Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20

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Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 C) 13.5 D) 15.9 B 0.75 C 1.00 B 21 [in] 1 Find the gauge pressure, at point 4, in pounds per inch squared. [pause] In this problem, a manometer is filled with, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 B 0.75 C 1.00 B 21 [in] 1 3 liquids, A, B, and C. The specific gravity of each of these 3 liquids, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 B 0.75 C 1.00 B 21 [in] 1 is provided in the data table. Point 1 is located on the open surface of --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 B 0.75 C 1.00 B 21 [in] 1 liquid A. And the absolute pressure, at point 1, equals, ---- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 B 0.75 C 1.00 B 21 [in] 1 14.7, pounds per inch squared. [pause] To solve this problem, we’ll use the equation, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] Liquid S.G. ΔP = ρ * g *Δh A 1.20 B 0.75 C 1.00 P1,abs= 14.7 [lb/in2] 21 [in] B 1 delta P, equals, rho, g, delta h. Which means, the pressure difference, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] Liquid S.G. ΔP = ρ * g *Δh pressure A 1.20 difference B 0.75 C 1.00 P1,abs= 14.7 [lb/in2] 21 [in] B 1 between 2 different locations in a fluid, equals, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] density Liquid S.G. ΔP = ρ * g *Δh pressure A 1.20 difference B 0.75 C 1.00 P1,abs= 14.7 [lb/in2] 21 [in] B 1 the density of the fluid, rho, times, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] density Liquid S.G. ΔP = ρ * g *Δh pressure A 1.20 difference B 0.75 acceleration C 1.00 P1,abs= 14.7 [lb/in2] 21 [in] B 1 the gravitational acceleration constant, g, times, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] density Liquid S.G. ΔP = ρ * g *Δh pressure change in A 1.20 difference height B 0.75 acceleration C 1.00 P1,abs= 14.7 [lb/in2] 21 [in] B 1 change in height within the fluid, delta h. For this equation, keep in mind that, the lower the height, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] density Liquid S.G. ΔP = ρ * g *Δh pressure change in A 1.20 difference height B 0.75 acceleration C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 in a fluid medium, the larger the pressure. For this reason, some authors add a minus sign, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] density Liquid S.G. ΔP = -ρ * g *Δh pressure change in A 1.20 difference height B 0.75 acceleration C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 to the equation. [pause] For example, if we want to know the pressure difference between, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] density Liquid S.G. ΔP = -ρ * g *Δh pressure change in A 1.20 difference height B 0.75 acceleration C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 point 1, and, a second location, point 2, --- A 15 [in] 4 10 [in] C

Find: P4,gauge [lb/in2] density Liquid S.G. ΔP = -ρ * g *Δh pressure change in A 1.20 difference height B 0.75 acceleration C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 [pause] then the pressure difference, delta P, 1, 2, --- A 15 [in] 4 10 [in] 2 C

Find: P4,gauge [lb/in2] Liquid S.G. ΔP12 = -ρA * g * (h2-h1) A 1.20 pressure B 0.75 difference C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 equals, negative rho A, times g, times the change in elevation between points 2 and 1. If we want to know the pressure at a point 3, --- A h1-h2 15 [in] 4 10 [in] 2 C

Find: P4,gauge [lb/in2] Liquid S.G. ΔP = -ρ * g * Δh A 1.20 pressure B 0.75 difference C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 which is located in liquid B, then we would have to evaluate the pressure difference from point A, --- 3 A h1-h2 15 [in] 4 10 [in] 2 C

