Coal Pulveriser Air Flow: 22-Mec-B3 Energy Conversion and Power Generation, December 2018, Q5

Question 5: Coal Pulveriser Air Flow (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Pulveriser capacity (as-received coal)	$\dot{M}_c$	18 Mg/h
Coal feed temperature	$T_c$	30 °C
Hot primary air from the air heater	$T_p$	250 °C
Cool tempering air	$T_t$	40 °C
Maximum permitted primary-air temperature	$T_{p,max}$	400 °C
Pulveriser outlet (coal–air mixture)	$T_o$	70 °C
Air–fuel ratio by mass	$AF$	2
Specific heat of air / of dry coal	$c_{p,a},\ c_c$	1.0 / 1.25 kJ/kg·K

Find. The coal, air and moisture flows; the heat-gain and heat-loss equations that close the mill as a heat exchanger; the maximum surface moisture the mill can dry under normal conditions; and the operating response when the coal is drier than that.

Figure 5.1 — the individual flow streams through the pulveriser, with the heat ledger the question asks for. Only the hot primary air loses heat; the tempering air, the dry coal and the surface moisture all gain heat, the moisture by far the most because it must be evaporated as well as warmed. The tempering-air stream is drawn dashed because at the limiting condition of part (c) it is shut off entirely.

Approach. Treat the mill as an adiabatic mixing heat exchanger: the only heat source is the hot primary air, and everything else in the control volume is a heat sink that must be brought to the 70 °C outlet temperature. Write the mass constraint on the two air streams and the energy balance, then solve the energy balance for the moisture flow at the limiting condition where no tempering air is admitted.

Part (a) — coal, air and moisture flows. The mill capacity converts directly: $$\dot{M}_c=\frac{18\times 1000}{3600}=5.0\ \text{kg/s},$$ and the stated air–fuel ratio by mass fixes the total air through the mill, $$\dot{M}_a=AF\times\dot{M}_c=2\times 5.0=10.0\ \text{kg/s}.$$ The as-received coal is dry coal plus surface moisture, so writing the moisture flow as $m$ and the surface-moisture fraction as $w$, $$\boxed{m=w\,\dot{M}_c=5.0\,w\ \text{kg/s},\qquad \dot{M}_{dry}=\dot{M}_c-m=(5.0-m)\ \text{kg/s}} .$$ The total air divides between the two supplies, $\dot{M}_p+\dot{M}_t=10.0$ kg/s.
Part (b) — the enthalpy a kilogram of moisture takes away. Surface moisture enters as liquid water at the coal-feed temperature and leaves as vapour in the air stream at the mill outlet. Because the vapour is at a low partial pressure, its enthalpy is essentially that of saturated vapour at the outlet temperature, so from the steam tables $$\Delta h_m=h_g(70\,{}^{\circ}\text{C})-h_f(30\,{}^{\circ}\text{C}) =2626.8-125.79=2501.0\ \text{kJ/kg}.$$ This single number dominates the whole question: evaporating one kilogram of water takes fifty times the heat needed to warm one kilogram of dry coal through the same 40 K.
Part (b) — the heat-loss and heat-gain equations. Only the hot primary air is above the outlet temperature, so it alone loses heat: $$\text{heat lost}=\dot{M}_p\,c_{p,a}\,(T_p-T_o)=180\,\dot{M}_p\ \text{kW}.$$ Everything else must be raised to 70 °C and therefore gains heat — the tempering air from 40 °C, the dry coal from 30 °C, and the moisture from liquid at 30 °C to vapour at 70 °C: $$\text{heat gained}=\dot{M}_t\,c_{p,a}\,(T_o-T_t) +(\dot{M}_c-m)\,c_c\,(T_o-T_c)+m\,\Delta h_m .$$ Substituting the numerical coefficients, the mill is closed by the pair $$\dot{M}_p+\dot{M}_t=10.0,\qquad 180\,\dot{M}_p=30\,\dot{M}_t+50\,(5.0-m)+2501.0\,m .$$ Two equations in the three unknowns $\dot{M}_p$, $\dot{M}_t$ and $m$: the system is one short, which is exactly why part (c) supplies the third condition.
Part (c) — the limiting condition. The mill is working hardest when every kilogram of air arrives at the full 250 °C, that is when $\dot{M}_t=0$ and $\dot{M}_p=\dot{M}_a=10.0$ kg/s. The heat then available is $$\dot{Q}_{avail}=10.0\times 1.0\times(250-70)=1800\ \text{kW},$$ and the balance becomes a single equation in $m$: $$1800=50\,(5.0-m)+2501.0\,m=250+2451.0\,m .$$
Part (c) — maximum surface moisture. Solving, $$\boxed{m=\frac{1800-250}{2451.0}=0.6324\ \text{kg/s}} ,$$ so the dry coal flow is $5.0-0.632=4.368$ kg/s and the surface moisture the mill can handle is $$\boxed{w=\frac{0.6324}{5.0}=0.1265=12.6\ \text{per cent as received}} .$$ Recomputing the balance from the answer confirms it: $50\times 4.368+2501.0\times 0.6324=218.4+1581.6=1800$ kW, matching the available heat exactly. A drying capacity in the region of 12 per cent surface moisture is typical of a vertical spindle mill on bituminous coal, and it is the number that decides whether a given coal can be burned at all without raising the primary-air temperature.
Part (d) — operating with drier coal. If the coal carries less than 12.6 per cent surface moisture, the evaporative load falls while the heat supplied by 10 kg/s of 250 °C air does not, so the mill outlet would drift above 70 °C. That is undesirable on three counts: it risks a mill fire or an explosion in the coal–air suspension, it drives off inherent as well as surface moisture and makes the pulverised coal dusty and hard to convey, and it takes the classifier away from its design operating point. The remedy is exactly what the mill is built for — admit tempering air, closing the hot-air damper and opening the cold-air damper so that the mixed air entering the throat is cooler while the total air flow, and therefore the air–fuel ratio and the carrying velocity through the classifier, stay unchanged. Quantitatively, at 8 per cent surface moisture ($m=0.40$ kg/s) the equations above give $$\dot{M}_p=7.29\ \text{kg/s},\qquad \dot{M}_t=2.71\ \text{kg/s},$$ i.e. about 27 per cent of the air bypasses the air heater. The mill therefore holds its 70 °C outlet across the whole range of coal moisture by trading hot air for cold at constant total flow.

