Question 2 of 8: Induced-draft cooling tower — air flow at the fan and make-up water

Question 2: Induced-draft cooling tower — air flow at the fan and make-up water (20 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.

Find. A flow sketch of the induced-draft tower, the volumetric air flow handled by the fan at its outlet in m³/s, and the make-up water mass flow in kg/s.

Approach. Two balances written across the dashed control volume close the problem: conservation of water substance links the evaporation rate to the air's humidity gain, and conservation of energy — with the fan work counted as an input — fixes the dry-air mass flow. The fan volume flow then follows from the specific volume at the saturated exit state.

1. Part (a) — the flow sketch. Warm water at 46 °C is sprayed over the fill; air is drawn in through the louvres at the base, rises counter-current to the falling water, and is pulled out of the top by the fan; cooled water collects in the basin and returns to the process. Because evaporation removes water from the circuit, make-up is admitted — here into the basin piping outside the tower. Figure 2.1 is the answer to part (a).
2. Fix the entering air state, reading “50 % saturation” literally. This paper family distinguishes percentage saturation μ = W/W_sat(t) from relative humidity φ = p_w/p_ws, so $$W_1 = \mu_1\,W_{sat}(15\,{}^{\circ}\text{C}) = 0.50 \times 0.010647 = \boxed{0.005324\ \text{kg/kg da}}$$ $$h_1 = 1.006\,t_1 + W_1\,(2501 + 1.86\,t_1) = 15.09 + 13.46 = 28.55\ \text{kJ/kg da}$$ For the record this state is 50.4 % relative humidity, so the distinction is worth 0.4 % here; it is quoted so the marker can see the convention was chosen deliberately.
3. Fix the leaving air state at saturation. $$W_2 = W_{sat}(25\,{}^{\circ}\text{C}) = 0.020081\ \text{kg/kg da}, \qquad h_2 = 25.15 + 51.16 = 76.31\ \text{kJ/kg da}$$ so each kilogram of dry air picks up ΔW = 0.014757 kg of water and Δh = 47.75 kJ of enthalpy on its way through.
4. Write the two balances. Water substance in equals water substance out: $$\dot m_{w1} + \dot m_a W_1 = \dot m_{w2} + \dot m_a W_2 \quad\Longrightarrow\quad \dot m_{w2} = \dot m_{w1} - \dot m_a\,\Delta W$$ and energy in equals energy out, with the fan work added: $$\dot m_a h_1 + \dot m_{w1} h_f(46) + \dot W_{fan} = \dot m_a h_2 + \dot m_{w2} h_f(25)$$ Substituting the mass balance into the energy balance eliminates ṁ_w2.
5. Solve for the dry-air mass flow. With h_f(46) = 192.62 and h_f(25) = 104.83 kJ/kg from the steam tables, $$\dot m_a = \frac{\dot m_{w1}\,[\,h_f(46) - h_f(25)\,] + \dot W_{fan}} {(h_2 - h_1) - \Delta W\,h_f(25)} = \frac{6(192.62 - 104.83) + 4}{47.75 - 0.014757 \times 104.83}$$ $$\dot m_a = \frac{530.7}{46.21} = \boxed{11.49\ \text{kg dry air/s}}$$ The numerator is the tower load: the water gives up 6 × 87.79 = 526.7 kW and the fan adds a further 4 kW, so 530.7 kW must leave in the air stream. The small correction in the denominator is the enthalpy the evaporated water carries out with it as liquid at the cold-water temperature.
6. Part (c) — the make-up water. Make-up exactly replaces what evaporates: $$\dot m_{mu} = \dot m_a\,\Delta W = 11.49 \times 0.014757 = \boxed{0.1695\ \text{kg/s}}$$ which is 610 litres per hour, or 2.8 % of the 6 kg/s circulating rate — a thoroughly typical evaporation loss for a 21 K range. In service a further blowdown stream of the same order is needed to hold the dissolved-solids concentration, so the real make-up demand would be roughly double this figure.
7. Part (b) — the volumetric flow at the fan outlet. The fan is in the leaving air, so its volume flow is evaluated at the saturated 25 °C state: $$v_2 = \frac{0.287042\,T_2\,(1 + 1.6079\,W_2)}{p} = \frac{0.287042 \times 298.15 \times 1.03228}{101.325} = 0.8719\ \text{m}^3/\text{kg da}$$ $$\dot V_2 = \dot m_a\,v_2 = 11.49 \times 0.8719 = \boxed{10.0\ \text{m}^3/\text{s}}$$ Strictly the fan itself raises the enthalpy by 4/11.49 = 0.348 kJ/kg, which at constant humidity ratio is 0.33 K of dry bulb and 0.11 % of volume — 10.02 m³/s. The paper states the leaving air is 25 °C saturated, so that is the state used, and the fan rise is reported as negligible rather than carried.
8. Check the result against the tower's own thermodynamics. The liquid-to-gas ratio is L/G = 6/11.49 = 0.52, at the low end of the 0.5–1.5 band real towers occupy, and the entering air wet bulb is 9.72 °C, so the cold water at 25 °C sits 15.3 K above the entering wet bulb across a 21 K range. Both figures say the same thing: this duty is easy for a tower, which is consistent with a modest air quantity. Substituting all four state points back into the energy balance reproduces 1487.65 kW on each side, so the arithmetic closes exactly.