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xeze [42]
3 years ago
11

Fresh cut potato strips with a moisture content of 50% (w/w) on wet basis are fried in peanut oil to produce French fries.

Engineering
1 answer:
Rufina [12.5K]3 years ago
4 0

Answer:

Percentage of oil uptake by the potato = 5.57% on a wet basis (including the moisture content)

Percentage of oil uptake by the potato = 7.24% on a dry basis (excluding the moisture content)

Explanation:

Starting with 1000kg of fresh potatoes,

It is given that this consists of 50% (w/w) of water.

Meaning that, 1000 kg of fresh potatoes consists of (1000 × 50%) of water = 500 kg of water.

If 1000 kg of potatoes consist of 500 kg of water, then the remaining 500 kg is pure potatoes (since it is stated that freshly cut potatoes have no oil content)

After frying, the weight of the French fries is 700 kg now.

But of this 700 kg, 23 % of its weight is moisture content, i.e. water,

23% of 700 kg = 161 kg.

This means that the amount of potatoes and oil in the French fries is 700 - 161 = 539 kg

But, recall, that the amount of potatoes in this process from the start is 500 kg. This amount doesn't change even on frying the fresh potatoes into French fries.

So, amount of oil in the French fries = 539 - 500 = 39 kg

Percentage of oil uptake by the potato = 100% × (amount of oil intake)/(total mass of potatoes) = 100% × 39/700 = 5.57% on a wet basis (including the moisture content)

And 100% × 39/539 = 7.24% on a dry basis (excluding the moisture content)

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Explanation:

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The height of section (1) above section (2), D₂ = 50 m

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Let 'v₂' represent the velocity at section (2)

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z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}

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g = The acceleration due to gravity = 9.8 m/s²

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z₂ = The reference = 0 m

By plugging in the values, we have;

50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

90.8777353 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

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