kcal/min
qra = (qs-(qbhp+ql+qle))					#Heat lost by radiation in kcal/min
nm = (bhp/ihp)*100					#Mechanical efficiency in percent
ith = ((ihp*4500)/(427*qs))*100					#Indicated thermal efficiency in percent

#Output
print '         HEAT BALANCE SHEET          ------------------  \
\nHeat supplied per minute is %3.1f kcal/min  \
\nHeat expenditure               kcal per minute  \
\n1.Heat equivalent of B.H.P is          %3.1f  \
\n2.Heat lost to jacket cooling water is %3.1f  \
\n3.Heat lost in exhaust gases is        %3.1f  \
\n4.Heat lost by radiation, etc, is     %3.1f \
\n--------                                       %3.1f                                      --------'%(qs,qbhp,ql,qle,qra,qs)

         HEAT BALANCE SHEET          ------------------
Heat supplied per minute is 352.7 kcal/min
Heat expenditure               kcal per minute
1.Heat equivalent of B.H.P is          88.2
2.Heat lost to jacket cooling water is 89.7
3.Heat lost in exhaust gases is        117.7
4.Heat lost by radiation, etc, is     57.1
--------                                       352.7                                      --------


## Example 23.12 Page no : 372¶

In [17]:
import math

#Input data
gu = 0.135					#Gas used in m**3/min at N.T.P
CV = 3990					#Calorific value of gas in kcal/m**3 at N.T.P
dg = 0.64					#Density of gas in kg/m**3 at N.T.P
au = 1.52					#Air used in kg/min
C = 0.24					#Specific heat of exhaust gases in kJ/kg.K
Te = 397+273					#Temperature of exhaust gases in K
Tr = 17+273					#Room temperature in K
cw = 6					#Cooling water per minute in kg
rT = 27.5					#Rise in temperature in degree C
ihp = 12.3					#Indicated horse power in h.p
bhp = 10.2					#Brake horse power in h.p

#Calculations
qs = (gu*CV*60)					#Heat supplied in kcal/hour
qbhp = ((bhp*4500*60)/427)					#Heat equivalent of B.H.P in kcal/hr
ql = (cw*60*rT)					#Heat lost in jacket cooling water in kcal/hr
mg = (gu*dg)					#Mass of gas used per minute in kg
me = (mg+au)					#Mass of exhaust gases per minute in kg
qe = (me*C*(Te-Tr)*60)					#Heat carried away by exhaust gases in kcal/hour
qun = (qs-(qbhp+ql+qe))					#Heat unaccounted in kcal/hour

#Output
print 'Heat supplied is %3.0f kcal/hour  \
\nHeat equivalent of B.H.P is %3.0f kcal/hr  \
\nHeat lost in jacket cooling water is %3.0f kcal/hr  \
\nHeat carried away by exhaust gases is %3.0f kcal/hour  \
\nHeat unaccounted is %3.0f kcal/hour'%(qs,qbhp,ql,qe,qun)

Heat supplied is 32319 kcal/hour
Heat equivalent of B.H.P is 6450 kcal/hr
Heat lost in jacket cooling water is 9900 kcal/hr
Heat carried away by exhaust gases is 8790 kcal/hour
Heat unaccounted is 7179 kcal/hour


## Example 23.13 Page no : 372¶

In [20]:
import math

#Input data
n = 4.					#Number of cylinders
r = 1.					#Radius in metre
N = 1400.					#Speed in r.p.m
bl = 14.5					#Net brake load in kg
P = [9.8,10.1,10.3,10]					#Loads on the brake in kg
d = 9.					#Bore in cm
l = 12.					#Stroke in cm

