please , give details to solve problem 9.11. i did it but the unit is m^2/s^2 in
ID: 1862761 • Letter: P
Question
please , give details to solve problem 9.11. i did it but the unit is m^2/s^2 instead of pa, it is wierd .
Determine the following conversion factors. The factor to multiply mm of Hg to get Pa (SHg = 13.6). The factor to multiply inches of H2O to get psi. The factor to multiply Pa to get m of H20. A vacuum pressure (below atmospheric) is applied to the column port of a well-type manometer. The well port is open to the atmosphere. The fluid is memory (S = 13.6). If the reading R is 130 mm, what is the applied pressure relative to atmospheric pressure? A vacuum pressure (below atmospheric) is applied to the column port of a well-type manometer. The well port is open to the atmosphere. The fluid is memory (S = 13.6). If the reading R is 25.0 in, what is the applied pressure relative to atmospheric pressure? A well-type inclined manometer has a gas pre anole nf7.5 cm of water column applied to its ports (higher pressure to well). If the manometer is at an angle of 7.5 degree and contains a fluid with a density of 0.8 g/cm3, what is the reading R? A well-type inclined manometer has a gas pressure difference of 3 in of water column applied to its ports (higher pressure to well). If the manometer is at an angle of 7 degree and contains a fluid with a density of 50 lbm/ft3, what is the reading R? An inclined manometer is used to measure the differential gas pressure across an orifice. The angle is 30 degree and the fluid used in the manometer has a specific gravity of 0.8. Calculate the sensitivity (cm length/Pa) and the resolution of the manometer (in Pa) if 0.5 mm (length) can be resolved with a naked eye. In Problem 9.11, calculate the length of the tube for measuring a maximum pressure of 10 cm of water. Determine the gas pressure exerted on the reservoir of an inclined manometer (P1 in Fig. 9.3) if it has a 15-degree angle, uses a fluid with specific gravity of 0.7, and reads 10.2 cm.Explanation / Answer
here a lenght of 0.5mm can resolved
so pressure difference due 0.5mm lenght = 800*5*10^-4*(sin 30)*9.8
= 1.96 Pa
sensitivity = 0.5mm/1.96Pa
= 0.05cm/1.96Pa
= 0.0255 cm/Pa
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