# Chapter 13: Fuel and Combustions¶

## Problem: 1, Page No: 350¶

In :
# Variables
C = 84                  # Percentage
S = 1.5                 # Percentage
N = 0.6                 # Percentage
H = 5.5                 # Percentage
O = 8.4                 # Percentage

# Solution
GCV = (8080 * C + 34500 * (H - O / 8) + 2240 * S) / 100
LCV = (GCV - 9 * H / 100 * 587)
print "Gross Calorific Value", int(GCV), "kcal / kg"
print "Net Calorific Value", "{:.2f}".format(LCV), "kcal / kg"

Gross Calorific Value 8356 kcal / kg
Net Calorific Value 8065.48 kcal / kg


## Problem: 2, Page No: 350¶

In :
# Variables
C = 90              # Percentage
O = 3.0             # Percentage
S = 0.5             # Percentage
N = 0.5             # Percentage
ash = 2.5           # Percentage
LCV = 8490.5        # kcal / kg

# Solution
print "HCV = LCV + 9 * H / 100 * 587"
print "HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)"
H = (8490.5 - 7754.8) / (345 - 52.8)
H = 4.575
print "The precentage of H is", H, "%"
HCV = LCV + 52.8 * H
print "Higeher calorific value of coal", "{:.1f}".format(HCV), "kcal / kg"

HCV = LCV + 9 * H / 100 * 587
HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)
The precentage of H is 4.575 %
Higeher calorific value of coal 8732.1 kcal / kg


## Problem: 3, Page No: 351¶

In :
# Variables
x = 0.72                # g
W = 250                 # g
w = 150                 # g
t1 = 27.3               # C
t2 = 29.1               # C

# Solution
HCV = ((W + w) * (t2 - t1)) / x
HCV *= 4185.0 / 10 ** 6
print "HCV of fuel is", "{:.3f}".format(HCV), "KJ / Kg"

HCV of fuel is 4.185 KJ / Kg


## Problem: 4, Page No: 351¶

In :
# Variables
x = 0.84                # g
W = 1060                # g
w = 135                 # g
delta_t = 2.5           # C

# Solution
HCV = ((W + w) * delta_t) / x
print "HCV of fuel is", "{:.2f}".format(HCV), "kcal / kg"

HCV of fuel is 3556.55 kcal / kg


## Problem: 5, Page No: 351¶

In :
# Variables
V = 0.1                 # m ^ 3
W = 25                  # kg
t1 = 20                 # C
t2 = 33                 # C
m = 0.025               # kg

# Solution
HCV = W * (t2 - t1) / V
LCV = HCV - (m / V) * 580
print "HCV is", HCV, "kcal / m^3"
print "LCV is", LCV, "kcal / m^3"

HCV is 3250.0 kcal / m^3
LCV is 3105.0 kcal / m^3


## Problem: 6, Page No: 351¶

In :
# Variables
w1 = 2.5                # g
w2 = 2.415              # g
r = 1.528               # g
ma = 0.245              # Mass of ash, g

# Solution
m = w1 - w2             # Mass of moisture in coal
mv = w2 - r             # Mass of volatile matter
moisp = m * 100 / w1
volp = mv * 100 / w1
ashp = ma * 100 / w1
carbp = 100 - (moisp + volp + ashp)
print "Percentage of moisture:", moisp, "%"
print "Percentage of volatile matter:", volp, "%"
print "Percentage of ash:", ashp, "%"
print "Percentage of fixed carbon:", carbp, "%"

Percentage of moisture: 3.4 %
Percentage of volatile matter: 35.48 %
Percentage of ash: 9.8 %
Percentage of fixed carbon: 51.32 %


## Problem: 7, Page No: 352¶

In :
# Variables
wt_coke = 2             # kg

# Solution
wt_O = 2 * 32 / 12.0
wt_air = wt_O * 100 / 23.2
Vol_air = wt_air / 28.94 * 22.4
print "Volume of air needed for the complete combustion of 2kg coke",
print "is", "{:.3f}".format(Vol_air), "litres at NTP"

Volume of air needed for the complete combustion of 2kg coke is 17.793 litres at NTP


## Problem: 8, Page No: 352¶

In :
# Variables
C = 86              # Percent
H = 4               # Percent
N = 1.3             # Percent
S = 3               # Percent
O = 4               # Percent
Ash = 1.7           # Percent
wt = 500            # g

# Solution
wt_C = C / 100.0
wt_S = S / 100.0
wt_H = H / 100.0
wt_O = O / 100.0

print "Nitrogen and ash are incombustible, so they do not require oxygen."
wt_O_C = 32 / 12.0 * wt_C
wt_O_S = 32 / 32.0 * wt_S
wt_O_H = 32 / 4.0 * wt_H

Twt_O = wt_O_H + wt_O_S + wt_O_C
wt_O_needed = Twt_O - wt_O
wt_air = (100.0 / 23.0 * wt_O_needed) * 500 / 1000.0
print "Minimum Wt. of air required by 500g of fuel", "{:.2f}".format(wt_air), "kg"

Nitrogen and ash are incombustible, so they do not require oxygen.
Minimum Wt. of air required by 500g of fuel 5.66 kg


