Upcoming SlideShare
×

# assignment 1 properties of fluids-Fluid mechanics

3,154 views

Published on

1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
3,154
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
103
0
Likes
1
Embeds 0
No embeds

No notes for slide

### assignment 1 properties of fluids-Fluid mechanics

1. 1. PHYS207 Assignment 1 Properties of FluidsQ1. Determine the gauge pressure at point A in the water pipe due to the deflection of mercury in theu-tube manometer. The liquid flowing in the pipe is water. The end of the manometer is open to theatmosphere.( H2O = 1,000 kg m-3Hg = 13,570 kg m-3)We can work from one side of the system (where the pressure is known to be atmospheric), to theother (unknown) end, adding static pressure as we progress downwards into depth and subtracting itupwards:“The pressure at D, plus 0.9 m of mercury, less 0.6 m of water, gives the pressure at A”Pressure changes are given by P = g Hso PA = PD + 13570(9.8)(3.9 – 3.0) – 1000(9.8)(3.6 – 3.0)PD is atmospheric pressure; the gauge pressure is what’s left after we subtract it:PA(Gauge) = 13,570(9.8)0.9 – 1000(9.8)0.6 = 113.8x103Pa = 113.8 kPaAB C 3.0 mD 3.9 m3.6 mX-section throughpressurised pipe
2. 2. Q2. The pressure in an automobile tyre depends on the temperature of the air in the tyre. When the airtemperature is 25 oC the pressure gauge reads 210 kPa. If the volume of the tyre is effectively constantat 0.025 m3, determine the pressure in the tyre when the air temperature inside the tyre rises to 50 oCdue to road heating. Also determine the mass (in kg) of air that must be bled off to restore pressure toits original value at this temperature. Assume atmospheric pressure to be 100 kPa.We can use the ideal gas law PV = nRT or its formulation in terms of density P = RT to relatepressure to temperature and amount of gas. It is important to recall that the ideal gas law uses absolutetemperatures and pressures, thus atmospheric pressure must be added to gauge pressures and 273 Kmust be added to centigrade temperatures:P1 = 210x103+ 100x103= 310x103PaT1 = 25 + 273 = 298 KSpecific Gas Constant R (Air) = 287 J/kg/K (Table page 15 Study Guide)We can now calculate the density of air in the tyre at 25 oC:P1 = RT1= P1 / RT1 = 310x103/ (287 x 298) = 3.625 kg/m3Since the volume of the tyre is effectively constant, the density does not change as the temperatureincreases to 50 oC:T2 = 50 + 273 = 323 KP2 = RT2 = 3.625(287)323 = 336x103Pa (absolute, or 236 x103Pa gauge)Restoring the pressure to 310x103Pa at 50 oC reduces the density:P3 = RT2 = 310x103P3 / RT2 = 310x103/(287x323) = 3.344 kg/m3The amount of gas removed can be calculated by expressing the density as mass by multiplying outthe volume:M =M1 = 3.625(0.025) = 90.6 x10-3kgM2 = 3.344(0.025) = 83.6 x10-3kgThe difference is 7 g of air.Q3. A tank consists of two chambers, one open to the atmosphere and another closed and filled with agas. The two chambers are connected by a sluice opening as shown. What is the residual pressure inthe gas, in gauge and absolute terms, if the barometric pressure reads 742 mm Hg?
3. 3. The barometric pressure is P = gH = 13570(9.8)0.742 = 98.68 kPaThe pressure at the outside water surface is atmospheric Po = 98.68 kPaThe absolute pressure at the bottom of the tank is P = 98680 + waterg(1.200 – 0.700) = 103.6 kPaThe absolute pressure at the gas is Pabs = 103600 – waterg(1.200) = 91.8 kPaThe gauge pressure in the gas subtracts the atmospheric pressure : Pgauge = 91.8 – 98.68 = – 6.88 kPaQ4. The viscosity of a fluid is to be measured by a viscometer constructed of two 75 cm longconcentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the twocylinders is 1 mm. With the outer cylinder fixed the inner cylinder is rotated at 300 RPM, and thetorque is measured by a spring gauge to be 0.8 Nm. Determine the viscosity of the fluid.The torque is produced by a force at the radius of the inner cylinder times the radius: T = F.roro = 0.15/2 = 0.075 mThe force F is a shear force due to viscous shear stress multiplied by the surface area of the innercylinder:F = AA = DoL = (0.15)0.75 = 0.3534 m2The shear stress is given by Newton’s law of viscositydydVdy is the gap between the cylinders: dy = 0.001 mdV is the linear velocity between the cylinders: this can be found from the angular velocity:= 300 RPM = 300x2 rads/minute = 300x2 rads/sec= 31.42 r/sThis gives a linear velocity of dV = r = 0.075(31.42) = 2.357 m/sdV/dy is thus 2.357/.001 = 2357 /sgaswaterBarometeratmosphere700 mmConnecting sluice1200 mm
4. 4. Assembling these terms:T = 0.8 N.m = Fro = F(0.075) = A(0.075) = dV/dy(0.3534)0.075 = (2357)(0.3534)(0.075)Thus = 0.8/(2357(0.3534)(0.075)) = 0.0128 Pa.sQ5. Molten rock 0.5 m thick flows down a 30oslope with an apparent velocity of 2 m/s. a) Assuminga linear velocity profile as a function of depth (V = ky) and a density of 2,500 kg/m3, calculate theviscosity of the lava.The apparent velocity Va is the surface velocity, since it is the only part of the flow that you can see.The linear velocity profile is V = Va(y/0.5) where y varies from 0 to 0.5, the depth of the flow.In steady flow the weight forces acting downslope are balanced by friction shear forces acting on theflow bed. Frictional shear is given by Newton’s law asdydVWhereYVdydV aA shear force is expressed over a given area of the flow bed:T = A = wL where w is the width (into the page) of a section of flowThe weight of that section of the flow is W = mg sin = Vg sin = wLYg sinThe shear force balances the weight force: T = Wie T = A = wL = W = wLYg sinor = Yg sinThus sinYgYVaSo 3089250250022sin..singVYa= 1531 Pa.sLmgV = f(y)=Va(y/Y)VaY = 0.5
5. 5. Q6. A droplet of water on a non-wetting surface forms a contact angle of 90owith the surface.Estimate the maximum thickness h of the droplet assuming static pressure, a density of 1000 kg/m3and a surface tension of 0.07 N/mThe maximum droplet depth is a minimum energy situation, ie a balance between potential energy andthe contributions of surface energy due to contact between the fluid, air and solid surfaces. Young’sequation (Lecture 4) gives a contact angle of 90oonly when the surface-vapour and surface-liquidtensions are equal, thus sv = sl.The minimum energy condition (also derived in Lecture 4) gives a formula in which these equal termscancel out:ggh lvlvsvsl 22(90ocontact angle only)ie h = √(2 g) = √(2(0.072)/9800) = 3.8 x 10-3= 3.8 mm.A droplet of water on glass is not far from this depth.