Tutorial 03 Conservation of Energy Question and Answer, Chemical Engineering Fluid Mechanics
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Descrição: Tutorial de Geoprocessamento tendo como base aplicativo ArcGis/ Desktop 9.3. Desenvolvido por: Osvaldo José Ribeiro Pereira Graduado em Geografia pela UNESP e mestre em Ciências do Solo pela USP.
proses desain database dan mengimplementasikan model REA dalam databaseFull description
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CBB 1033 Chem. Eng. Fluid Mechanics
January 2011
Chapter 2 Solution to Tutorial Question
1.
2
The pressure in a steam pipeline is measured to be 500 kN/m . Express this pressure as a head of water. If the atmospheric pressure is 1.013 bar, what is the absolute pressure of the steam in 2 kN/m . Solution: Pg
= 500 kPa = gh 500 10
h=
3
= 50.97 m
1000 9.81
101.3kPa = 601.3 kPa Pabs = 500 kPa 1.013bar ba r 1.013bar
2.
A mercury barometer reads 760 mm at sea level and 750 mm on top of a mountain. How high is 3 the mountain above the sea level? Assume a constant density of air of 1.2 kg/m and the specific gravity of mercury to be 13.6. Solution: P2 – P1 = – g h
h =
3.
760 750 103 13.6 1000 = 113.3 m 1.2
The hydraulic press shown in the figure below is expected to give a load M of 2 tonnes at the end of a 120 mm piston. What force F will be required at the plunger, which is of 20 mm? (SG for oil used in the hydraulic press is 0.8)
F
120 mm
1m W
20 mm
1
CBB 1033 Chem. Eng. Fluid Mechanics
January 2011
Solution: Taking P1 at the entry of 20 mm piston, and P 2 at the end of 120 mm piston. P1 - gh = P 2 F W ρ oilgh Aφ20mm Aφ120mm
The hydraulic brakes in a motorcar are worked by a plunger pump of diameter 2.0 cm. The fourwheel cylinders are of diameter 3.0 cm each. If the operator applies a force of 35 N at the plunger, what will be the force exerted at each wheel cylinder Assume that all components are at the same horizontal plane. Solution: FP1
FP2
FP3
FP4
35 N
B
A
Since all the cylinders are at horizontal position, h = 0 And PP1 = PP2 = PP3 = PP4
F Aφ2cm
FP Aφ3 cm
33 FP 35 2 78.75N 2
5.
Water contained in a sealed tank with air occupying the space above the water, as shown in the figure below. A mercury U-tube manometer is connected to the air space as shown. A pressure gage is connected 2 m below the surface of the water. If the manometer level in the U-tube reads 0.2 m, what pressure will the pressure gage read in kPa? (SG for mercury = 13.6) 2
CBB 1033 Chem. Eng. Fluid Mechanics
January 2011
0.2 m
Pressure gage
2m
Solution: Assumption: All liquids are incompressible fluids. Pressure of air in the tank acts in all direction and equal to the pressure at the surface of water. Pgage P – watergh = P surface = – Hggh + P atm Pgage P = watergh – Hggh + P atm = (1000)(9.81)(2) – (13.6)(1000)(9.81)(0.2) + 0 = – 7.063 kPa 6.
A mercury manometer is connected to a closed oil tank which has a layer of water at the bottom. Calculate the pressure of the air above the oil.
Air 2m
Oil (SG = 0.8)
Water
5m 0.5 m
Mercur
SG = 13.6
Solution: Assumption: All liquids are incompressible fluids so that the densities are constant. Pair + Poil + Pwater – PHg = Patm Pair = PHg – Pwater – Poil + P atm = (1000)(9.81)[(13.6)(0.5) – (5) – (0.8)(2)] = 1962 Pa 3
CBB 1033 Chem. Eng. Fluid Mechanics
7.
January 2011
Oil of specific gravity 0.85 is flowing in a pipeline as shown below. A manometer containing water is connected as shown. Calculate the pressure in the pipeline.
