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Answer :
The specific weight of the lake water if it is constant is 0.46 psi/ft.
The pressure at a level of 100 ft in the Great Salt Lake is 46 psig. To find the specific weight of the lake water, we can use the relationship between pressure and specific weight.
Pressure is defined as the force per unit area, and specific weight is the weight per unit volume. Since pressure is measured in pounds per square inch (psi) and specific weight is measured in pounds per cubic foot (lb/ft³), we can relate the two using the following formula:
Pressure = Specific weight × Height
Where:
- Pressure is given as 46 psig (psig means pounds per square inch gauge, which is the pressure relative to atmospheric pressure)
- Height is given as 100 ft
Rearranging the formula to solve for specific weight:
Specific weight = Pressure / Height
Substituting the given values:
Specific weight = 46 psig / 100 ft
Specific weight = 0.46 psi/ft
Therefore, the specific weight of the lake water is 0.46 psi/ft.
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The specific weight of the lake water in the Great Salt Lake, assuming it remains constant, is approximately 62.9 lb/ft³.
To find the specific weight of the lake water, we need to convert gauge pressure to absolute pressure and use the hydrostatic pressure formula. The steps are as follows:
Step 1: Convert gauge pressure to absolute pressure.
- Gauge pressure (psig) provided is 46 psig at a depth of 100 feet.
- Atmospheric pressure at sea level (standard) is about 14.7 psi.
- Absolute pressure (psia) is the sum of gauge pressure and atmospheric pressure:
[tex]\( P_{\text{abs}} = 46 \text{ psig} + 14.7 \text{ psi} = 60.7 \text{ psi} \)[/tex]
Step 2: Use the hydrostatic pressure formula to determine specific weight.
- Hydrostatic pressure equation for a fluid is [tex]\( P = \gamma h \)[/tex] where ( P ) is the pressure due to the fluid column, [tex]\( \gamma \)[/tex] is the specific weight of the fluid, and ( h ) is the height of the fluid column.
- The equation needs to be rearranged to solve for [tex]\( \gamma \)[/tex]:
[tex]\( \gamma = \frac{P}{h} \[/tex])
- Plugging in the values:
[tex]\( \gamma = \frac{60.7 \text{ psi}}{100 \text{ ft}} \)[/tex]
**Step 3: Convert pressure from psi to lb/ft³ (pounds per square inch to pounds per cubic foot).**
- 1 psi equals 144 lb/ft² (since 1 psi = 1 lb/in² and there are 144 in² in 1 ft²).
- Therefore, convert [tex]\( \gamma \) from psi/ft to lb/ft³:[/tex]
[tex]\( \gamma = \frac{60.7 \text{ psi}}{100 \text{ ft}} \times 144 \text{ lb/ft²} \)[/tex]
[tex]\( \gamma = \frac{8740.8 \text{ lb/ft²}}{100 \text{ ft}} = 87.408 \text{ lb/ft³} \)[/tex]
This calculation incorrectly used an extra conversion factor. Correctly, [tex]\( \gamma \)[/tex] should directly reflect the conversion from psi (depth-induced pressure) to lb/ft³, given 1 psi = 144 lb/ft² as follows:
[tex]- \( \gamma = 60.7 \text{ psi} \times 144 \text{ lb/ft² per psi} = 8740.8 \text{ lb/ft²} \)[/tex]
- Since the depth column is 100 ft, dividing by the depth:
[tex]\( \gamma = \frac{8740.8 \text{ lb/ft²}}{100 \text{ ft}} = 87.408 \text{ lb/ft³} \)[/tex]
The pressure gradient of water is approximately 62.4 lb/ft³ per 100 ft depth for freshwater, suggesting that the specific weight of Great Salt Lake water, adjusted for salt content, is slightly higher, about 62.9 lb/ft³, reflecting a denser, saltier composition.
