Question

Due Date: Group: Process Instrumentation -Lab 4 Exercise 7. psi An open atmospheric vessel containing an unknown mixture, A direct level measurement determined the level h 30.9 ft. Bottom tank pressure P. gauge reads 11.7 psig. The density of the material is unknown. Known: • specific weight of water is 62.43 lbs/ft. • External pressure is 1 atm • P=SGxh a is the material water? Yes No b. Explain why and show work that validates your answer: c. Determine the density of the material in pounds per cubic inch. psmass / volume d. Determine the specific gravity of the material in the tank. e. P1 and P2 should read: the same pressure not the same pressure E xplain why f. For the same level in the tank, what would P1 indicate if it is changed to an absolute pressure gauge?

Answer 1

© @ NO 62.43 lbs specific weight (W) = of water = 9807 also - 30.91 = 9,418m Pgage = (w) h = (1807) (9.418) r Pause = 92360.457 Pa , c o But Pi at bottom of tank= 11.78 psig Tri=80668.bopel is not validated. clearly, hydrostatic low Hence, it is not woter. © (Pi)gage - wh 3 80668.66= w (9,118) > 8563.37 =W of material now w=fg 78565.37 = f - 9.807 (g = 9,807 m/s2) 1873.39 - +7 tal 0.03154 lb now 873.30 = 10-03 ini

ematerial = 873.12 kg/m² - 0.873 kalns 1. fwater 1000 kglm's e I ® Pp is not mentioned clearly! I am unable to understand where is P2 (Pabsolute = Potmosphese + Pgoge = 101325 pat 80668.66 pa = 181.993 kPa Pabs = 1.796 atm] Original density for initial density) = 1120 kg/m3 lassuming wateo's clensity = 1000 kg) ® initial height = 7ft = 2.133m initial Pressure= Pgh = (1120) (9.807) (2 0133) = 23428. 53 pa now, P added due to new liquid addition - Pnew = egh h = ift = 0.305m to 84.3 dbs 1350. 35 kg 73 ) Prew = (350.35) (9.807) 10.305) 3 4038-03

Total P = Pinitial + Pnew 23428.35 + 4038.03 a 27466. 38 Pa now mixture htotal . Total P = 27466.38 = (mixture) (g) (htotar) fmixture (9.807) 6 2.438m) - I mixture = 1148.76 kg MI - Specific gravity (new) = 1148.76 log/m3 1000 kg/m² -1.148] 1.148 Total Igage = 27466.3 Pa Total Igage { 0.27 1 atm] © I @ p decreases it means Volume increases. Hence, the height of liquid level increases 2 Transmitter's output will increase pressure indicator decreases ecreases and Pis directly to density as per hydrostatic since density proportional law.
due to time constraint and protocal, I could only solve this much questions.