# Question 5.2: The composition of a sample of waste material to be incinera......

he composition of a sample of waste material to be incinerated is as follows (in wt %)

Moisture – 24.0
Carbon – 28.0
Hydrogen – 3.5
Oxygen – 22.0
Nitrogen – 0.33
Sulfur – 0.16
Noncombustible matter (ash, metals, glass, etc.) –23.10
Determine (on 1 kg basis)
1. Heat required to be externally supplied for removal of moisture
2. Heat taken up by noncombustible matter
3. Theoretical and empirical calorific value
4. Air required for combustion
5. Volume and composition of combustion products
6. Theoretical flame temperature
7. Overall heat balance
8. Net heat available for external use

Step-by-Step
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The sample contains 25% moisture. In incineration the moisture will first be expelled.

Assuming initial temperature 25°C, on 1 kg basis, the heat required to raise the temperature of moisture to 100°C

= $W \times C_{w} \times (t_2 – t_1)$

where W = weight (=1 kg)

$C_w$ = specific heat of water (J/kg) ≈ 4.2 kJ/kg

$t_2 – t_1$ =100 – 25 = 75

substituting

Heat in water = 1 × 4.2 × 75

= 315 kJ/kg

Latent heat of evaporation is 2270 kJ/kg. The vapor will absorb heat to reach combustion temperature, which is unknown. Assuming the combustion temperature as 1500°C, the heat absorbed will be (specific heat being 2.0 kJ/kg °C)

$= W \times C_S (t_3 – 100) \\ = 1 \times 2 (1500 – 100) \\ = 2800 \ kJ/kg$

Total heat absorbed by the moisture for heating from 25 to 1500°C is

2270 + 315 + 2800 = 5385  kJ/kg

1 kg of sample will contain 0.25 kg moisture which will absorb

5385 × 0.25 = 1346.25 kJ = He

Calculating the calorific value from Mendeleev’s formula

$Q^h=4.187[81 \times \% \mathrm{C}+300 \times \% \mathrm{H}-26(\% \mathrm{O}-\% \mathrm{~S})]$

substituting

$=4.187[81 \times 0.28-300 \times 0.035-26(22-0.16)]$

we get

$Q^h$ = 10,911 KJ/kg

As expected, this value is lower and more realistic. We will use this value of $Q^h$ for further calculations.

Air required for combustion, volume, and contents of combustion products is calculated and given in the following Table (5.5).

This shows that
For 0% excess air

Air required – 2.70 m³/kg
Combustion products – 3.93 m³/kg
Contents of combustion products (%)

C$O_2$ – 13.3
$H_2$O – 17.6
S$O_2$ – Trace
$N_2$ – 69.1
$Q^h$ – calorific value – 10911 kJ

Assuming specific heat of combustion products $C_c$ = 1.6

\begin{aligned}Q_h & =C_c V_c t \\t & =\frac{10,911}{3.93 \times 1.6} \\& =1,735^{\circ} \mathrm{C}\end{aligned}

For 50% excess air
Air required – 4.04 m³/kg
Combustion products – 4.68 m³/kg
Contents of combustion products (%)

C$O_2$ – 11.12
$H_2$O – 14.55
$O_2$ – 6.10
$N_2$ – 68.23
S$O_2$ – Trace

The sample contains 23% noncombustible matter and ash. Assuming specific heat of this matter as 0.8 kJ/kg°C and an average temperature of 600°C the heat absorbed by 1 kg of noncombustible is

= 1 × 0.8 × (600 – 25)

= 460 kJ/kg

At 23% ash the heat absorbed by ash in 1 kg of garbage is

460 × 0.23 = 106 kJ

Heat evolved by carbon (28%) in the waste,

C + $O_2$ → C$O_2$ + 34070 kJ/kg °C

Thus for 0.28 carbon the heat evolved is

34,070 × 0.28  = 9,540 kJ

Heat evolved by hydrogen (3.5%)

$\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}+121,025 \ \mathrm{~kJ} / \mathrm{kg} .$

For 0.035 kg hydrogen the heat is

121,025 × 0.035 = 4,236 kJ

Sample contains 0.16% sulfur.
Assuming that it is in elemental form

$\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2+9,131 \ \mathrm{~kJ}$

For 0.0016 kg sulfur the heat evolved is

9,131 × 0.0016 = 14.6 kJ

The total heat evolved on combustion of 1 kg waste is

9,540 + 4,236 + 14.6 ≅ 13,790 kJ = $Q^h$

The actual combustion process is not likely to be simple elemental oxidation. The heat evolved may be less than the above calculated value.
The combustion temperature is

\begin{aligned}& =\frac{10,911}{4.68 \times 1.65} \\& =1,457^{\circ} \mathrm{C}\end{aligned}

Assume that the combustion products leave the incineration chamber at 600°C. To bring about complete combustion and remove odors, this gas will have to be heated to about 1000°C in the mixing chamber or the flue.
Quantity of combustion products is about 4–5 m³/kg and has a specific heat 1.6 kJ/m³ °C.

\begin{aligned}H_A & =5 \times 1.6(100-600) \\& =3200 \ \mathrm{~kJ}\end{aligned}

We can now calculate heat balance.

\begin{aligned}\text { Heat required externally } & =H_e+H_A \\& =1346+3200 \\& =4546 \ \mathrm{~kJ} / \mathrm{kg}\end{aligned}

Heat available from combustion

$Q^h$ = 10911 kJ/kg

Heat available from combustion

= 10911 – 4546 = 6365 kJ/kg

Assume 10% loss to walls, etc., and 106 kJ as heat taken by ash.
Net heat available ≅ 5600 kJ/kg waste. This heat can be utilized for water heating and other purposes.

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