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Mrac [35]
3 years ago
7

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.
Engineering
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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The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
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Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

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For cylinder BC:

Let Length of BC = 18 in

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2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

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