#Calculations
bhp = (bl*2*3.14*r*N)/4500					#Brake horse power in h.p
bhp1 = (P[0]*2*3.14*r*N)/4500					#Brake horse power in h.p
bhp2 = (P[1]*2*3.14*r*N)/4500					#Brake horse power in h.p
bhp3 = (P[2]*2*3.14*r*N)/4500					#Brake horse power in h.p
bhp4 = (P[3]*2*3.14*r*N)/4500					#Brake horse power in h.p
ihp1 = bhp-bhp1					#Indicated horse power in h.p
ihp2 = bhp-bhp2					#Indicated horse power in h.p
ihp3 = bhp-bhp3					#Indicated horse power in h.p
ihp4 = bhp-bhp4					#Indicated horse power in h.p
ihp = (ihp1+ihp2+ihp3+ihp4)					#Indicated horse power in h.p
nm = (bhp/ihp)*100					#Mechanical efficiency in percent
pm = ((4500*bhp)/((l/100)*(3.14/4)*d**2*(N/2)))					#Brake mean effective pressure in kg/cm**2

#Output
print 'I.H.P is %3.1f h.p  \
\nMechanical efficiency is %3.1f percent  \
\nBrake mean effective pressure is %3.0f kg/cm**2'%(ihp,nm,pm)

I.H.P is 34.8 h.p
Mechanical efficiency is 81.5 percent
Brake mean effective pressure is  24 kg/cm**2


## Example 23.14 Page no : 374¶

In [23]:
import math

#Input data
N = 350					#Speed in r.p.m
L = 60					#Net brake load in kg
mep = 2.75					#Mean effective pressure in kg/cm**2
oc = 4.25					#Oil consumption in kg/hour
jcw = 490					#Jacket cooling water in kg/hour
Tw = [20+273,45+273]					#Temperature of jacket water at inlet and outlet in K
au = 31.5					#Air used per kg of oil in kg
Ta = 20+273					#Temperature of air in the test room in K
Te = 390+273					#Temperature of exhaust gases in K
d = 22					#Cylinder diameter in cm
l = 28					#Stroke in cm
bd = 1					#Effective brake diameter in m
CV = 10500					#Calorific value of oil in kcal/kg
pH2 = 15					#Proportion of hydrogen in fuel oil in percent
C = 0.24					#Mean specific heat of dry exhaust gases
Cs = 9.5					#Specific heat of steam in kJ/kg.K

#Calculations
ibp = (mep*(l/100)*(3.14/4)*d**2*N)/4500					#Indicated brake power in h.p
bhp = (L*3.14*N*bd)/4500					#Brake horse power in h.p
qs = (oc*CV)/60					#Heat supplied per minute in kcal
qbhp = (bhp*4500)/427					#Heat equivalent of B.H.P in kcal/min
pqbhp = (qbhp/qs)*100					#Percenatge of  heat
ql = (jcw/60)*(Tw[1]-Tw[0])					#Heat lost to cooling water in kcal/min
pql = (ql/qs)*100					#Percenatge of  heat
wH2O = (9*(pH2/100)*(oc/60))					#Weight of H2O produced per kg of fuel burnt in kg/min
twe = (oc*(au+1))/60					#Total weight of wet exhaust gases per minute in kg
twd = (twe-wH2O)					#Weight of dry exhaust gases per minute in kg
qle = (twd*C*(Te-Ta))					#Heat lost to dry exhaust gases/min in kcal
pqle = (qle/qs)*100					#Percenatge of  heat
qx = (100+538.9+0.5*(Te-373))					#Heat in kcal/kg
qst = (wH2O*qx)					#Heat to steam in kcal/min
pqst = (qst/qs)*100					#Percenatge of  heat
qra = (qs-(qbhp+ql+qle+qst))					#Heat lost by radiation in kcal/min
pqra = (qra/qs)*100					#Percenatge of  heat

