If the pressure in a tank of oil (sp.gr.=0.80) is 400 kPa, what is the equivalent head in meters of oil?
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- 1. If the pressure in a tank of oil (s = 0.80) is 60 lb per sq.in., what is the equivalent head: (a) in feet of the oil; (b) in feet of water; (c) in inches of mercury?QUESTIONS AND PROBLEMS: 1. If 5.6 mm3 of oil weighs 46,800 N. Calculate its density and its specific gravity. 2. Determine the pressure at A in bar gage due to the deflection of the mercury (sp. gr. = 13.6) in the U-tube gage shown in the figure. %3D Water 0.3 m A 0.6 m Mercury 3. For a gage pressure at A of 11,000 Pa vacuum, find the specific gravity of the gage liquid Y in the figure. Air- Liquid X Sp. gr. 1.6 0.4 m 0.50 m 0.3 m Liquid Y 4. a.) convert a pressure head of 15 m of water to meters of oil (sp. gr. =0.75). b.) Convert a pressure head of 600 mm of mercury to meters of oil (sp. gr.= 0.75)If the atmospheric pressure is 101.30 kPa and the absolute pressure at the bottom of the tank as shown is 231.30kPa. a. What is the absolute pressure at the interface of the olive oil and the mercury? b. What is the gage pressure at the interface of the olive oil and the mercury? 1.5 m sea oil, s = 0.89 2.5 m water 2.9 m olive oil 0.4 m mercury
- A gas expands from 0.03 m3 to 0.05 m3 at constant pressure of 3.5 atm. Determine the work done.The closed cylinder shown is filled with water to a depth of 100mm and rest with air. What is the pressure at the bottom of the periphery of the tank? 0.02 0.10 AIR 0.202. The pressure on a closed tank reads 58.86 kPa. (a) What is the equivalent height in water? (b) What is the equivalent height in terms of oil having sg = 0.85? (c) What is the equivalent height in terms of mercury having sg = 13.6?
- A glass 12 cm tall filled with water is invert the pressure at the closed end? Barometric pressure is 101.325 kPa. What is Ans. 100.15 kPaa Problem 2- 51 In Figure 13, in which fluid will a pressure of 700 kPa first be achieved? Ans: glycerin Proble Po 90 kPa A diffe water ethyl alcohol p= 773.3 kg/m³ mm, 1 betwe 60 m ol p= 899.6 kg/m 10 m water P= 979 kg/m' glycerin p = 1236 kg/m 5 m 5 m 2021/04/16 18:27A pressure in a given tank reads 265 mm of Hg. Determine the equivalentheight of a certain fluid in m with sp. gr = 3.44.2. If the atmospheric pressure is 101.33 kPa and the absolute pressure at the bottom of the tank as shown in the figure is 215 kPa. a.) What is the specific gravity of olive oil? b. ) What is the absolute pressure at the interface of the olive oil and the mercury c.) What is the pressure at the bottom of the tank Sea Oil (sg-0.89) Water Olive oil Mercury (sg= 13.6) 1.5 m 2.5 m 3.0m 0.5 m
- A tank contains oil sg=0.80, gasoline sg=0.90, and sea water sg=1.05. What is the pressure at a depth of 1.20 m If the depth of liquids are, 0.50, 0.60, and 0.80m respectively. (Draw the figure) a. 6318 b. 1045 c. 742 d. 4812Q2: A pressure gage 19.0 ft above the bottom of a tank containing a liquid reads 13.19 psi,another gage at height 14 .0 ft reads 15.12 psi. The density of * :the liquid is 1. 73 slug/ft O 3.10 slug/ft O 0.0216 slug/ft O 4.83 slug/ft OIn which fluid will a pressure of 680 kPa first be achieved? Po = 90 kPa ethyl alcohol p = 773.3 kg/m 60 m oil 10 m p = 899.6 kg/m³ water 5 m p = 979 kg/m3 glycerin p = 1236 kg/m3 5 m Answer: