In [1]:

```
#pg 589
print('Example 17.1');
# aim : To determine
# the indicated and brake output and the mechanicl efficiency
# draw up an overall energy balance and as % age
import math
# given values
h = 21;# height of indicator diagram, [mm]
ic = 27;# indicator calibration, [kN/m**2 per mm]
sv = 14*10**-3;# swept volume of the cylinder;,[m**3]
N = 6.6;# speed of engine, [rev/s]
ebl = 77;# effective brake load, [kg]
ebr = .7;# effective brake radious, [m]
fc = .002;# fuel consumption, [kg/s]
CV = 44000;# calorific value of fuel, [kJ/kg]
cwc = .15;# cooling water circulation, [kg/s]
Ti = 38;# cooling water inlet temperature, [C]
To = 71;# cooling water outlet temperature, [C]
c = 4.18;# specific heat capacity of water, [kJ/kg]
eeg = 33.6;# energy to exhaust gases, [kJ/s]
g = 9.81;# gravitational acceleration, [m/s**2]
# solution
PM = ic*h;# mean effective pressure, [kN/m**2]
LA = sv;# swept volume of the cylinder, [m**3]
ip = PM*LA*N/2;# indicated power,[kW]
T = ebl*g*ebr;# torque, [N*m]
bp = 2*math.pi*N*T;# brake power, [W]
n_mech = bp/ip;# mechanical efficiency
print ' The Indicated power is (kW) = ',round(ip,2)
print ' The Brake power is (kW) = ',round(bp*10**-3)
print ' The mechanical efficiency is (percent) = ',round(n_mech)
ef = CV*fc;# energy from fuel, [kJ/s]
eb = bp*10**-3;# energy to brake power,[kJ/s]
ec = cwc*c*(To-Ti);# energy to coolant,[kJ/s]
es = ef-(eb+ec+eeg);# energy to surrounding,[kJ/s]
print('Energy can be tabulated as :-');
print('----------------------------------------------------------------------------------------------------');
print(' kJ/s Percentage ')
print('----------------------------------------------------------------------------------------------------');
print ' Energy from fuel ',ef,' ',ef/ef*100,'\n Energy to brake power ',round(eb),' ',round(eb/ef*100),'\n Energy to coolant ',round(ec,1),' ',round(ec/ef*100,1),' \n Energy to exhaust ',eeg,' ',round(eeg/ef*100,1),'\n Energy to suroundings,etc. ',round(es,1),' ',round(es/ef*100,1)
# End
```

In [2]:

```
#pg 591
print('Example 17.2');
import math
# aim : To determine
# (a) bp
# (b) ip
# (c) mechanical efficiency
# (d) indicated thermal efficiency
# (e) brake specific steam consumption
# (f) draw up complete energy account for the test one-minute basis taking 0 C as datum
# given values
d = 200.*10**-3;# cylinder diameter, [mm]
L = 250.*10**-3;# stroke, [mm]
N = 5.;# speed, [rev/s]
r = .75/2;# effective radious of brake wheel, [m]
Ps = 800.;# stop valve pressure, [kN/m**2]
x = .97;# dryness fraction of steam
BL = 136.;# brake load, [kg]
SL = 90.;# spring balance load, [N]
PM = 232.;# mean effective pressure, [kN/m**2]
Pc = 10.;# condenser pressure, [kN/m**2]
m_dot = 3.36;# steam consumption, [kg/min]
CC = 113.;# condenser cooling water, [kg/min]
Tr = 11.;# temperature rise of condenser cooling water, [K]
Tc = 38.;# condensate temperature, [C]
C = 4.18;# heat capacity of water, [kJ/kg K]
g = 9.81;# gravitational acceleration, [m/s**2]
# solution
# from steam table
# at 800 kN/m**2
tf1 = 170.4;# saturation temperature, [C]
hf1 = 720.9;# [kJ/kg]
hfg1 = 2046.5;# [kJ/kg]
hg1 = 2767.5;# [kJ/kg]
vg1 = .2403;# [m**3/kg]
# at 10 kN/m**2
tf2 = 45.8;# saturation temperature, [C]
hf2 = 191.8;# [kJ/kg]
hfg2 = 2392.9;# [kJ/kg]
hg2 = 2584.8;# [kJ/kg]
vg2 = 14.67;# [m**3/kg]
# (a)
T = (BL*g-SL)*r;# torque, [Nm]
bp = 2*math.pi*N*T*10**-3;# brake power,[W]
print ' (a) The brake power is (kW) = ',round(bp,3)
# (b)
A = math.pi*d**2/4;# area, [m**2]
ip = PM*L*A*N*2;# double-acting so*2, [kW]
print ' (b) The indicated power is (kW) = ',round(ip,1)
# (c)
n_mec = bp/ip;# mechanical efficiency
print ' (c) The mechanical efficiency is (percent) = ',round(n_mec*100,1)
# (d)
h = hf1+x*hfg1;# [kJ/kg]
hf = hf2;
ITE = ip/((m_dot/60)*(h-hf));# indicated thermal efficiency
print ' (d) The indicated thermal efficiency is (percent) = ',round(ITE*100,2)
# (e)
Bsc=m_dot*60/bp;# brake specific steam consumption, [kg/kWh]
print ' (e) The brake steam consumption is (kg/kWh) = ',round(Bsc,2)
# (f)
# energy balanvce reckoned from 0 C
Es = m_dot*h;# energy supplied, [kJ]
Eb = bp*60;# energy to bp, [kJ]
Ecc = CC*C*Tr;# energy to condensate cooling water, [kJ]
Ec = m_dot*C*Tc;# energy to condensate, [kJ]
Ese = Es-Eb-Ecc-Ec;# energy to surrounding,etc, [kJ]
print ' (f) Energy supplied/min is (kJ) = ',round(Es)
print ' Energy to bp/min is (kJ) = ',round(Eb)
print ' Energy to condenser cooling water/min is (kJ) = ',round(Ecc)
print ' Energy to condensate/min is (kJ) = ',round(Ec)
print ' Energy to surrounding, etc/min is (kJ) = ',round(Ese)
print 'answer in the book is misprinted for Es'
# End
```

