Unless otherwise stated, take the density of water to be Pw 1000 kg/m² and its specific weight to be w 62.4 lb/ft³. Also, assume all pressures are gage pressures. *2-56. The Morgan Company manufactures a micro- manometer that works on the principles shown. Here there are two reservoirs filled with kerosene, each having a cross- sectional area of 300 mm². The connecting tube has a cross- sectional area of 15 mm² and contains mercury. Determine h if the pressure difference PA PB = 40 Pa. What would h be if water were substituted for mercury? PHg = 13550 kg/m³, Pke = 814 kg/m³. SOLUTION Referring to Fig. a, write the manometer equation starting at A and ending at B. PA + Pigh-Pagh-Pig(h₁-h+h₂) = PB PA PB Pagh Pigh + Pigh₂ (1) Since the same amount of liquidlevingne teservoir will enter the left tube AR= Substitute this result into Eq. (1) Mercury: h When P₁ = Pke When p₁ Pke= 40 N/m² = [ (13550 kg/m³) (9.81 m/s²) - (1. = 0.3191 (10) m = 0.319 mm = = A PA PB P2gh pigh + pig| PA PB = = [P8 -(1-4)P₁8] (+) h 814 kg/m³, p2 = PHg = 13550 kg/m² and PA - PB = 40 Pa, 814 kg/m², P₂ 15 300 (2) (814 kg/m³) (9.81 m/s²) = Pw 1000 kg/m³ and PA PB = 40 Pa Ans. 15 40 N/m² = [ (1000 kg/m³) (9.81 m/s²) - (1 - 355) (814 kg/m³) (9.81 m/s²)] 300 Water: h = 0.01799 m = 18.0 mm Ans. From the results, we notice that if p2p₁, h will be too small to be read. Hence, when choosing the liquid to be used, p2 should be slightly larger than p, so that the sensitivity of the micromanometer is increased. Liquid 1 initial level h₂ +IN Liquid 2 -Liquid 1 Liquid 2 initial level
Unless otherwise stated, take the density of water to be Pw 1000 kg/m² and its specific weight to be w 62.4 lb/ft³. Also, assume all pressures are gage pressures. *2-56. The Morgan Company manufactures a micro- manometer that works on the principles shown. Here there are two reservoirs filled with kerosene, each having a cross- sectional area of 300 mm². The connecting tube has a cross- sectional area of 15 mm² and contains mercury. Determine h if the pressure difference PA PB = 40 Pa. What would h be if water were substituted for mercury? PHg = 13550 kg/m³, Pke = 814 kg/m³. SOLUTION Referring to Fig. a, write the manometer equation starting at A and ending at B. PA + Pigh-Pagh-Pig(h₁-h+h₂) = PB PA PB Pagh Pigh + Pigh₂ (1) Since the same amount of liquidlevingne teservoir will enter the left tube AR= Substitute this result into Eq. (1) Mercury: h When P₁ = Pke When p₁ Pke= 40 N/m² = [ (13550 kg/m³) (9.81 m/s²) - (1. = 0.3191 (10) m = 0.319 mm = = A PA PB P2gh pigh + pig| PA PB = = [P8 -(1-4)P₁8] (+) h 814 kg/m³, p2 = PHg = 13550 kg/m² and PA - PB = 40 Pa, 814 kg/m², P₂ 15 300 (2) (814 kg/m³) (9.81 m/s²) = Pw 1000 kg/m³ and PA PB = 40 Pa Ans. 15 40 N/m² = [ (1000 kg/m³) (9.81 m/s²) - (1 - 355) (814 kg/m³) (9.81 m/s²)] 300 Water: h = 0.01799 m = 18.0 mm Ans. From the results, we notice that if p2p₁, h will be too small to be read. Hence, when choosing the liquid to be used, p2 should be slightly larger than p, so that the sensitivity of the micromanometer is increased. Liquid 1 initial level h₂ +IN Liquid 2 -Liquid 1 Liquid 2 initial level
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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I don't get this part, please explain
![Unless otherwise stated, take the density of water to be
Pw = 1000 kg/m³ and its specific weight to be yw =
62.4 lb/ft³. Also, assume all pressures are gage pressures.
