What is the minimal mass of helium (density 0.18 kg/m^3) needed to lift a balloo

Question

What is the minimal mass of helium (density 0.18 kg/m^3) needed to lift a balloon carrying two people in a basket, if the total mass of people, basket, and balloon (but not gas) is 320 kg? m,min=___kg Repeat for a hot-air balloon whose air density is 10% less than that of the surrounding atmosphere. m',min=___kg

Explanation / Answer

M[min] = M[basket+people+ balloon, not gas] * ?R/R[b] ?R is the difference in density between the gas inside and surrounding the balloon. R[b] is the density of gas inside the baloon. ==================================== Let V be the volume of helium required. Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V U = 1.225gV newtons ---- Weight of Helium = Volume of Helium * Density of Helium * g W[h] = 0.18gV N Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N ----- Weight of 260kg = 2549.7 N Then to lift the whole thing, F > 2549.7 So minimal F would be 2549.7 ---- 1.045gV = 2549.7 V = 248.8 m^3 Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf) ===== Let the density of the surroundings be R Then U-W = (1-0.9)RgV = 0.1RgV So 0.1RgV = 2549.7 N V = 2549.7 / 0.1Rg Assuming that R is again 1.255, V = 2071.7 m^3 Then mass of hot air required = 230.2 * 0.9R = 2340 kg Notice from this that M = 2549.7/0.9Rg * 0.1R so M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon) M[min] = M[basket] * ?R/R[b]

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