Fluid Mechanics Physics

Question 1. A piece of metal has a mass of 5.0 kg and a volume of 1000 cm3. [3 marks] a) Find the density of the metal. b) Would a 2000 cm3 piece of the same metal have the same density? c) What would the mass of the 2000 cm3 piece be? 2. Water expands when heated. If a beaker of water is heated from 10 C to 90 C, does the pressure at the bottom of the beaker increase, decrease, or stay the same? Explain your reasoning. 3. Your friend who hasn't studied fluids wants to explore the bottom of a deep lake but can't afford SCUBA gear. He proposes that you hold one end of a garden hose in a boat while he fills his pockets with rocks and uses the hose to breathe. Explain to your friend why this is a bad idea. 4. A 2000 kg car rests on a platform that sits on a container of fluid. The platform has an area of 4 m2. At the other end of the container has a plunger of area 100 cm2. a) What pressure must be exerted by the fluid on the platform to keep the car at rest? b) What force must be exerted on the plunger to supply this pressure? 5. Water flows at a rate of 2.0 L/s through a pipe with a diameter of 10 cm. The pipe then goes up a height of 5m while narrowing to a diameter of 5 cm. What is the pressure gradient (the change in pressure) from the top to the bottom of the pipe? [hint: draw this and label with P1, v1, P2, v2 etc.) 6. A hurricane wind blows across the top of a flat roof measuring 12 m by 5m at a speed of 120 km/h. a) Is the air pressure above the roof higher or lower than the pressure inside the house? Explain. b) Find the pressure difference. c) How much force is exerted on the roof? If the roof's structure fails, will in blow in or blow out? 7. Astronauts are protected from the vacuum of space (P ~ 0) by either a thin suit or the walls of the spaceship. Movies often show the effects of a hull breach -- but people or objects are not sucked out -- they are blown out by the air inside the ship (at about 100 kPa of pressure). [4 marks] a) Estimate the force exerted by the air if the hole in the hull has a diameter of 1.0 m. b) How fast would the air exit the hole (consult Table 13.1 for density of air)? c) The International Space Station has 916 m3 of pressurized air. How long do the astronauts have to plug the hole before the station runs out of air? Answer: 1. We know that density = Mass/Volume In the given case, the mass of the metal = 5 kg Volume of the metal = 1000cm3 = 10-3 m3 Hence density = 5/10-3 = 5000 kg/m3 Density essentially depends on the material and since the 2000 cm3 piece is of the same metal, hence the density would be the same as calculated above. Volume of the given piece of metal = 2000cm3 = 2*10-3 m3 Density of the metal = 5000 kg/m3 Let the mass of the given piece be M kg Hence 5000 = M/2*10-3 which implies M= 10kg 2. In the given case, the pressure at the bottom of the beaker would stay the same since there is no change in the overall mass of the water and/or the depth which are the essential factors which would impact the pressure. 3. Clearly this is a bad idea since human lungs tend to get compressed when we go under water due to which air also becomes compressed and hence unable to automatically move air from a region of low pressure (i.e. at the end of the hose in air) to the region of high pressure (i.e. to the lungs submerged under the water). Normally this problem is resolved through diaphragm which tends to lower the lung pressure but this capacity is severely limited at depths of greater than 1m. In the given case, since my friend intends to go to the bottom of the case, clearly this is a bad idea since he will not be able to suck the outside air and thus this arrangement can prove to be fatal. 4. To keep the car at rest and maintain the equilibrium the pressure at the two points A1 and point A2 must be the same. Mass of the car = 2000 kg Total force exerted by the car = Mg = 2000*9.81 = 19620 N Total area of the platform = 4m2 Hence pressure at point A1 = 19620/4 = 4905 Pascal Hence this is the total pressure required to keep the car at rest. Let the requisite force exerted on the plunger be F The pressure at point A2 needs to be 4905 Pascal Area of contact at the plunger = 100cm2 = 10-2m2 Hence 4905 = F/10-2 which implies that F = 49.05 N 5. Since the volume is conserved hence volume at the bottom should be equal to the volume at the top. Let the speed at the bottom be Vb . Let the speed at the top be Vt Then *(100/4)*10-4 * Vb = 0.002 which implies Vb = 0.25m/s Further *(25/4)*10-4 * Vt = 0.002 which implies Vt = 1.02m/s Further applying the bernoullis equation we get the following Pt + ght + vt2 = Pb + ghb + vb2 Let us call Pb Pt = P or the pressure difference which is to be determined Hence P = g(ht - hb) + (vt2 - vb2) We are given that (ht - hb) = 5m , g assumed to be 9.81m/s2, for water = 1000kg/m3 P = 1000*9.81*5 + 500*(1.022 0.252) = 49.53 KPa 6. As per the Bernoullis theorem, since the air velocity above the roof is considerably higher than the velocity of air below the roof, hence it may be concluded that the pressure is higher inside the house. The pressure difference (P) can be calculated as shown below. p = (1/2)(d)(v2) Here d is the density of the air while v is the speed of the wind (m/s) In the given case d = 1.29 kg/m3 and v= 120kmph = 33.33m/s Putting these values in the above formula, we get p = 716 Pa If the roof structure fails, clearly the roof will flow out since the pressure outside the house is lower than that inside the house. We know Pressure = Force/Area Hence Force = Pressure *Area Area of the roof = 12*5 = 60m2 Force on the roof = 716*60 = 42960 N 7 Diameter of the hull = 1m Hence radius of the hull = 0.5m Area of the hull = (0.5)2 = 0.786 m2 Let the force be F We know F= Pressure * Area = 100000*0.785714 = 78571.4 N We use the following formula. p = (1/2)(d)(v2) Here d is the density of the air while v is the speed of the wind (m/s) In the given case d = 1.29 kg/m3 100000 = 0.5*1.29*v2 Solving the above, we get v= 393.75m/s Speed of air exiting the space station = 393.75m/s Area of the hull = 0.786 m2 Volume of air exiting the space station per second = 393.75 * 0.786 = 309.38m3 Total air in the space station = 916m3 Time available to astronauts to plug the hole = 916/309.38 = 2.96 seconds