Find: P4,gauge [lb/in2] ΔP13=-ρA* g * (hAB-h1)+… Liquid S.G. A 1.20 pressure B 0.75 difference C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 to the interface between fluids A and B, and combine that to the pressure difference, --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] -ρB* g * (h3-hAB) ΔP13=-ρA* g * (hAB-h1) Liquid S.G. -ρB* g * (h3-hAB) A 1.20 pressure B 0.75 difference C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 from the fluid interface, AB, to point 3. [pause] In this problem, we’re asked to find --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] -ρB* g * (h3-hAB) ΔP13=-ρA* g * (hAB-h1) Liquid S.G. -ρB* g * (h3-hAB) A 1.20 pressure B 0.75 difference C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 the gauge pressure, at point 4. Gauge pressure, equals, --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] S.G. P4,gauge = P4,abs-Patm A 1.20 B 0.75 C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 the absolute pressure, minus the atmospheric pressure. The atmostpheric pressure equals, --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] S.G. P4,gauge = P4,abs-Patm A 1.20 Patm= 14.7 [lb/in2] B 0.75 C 1.00 lower height = higher P1,abs= 14.7 [lb/in2] pressure 21 [in] B 1 14.7 pounds per inch squared, and the absolute pressure, at point 4, --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] P4,abs = P1,abs+ΔP14 S.G. P4,gauge = P4,abs-Patm A 1.20 Patm= 14.7 [lb/in2] B 0.75 P4,abs = P1,abs+ΔP14 C 1.00 P1,abs= 14.7 [lb/in2] 21 [in] B 1 equals, the absolute pressure at point 1, plus the pressure difference between points 4 and 1. And since we know, --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] P4,abs = P1,abs+ΔP14 S.G. P4,gauge = P4,abs-Patm A 1.20 Patm= 14.7 [lb/in2] B 0.75 P4,abs = P1,abs+ΔP14 C 1.00 P1,abs= 14.7 [lb/in2] 21 [in] B 1 the absolute pressure at point 1, is also, 14.7 pounds per square inch, then our equation for gauge pressure at point 4, --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] P4,abs = P1,abs+ΔP14 S.G. P4,gauge = P4,abs-Patm A 1.20 Patm= 14.7 [lb/in2] B 0.75 P4,abs = P1,abs+ΔP14 C 1.00 P4,gauge = ΔP14 P1,abs= 14.7 [lb/in2] 21 [in] B 1 simplifies, to the change in pressure between points 4 and 1. Therefore, for notational convenience, we’ll reposition our points, 2 and 3, --- 3 A h1-h2 15 [in] AB 4 10 [in] 2 C

Find: P4,gauge [lb/in2] P4,abs = P1,abs+ΔP14 S.G. P4,gauge = P4,abs-Patm A 1.20 Patm= 14.7 [lb/in2] B 0.75 P4,abs = P1,abs+ΔP14 C 1.00 P4,gauge = ΔP14 P1,abs= 14.7 [lb/in2] 21 [in] B 1 at the fluid boundaries A, B, and, B, C, respectively. [pause] Next, we’ll write out our equation for --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge [lb/in2] -ρB* g * (h3-h2) -ρC* g * (h4-h3) S.G. P4,gauge = ΔP14 A 1.20 ΔP14=-ρA* g * (h2-h1) B 0.75 -ρB* g * (h3-h2) C 1.00 -ρC* g * (h4-h3) 21 [in] B 1 the change in pressure between points 4 and 1. The problem statement provides the change in height, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge [lb/in2] -ρB* g * (h3-h2) -ρC* g * (h4-h3) S.G. P4,gauge = ΔP14 A 1.20 ΔP14=-ρA* g * (h2-h1) B 0.75 -ρB* g * (h3-h2) C 1.00 -ρC* g * (h4-h3) 21 [in] B 1 between consecutive points, as, negative 21 inches, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge [lb/in2] -ρB* g * (h3-h2) -ρC* g * (h4-h3) -21 [in] S.G. P4,gauge = ΔP14 -21 [in] A 1.20 ΔP14=-ρA* g * (h2-h1) 10 [in] B 0.75 -ρB* g * (h3-h2) C 1.00 -ρC* g * (h4-h3) -15 [in] 21 [in] B 1 32.2 feet per second squared. [pause] The gravitational acceleration term, g, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge [lb/in2] -ρB* g * (h3-h2) -ρC* g * (h4-h3) -21 [in] S.G. P4,gauge = ΔP14 -21 [in] A 1.20 ΔP14=-ρA* g * (h2-h1) 10 [in] B 0.75 -ρB* g * (h3-h2) C 1.00 -ρC* g * (h4-h3) -15 [in] 21 [in] g=32.2 [ft/s2] 1 B equals, 32.2 feet per second squared. [pause] And the density of a fluid, equals, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge [lb/in2] ρ = ρH2O * S.G. -ρB* g * (h3-h2) P4,gauge = ΔP14 -21 [in] A 1.20 ΔP14=-ρA* g * (h2-h1) B 0.75 -ρB* g * (h3-h2) 10 [in] C 1.00 -ρC* g * (h4-h3) -15 [in] 21 [in] g=32.2 [ft/s2] 1 B the density of water, times the specific gravity of the fluid, where the density of water equals, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge ρH2O= 62.4 [lb/ft3] ρ = ρH2O * S.G. -ρB* g * (h3-h2) P4,gauge = ΔP14 ρ = ρH2O * S.G. -21 [in] A 1.20 ΔP14=-ρA* g * (h2-h1) B 0.75 -ρB* g * (h3-h2) 10 [in] C 1.00 -ρC* g * (h4-h3) -15 [in] 21 [in] g=32.2 [ft/s2] 1 B 62.4 pounds per cubic foot, and the specific gravity of each fluid, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge ρH2O= 62.4 [lb/ft3] ρ = ρH2O * S.G. -ρB* g * (h3-h2) P4,gauge = ΔP14 ρ = ρH2O * S.G. -21 [in] A 1.20 ΔP14=-ρA* g * (h2-h1) B 0.75 -ρB* g * (h3-h2) 10 [in] C 1.00 -ρC* g * (h4-h3) -15 [in] 21 [in] g=32.2 [ft/s2] 1 B is identified in the table. [pause] After writing out this equation, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) 1 the change in pressure, between points 4 and 1, equals, --- 3 A 15 [in] 2 4 10 [in] C

