Tunnel Ventilation Conversion of a 12,000-Bird Broiler House: Fan Count, Static Pressure and Hot-End Heat Build-Up
A 12 m × 100 m (1,200 m²) broiler house in Batangas holds 12,000 broilers at 2.5 kg market weight. It is being converted to tunnel ventilation; the effective cross-section at the fan end is 30 m². Design tunnel velocity is 2.5 m/s. The supplier quotes 1.4 m (50 in) fans rated 40,000 m³/h at 25 Pa. The owner has installed five fans and a single evaporative cooling pad, 12 m wide × 3.5 m high, at the inlet end. On a 34°C dry-bulb, 24°C wet-bulb afternoon, the inner house temperature reads 31°C near the pad and 37°C at the fan end, inside relative humidity is 78%, and birds at the fan end are panting with wings spread while those near the pad look comfortable. A hand-held anemometer reads 1.6, 1.7 and 1.8 m/s at three points down the house. Static pressure at the fan wall is 45 Pa.
1. Compute the airflow required for 2.5 m/s tunnel velocity and the number of fans needed. Compare that with what is installed and with the measured air speed. What do the three numbers tell you?
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Required airflow. Q = v × A = 2.5 m/s × 30 m² = 75 m³/s = 270,000 m³/h.
Fans required at rated output. 270,000 / 40,000 = 6.75, so 7 duty fans, plus one standby for bird welfare during a failure - 8 installed in total.
Installed capacity. 5 × 40,000 = 200,000 m³/h = 55.6 m³/s, giving 55.6 / 30 = 1.85 m/s. That is 74% of design airflow and well below the 2.5 m/s needed to remove bird heat at market weight.
Measured speed. The anemometer readings of 1.6–1.8 m/s sit close to the 1.85 m/s predicted from five fans. That agreement is important: it says the fans are moving roughly what they should for their number, and the problem is the count, not a gross mechanical failure.
But add the static pressure. The fans are rated at 25 Pa and the system is running at 45 Pa. Fan output falls steeply as static pressure rises, so the true per-fan delivery at 45 Pa is below 40,000 m³/h. The measured 1.7 m/s implies about 51 m³/s (roughly 10,200 m³/h per fan), which means each fan is delivering about 25% below nameplate and even 7 fans may not reach 2.5 m/s until the pressure is brought down. The shortfall is therefore both a count problem and a system-restriction problem, and fixing only one of them will not solve it.
2. The house shows a 6°C gradient from pad end to fan end. Using an air-mass balance, estimate the temperature rise the airflow should produce, and use the gap between prediction and observation to structure your diagnostic.
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Heat load. Sensible heat production for 2.5 kg broilers at market weight is roughly 7 W/kg: 12,000 × 2.5 × 7 = 210 kW.
Air-mass balance. At about 51 m³/s and 1.2 kg/m³, mass flow = 61 kg/s. ΔT = Q / (m × cp) = 210 / (61 × 1.006) ≈ 3.4 K. At the design 75 m³/s (90 kg/s) it would be about 2.3 K.
The gap. About 3.4 K is predicted but 6 K is measured. A gradient roughly double the prediction means the house is not behaving as a simple plug-flow tunnel. Candidate causes, in order of likelihood: (1) air leaking in at the fan end through curtains, doors, feed-line penetrations and the fan wall itself - leak air bypasses the birds and destroys the longitudinal flow; (2) short-circuiting between pad and fans along the sidewalls; (3) internal obstructions (feeders, brooders, chick screens) creating dead zones and channeling; (4) the pad not fully wetted, so inlet air is not being cooled as designed; (5) instrument error or a sensor in direct radiant heat from the roof.
Diagnostic sequence. Smoke or a fogger at the pad to visualize flow; a hand traverse at several cross-sections, not just three points on the centreline; a leak survey with a thermal camera or by hand at the fan wall; static pressure at the pad and at the fans to locate the restriction; check belt tension and shutter opening on each fan; check pad water distribution and scale; then check roof insulation and radiant heat at bird level. Correct the envelope and the restriction first, then add the missing fans - adding fans to a leaking house only buys a higher electricity bill.
3. Evaluate the evaporative cooling pad. Using a pad saturation efficiency of 80%, what inlet temperature should the house see, and does the observed 31°C near the pad indicate a pad problem?
Read the worked answer
Expected pad performance. Outlet temperature = dry bulb − efficiency × (dry bulb − wet bulb) = 34 − 0.8 × (34 − 24) = 34 − 8 = 26°C. The air entering the birds should therefore be about 26°C, with the same moisture content as the outside air.
Resulting humidity. The outside air at 34°C and 24°C wet bulb carries roughly 15.5 g of water per kg of dry air. At 26°C, saturation is about 21.5 g/kg, so relative humidity after cooling is roughly 70–72%. That is acceptable for tunnel ventilation and is close to the measured 78%, so the humidity reading is not itself alarming.
Pad face velocity. Pad area = 12 × 3.5 = 42 m². At the installed airflow of about 51 m³/s, face velocity = 1.21 m/s; at the design 75 m³/s it would be 1.79 m/s. Both are within the usual 1.5–2.0 m/s band, so the pad is not obviously undersized for the design airflow.
So what explains 31°C near the pad instead of 26°C? The pad is not delivering its rated 80% saturation. Likely causes: insufficient water flow over the pad face or dry streaks, scale or algae blocking the flutes, water distributor holes plugged, the pump cycling, or the pad drawing air around its edges rather than through the medium. Also confirm the temperature sensor is not reading radiant heat from the roof or a nearby bird.
Actions. Measure the water flow rate against the pad manufacturer's figure, inspect the pad face for dry bands, descale or replace the medium if it is more than a few seasons old, and check the pump and distribution header. Only after the pad is verified should the extra fans be added; a wet-bulb-limited system cannot be fixed by airflow alone, and if the wet bulb is high the honest answer is to reduce stocking density or improve roof insulation.