## Problem: 9, Page No: 353¶

In :
# Variables
wt_C = 3                # kg

# Solution
wt_air = wt_C * 32 * 100 / 12.0 / 23.0
vol_air = wt_air * 1000 * 22.4 / 28.94

print "H2(g)  +  1/2 O2(g)  --> H20(l)"
print " 1            0.5         1\t\t(By Vol.)"
print "CO(g)  +  1/2 O2(g)  --> CO2(g)"
print " 1             0.5        1\t\t(By Vol.)"
print "CH4(g) +  2   O2(g)  --> CO2(g) + 2H2O(l)"
print " 1             2          1\t\t(By Vol.)"

print "Weight of air for the combustion of 3kg carbon",
print "{:.3f}".format(wt_air), "kg"
print "Vol. pf air required for combustion of 3kg carbon",
print "{:.3e}".format(vol_air), "L",
print "or", "{:.2f}".format(vol_air / 1000), "m^3"

H2(g)  +  1/2 O2(g)  --> H20(l)
1            0.5         1		(By Vol.)
CO(g)  +  1/2 O2(g)  --> CO2(g)
1             0.5        1		(By Vol.)
CH4(g) +  2   O2(g)  --> CO2(g) + 2H2O(l)
1             2          1		(By Vol.)
Weight of air for the combustion of 3kg carbon 34.783 kg
Vol. pf air required for combustion of 3kg carbon 2.692e+04 L or 26.92 m^3


## Problem: 10, Page No: 353¶

In :
# Variables
H = 0.30                # m^3
CO = 0.10               # m^3
CH4 = 0.04              # m^3
N2 = 0.56               # m^3

# Soution
vol_oxygen = H * 0.5 + CO * 0.5 + CH4 * 2
vol_air = vol_oxygen * 100 / 21
print "Volumer of air required for complete combustion of 1 m^3 of",
print "producer gas:", "{:.3f}".format(vol_air), "m^3"

Volumer of air required for complete combustion of 1 m^3 of producer gas: 1.333 m^3


## Problem: 11, Page No: 354¶

In :
# Variables
H = 15.4                # Percentage
C = 84.6                # Percentage
wt_fuel = 1             # kg
wt_C = 0.846            # kg
wt_H = 0.154            # kg

# Solution
print "The combustion reactions are,"
print "C + O2 --> CO2"
print "12  32  \t(by Weight)"
print "2H2 + O2 --> H20"
print " 4    32\t(by Weight)"

wt_O = 32 / 12.0 * wt_C
wt_O_H = 32 / 4.0 * wt_H
Twt_O = wt_O + wt_O_H
print "Because 32 gm of O2 occupies a volume of 22.4 liters at NTP"
print "3.488 * 1000 gm of O2 will occupy",
print "{:.1f}".format(22.4 / 32 * Twt_O * 1000), "liters"

The combustion reactions are,
C + O2 --> CO2
12  32  	(by Weight)
2H2 + O2 --> H20
4    32	(by Weight)
Because 32 gm of O2 occupies a volume of 22.4 liters at NTP
3.488 * 1000 gm of O2 will occupy 2441.6 liters


## Problem: 12, Page No: 354¶

In :
# Variables
C = 750                 # g
H = 52                  # g
O = 121                 # g
N = 32                  # g
ash = 45                # g

# Solution
min_wt_air = (C * 32 / 12. + H * 16 / 2. - O) * 100 / 23.
HCV = 1 / 1000. * (8080 * C + 34500 * (H - O / 8.) + 2240 * 0)
LCV = HCV - 0.09 * H * 587 / 10.0

print "HCV is", int(HCV), "kcal/kg"
print "LCV is", int(LCV), "kcal/kg"

HCV is 7332 kcal/kg
LCV is 7057 kcal/kg


## Problem: 13, Page No: 355¶

In :
# Variables
C = 81              # Percent
H = 8               # Percent
N = 2               # Percent
O = 5               # Percent

# Solution
print "In 1kg coal,"

wt_C = C * 10
wt_H = H * 10
wt_N = N * 10
wt_O = O * 10
wt_ash = 100 - (wt_O + wt_N + wt_H + wt_C)

wt_air = ((wt_C * 32 / 12. + wt_H * 16 / 2. - wt_O) * 100 / 23.) / 1000.

print "Weight of air required for complete combustion of 10kg coal",
print "=", "{:.2f}".format(wt_air * 10), "kg"

HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.))
LCV = HCV - 0.09 * H * 587

print "HCV is", int(HCV), "kcal/kg"
print "LCV is", int(LCV), "kcal/kg"

In 1kg coal,
Weight of air required for complete combustion of 10kg coal = 119.57 kg
HCV is 9089 kcal/kg
LCV is 8666 kcal/kg


## Problem: 14, Page No: 355¶

In :
# Variables
C = 80              # Percent
H = 7               # Percent
N = 2.1             # Percent
O = 3               # Percent
S = 3.5             # Percent
Ash = 4.4           # Percent

# Solution
HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.) + 2240 * S)
LCV = HCV - 0.09 * H * 587

print "HCV is", int(HCV), "kcal/kg"
print "LCV is", int(LCV), "kcal/kg"

HCV is 8828 kcal/kg
LCV is 8458 kcal/kg