In [3]:

```
#pg 593
print('Example 17.3');
# aim : To determine
# (a) the brake power
# (b) the brake specific fuel consumption
# (c) the indicated thermal efficiency
# (d) the energy balance, expressing the various items
import math
# given values
t = 30.;# duration of trial, [min]
N = 1750.;# speed of engine, [rev/min]
T = 330.;# brake torque, [Nm]
mf = 9.35;# fuel consumption, [kg]
CV = 42300.;# calorific value of fuel, [kJ/kg]
cwc = 483.;# jacket cooling water circulation, [kg]
Ti = 17.;# inlet temperature, [C]
To = 77.;# outlet temperature, [C]
ma = 182.;# air consumption, [kg]
Te = 486.;# exhaust temperature, [C]
Ta = 17.;# atmospheric temperature, [C]
n_mec = .83;# mechanical efficiency
c = 1.25;# mean specific heat capacity of exhaust gas, [kJ/kg K]
C = 4.18;# specific heat capacity, [kJ/kg K]
# solution
# (a)
bp = 2*math.pi*N*T/60*10**-3;# brake power, [kW]
print ' (a) The Brake power is (kW) = ',round(bp,1)
# (b)
bsf = mf*2/bp;#brake specific fuel consumption, [kg/kWh]
print ' (b) The brake specific fuel consumption is (kg/kWh) = ',round(bsf,3)
# (c)
ip = bp/n_mec;# indicated power, [kW]
ITE = ip/(2*mf*CV/3600);# indicated thermal efficiency
print ' (c) The indicated thermal efficiency is (percent) = ',round(ITE*100,1)
# (d)
# taking basis one minute
ef = CV*mf/30;# energy from fuel, [kJ]
eb = bp*60;# energy to brake power,[kJ]
ec = cwc/30*C*(To-Ti);# energy to cooling water,[kJ]
ee = (ma+mf)/30*c*(Te-Ta);# energy to exhaust, [kJ]
es = ef-(eb+ec+ee);# energy to surrounding,etc,[kJ]
print ' (d) Energy from fuel is (kJ) = ',round(ef)
print ' Energy to brake power is (kJ) = ',round(eb)
print ' Energy to cooling water is (kJ) = ',round(ec)
print ' Energy to exhaust is (kJ) = ',round(ee)
print ' Energy to surrounding, etc is (kJ) = ',round(es)
print 'The answer is a bit different due to rounding off error in textbook'
# End
```

In [4]:

```
#pg 594
print('Example 17.4');
# aim : To determine
# (a) the indicated power of the engine
# (b) the mechanical efficiency of the engine
# given values
bp = 52;# brake power output, [kW]
bp1 = 40.5;# brake power of cylinder cut1, [kW]
bp2 = 40.2;# brake power of cylinder cut2, [kW]
bp3 = 40.1;# brake power of cylinder cut3, [kW]
bp4 = 40.6;# brake power of cylinder cut4, [kW]
bp5 = 40.7;# brake power of cylinder cut5, [kW]
bp6 = 40.0;# brake power of cylinder cut6, [kW]
# sollution
ip1 = bp-bp1;# indicated power of cylinder cut1, [kW]
ip2 = bp-bp2;# indicated power of cylinder cut2, [kW]
ip3 = bp-bp3;# indicated power of cylinder cut3, [kW]
ip4 = bp-bp4;# indicated power of cylinder cut4, [kW]
ip5 = bp-bp5;# indicated power of cylinder cut5, [kW]
ip6 = bp-bp6;# indicated power of cylinder cut6, [kW]
ip = ip1+ip2+ip3+ip4+ip5+ip6;# indicated power of engine,[kW]
print ' (a) The indicated power of the engine is (kW) = ',ip
# (b)
n_mec = bp/ip;# mechanical efficiency
print ' (b) The mechanical efficiency of the engine is (percent) = ',round(n_mec*100,1)
# End
```

In [5]:

```
#pg 595
print('Example 17.5');
# aim : To determine
# the brake power,indicated power and mechanicl efficiency
# draw up an energy balance and as % age of the energy supplied
# given values
N = 50.;# speed, [rev/s]
BL = 267.;# break load.,[N]
BL1 = 178.;# break load of cylinder cut1, [N]
BL2 = 187.;# break load of cylinder cut2, [N]
BL3 = 182.;# break load of cylinder cut3, [N]
BL4 = 182.;# break load of cylinder cut4, [N]
FC = .568/130;# fuel consumption, [L/s]
s = .72;# specific gravity of fuel
CV = 43000;# calorific value of fuel, [kJ/kg]
Te = 760;# exhaust temperature, [C]
c = 1.015;# specific heat capacity of exhaust gas, [kJ/kg K]
Ti = 18;# cooling water inlet temperature, [C]
To = 56;# cooling water outlet temperature, [C]
mw = .28;# cooling water flow rate, [kg/s]
Ta = 21;# ambient tempearture, [C]
C = 4.18;# specific heat capacity of cooling water, [kJ/kg K]
# solution
bp = BL*N/455;# brake power of engine, [kW]
bp1 = BL1*N/455;# brake power of cylinder cut1, [kW]
i1 = bp-bp1;# indicated power of cylinder cut1, [kW]
bp2 = BL2*N/455;# brake power of cylinder cut2, [kW]
i2 = bp-bp2;# indicated power of cylinder cut2, [kW]
bp3 = BL3*N/455;# brake power of cylinder cut3, [kW]
i3 = bp-bp3;# indicated power of cylinder cut3, [kW]
bp4 = BL4*N/455;# brake power of cylinder cut4, [kW]
i4 = bp-bp4;# indicated power of cylinder cut4, [kW]
ip = i1+i2+i3+i4;# indicated power of engine, [kW]
n_mec = bp/ip;# mechanical efficiency
print ' The Brake power is (kW) = ',round(bp,1)
print ' The Indicated power is (kW) = ',round(ip,1)
print ' The mechanical efficiency is (percent) = ',round(n_mec*100,1)
mf = FC*s;# mass of fuel/s, [kg]
ef = CV*mf;# energy from fuel/s, [kJ]
me = 15*mf;# mass of exhaust/s,[kg],(given in condition)
ee = me*c*(Te-Ta);# energy to exhaust/s,[kJ]
ec = mw*C*(To-Ti);# energy to cooling water/s,[kJ]
es = ef-(ee+ec+bp);# energy to surrounding,etc/s,[kJ]
print('Energy can be tabulated as :-');
print('----------------------------------------------------------------------------------------------------');
print(' kJ/s Percentage ')
print('----------------------------------------------------------------------------------------------------');
print ' Energy from fuel ',round(ef,1),' ',ef/ef*100,'\n Energy to brake power ',round(bp,1),' ',round(bp/ef*100.,1),'\n Energy to exhaust ',round(ee,1),' ',round(ee/ef*100),'\n Energy to coolant ',round(ec,1),' ',round(ec/ef*100,1),'\n Energy to suroundings,etc. ',round(es,1),' ',round(es/ef*100,1)
print 'there is minor variation in the result reported in the book due to rounding off error'
# End
```

In [6]:

```
#pg 596
print('Example 17.6');
# aim : To determine
# (a) the break power of engine
# (b) the fuel consumption of the engine
# (c) the brake thermal efficiency of the engine
import math
# given values
d = 850*10**-3;# bore , [m]
L = 2200*10**-3;# stroke, [m]
PMb = 15;# BMEP of cylinder, [bar]
N = 95./60;# speed of engine, [rev/s]
sfc = .2;# specific fuel oil consumption, [kg/kWh]
CV = 43000;# calorific value of the fuel oil, [kJ/kg]
# solution
# (a)
A = math.pi*d**2/4;# area, [m**2]
bp = PMb*L*A*N*8/10;# brake power,[MW]
print ' (a) The brake power is (MW) = ',round(bp,3)
# (b)
FC = bp*sfc;# fuel consumption, [kg/h]
print ' (b) The fuel consumption is (tonne/h) = ',round(FC,2)
# (c)
mf = FC/3600;# fuel used, [kg/s]
n_the = bp/(mf*CV);# brake thermal efficiency
print ' (c) The brake thermal efficiency is (percent) = ',round(n_the*100)
# End
```