Homework #2ME 363 - Fluid Mechanics

Due Feb 6, 2008Spring Semester 2008

1]The viscosity of a fluid is to be measured by a viscometer constructed of two 75-mm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinders is 0.12 cm. The inner cylinder is rotated at 200 rpm, and the torque is measured to be 0.8 N-m. Determine the viscosity of the fluid. What kind of fluid might have this viscosity?

2]When measuring small pressure differences with a manometer, one arm of the manometer may be inclined to improve the accuracy of the reading. (The pressure difference is still proportional to the vertical distance and not the actual length of the fluid along the tube.) The air pressure in a circular duct is to be measured using a manometer whose open arm is inclined 35 degrees from the horizontal. The density of the liquid in the manometer is 0.81 kg/L, and the vertical distance between the fluid levels in the two arms of the manometer is 8 cm. Determine the gage pressure of air in the duct and the length of the fluid column in the inclined arm above the fluid level in the vertical arm.

3]A multifluid container is connected to a U-tube. For the given specific gravities and fluid column heights, determine the gage pressure at A. Also determine the height of a mercury column that would create the same pressure at A.

4]A glass that is fully filled with water and covered with a thin paper is inverted. Determine the pressure at the bottom (narrow end) of the glass, and explain why water does not fall out.

5]A retaining wall against a mud slide is to be constructed by placing 0.8-m-high by 0.2-m-wide rectangular concrete blocks (ρ=2700 kg/m3) side-by-side. The friction coefficient between the ground and the concrete blocks is f = 0.3, and the density of the mud is 1800 kg/m3. Determine the mud height at which (a) the blocks will overcome friction and begin sliding and (b) the blocks will tip over.

6]A water trough of semicircular cross-section of radius 0.5 m consists of two symmetric parts hinged to each other at the bottom. The two parts are held together by a cable and turnbuckle placed every 3 m along the length of the trough (the dimension going into the page). Calculate the tension in each cable when the trough is filled to the rim.

Problem 1:

SolutionThe torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be determined.

Assumptions1 The inner cylinder is completely submerged in oil. 2 The viscous effects on the two ends of the inner cylinder are negligible. 3 The fluid is Newtonian.

AnalysisSubstituting the given values, the viscosity of the fluid is determined to be

DiscussionThis is the viscosity value at the temperature that existed during the experiment. Viscosity is a strong function of temperature, and the values can be significantly different at different temperatures. This could be SAE 10W oil at a high temperature ~ 50 degrees C.

Problem 2:

SolutionThe air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined.

AssumptionsThe manometer fluid is an incompressible substance.

PropertiesThe density of the liquid is given to be  = 0.81 kg/L = 810 kg/m3.

AnalysisThe gage pressure in the duct is determined from

The length of the differential fluid column is

DiscussionNote that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability (and therefore higher precision).

Problem 3:

SolutionA multi-fluid container is connected to a U-tube. For the given specific gravities and fluid column heights, the gage pressure at A and the height of a mercury column that would create the same pressure at A are to be determined.

Assumptions1 All the liquids are incompressible. 2 The multi-fluid container is open to the atmosphere.

PropertiesThe specific gravities are given to be 1.26 for glycerin and 0.90 for oil. We take the standard density of water to be w=1000 kg/m3, and the specific gravity of mercury to be 13.6.

AnalysisStarting with the atmospheric pressure on the top surface of the container and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach point A, and setting the result equal to PA give

Rearranging and using the definition of specific gravity,

or

Substituting,

The equivalent mercury column height is

DiscussionNote that the high density of mercury makes it a very suitable fluid for measuring high pressures in manometers.

Problem 4

SolutionA glass filled with water and covered with a thin paper is inverted. The pressure at the bottom of the glass is to be determined.

Assumptions1 Water is an incompressible substance. 2 The weight of the paper is negligible. 3 The atmospheric pressure is 100 kPa.

PropertiesWe take the density of water to be  =1000 kg/m3.

AnalysisThe paper is in equilibrium, and thus the net force acting on the paper must be zero. A vertical force balance on the paper involves the pressure forces on both sides, and yields



That is, the pressures on both sides of the paper must be the same.

The pressure at the bottom of the glass is determined from the hydrostatic

pressure relation to be



Substituting,

DiscussionNote that there is a vacuum of 1 kPa at the bottom of the glass, and thus there is an upward pressure force acting on the water body, which balanced by the weight of water. As a result, the net downward force on water is zero, and thus water does not flow down.

Problem 5

SolutionA retaining wall against mud slide is to be constructed by rectangular concrete blocks. The mud height at which the blocks will start sliding, and the blocks will tip over are to be determined.

AssumptionsAtmospheric pressure acts on both sides of the wall, and thus it can be ignored in calculations for convenience.

PropertiesThe density is given to be 1800 kg/m3 for the mud, and 2700 kg/m3 for concrete blocks.

Analysis(a) The weight of the concrete wall per unit length (L = 1 m) and the friction force between the wall and the ground are

The hydrostatic force exerted by the mud to the wall is

Setting the hydrostatic and friction forces equal to each other gives

(b) The line of action of the hydrostatic force passes through the pressure center, which is 2h/3 from the free surface. The line of action of the weight of the wall passes through the midplane of the wall. Taking the moment about point A and setting it equal to zero gives

Solving for h and substituting, the mud height for tip over is determined to be

DiscussionThe concrete wall will slide before tipping. Therefore, sliding is more critical than tipping in this case.

Problem 6

SolutionTwo parts of a water trough of semi-circular cross-section are held together by cables placed along the length of the trough. The tension Tin each cable when the trough is full is to be determined.

Assumptions1 Atmospheric pressure acts on both sides of the trough wall, and thus it can be ignored in calculations for convenience. 2 The weight of the trough is negligible.

PropertiesWe take the density of water to be 1000 kg/m3 throughout.

AnalysisTo expose the cable tension, we consider half of the trough whose cross-section is quarter-circle. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are:

Horizontal force on vertical surface:

The vertical force on the horizontal surface is zero, since it coincides with the free surface of water. The weight of fluid block per 3-m lengthis

Then the magnitude and direction of the hydrostatic force acting on the surface of the 3-m long section of the trough become

Therefore, the line of action passes through the center of the curvature of the trough, making 57.5 downwards from the horizontal. Taking the moment about point A where the two parts are hinged and setting it equal to zero gives

Solving for T and substituting, the tension in the cable is determined to be

DiscussionThis problem can also be solved without finding FR by finding the lines of action of the horizontal hydrostatic force and the weight.