Cooling a Hot Flue Gas by Spraying Liquid Water: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2015, Q1

Question 1: Cooling a Hot Flue Gas by Spraying Liquid Water (Part A — 15 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. Hot combustion gas cooled from 600 °C to 450 °C in an adiabatic spray cooler; liquid water fed at 20 °C leaves as vapour at 450 °C. Basis: 1000 mol dry gas/h. Heat-capacity data (Felder Table B.2, $C_p=a+bT+cT^2+dT^3$ in kJ/mol·°C); water latent heat $\lambda(100^\circ\text{C})=40.66$ kJ/mol and $C_{p,\text{liq}}=0.0754$ kJ/mol·°C.

Quantity	Value
Dry-gas basis	1000 mol/h (8.5% CO, 10.5% CO₂, 0.5% O₂, 80.5% N₂)
Inlet / outlet gas temperature	600 °C → 450 °C, 1.0 atm
Spray-water inlet temperature	20 °C (liquid)
Gas dew point	57 °C ⇒ $p_w=P^{sat}(57^\circ\text{C})=129.9$ mm Hg

Find. The mass flow rate (kg/h) of spray water needed to cool the gas to 450 °C.

Figure 1 — Adiabatic spray cooler: cold water injected into the hot gas evaporates and cools it to 450 C.

Approach. Treat the cooler as adiabatic: the sensible heat given up by the hot gas (dry gas plus the moisture it already carries) exactly supplies the heat needed to take the spray water from liquid at 20 °C to vapour at 450 °C. Because 450 °C is far above the 57 °C dew point, all water remains vapour, so no condensation term appears.

Moisture already in the gas. The dew point fixes the water partial pressure, so the moles of water per mole of dry gas follow from Raoult/Dalton at saturation: $$\dot n_{w,0}=\dot n_{dry}\,\frac{p_w}{P-p_w}=1000\cdot\frac{129.9}{760-129.9}=\boxed{206\ \text{mol/h}}.$$ This moisture is cooled with the gas from 600 °C to 450 °C.
Heat released by cooling the incoming gas. Integrating each $C_p$ from 450 °C to 600 °C and summing (dry gas + its moisture) gives the sensible heat available: $$Q_{rel}=\sum_i \dot n_i\!\int_{450}^{600}\!C_{p,i}\,dT=1000\!\sum_i y_i\overline{C}_{p,i}\Delta T+\dot n_{w,0}\!\int_{450}^{600}\!C_{p,\text{H}_2\text{O}}\,dT=6217\ \text{kJ/h}.$$
Heat to raise one mole of spray water to vapour at 450 °C. Sum the liquid sensible heat (20→100 °C), the latent heat at 100 °C, and the vapour sensible heat (100→450 °C): $$\Delta\hat H_{spray}=C_{p,\text{liq}}(100-20)+\lambda_{100}+\int_{100}^{450}\!C_{p,\text{H}_2\text{O}}\,dT=6.03+40.66+12.57=59.26\ \text{kJ/mol}.$$
Water feed rate. An adiabatic energy balance sets released heat equal to absorbed heat: $$\dot n_{spray}=\frac{Q_{rel}}{\Delta\hat H_{spray}}=\frac{6217}{59.26}=104.9\ \text{mol/h}\;\Rightarrow\;\dot m_{spray}=104.9\times0.01802=\boxed{1.89\ \text{kg/h}}.$$

Quantity	Result
Moisture carried by the feed gas	206 mol H₂O/h
Sensible heat released (600→450 °C)	6217 kJ/h
Enthalpy to vaporise spray water (20 °C liq → 450 °C vap)	59.26 kJ/mol
Cooling-water feed rate	≈ 1.89 kg/h (105 mol/h)

Concept & Theory

Topic. Adiabatic direct-contact (spray) cooling — a combined mass/energy balance with evaporative cooling.

Key relations. Dew-point/humidity relation $\dot n_w=\dot n_{dry}\,p_w/(P-p_w)$; sensible heat $\int C_p\,dT$; latent heat $\lambda$; adiabatic balance $Q_{released}=Q_{absorbed}$.

Why this works. The spray water cools the gas by absorbing energy as it warms and evaporates; with no external heat, the gas's sensible-heat loss is the only source. Choosing a path (heat liquid → vaporise → superheat) makes the enthalpy change of the water easy to tabulate, and the incoming moisture must be included because it too gives up sensible heat.

Common pitfall. Ignoring the water already present in the feed (the dew point is not decoration — it sets 206 mol/h of extra heat capacity), or adding a condensation term even though 450 °C is far above the dew point so nothing condenses.

Question 1 of 7: Cooling a Hot Flue Gas by Spraying Liquid Water

Question 1: Cooling a Hot Flue Gas by Spraying Liquid Water (Part A — 15 marks)