*2-56. The Morgan Company manufactures a micro-
manometer that works on the principles shown. Here there
are two reservoirs filled with kerosene, each having a cross-
sectional area of 300 mm². The connecting tube has a cross-
sectional area of 15 mm² and contains mercury. Determine
h if the pressure difference PA PB = 40 Pa. What would h
be if water were substituted for mercury? PHg = 13550 kg/m³,
Pke = 814 kg/m³.
SOLUTION
Referring to Fig. a, write the manometer equation starting at A and ending at B.
PA + Pigh₁ pagh - Pig(h₁h+h₂) = PB
PA PB Pagh - Pigh + Pigh₂
(1)
Since the same amount of liquidering the teservoir will enter the left tube
h₂
AR =
Substitute this result into Eq. (1)
h₂ =
PA
= A
PA PB = P2gh - pigh+ pig|
8 (4)
- PB = h[P28 - (1 - A)P₁8]
AR
When P₁ = Pke = 814 kg/m², P2 = PHg = 13550 kg/m³ and PA PB = 40 Pa,
40 N/m² = [ (13550 kg/m³) (9.81 m/s²) – (1
1-
Mercury: h= 0.3191 (10³) m = 0.319 mm
When P₁ = Pie
15
300
(2)
(814 kg/m²) (9.81 m/s²)]
Ans.
814 kg/m³, P₂ = Pw = 1000 kg/m³ and PA - PB = 40 Pa
15
40 N/m² = [ (1000 kg/m³) (9.81 m/s²) - (1 - 350) (814 kg/m³) (9.81 m/s²)]
300,
Water: h = 0.01799 m = 18.0 mm
Ans.
From the results, we notice that if p₂p₁, h will be too small to be read. Hence,
when choosing the liquid to be used, p2 should be slightly larger than p₁ so that the
sensitivity of the micromanometer is increased.
Liquid 1
initial level
h₂
h₂
T
Liquid 2
B
h₂
Liquid 1
-Liquid 2
initial level](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2ccf4a7-576c-462f-b8c3-aa1d4f61721d%2F03f58bfa-9b51-4d3b-9446-c3c2a7d9fe69%2Fkoryl69_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Unless otherwise stated, take the density of water to be
Pw = 1000 kg/m³ and its specific weight to be yw =
62.4 lb/ft³. Also, assume all pressures are gage pressures.
*2-56. The Morgan Company manufactures a micro-
manometer that works on the principles shown. Here there
are two reservoirs filled with kerosene, each having a cross-
sectional area of 300 mm². The connecting tube has a cross-
sectional area of 15 mm² and contains mercury. Determine
h if the pressure difference PA PB = 40 Pa. What would h
be if water were substituted for mercury? PHg = 13550 kg/m³,
Pke = 814 kg/m³.
SOLUTION
Referring to Fig. a, write the manometer equation starting at A and ending at B.
PA + Pigh₁ pagh - Pig(h₁h+h₂) = PB
PA PB Pagh - Pigh + Pigh₂
(1)
Since the same amount of liquidering the teservoir will enter the left tube
h₂
AR =
Substitute this result into Eq. (1)
h₂ =
PA
= A
PA PB = P2gh - pigh+ pig|
8 (4)
- PB = h[P28 - (1 - A)P₁8]
AR
When P₁ = Pke = 814 kg/m², P2 = PHg = 13550 kg/m³ and PA PB = 40 Pa,
40 N/m² = [ (13550 kg/m³) (9.81 m/s²) – (1
1-
Mercury: h= 0.3191 (10³) m = 0.319 mm
When P₁ = Pie
15
300
(2)
(814 kg/m²) (9.81 m/s²)]
Ans.
814 kg/m³, P₂ = Pw = 1000 kg/m³ and PA - PB = 40 Pa
15
40 N/m² = [ (1000 kg/m³) (9.81 m/s²) - (1 - 350) (814 kg/m³) (9.81 m/s²)]
300,
Water: h = 0.01799 m = 18.0 mm
Ans.
From the results, we notice that if p₂p₁, h will be too small to be read. Hence,
when choosing the liquid to be used, p2 should be slightly larger than p₁ so that the
sensitivity of the micromanometer is increased.
Liquid 1
initial level
h₂
h₂
T
Liquid 2
B
h₂
Liquid 1
-Liquid 2
initial level
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