If you measured h=106.7 mm Hg using simple manometer, then the pressure (in kpa) will be:(Given: Mercury specific gravity is 13.55) Select one: a. 15.25 O b. 14.18 c. 13.12 d. 16.32 e. 12.05
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- Using Boyle's Law Demonstrator, the absolute pressure and the volume of air are respectively 61.15 kPa and 0.5 L. When the oil height is decreased by 2 cm, the new absolute pressure becomes 50.95 kPa. So, the new air volume will be approximately equal to: Select one: a. 0.6 L b. 0.4 L c. It can't be found because there are missed given data. d. 0.2 LA U-tube differential manometer connects two pressure pipes A and B. Pipe A contains carbon tetrachloride having specific gravity 1.594 under a pressure of 11.772 kPa and pipe B contains oil of sp. gr. 0.8 under a pressure of 11.772 kPa. The pipe lies 2.5 m above pipe B. Find the difference of pressure measured by mercury as fluid filling U-tube. a. 31.36 cm of mercury b. 25.11 cm of mercury c. 64.74 cm of mercury d. 13.62 cm of mercuryWith the manometer reading as shown in Fig. 2.33, calculate the difference of pressures in the two tubes A and B containing water. [Ans. 9.07 kPa] 1.65 m 80.25 m T 0.5,m & -Oil sp. gr. 0.9 Water-- Fig. 2.33. B
- If you measured h=890.9 mm Hg using simple manometer, then the pressure (in kpa) will be:(Given: Mercury specific gravity is 13.55) Select one: a. 127.33 b. 100.61 O c. 109.51 O d. 136.24 e. 118.4286 N/cm"] 11. A single column vertical manometer (ie., micrometer) is connected to a pipe containing oil of sp. gr. 0.9. The area of the reservoir is 80 times the area of the manometer tube. The reservoir contains mercury of sp. gr. 13.6. The level of mercury in the reservoir is at a height of 30 cm below the centre of the pipe and difference of mercury levels in the reservoir and right linb is 50 cm. Find the pressure in the pipe. (Ans. 6.474 N/em) 12. A pipe contains an oil of sp. gr. 0.8. A differential manometer connected at the two points A and B of the pipe shows a difference in mercury level as 20 cm. Find the difference of pressure at the two paints. [Ans. 25113.6 N/m³) 13. A U-tube differential manometer connects two pressure pipes A and B. Pipe A contains carbon tetrachlo- ride having a specific gravity 1.594 under a pressure of 11.772 N/cm and pipe B contains oil of sp. gr. 0.8 under a pressure of 11.772 N/cm. The pipe A lies 2.5 m above pipe B. Find the difference of…[Hint: Pressure in the pipe B is more than that in pipe A.] An inverted differential manometer containing an oil of sp. gr. 0.8 is connected to find the difference Water- -Oil of sp. gr. 0.8 3.5 m B T A 300 B Fig. 2.31. Fig. 2.32. of pressure at two points of a pipe containing water as shown in Fig. 2.32. Find the difference of pressures, if the manometer reading be 300 mm.
- If you measured h=851.1 mm Hg using simple manometer, then the pressure (in kpa) will be:(Given: Mercury specific gravity is 13.55) uor Select one: O a. 104.62 Ob. 96.11 O c. 130.15 d. 121.64 Oe. 113.13 Inviscid water Flows as shown, The manometers helghts are h1-59.3 cm, and h2=20 em, what is the water acceleration (m/s2)?2)What is the maximum gauge pressure of water that can be measured using a piezometer 2.5 m high? almosphere? (Ans. 24.5 kN/m2)12. Examine Figure 1. Given that a=7.5 in. and b=12 in., what is the height of the water in the ight hand vessel? Density ofthe CCI4 is 1.6 g/cm³. Manometer oil,p = 0.7g/cm3 Open to atmosphere Vapor 6.6 psig Liquid 10 in. Figure CS CamScanner ile tarandı
- Convert the reading of pressure to kpa. Assuming that borometer reads 760 mm of Hg. (a). 4.2 bar31. Consider a double-fluid manometer attached to an air pipe shown below. If the specific gravity of one fluid is 13.55, determine the specific gravity of the other fluid for the indicated absolute pressure of air. Take the atmospheric pressure to be 100 kPa. Air P= 76 APa 40 em 22 em SG, = 13.5A closed vessel contains air at a pressure of 140 kPag and a temperature of 20°C. Find the final absolute pressure in KPA if the air is heated at a constant volume to 100°C. The atmospheric pressure is 28 in. Hg. a. 298.93 b. 300.93 c. 278.93 d. 288.93