# Chapter 17 - Engine and plant trails¶

## Example 1: pg 589¶

In :
#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

Example 17.1
The Indicated power is (kW) =  26.2
The Brake power is (kW) =  22.0
The mechanical efficiency is (percent) =  837.0
Energy can be tabulated as :-
----------------------------------------------------------------------------------------------------
kJ/s                           Percentage
----------------------------------------------------------------------------------------------------
Energy from fuel                          88.0                     100.0
Energy to brake power                 22.0                      25.0
Energy to coolant                        20.7                     23.5
Energy to exhaust                       33.6                      38.2
Energy to suroundings,etc.          11.8                      13.4


## Example 2: pg 591¶

In :
#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

Example 17.2
(a) The brake power is (kW) =  14.657
(b) The indicated power is (kW) =  18.2
(c) The mechanical efficiency is (percent) =  80.4
(d) The indicated thermal efficiency is (percent) =  12.94
(e) The brake steam consumption is (kg/kWh) =  13.75
(f) Energy supplied/min is (kJ) =  9092.0
Energy to bp/min is (kJ) =  879.0
Energy to condenser cooling water/min is (kJ) =  5196.0
Energy to condensate/min is (kJ) =  534.0
Energy to surrounding, etc/min is (kJ) =  2483.0
answer in the book is misprinted for Es


## Example 3: pg 593¶

In :
#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

Example 17.3
(a) The Brake power is (kW) =  60.5
(b) The brake specific fuel consumption is (kg/kWh) =  0.309
(c) The indicated thermal efficiency is (percent) =  33.2
(d) Energy from fuel is (kJ) =  13184.0
Energy to brake power is  (kJ) =  3629.0
Energy to cooling water is (kJ) =  4038.0
Energy to exhaust is (kJ) =  3739.0
Energy to surrounding, etc is (kJ) =  1778.0
The answer is a bit different due to rounding off error in textbook


## Example 4: pg 594¶

In :
#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

Example 17.4
(a) The indicated power of the engine is (kW) =  69.9
(b) The mechanical  efficiency of the engine is (percent) =  74.4


## Example 5: pg 595¶

In :
#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

Example 17.5
The Brake power is (kW) =  29.3
The Indicated power is (kW) =  37.3
The mechanical efficiency is (percent) =  78.8
Energy can be tabulated as :-
----------------------------------------------------------------------------------------------------
kJ/s                           Percentage
----------------------------------------------------------------------------------------------------
Energy from fuel                          135.3                   100.0
Energy to brake power                  29.3                     21.7
Energy to exhaust                        35.4                     26.0
Energy to coolant                         44.5                     32.9
Energy to suroundings,etc.            26.1                    19.3
there is minor variation in the result reported in the book due to rounding off error


## Example 6: pg 596¶

In :
#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

Example 17.6
(a) The brake power is (MW) =  23.719
(b) The fuel consumption is (tonne/h) =  4.74
(c) The brake thermal efficiency is (percent) =  42.0