1. 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

2. Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 B 0.75C
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

3. Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 B 0.75C
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

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

5. Find: P4,gauge [lb/in2] Liquid S.G. P1,abs=14.7 [lb/in2] A 1.20 B 0.75C
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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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

35. 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

36. 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

37. 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]

38. 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]

39. 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]

40. 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]

41. 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

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