$ Water P2 Pa = 101 kPa 3.54 For the pipe-flow reducing section of Fig. P3.54, D1 = 8 cm, D2 = 5 cm, and p2 = 1 atm. All fluids are at 20°C. If V1 = 5 m/s and the manometer reading is h = 58 cm, estimate the total horizontal force resisted by the flange bolts. h Mercury Fig. P3.54 Solution: Let the CV cut through the bolts and through section 2. For the given manometer reading, we may compute the upstream pressure: P₁ P2 = (merc - water )h = (132800-9790)(0.58 m) ≈ 71300 Pa (gage) Now apply conservation of mass to determine the exit velocity: Q₁ = Q₂, or (5 m/s)(π/4)(0.08 m)² = V₂(π/4)(0.05)², solve for V₂ ~ 12.8 m/s Finally, write the balance of horizontal forces: ΣF、 =-Fbolts +P1,gageA₁ €m(V₂ − V₁), T T or: Fbolts (71300) (0.08)²-(998) (0.08)² (5.0)[12.8-5.0] ≈ 163 N_Ans. = 4
$ Water P2 Pa = 101 kPa 3.54 For the pipe-flow reducing section of Fig. P3.54, D1 = 8 cm, D2 = 5 cm, and p2 = 1 atm. All fluids are at 20°C. If V1 = 5 m/s and the manometer reading is h = 58 cm, estimate the total horizontal force resisted by the flange bolts. h Mercury Fig. P3.54 Solution: Let the CV cut through the bolts and through section 2. For the given manometer reading, we may compute the upstream pressure: P₁ P2 = (merc - water )h = (132800-9790)(0.58 m) ≈ 71300 Pa (gage) Now apply conservation of mass to determine the exit velocity: Q₁ = Q₂, or (5 m/s)(π/4)(0.08 m)² = V₂(π/4)(0.05)², solve for V₂ ~ 12.8 m/s Finally, write the balance of horizontal forces: ΣF、 =-Fbolts +P1,gageA₁ €m(V₂ − V₁), T T or: Fbolts (71300) (0.08)²-(998) (0.08)² (5.0)[12.8-5.0] ≈ 163 N_Ans. = 4
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![Water
da ne
P₂ Pa = 101 kPa
=
3.54 For the pipe-flow reducing section
of Fig. P3.54, D1 = 8 cm, D2 = 5 cm, and p2
1 atm. All fluids are at 20°C. If V1 = 5
m/s and the manometer reading is h = 58
cm, estimate the total horizontal force
resisted by the flange bolts.
h
Mercury
Fig. P3.54
Solution: Let the CV cut through the bolts and through section 2. For the given
manometer reading, we may compute the upstream pressure:
P₁ P2 = (merc - / water )h = (132800-9790)(0.58 m)≈ 71300 Pa (gage)
Now apply conservation of mass to determine the exit velocity:
Q₁ = Q₂, or (5 m/s)(π/4)(0.08 m)² = V₂(π/4)(0.05)², solve for V₂ ≈ 12.8 m/s
Finally, write the balance of horizontal forces:
ΣFx = -Fbolts +P1,gage A₁ m(V₂ - V₁),
T
T
or: Folts (71300) (0.08)²-(998) (0.08)² (5.0)[12.8–5.0] ≈ 163 N Ans.
-
0000000000
seves
in
S](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e361db2-2af2-4ae7-a51c-edcbfca125c1%2Fdb62092e-8ddb-45f5-954a-4db363bc2ac4%2F9vca6lb_processed.png&w=3840&q=75)
Transcribed Image Text:Water
da ne
P₂ Pa = 101 kPa
=
3.54 For the pipe-flow reducing section
of Fig. P3.54, D1 = 8 cm, D2 = 5 cm, and p2
1 atm. All fluids are at 20°C. If V1 = 5
m/s and the manometer reading is h = 58
cm, estimate the total horizontal force
resisted by the flange bolts.
h
Mercury
Fig. P3.54
Solution: Let the CV cut through the bolts and through section 2. For the given
manometer reading, we may compute the upstream pressure:
P₁ P2 = (merc - / water )h = (132800-9790)(0.58 m)≈ 71300 Pa (gage)
Now apply conservation of mass to determine the exit velocity:
Q₁ = Q₂, or (5 m/s)(π/4)(0.08 m)² = V₂(π/4)(0.05)², solve for V₂ ≈ 12.8 m/s
Finally, write the balance of horizontal forces:
ΣFx = -Fbolts +P1,gage A₁ m(V₂ - V₁),
T
T
or: Folts (71300) (0.08)²-(998) (0.08)² (5.0)[12.8–5.0] ≈ 163 N Ans.
-
0000000000
seves
in
S
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