Sunday, January 17, 2010

HIDROSTATIC (TUTORIAL)

1. Find the pressure due to the fluid at a depth of 76cm in still,
a) Water (Pω= 1.00g/cm )
b) Mercury (P= 13.6 g/cm )

water
P = ρgh
= ( 0.001 kg/m³) (9.81m/s²) ( 0.76m)
= 7.4556 kPa
= 7.5 kPa

Mercury
( pm= 13.6g / cm³)
P = ρgh
= ( 0.0136 kg / m³) ( 9.81m/s²) (0.76m)
= 101.3962 kPa
= 101.4 kPa


2. A weighted piston confines a fluid density P in a closed container. The combined weight of piston and weight is 200N, and the cross- sectional area of the piston is A= 8.0 cm . Find the total pressure at point B if the fluid is mercury and h= 25cm. What would aan ordinary pressure gauge read mercury and h= 25cm. What would an ordinary pressure gauge read at B?

P = F / A + Pgh
= [200N/8 x 10 ̄4] + [(13600)(9.8m/s²)(0.25)]
= 250000+33354
=283,354 @ 283.4 kPa .
Pm =(13.6g/cm³)(9.81)(76)
= 0.136 kg/m (9.81)(76)
= 1.01 x 10 5 Nm
P = 283.4 kPa + Pm
= (283.4kPa) + (1.01 x 10 5 Nm)
= 384.4 kPa.


3. A vertical test tube has 2.0cm of oil (P=0.8g/cm ) floating on 8.0cm of water. What is the pressure at the bottom of the tube due to the fluid in it?

P 1 = ρgh
= 800 kg/m ( 9.8 m/s ̄ ² ) (2 x 10 ̄ ²)
= 156.96
P 2 = ρgh
= 1000 ( 9.8 m/s ²) ( 8 x 10 ̄ ²)
= 784
P1 + P2 = 156.96 + 784
= 0.94 kpa


4. The U-tube device conneted to the tank. What is the pressure in the tank it atmospheric pressure is 76cm of mercury? The density of mercury is 13.6 g/cm .


ρ = 76 cm – 13.6 g/cm³ (9.81 mg²)(5 x 10 ̄ ²)
= 76 cm – 13600 kg /cm³ ( 9.81 ms²)(5 x 10 ̄ ²)
= 101.325 kpa – 66708
= 95 kpa


5. The mass of a block of aluminium is 25.0g.
a) What is it’s volume?
b) What will be tension in a string that suspends the block when the block is totally submerged in water?
The density of aluminium is 2700 kg/m .

F = vg( e- pai)
= (9.26 x 10̄̄ ̄6 x 9.8 ) ( -1700 )
=9.0748 x 10 ̄ 5 (-1700)
= 0.154N
V = m/p
= 0.025/ 2700
= 9.259 x 10 ̄ 6
= 9.26 X 10 ̄ 6


6. A solid aluminium cylinder with P= 2700kg/mpadu has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

FB = Ví , Pí , a
PT = FB / Vi a
FB + Fr = mg
FB = mg – Fr
= ( 0.067 x 9.81 ) – ( 0.045 x 9.81 )
= 0.65727 – 0.44145
= 0.21582
V1 = M / P = 0.067 / 2700
= 2.481 x 10 ̄ 5
P1 = 0.21582
= ( 2.481 x 10 ̄ 5 ) ( 9.81 m / s²)
= 8.9 x 10 ² kg / m³

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