Chapter 3 Problem Definition Situation Regarding The Hydraulic Jack

Unlock access to all the studying documents.

View Full Document

Assumptions:

Density of air is constant.

Properties:

Air, ρ=1.1kg/m3.

Solution:

Pressure at summit:

d.)

Situation:

Pressure increases with depth in a lake.

∆z=350m.

Find:

Pressure (MPa).

Properties:

Water, γ=9810N/m3.

Solution:

e.)

Situation:

Pressure increase with water depth in a standpipe.

∆z=70m.

Find:

Pressure (kPa).

Properties:

Water, γ=9810N/m3.

Solution:

3.15: PROBLEM DEFINITION

Situation:

Airabovealongtubeispressurized.

Initial state: pair1 =50kPa-vacuum

Final state: pair2 =25kPa-vacuum.

Find:

Will hincrease or decrease?

The change in water column height (∆h)in meters.

Assumptions:

PLAN

Since pressure increases, the water column height will decrease. Use absolute pressure

in the hydrostatic equation.

1. Find h(initial state)by applying the hydrostatic equation.

SOLUTION

1. Initial State. Locate point 1 on the reservoir surface; point 2 on the water surface

inside the tube:

2. Final State:

REVIEW

Tip! In the hydrostatic equation, use gage pressure or absolute pressure. Using

vacuum pressure will give a wrong answer.

3.16: PROBLEM DEFINITION

Situation:

A closed tank contains air, oil, and water.

Find:

Speciﬁc gravity of oil.

Pressure at C (kPa-gage).

Sketch:

0.5 m

1.0 m

Air

= 50.0 kPa

Oil

Properties:

Water (10 ◦C), Table A.5, γ=9810N/m3.

PLAN

1. Find the oil speciﬁc gravity by applying the hydrostatic equation from A to B.

2. Apply the hydrostatic equation to the water.

SOLUTION

1. Hydrostatic equation (from oil surface to elevation B):

2. Hydrostatic equation (in water):

3. Hydrostatic equation (in oil):

3.17: PROBLEM DEFINITION

Situation:

A manometer is described in the problem statement.

dleft =1mm,d

right =3mm.

Find:

Water surface level in the left tube as compared to the right tube.

SOLUTION

3.18: PROBLEM DEFINITION

Situation:

A force is applied to a piston.

F1=200N,d1=4cm,d2=10cm.

Find:

Force resisted by piston.

Assumptions:

Neglect piston weight.

PLAN

Apply the hydrostatic equation and equilibrium.

SOLUTION

1. Equilibrium (piston 1)

2. Hydrostatic equation

3. Equilibrium (piston 2)

3.19: PROBLEM DEFINITION

Situation:

Regarding the hydraulic jack in Problem 3.18 (EFM 10e), which ideas were used

to analyze the jack? (select all that apply)

a. pressure = (force)(area)

b. pressure increases linearly with depth in a hydrostatic ﬂuid

c. the pressure at the very bottom of the 4-cm chamber is larger than the pressure

at the very bottom of the 10-cm chamber

d. when a body is stationary, the sum of forces on the object is zero

e. when a body is stationary, the sum of moments on the object is zero

f. pressure = (weight/volume)(change in elevation)

SOLUTION

Correct answers are a, b, d, e and f.

3.20: PROBLEM DEFINITION

Situation:

A diver goes underwater.

∆z=50m.

Find:

Gage pressure (kPa).

Ratio of pressure to normal atmospheric pressure.

Properties:

Water (20 ◦C), Table A.5, γ=9790N/m3.

PLAN

1. Apply the hydrostatic equation.

2. Calculate the pressure ratio (use absolute pressure values).

SOLUTION

1. Hydrostatic equation

2. Calculate pressure ratio

3.21: PROBLEM DEFINITION

Situation:

Water and kerosene are in a tank.

zwater =0.8m,zkerosene =0.3m.

Find:

Gage pressure at bottom of tank (kPa-gage).

Properties:

Water (20 ◦C),TableA.5,γw= 9790 N/m3.

Kerosene (20 ◦C) ,Table A.4, γk=8010N/m3.

SOLUTION

Manometer equation (add up pressure from the top of the tank to the bottom of the

3.22: PROBLEM DEFINITION

Situation:

A hydraulic lift is being designed.