Check: two assumptions stated under the paper's Note 6. First, the 18 Mg/h capacity is taken as as-received coal, so the moisture is part of the 5.0 kg/s rather than additional to it, and the air–fuel ratio of 2 is likewise referred to the as-received coal; reading the capacity as dry coal instead would give 0.620 kg/s of moisture, a change of under two per cent in the answer. Second, the evaporated moisture is assumed to leave as low-partial-pressure vapour at the mill outlet temperature, so its enthalpy is $h_g$ at 70 °C; the mill is also assumed adiabatic to atmosphere and the grinding work is neglected, both standard for this calculation. Note finally that the 400 °C ceiling on primary-air temperature is the mill's upset limit, not a normal condition: pushing the air heater outlet to 400 °C would raise the available heat to 3300 kW and the drying capacity to 1.244 kg/s, or 24.9 per cent surface moisture, which is how a plant copes with wet coal after rain.

Part	Quantity	Result
(a)	Coal mass flow (as received)	5.0 kg/s
(a)	Total air mass flow	10.0 kg/s
(a)	Moisture flow in terms of coal flow	$m=5.0\,w$ kg/s
(b)	Mass constraint	$\dot{M}_p+\dot{M}_t=10.0$ kg/s
(b)	Energy balance	$180\,\dot{M}_p=30\,\dot{M}_t+50(5.0-m)+2501\,m$
(b)	Moisture evaporation enthalpy	2501.0 kJ/kg
(c)	Heat available (no tempering air)	1800 kW
(c)	Maximum moisture flow	0.6324 kg/s
(c)	Maximum surface moisture	12.6 per cent as received
(d)	Response to drier coal	admit tempering air at constant total flow (7.29 / 2.71 kg/s at 8 per cent moisture)

Topic. The pulverised-fuel mill as a simultaneous grinder, classifier and dryer: primary and tempering air, mill outlet temperature and drying capacity.

Key relations. Mill throughput $\dot{M}_c=\text{capacity}/3600$; air–fuel ratio $\dot{M}_a=AF\,\dot{M}_c$; air mass split $\dot{M}_p+\dot{M}_t=\dot{M}_a$; adiabatic energy balance $\dot{M}_p c_{p,a}(T_p-T_o)=\dot{M}_t c_{p,a}(T_o-T_t)+\dot{M}_{dry}c_c(T_o-T_c)+m\,[\,h_g(T_o)-h_f(T_c)\,]$.

Why this works. The mill is a well-insulated box with three streams in and one out, so a first-law balance over it must close with no heat crossing the boundary. Choosing the outlet temperature as the datum turns the balance into a simple ledger: anything entering above 70 °C is a source, anything entering below is a sink, and the phase change of the surface moisture is a sink of enormous size because the latent heat of water at these conditions is around 2350 kJ/kg. Fixing the outlet temperature at 70 °C is what makes the problem determinate at all: the mill has one degree of freedom, the split between hot and tempering air, and the outlet temperature is the controlled variable that consumes it. That is also the physical control loop on a real mill — a temperature element in the mill outlet duct modulates the hot and cold dampers in opposition, holding the outlet constant while the total primary air is set independently by the coal flow.

Common pitfall. Forgetting that the tempering air is a heat sink, not a second source: it enters at 40 °C, below the 70 °C outlet, so it must be warmed. Treating it as a source reverses a sign and makes the mill appear to have unlimited drying capacity. The second pitfall is using the latent heat alone for the moisture instead of the full liquid-to-vapour enthalpy change across the temperature difference, which understates the moisture duty by about seven per cent ($h_{fg}$ at 70 °C is 2333.8 kJ/kg against the full 2501.0 kJ/kg).

Source. Rayaprolu, Boilers for Power and Process, chapters on pulverised-fuel firing and mills (vertical spindle mills, primary air systems and mill drying capacity); Babcock & Wilcox, Steam: Its Generation and Use, 41st ed., chapter on coal pulverisation.

Question 5 of 8: Coal Pulveriser Air Flow

Question 5: Coal Pulveriser Air Flow (10 marks)