Frost risk in Frost Hollow vineyard
Frost Hollow is a valley vineyard district. A clear, calm night follows a warm spring day. At 02:00 the temperature at the valley floor (35 m) is –3.2 °C, at mid-slope (60 m) –0.8 °C and on the plateau (95 m) –1.4 °C. Wind speed is 2 km/h and relative humidity is 78%. The vines are at flowering, when 30 minutes below –1.5 °C can destroy the crop. The farmer owns an 8 ha plot on the mid-slope and wants to expand onto one of three 5 ha sites: the valley floor, the mid-slope, or the plateau. A frost fan costs £18,000 and mixes air; overhead sprinklers cost £30,000 and use 4,000 L per hour per hectare. The farm's water storage is limited to 600 m³ and the local river is licensed for only 1,200 m³ per week in spring. Soil on the valley floor is wet and organic; the plateau is thin and stony. The forecast is for similar conditions for three nights.
1. Explain the temperature differences at 02:00 using radiation and air movement.
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Under clear skies and light wind, the ground loses heat by long-wave radiation faster than it gains heat, so air near the surface cools. On the valley floor, cold, dense air drains down the slopes (katabatic flow) and collects because it cannot escape; this is why the floor is coldest at –3.2 °C. The mid-slope is warmest at –0.8 °C because cold air drains away downslope and some mixing with warmer air above occurs; it is not a warm 'thermal belt' by heating, but by drainage. The plateau is cooler than the mid-slope (–1.4 °C) because it is exposed to the open sky and has no slope above to supply drained air, but it is warmer than the floor because cold air does not pond there. Wet organic soil on the floor may hold more water and have different heat capacity, but the dominant control is drainage. The 2 km/h wind is too weak to mix the valley inversion fully. Humidity of 78% means some condensation may release latent heat, but not enough to offset the inversion.
2. Choose the expansion site and one protection method. Show a calculation to support your choice.
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Choose the mid-slope. It is already the warmest site (–0.8 °C), above the –1.5 °C critical threshold on the measured night, and cold air drains away rather than ponding. The valley floor is a frost trap at –3.2 °C and would need protection on most frost nights; the plateau is intermediate but more exposed. Use frost fans rather than sprinklers. Fans cost £18,000 and mix warmer air downward; sprinklers cost £30,000 and need water. Calculation: at 4,000 L/hour/ha, a 5 ha block needs 20,000 L/hour = 20 m³/hour. For 6 hours per frost night, that is 120 m³; for three nights, 360 m³. The 600 m³ store could cover it, but the river licence of 1,200 m³/week would be close if other users also abstract, and refilling is uncertain. Fans use no water and are cheaper, but they work best when a temperature inversion exists. Install fans on the mid-slope as insurance, and keep the valley floor for non-frost-sensitive uses.
3. How reliable is a single night's data for the farmer's decision? Suggest one improvement.
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One night shows the pattern but not its frequency. The farmer needs to know how often the mid-slope falls below –1.5 °C, not just that it was –0.8 °C once. A single sensor may also be in a hollow or near a hedge, so the reading may not represent a 5 ha block. The forecast is for similar conditions for three nights, but 'similar' can change with cloud, wind and soil moisture; if wind rises above about 8 km/h, the inversion may mix out and the valley floor could be less cold, while the mid-slope may cool more. Improvement: place three shielded loggers per site at 1.5 m, record every 15 minutes for at least one full frost season, and compare with a standard screen. Then calculate the number of nights below the critical threshold and the duration. Use that frequency to compare the cost of fans (£18,000) with expected crop loss, rather than deciding from one night.