Fluid dynamics.. does anyone know the answers or how to solve these problems?
At low pressures there is a significant difference between the densities of liquids and of gases. For example, at 1 at, the densest gas known to the author is uranium hexaflouride, which has M = 352g/mol; its normal boiling point is 56.2 C, assuming that it obeys the ideal gas law. The least dense liquid known to the author is liquid hydrogen, which at its normal boiling point, 20 K, had a density of 0.071g/cm^3. Liquid helium also has a very low density, about .125g/cm^3 (at 4 K). Excluding these remarkable materials, make a list of liquids which at 1 atm can exist at densities of less than 0.5g/cm^3.
14) For some oil and gas drilling operations we need a high-density drilling fluid (called "drilling mud"). What is the density of a mud that is 50% wt. water, 50% wt. BaSO4 (barite), SG of barite = 4.49.
I know the answer is 102 lbm/ft^3, but I don’t know how to go about solving the problem.
15) Why are specific gravities most often referred to the density of water at a temperature of 4 C instead of at 0 C?
19) A cubic foot of water at a temp of 68 F (20 C) weighs 62.3 lbf on earth…
a) What is its density?
b) What does it weigh on the moon (g = 6ft/s^2)?
c) What is its density on the moon?
27) The slug and the poundal were invented to make the conversion factor (mass/length) / (force/time^2) have a coefficient of 1. A new unit of length or a new unit of time could just as logically have been invented for this. Let us name those units the "toof" and the "dnoces". What are the valies of the toof and the dnoces in terms of the foot and the second.
I also know the answer to this is the toof = 32.2 ft, and the dnoces = s/sqrt 32.2, but I don’t know how to show this.
Any help on these problems would be awesome, thanks a bunch! =]
April 23rd, 2010 at 8:08 pm
I’ll give you two of the five…
14) Specific gravity of a substance is the ratio of the density of the substance to that of water. SG = 4.49 means that BaSO4 is 4.49 times as dense as water.
To solve the problem, first calculate the density BaSO4. Knowing water is 62.4 lb/ft^3, the density of BaSO4 is 4.49 x 62.4, which equals 280.18 lb/ft^3.
The problem states that the drilling mud is 50% by wt H2O and 50% by wt BaSO4. To achieve the weight balance, you would need 4.49 times the volume of water.
the solution is: (4.49 ft^3 H20 * 62.4 lb/ft^3)+(1ft^3 BaSO4 * 280.18 lb/ft^3) / (4.49 ft^3 +1 ft^3)
…in words: total mass divided by total volume
15) The density of water at 0 C (ice) is lower than that of water at 4 C. So using 0 C water would make SG higher for other substances.