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Stels [109]
3 years ago
5

A corn ethanol production plant receives 500,000.0 kg/day corn feedstock at a moisture content of 15.5% (wet basis). If all of t

his corn is converted into ethanol what is the theoretical volume of ethanol that this facility can produce per day? Assume that the starch content of corn grain is 68.5%. The density of ethanol is 789 kg/m3.
Engineering
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

207 m³/day

Explanation:

Dry corn feed stock = 500000 × ( 100 - 15.5%) = 500000 × 84.5% = 500000× 0.845 = 422500

Starch yield = 68.5% × 422500 = 289412.5

Glucose yield = 1.11 × 289412.5 = 321247.875 where 1.11 is the scarification factor of starch to glucose

Ethanol yield = 0.51 × 321247.875 = 163836.416 where 0.51 is theoretical yield of ethanol from one mole of glucose

density = mass / volume

volume = mass / density = 163836.416 / 789 = 207 m³ / day

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Simora [160]

Answer:

1064.8 kg/m³

Explanation:

Weight of the hydrometer = ρghA where ρ is the density, g is acceleration due to gravity, h is the submerged height and A is the cross sectional area.

W in water = ρwghwA

W in liquid = (ρliq)g hliq A where the cross sectional area is constant

W in water = W in liquid

(ρw)ghwA = (ρliq)g hliq A  where ρw is density of water, ρliq is the density of liquid and hw and hliq are the heights of the liquid and that water. g acceleration due to gravity cancel on both sides as well as the constant A

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8 0
3 years ago
Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: L
Travka [436]

Answer:

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Explanation:

Load 1: Active power P_1 = 20 HP = 14.91kW;

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                               = 14.91\times tan(cos^{-}0.8) = 11.18 kvar


Load 2: Active power P_2 = 20 kW;

Reactive power Q2 = 0 since the load is purely resistive.

Load 3: Active power P_3 = 0 due to purely capacitiveload

           Reactive power Q_3 = -20 Var

a) since all three loads are connected in parallel therefore

    The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar

Since Q = 0, the power factor is unity.

Supply current per phase is given by

I = \frac{P}{\sqrt{3}V_{L}}

= \frac{34910}{\sqrt{3}\times 480} = 41.99 A

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