Wmax =10ton = 20000 lbf,Wparts =1000lbf.

∆L=6ft,∆t=20s.

Diameter range: 2−8in.

Pressure range: 200 −3000 psig.

Available pumping capacity: 5,10,15 gpm.

Find:

Select a hydraulic pump capacity (gpm).

Select a cylinder diameter (D).

PLAN

Apply equilibrium to ﬁnd the smallest bore diameter (D) that works. Then ﬁnd the

SOLUTION

Equilibrium (piston)

Corresponding minimum bore diameter is

Conversion from gallons to cubic feet ¡ft3¢:7.48gal=1ft

3.Thus, the maximum

bore diameter for three pumps (to meet the lift speed speciﬁcation) is given in the

table below.

pump (gpm) pump (cfm) A (ft2)D

max (in)

1.) The 10 gpm pump will work with a bore diameter between 3.0 and 3.6 inches.

2.) The 15 gpm pump will work with a bore diameter between 3.0 and 4.6 inches.

REVIEW

1. These are preliminary design values. Other issues such as pressure drop in the

hydraulic lines and valves would have to be considered.

2. We recommend selecting the 15 gpm pump and a 4.5 inch bore to provide

latitude to handle pressure losses, and to reduce the maximum system pressure.

3.23: PROBLEM DEFINITION

Situation:

Initial State: Water levels as shown. Valve in open.

Final State: Water is added to the tank with the valve closed.

Find:

Increase of water level ∆in manometer (in meters).

Properties:

Water (20 ◦C),TableA.5,γw=9790 N/m3.

patm =100kPa.

Assumptions:Idealgas.

PLAN

Apply the hydrostatic equation and the ideal gas law.

SOLUTION

Ideal gas law (mole form; apply to air in the manometer tube)

State 1 (before air is compressed)

3.24: PROBLEM DEFINITION

Situation:

Atankisﬁtted with a manometer.

S=3,z1=0.15 m.

Properties:

γwater=9810 N/m3.

PLAN

Apply the hydrostatic principle to the water and then to the manometer ﬂuid.

SOLUTION

1. Hydrostatic equation (location 1 is on the free surface of the water; location 2 is

the interface)

2. Hydrostatic equation (manometer ﬂuid; let location 3 be on the free surface)

3. Solve for ∆h

3.25: PROBLEM DEFINITION

Situation:

A mass sits on top of a piston situated above a reservoir of oil.

Weight

Piston

Find:

Derive an equation for h2in terms of the speciﬁed parameters.

Assumptions:

Neglect the mass of the piston.

Neglect friction between the piston and the cylinder wall.

The pressure at the top of the oil column is 0 kPa-gage.

PLAN

1. Relate wto pressure acting on the bottom of the piston using equilibrium.

2. Related pressure on the bottom of the piston to the oil column height using the

hydrostatic equation.

3. Find h2bycombiningsteps1and2.

SOLUTION

1. Equilibrium (piston):

2. Hydrostatic equation. (point 1 at btm of piston; point 2 at top of oil column):

water h2(2)

3. Combine Eqs. (1) and (2):

REVIEW

1. Notice. Column height h2increases linearly with increasing weight w.Similarly, h2

decreases linearly with Sand decreases quadratically with D1.

2. Notice. The apparatus involved in the problem could be used to create an instru-

ment for weighing an object.

3.26: PROBLEM DEFINITION

Situation:

A mass sits on top of a piston situated above a reservoir of oil.

m=5kg,S=0.8,h1=42mm.

D1=120mm,D2=5mm.

Weight

Piston

Find:

Calculate h2(m).

Assumptions:

Neglect the mass of the piston.

Neglect friction between the piston and the cylinder wall.

The pressure at the top of the oil column is 0 kPa-gage.

PLAN

1. Relate mass mto pressure acting on the bottom of the piston using equilibrium.

2. Related pressure on the bottom of the piston to the oil column height using the

hydrostatic equation.

3. Find h2bycombiningsteps1and2.

SOLUTION

1. Equilibrium (piston):

2. Hydrostatic equation. (point 1 at btm of piston; point 2 at top of oil column):

3. Combine Eqs. (1) and (2):