Find: Wcone [lbf] D D = 3 [ft] air SGo = 0.89 d3 SGw= 1.00 oil d2

Find: Wcone [lbf] D D = 3 [ft] air SGo = 0.89 d3 SGw= 1.00 oil d2
water 457 476 495 514 Find the weight of the cone, in pounds force. [pause] In this problem, --- d1 d3 = 2 [ft] d2 = 2 [ft] d1 = 4 [ft]

water 457 476 495 514 a cone rests at equilibrium and is in contact with 3 separate fluids, --- d1 d3 = 2 [ft] d2 = 2 [ft] d1 = 4 [ft]

water 457 476 495 514 air, oil, and water. The specific gravity of the oil, and water, ---- d1 d3 = 2 [ft] d2 = 2 [ft] d1 = 4 [ft]

water 457 476 495 514 are provided. The depth of the cone, lower-case d, within each of these three fluids, --- d1 d3 = 2 [ft] d2 = 2 [ft] d1 = 4 [ft]

water 457 476 495 514 is identified using variables d1, d2 and d3. Also, the diameter of the cone, is provided. --- d1 d3 = 2 [ft] d2 = 2 [ft] d1 = 4 [ft]

water 457 476 495 514 [pause] To solve this problem, we’ll begin by summing the forces, --- d1 d3 = 2 [ft] d2 = 2 [ft] d1 = 4 [ft]

Find: Wcone [lbf] ΣFy = 0 D = 3 [ft] air d3 d2 oil water d1
in the vertical direction, which equals, the bouyant force, --- d1

Find: Wcone [lbf] ΣFy = 0 = Fb - Wcone D = 3 [ft] air d3 Wcone d2 oil
water F b, minus the weight of the cone, W cone. --- d1 Fb

Find: Wcone [lbf] ΣFy = 0 = Fb - Wcone D = 3 [ft] bouyant air cone
force weight d3 Wcone d2 oil water [pause] Therefore, the weight of the cone, equals, --- d1 Fb

force weight d3 Wcone Wcone= Fb d2 oil water the bouyant force. [pause] The bouyant force, equals, --- d1 Fb

force weight d3 Wcone Wcone= Fb d2 oil Fb= Σ ρ * g * V water the sum of rho, g, V, for the volume of all fluids displaced by the cone. This concept is sometimes referred to as, --- d1 Fb

Find: Wcone [lbf] D = 3 [ft] “Archimedes’ air Principle” d3 Wcone
Wcone= Fb d2 oil Fb= Σ ρ * g * V water Archimedes' Principle. [pause] Therefore the bouyant force, incudes, ---- d1 Fb

Wcone= Fb d2 oil Fb= Σ ρ * g * V water a contribution from the water displaced --- d1 Fb= ρw * g * Vw+ ρo * g * Vo Fb

Wcone= Fb d2 oil Fb= Σ ρ * g * V water by the cone, and a contribution from the oil displaced, --- d1 Fb= ρw * g * Vw+ ρo * g * Vo water Fb contribution

Wcone= Fb d2 oil Fb= Σ ρ * g * V water by the cone. [pause] Also, the density of a fluid, rho i, equals, ---- d1 Fb= ρw * g * Vw+ ρo * g * Vo water oil Fb contribution contribution

Find: Wcone [lbf] Wcone= Fb D = 3 [ft] Fb= ρw * g * Vw+ ρo * g * Vo d3
ρi = SGi * ρw d2 oil water the specific gravity of that fluid, SG i, times, --- d1 Fb

ρi = SGi * ρw d2 oil specific density water gravity the density of water, rho w. [pause] After making this substitution, ---- of water d1 Fb

ρi = SGi * ρw d2 oil specific density water gravity for the densities of water and oil, our revised equation, ---- of water d1 Fb

ρi = SGi * ρw d2 oil specific density water gravity for the bouyant force is rewritten, and simplified of water d1 Fb= SGw * ρw * g * Vw+ SGo * ρw * g * Vo Fb

ρi = SGi * ρw d2 oil specific density water gravity [pause] The density of water equals, --- of water d1 Fb= SGw * ρw * g * Vw+ SGo * ρw * g * Vo Fb= ρw * g * (SGw *Vw + SGo *Vo) Fb

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water 62.3 pounds mass per cubic foot. The gravitational acceleration constant, --- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) Fb

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water equals 32.2 feet per second squared, and the problem statement provided, --- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) g= 32.2 [ft/s2] Fb

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water the specific gravitites of water and oil. [pause] This means we’re left to compute --- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) g= 32.2 [ft/s2] SGo = 0.89 SGw= 1.00

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water the volume of the cone submerged in water, V w, and, the volume of the cone, ---- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) g= 32.2 [ft/s2] SGo = 0.89 SGw= 1.00

