Why is it that the P2 = 101.325 is added by 9.81h? Problem 2-44A bottle consisting of a cylinder 15 cm in diameter and 25 cm high, has a neckhich is 5 cm diameter and 25 cm long. The bottle is inserted vertically inater, with the open end down, such that the neck is completely filled withwater. Pind the depth to which the open end is submerged. Assume normalharometric pressure and neglect vapor pressure.BolutionApplying Boyle's Lawn Vi-pa Va15 cmoefore the bottle was insertedVolume of air25 amV - (15P (25) + (5 (25)V = 4,908.74 cmWaterAbsolute pressure in airP- 101.325Water25 amWhen the bottle is inserted:Volume of air:5 omVz = (15) (25)V=4,417.9 cmPressure in air:P= 101.325 + 981Ip V - pa Val101.325(4,908.74) (101.325 - 981 4,417 9)101.325 + 9.81 - 112.58=1.15 cm*-h + 25 = 26.15 cm

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Engineering

Civil Engineering

Why is it that the P2 = 101.325 is added by 9.81h?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Why is it that the P2 = 101.325 is added by 9.81h?

Problem 2-44
A bottle consisting of a cylinder 15 cm in diameter and 25 cm high, has a neck
hich is 5 cm diameter and 25 cm long. The bottle is inserted vertically in
ater, with the open end down, such that the neck is completely filled with
water. Pind the depth to which the open end is submerged. Assume normal
harometric pressure and neglect vapor pressure.
Bolution
Applying Boyle's Law
n Vi-pa Va
15 cmo
efore the bottle was inserted
Volume of air
25 am
V - (15P (25) + (5 (25)
V = 4,908.74 cm
Water
Absolute pressure in air
P- 101.325
Water
25 am
When the bottle is inserted:
Volume of air:
5 om
Vz = (15) (25)
V=4,417.9 cm
Pressure in air:
P= 101.325 + 981
Ip V - pa Val
101.325(4,908.74) (101.325 - 981 4,417 9)
101.325 + 9.81 - 112.58
=1.15 cm
*-h + 25 = 26.15 cm

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