#Output
print '         HEAT BALANCE SHEET          ------------------  \
\nHeat supplied per minute is        %3.0f kcal/min  100 percent  \
\nHeat expenditure                  kcal per minute      percent  \
\n1.Heat equivalent of B.H.P is              %3.1f      %3.1f  \
\n2.Heat lost to cooling water is            %3.0f        %3.1f  \
\n3.Heat lost to dry exhaust gases is        %3.1f      %3.1f \
\n4.Heat lost of steam in exhaust gases is  %3.0f         %3.1f \
\n5.Heat lost by radiation, etc., is         %3.0f        %3.1f \
\n ---------- Total                         %3.0f        %3.0f  ------------------'\
%(qs,qbhp,pqbhp,ql,pql,qle,pqle,qst,pqst,qra,pqra,qs,(pqbhp+pql+pqle+pqst+pqra))

         HEAT BALANCE SHEET          ------------------
Heat supplied per minute is        744 kcal/min  100 percent
Heat expenditure                  kcal per minute      percent
1.Heat equivalent of B.H.P is              154.4      20.8
2.Heat lost to cooling water is            200        26.9
3.Heat lost to dry exhaust gases is        204.4      27.5
4.Heat lost of steam in exhaust gases is    0         0.0
5.Heat lost by radiation, etc., is         185        24.9
---------- Total                         744        100  ------------------


## Example 23.15 Page no : 376¶

In [24]:
import math

#Input data
d = 20					#Diameter in cm
l = 40					#Stroke in cm
mep = 5.95					#Mean effective pressure in kg/cm**2
bt = 41.5					#Brake torque in kg.m
N = 250					#Speed in r.p.m
oc = 4.2					#Oil consumption in kg per hour
CV = 11300					#Calorific value of fuel in kcal/kg
jcw = 4.5					#Jacket cooling water in kg/min
rT = 45					#Rise in temperature in degree C
au = 31					#Air used in kg
Te = 420					#Temperature of exhaust gases in degree C
Tr = 20					#Room temperature in degree C
Cm = 0.24					#Mean specific heat of exhaust gases in kJ/kg.K

#Calculations
ihp = (mep*(l/100)*(3.14/4)*d**2*(N/2))/4500					#Indicated horse power in h.p
bhp = (bt*2*3.14*N)/4500					#Brake horse power in h.p
q = (oc*CV)					#Heat supplied in kcal/hour
qbhp = (bhp*4500*60)/427					#Heat equivalent of B.H.P in kcal/hour
qfhp = ((ihp-bhp)*4500*60)/427					#Heat equivalent F.H.P in kcal/hour
qc = (jcw*rT*60)					#Heat lost in cooling water in kcal/hour
qe = (oc*32*Cm*(Te-Tr))					#Heat lost in exhaust gases in kcal/hour
hu = (q-(qbhp+qfhp+qc+qe))					#Heat unaccounted in kcal/hour

#Output
print 'Indicated horse power is %3.1f h.p  \
\nBrake horse power is %3.2f h.p  \
\nHeat supplied is %3.0f kcal/hour  \
\nHeat equivalent of B.H.P is %3.0f kcal/hour  \
\nHeat equivalent of F.H.P is %3.0f kcal/hour  \
\nHeat lost in cooling water is %3.0f kcal/hour  \
\nHeat lost in exhaust gases is %3.0f kcal/hour  \
\nHeat unaccounted is %3.0f kcal/hour'%(ihp,bhp,q,qbhp,qfhp,qc,qe,hu)

Indicated horse power is 0.0 h.p
Brake horse power is 14.48 h.p
Heat supplied is 47460 kcal/hour
Heat equivalent of B.H.P is 9155 kcal/hour
Heat equivalent of F.H.P is -9155 kcal/hour
Heat lost in cooling water is 12150 kcal/hour
Heat lost in exhaust gases is 12902 kcal/hour
Heat unaccounted is 22408 kcal/hour