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water submerged in oil, V o. [pause] The volume of the cone submerged in water, V w, --- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) g= 32.2 [ft/s2] SGo = 0.89 SGw= 1.00

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water equals, the volume of a smaller cone having a height of, --- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) g= 32.2 [ft/s2] SGo = 0.89 SGw= 1.00

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water lower-case d 1, and having a diameter, we’ll define as, upper-case, --- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) g= 32.2 [ft/s2] SGo = 0.89 SGw= 1.00

ρi = SGi * ρw d2 oil ρw = 62.3 [lbm/ft3] water D 1. [pause] The volume of this cone, equals, --- d1 Fb= ρw * g * (SGw *Vw + SGo *Vo) g= 32.2 [ft/s2] SGo = 0.89 SGw= 1.00

Find: Wcone [lbf] Fb= ρw * g * (SGw *Vw + SGo *Vo) D = 3 [ft]
Vw= (1/12) * π * D1 * d1 2 d3 D1 d2 oil water one twelvth, PI, D1 squared, d1. [pause] From the problem statement, we know, lower-case, --- d1

Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 oil water d 1, equals, 4 feet. Upper-case D 1, --- d1

Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 oil water the diameter, can be computed by using similar traingles, --- d1

Find: Wcone [lbf] = Fb= ρw * g * (SGw *Vw + SGo *Vo) D = 3 [ft]
Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 D1 d1 oil = D d1+d2+d3 water and noticing the diameter and height, of the cone surbmerged in water, ---- d1

Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 D1 d1 oil = D d1+d2+d3 water is proportional to the diameter and height of the entire cone. [pause] After solving for diameter D 1, --- d1

Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 D1 d1 oil = D d1+d2+d3 water and plugging in the given values, --- d1 d1 D1 = D * d1+d2+d3

Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 D1 d1 oil = D d1+d2+d3 water the diameter of this smaller cone, equals, --- d1 d1 d1 = 4 [ft] D1 = D * d1+d2+d3 d2 = 2 [ft] d3 = 2 [ft]

Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 D1 d1 oil = D d1+d2+d3 water 1.5 feet. [pause] Now we can solve for the volume of the cone, --- d1 d1 d1 = 4 [ft] D1 = D * d1+d2+d3 d2 = 2 [ft] D1 = 1.5 [ft] d3 = 2 [ft]

Vw= (1/12) * π * D1 * d1 2 d3 d1 = 4 [ft] D1 d2 D1 d1 oil = D d1+d2+d3 water submerged in water, which equals, --- d1 d1 d1 = 4 [ft] D1 = D * d1+d2+d3 d2 = 2 [ft] D1 = 1.5 [ft] d3 = 2 [ft]

Vw= [ft3] Vw= (1/12) * π * D1 * d1 2 d3 D1 d1 = 4 [ft] d2 oil water 2.356 feet cubed. [pause] At this point, the last unknown variable, ---- d1 d1 d1 = 4 [ft] D1 = D * d1+d2+d3 d2 = 2 [ft] D1 = 1.5 [ft] d3 = 2 [ft]

Vw= [ft3] Vw= (1/12) * π * D1 * d1 2 d3 D1 d1 = 4 [ft] d2 oil water is the volume of the cone submerged in oil. [pause] To solve for this volume, we’ll define variable, --- d1 d1 d1 = 4 [ft] D1 = D * d1+d2+d3 d2 = 2 [ft] D1 = 1.5 [ft] d3 = 2 [ft]

Vw= [ft3] Vw= (1/12) * π * D1 * d1 D2 2 d3 d1 = 4 [ft] d2 oil water upper-case D 2, for the diameter of the cone, at the oil-air interface. [pause] Now, we’ll define --- d1 d1 d1 = 4 [ft] D1 = D * d1+d2+d3 d2 = 2 [ft] D1 = 1.5 [ft] d3 = 2 [ft]

Find: Wcone [lbf] -Vw Fb= ρw * g * (SGw *Vw + SGo *Vo) D = 3 [ft]
Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 oil water V o, as, the volume of the cone, with a diameter upper-case D 2, --- d1

Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 oil water MINUS, the volume of the cone which is submerged in water, ---- d1

Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 Vw= [ft3] oil water V w, which we just calculated. [pause] The only unknown variable is the diameter, ---- d1

Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 Vw= [ft3] oil water upper-case D 2, which we can solve, again, by recognizing the proportionality between ---- d1

Find: Wcone [lbf] -Vw = Fb= ρw * g * (SGw *Vw + SGo *Vo) D = 3 [ft]
Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 D2 d1+d2 oil = D d1+d2+d3 water this cone, and the entire cone. After solving for the diameter, D 2, --- d1

Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 D2 d1+d2 oil = D d1+d2+d3 water and plugging in the known values, diameter D 2, --- d1 d1+d2 D2 = D * d1+d2+d3

Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 D2 d1+d2 oil = D d1+d2+d3 water equals, 2.25 feet. [pause] Now if we plug in diameter 2, --- d1 d1+d2 d1 = 4 [ft] D2 = D * d1+d2+d3 d2 = 2 [ft] D2 = [ft] d3 = 2 [ft]

Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 oil Vw= [ft3] water depth 1, depth 2, and V w, this makes the volume V o, equal to, --- d1 d1+d2 d1 = 4 [ft] D2 = D * d1+d2+d3 d2 = 2 [ft] D2 = [ft] d3 = 2 [ft]

Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 oil Vw= [ft3] water Vo= [ft3] 5.596 cubic feet. [pause] Lastly, we’ll return to our equation for --- d1 d1+d2 d1 = 4 [ft] D2 = D * d1+d2+d3 d2 = 2 [ft] D2 = [ft] d3 = 2 [ft]

Find: Wcone [lbf] -Vw Fb= ρw * g * (SGw *Vw + SGo *Vo)
Vo= (1/12) * π * D2 * (d1+d2) 2 D2 -Vw d3 d2 oil Vw= [ft3] water Vo= [ft3] the bouyant force, F b, and after plugging in, --- d1 d1+d2 d1 = 4 [ft] D2 = D * d1+d2+d3 d2 = 2 [ft] D2 = [ft] d3 = 2 [ft]

Find: Wcone [lbf] SGw= 1.00 SGo = 0.89
Fb= ρw * g * (SGw *Vw + SGo *Vo) ρw = 62.3 [lbm/ft3] Vw= [ft3] g= 32.2 [ft/s2] Vo= [ft3] the known variables, the bouyant force, equals, ----

Fb= ρw * g * (SGw *Vw + SGo *Vo) ρw = 62.3 [lbm/ft3] Vw= [ft3] g= 32.2 [ft/s2] Vo= [ft3] lbm* ft Fb= 14,717 14,717, pounds mass, feet, per second squared. [pause] After multiping this value by --- s2

Fb= ρw * g * (SGw *Vw + SGo *Vo) ρw = 62.3 [lbm/ft3] Vw= [ft3] g= 32.2 [ft/s2] Vo= [ft3] lbm* ft lbf* s2 1 Fb= 14,717 * s2 32.2 a unit conversion, the bouyant force, equals, --- lbm*ft

Fb= ρw * g * (SGw *Vw + SGo *Vo) ρw = 62.3 [lbm/ft3] Vw= [ft3] g= 32.2 [ft/s2] Vo= [ft3] lbm* ft lbf* s2 1 Fb= 14,717 * s2 32.2 457 pounds force. [pause] Which we previously determined was, --- lbm*ft Fb= 457 [lbf]

Fb= ρw * g * (SGw *Vw + SGo *Vo) ρw = 62.3 [lbm/ft3] Vw= [ft3] g= 32.2 [ft/s2] Vo= [ft3] lbm* ft lbf* s2 1 Fb= 14,717 * the weight of the cone. [pause] s2 32.2 lbm*ft Fb= 457 [lbf] Wcone= Fb Wcone= 457 [lbf]

Fb= ρw * g * (SGw *Vw + SGo *Vo) ρw = 62.3 [lbm/ft3] Vw= [ft3] g= 32.2 [ft/s2] Vo= [ft3] lbm* ft lbf* s2 1 457 476 495 514 Fb= 14,717 * s2 32.2 When reviewing the possible solutions, --- lbm*ft Fb= 457 [lbf] Wcone= Fb Wcone= 457 [lbf]

Fb= ρw * g * (SGw *Vw + SGo *Vo) ρw = 62.3 [lbm/ft3] Vw= [ft3] g= 32.2 [ft/s2] Vo= [ft3] lbm* ft lbf* s2 1 457 476 495 514 answerA Fb= 14,717 * 32.2 the correct answer is A. [fin] s2 lbm*ft Fb= 457 [lbf] Wcone= Fb Wcone= 457 [lbf]

* ΔP13=-ρA* g * (hAB-h1)+… depth to centroid kg*cm3 1,000 g * m3
a monometer contains 5 different fluids, which include, --- kg*cm3 1,000 * g * m3

Find: Wcone [lbf] D D = 3 [ft] air SGo = 0.89 d3 SGw= 1.00 oil d2

Similar presentations

Presentation on theme: "Find: Wcone [lbf] D D = 3 [ft] air SGo = 0.89 d3 SGw= 1.00 oil d2"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Find: Wcone [lbf] D D = 3 [ft] air SGo = 0.89 d3 SGw= 1.00 oil d2

Similar presentations

Presentation on theme: "Find: Wcone [lbf] D D = 3 [ft] air SGo = 0.89 d3 SGw= 1.00 oil d2"— Presentation transcript:

Similar presentations

About project

Feedback