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in ΔP14= 65,703 65,703 pounds inches, per feet square per second squared. [pause] After converting, ---- ft2*s2

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in 1 lbf*s2 ΔP14= 65,688 pounds mass to pounds force, and after convering, feet to inches, --- * ft2*s2 32.2 lbm*ft

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in 3 1 lbf*s2 1 ft ΔP14=65,688 the pressure difference between points 4, and 1, equals, --- * * ft2*s2 32.2 lbm*ft 12 in

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in 3 1 lbf*s2 1 ft ΔP14=65,688 1.18 pounds force per inch square. [pause] This means, --- * * ft2*s2 32.2 lbm*ft 12 in Δ P14 = 1.18 [lb/in2]

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in 3 1 lbf*s2 1 ft ΔP14=65,688 the gauge pressure at point 4, equals, --- * * ft2*s2 32.2 lbm*ft 12 in Δ P14 = 1.18 [lb/in2]

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 P4,gauge=1.18 [lb/in2] ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in 3 1 lbf*s2 1 ft ΔP14=65,688 1.18 pounds force per inch square. [pause] * * ft2*s2 32.2 lbm*ft 12 in Δ P14 = 1.18 [lb/in2]

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 P4,gauge=1.18 [lb/in2] ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in A) -1.2 B) 1.2 C) 13.5 D) 15.9 ΔP14=65,688 When reviewing the possible solutions, ft2*s2 Δ P14 = 1.18 [lb/in2]

Find: P4,gauge [lb/in2] -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) P4,gauge = Δ P14 P4,gauge=1.18 [lb/in2] ΔP14=-ρA* g *Δh12-ρB* g*Δh23-ρC * g *Δh23 =-62.4 [lb/ft3]*1.20 *32.2[ft/s2]*(-21[in]) -62.4 [lb/ft3]*0.75 *32.2[ft/s2]*(10 [in]) -62.4 [lb/ft3]*1.00 *32.2[ft/s2]*(-15 [in]) lbm*in A) -1.2 B) 1.2 C) 13.5 D) 15.9 ΔP14=65,688 the answer is B. [fin] ft2*s2 Δ P14 = 1.18 [lb/in2] answerB

Find the mass of grain, in kilograms, in the truck, when the scale first reads 675 kilograms. [pause] In this problem, ---