## Example 23.16 Page no : 381¶

In [26]:
import math

#Input data
ihp = 45.					#Indicated horse power in h.p
bhp = 37.					#Brake horse power in h.p
fu = 8.4					#Fuel used in kg/hour
CV = 10000.					#Calorific value in kcal/kg
Tc = [15.,70.]					#Inlet and outlet temperatures of cylinders in degree C
cj = 7.					#Rate of flow of cylinder jacket in kg/min
Tw = [15.,55.]					#Inlet and outlet temperatures of water in degree C
rw = 12.5					#Rate of water flow in kg per minute
Te = 82.					#Final temperature of exhaust gases in degree C
Tr = 17.					#Room temperature in degree C
af = 20.					#Air fuel ratio
Cm = 0.24					#Mean specific heat of exhaust gases in kJ/kg.K

#Calculations
q = (fu/60)*CV					#Heat supplied in kcal/min
qbhp = (bhp*4500)/427					#Heat equivalent of B.H.P in kcal/min
ql = (cj*(Tc[1]-Tc[0]))					#Heat lost to cylinder jacket cooling water in kcal/min
qe = (rw*(Tw[1]-Tw[0]))					#Heat lost by exhaust gases in kcal/min
qee = (Te-Tr)*Cm*(af+1)*fu/60					#Heat of exhaust gas in kcal/min
te = (qe+qee)					#Total heat lost to exhaust gases in kcal/min
hra = (q-(qbhp+ql+te))					#Heat lost to radiation in kcal/min
ith = ((ihp*4500)/(427*q))*100					#Indicated thermal efficiency in percent
bth = ((bhp*4500)/(427*q))*100					#Brake thermal efficiency in percent
nm = (bhp/ihp)*100					#Mechanical efficiency in percent

#Output
print 'Heat supplied is %3.0f kcal/min  \
\nHeat equivalent of B.H.P is %3.0f kcal/min  \
\nHeat lost to cylinder jacket cooling water is %3.0f kcal/min  \
\nTotal heat lost to exhaust gases is %3.1f kcal/min  \
\nHeat lost to radiation is %3.1f kcal/min  \
\nIndicated thermal efficiency is %3.1f percent  \
\nBrake thermal efficiency is %3.1f percent  \
\nMechanical efficiency is %3.1f percent'%(q,qbhp,ql,te,hra,ith,bth,nm)

Heat supplied is 1400 kcal/min
Heat equivalent of B.H.P is 390 kcal/min
Heat lost to cylinder jacket cooling water is 385 kcal/min
Total heat lost to exhaust gases is 545.9 kcal/min
Heat lost to radiation is 79.2 kcal/min
Indicated thermal efficiency is 33.9 percent
Brake thermal efficiency is 27.9 percent
Mechanical efficiency is 82.2 percent


## Example 23.17 Page no : 382¶

In [28]:
import math

#Input data
Vs = 0.0015					#Stroke volume in cu.m
rc = 5.5					#Volume compression ratio
p2 = 8.					#Pressure at the end of compression stroke in kg/cm**2
T2 = 350.+273					#Temperature at the end of compression stroke in K
p3 = 25.					#Pressure in kg/cm**2
x = (1./30)					#Fraction of dismath.tance travelled by piston
pa = 1./16					#Petrol air mixture ratio
R = 29.45					#Characteristic gas constant in kg.m/kg degree C
CV = 10000.					#Calorific value of fuel in kcal per kg
Cv = 0.23					#Specific heat in kJ/kg.K

#Calculations
V2 = (Vs*10**6)/(rc-1)					#Volume in c.c
V3 = (Vs*10**6)*x+V2					#Volume in c.c
T3 = (T2*p3*V3)/(p2*V2)					#Temperature in K
W = ((p3+p2)/2)*(V3-V2)					#Workdone in kg.cm
mM = ((p2*V2)/(T2*R*100))					#Mass of mixture present in kg
dE = (mM*Cv*(T3-T2))					#Change in energy in kcal
q = (dE+(W/(427*100)))					#Heat in kcal
qc = (1/(1+(1/pa)))*mM*CV 					#Heat in kcal
ql = (qc-q)/mM					#Heat lost in kcal per kg of charge

#Output
print 'Heat lost per kg of charge during explosion is %3.0f kcal'%(ql)

Heat lost per kg of charge during explosion is 203 kcal