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3 years ago

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The water requirement for Class H cement is 38% (i.e.,water (%) by weight of cement),whereas the water requirement for barite is

2.4 gal/100 lbm barite. Compute the amount of barite that should be blended with each sack of cement for obtaining a Class H cement slurry of 15.7 lbm/gal.

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

weight of barite = 398.4355 kg

Explanation:

Solution:- The values given in the question are as follows:

water requirement for H class cement = 38% by weight of cement

water requirement for barite = 2.4 gal / 100 lbm

H class cement slurry = 15.7 lbm/gal

one sack of cement = 50 kg or 110.231 lbm

one sack of cement require water = (38/100)*110.231

one sack of cement require water = 41.8877 gal

water required 100 lbm barite = 2.4 gal

or water required barite = 2.4% by weight of barite

H class cement slurry = (weight of cement + weight of barite)/total weight of water

15.7 =(110.231 + weight of barite)/(water required one sack of cement + 2.4%*weight of barite)

15.7 = 110.231 + (weight of barite)/(41.887 + 0.024*weight of barite)

15.7*41.8877 + 15.7*0.024*weight of barite = 110.231 + weight of barite

657.6368 + 0.3768*weight of barite = 110.231 + weight of barite

547.4058 =0.6232*weight of barite

weight of barite = 878.379 lbm or 878.4 lbm

weight of barite = 398.4355 kg

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Answer:

Explanation:

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Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

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V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

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V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

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$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

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