Water
150 mm 100 mm
Oil
Solution: Assumption: All liquids are incompressible fluids so that the densities are constant. Taking P2 at the manometer entry at the pipe wall, and P 1 = Patm P2
In the system shown on the next page, the pressure gage reading at A is 2 kPa. Specific gravity of oil is 0.8. Calculate: a. The length of the water column Y in the open piezometer tube B. b. The difference of the mercury columns, h, in the U-tube manometer. (SG mercury = 13.6)
Solution: a.
Assumption: Air in the tank acts in all directions. All liquids are incompressible fluid with constant densities. Pgage + Poil + Pwater – Pwater = Patm 2000 + (0.8)(1000)(9.81)(0.5) + (1000)(9.81)(0.8) – (1000)(9.81)(Y) = 0 Y = 1.4 m
b.
Taking the opening of tube as P 1 and opening of piezometer tube a s P 2 P1 + (13.6)(1000)(9.81)(h) – (1000)(9.81)(1.4+ 0.6) = P 2 h = 0.147 m
4
CBB 1033 Chem. Eng. Fluid Mechanics
January 2011
A B
Air
0.5 m
Oil Y Water
0.8 m
h
9.
0.6 m
Two pipes A and B carry water and oil, respectively. A differential manometer is connected as shown. The relevant measurements are shown in the figure. Find the difference of pressure in the pipes A and B. Specific gravity of oil is 0.6 and that of mercury is 13.6.
A U-tube manometer containing water is connected to two tanks. Find the pressure difference in the air in the two tanks.
0.8 m
Air
Air
Oil (SG = 0.9)
Liquid A (SG = 0.8)
0.5 m
1.2 m Tank X
Tank Y 0.8 m
Solution: Px + (0.9)(1000)(9.81)(0.8+1.2) – (1000)(9.81)(0.8) – (0.8)(1000)(9.81)(0.5 + (1.2 – 0.8)) = P Y PY – PX = 2746.8 Pa 11.
An inverted U-tube manometer is used to measure the difference of water pressure between two points in a pipe. a. Find the difference of pressure between the points. b. If the pressure at one point is reduced by 2.5 kPa, what will be the new difference of levels of water in the manometer? Solution: a.
PA + (1000)(9.81)(x) – (1000)(9.81)(0.45 + x + 0.23) = P B PB – PA = (1000)(9.81)(0.45 + 0.23) = 6670.8 Pa
b.
Say New PA = PA – 2500 Pa PA – 2500 Pa + (1000)(9.81)(x) – (1000)(9.81)(y + x + 0.23) = P B PB – PA + 2500 Pa = (1000)(9.81)(y + 0.23) 6670.8 + 2500 = (1000)(9.81) y + 2256.3 y = 0.705 m Say new PB = PB – 2500 PA + (1000)(9.81)(x) – (1000)(9.81)(y + x + 0.23) = P B – 2500 Pa PB – PA – 2500 Pa = (1000)(9.81)(y + 0.23) 6670.8 – 2500 = (1000)(9.81) y + 2256.3 y = 0.195 m
6
CBB 1033 Chem. Eng. Fluid Mechanics
January 2011
Air
0.45 m
Water 0.23 m Water
Pipe A
12.
Pipe B
X and Y are two pipes carrying water. Inverted U-tube manometer containing oil is connected as shown. Find the difference of pressure, p x – py.
Oil (SG = 0.85)
200 mm
+
600 mm
Pipe Y
+
Pipe X Solution: Px – (1000)(9.81)(0.6 + a + 200) + (0.85)(1000)(9.81)(0.2) + (1000)(9.81)(a) = P Y PX – PX = (1000)(9.81)(0.6 + a + 0.2) – (0.85)(1000)(9.81)(0.2) – (1000)(9.81)(a) = 6180.3 Pa
7
CBB 1033 Chem. Eng. Fluid Mechanics
13.
January 2011
In the compound U-tube manometer shown, find the specific gravity of the gauge liquid B. The gage pressure at A is –10 kPa and the liquid in pipe A is water.
Air
400 mm
Pipe A
+
500 mm
Liquid B
Solution: Assumption: density of air is very low compared to manometer liquids, so the pressure exerted by air in the tube is relatively negligible compared to the pressure exerted by the manometer liquids. PA + wgh500mm - airgh + Bgh400mm = Patm 3 -3 -3 – 10 10 + (1000)(9.81)(500 10 ) + B (9.81)(400 10 ) = 0
