Homework Problem 1 Tall buildings can develop a significant difference in pressure between the inside and the outside of the build lock entrances are used at the ground level so the pressure at the ground floor is not the same as atmospheric pressure. Assume that a skyscraper doesn't use an airlock entrance and that the temperature inside a 300 m skyscraper is 1 C and the temperature outside is a uniform 0 C. Assume standard atmospheric conditions at base of the building and that the pressure is the same inside and outside at ground level. Is the compressibility of air important in this problem? Derive an equation for the pressure difference between the inside and the outside of the building as a func of height taking compressibility into account. What is the pressure differential at the top of the building in Pa? What is the equivalent height of a water column? (.7 cm) If someone opens a window on the top floor will air go in or out? Solution 1 To determine if compressibility is important, calculate the change in pressure between top and bottom of building neglecting compressibility. If p is small relative to p atm, then compressibility isn't significant h := 300m ρ air := 1. kg p := ρ air gh kpa := 1000Pa p atm := 101kPa 9 kg M gas 1000 mol := i is inside o is outside effects are small. R gas := 8.31447 joule p = 0.036 Compressibility p atm molek T i := 73.15K + 1K T i = 94.15 K T o := 73.15K + 0K T o = 73.15 K z := 0, 1m.. 300m Find the pressure inside the building and outside the building as functions of elevation. The problem states th temperature is uniform both inside and outside the building M gas gz := pressure outside of the building R gas T isothermal o p o ( z) p atm exp p i ( z) := p atm exp M gas gz R gas T i pressure inside the building
Figure 1. Pressure in atmospheres vs height in the building. The pressure outside the building drops faster than the pressure inside the building because cold air is more dense. Air will go out an open window. 1 p o ( z) 0.99 p atm 0.98 p i ( z) p atm 0.97 0.96 0 60 10 180 40 300 The pressure inside the building doesn't decrease as rapidly because the air density is lower inside. Therefore the pressure at the top of the building is higher inside the building than it is outside. M gas gz M gas gz := exp Pressure difference between inside and outside R gas T i R gas T o p( z) p atm exp 300 z m 40 180 p( z) Pa 10 60 0 0 60 10 180 40 300 z m Figure. Increased pressure in the inside of the building as a function of elevation. The pressure differential at 300 m is: z := 300m p( z) := p atm exp M gas gz R gas T i exp M gas gz R gas T o equivalent height of water ρ water 1000 kg h := p( z) ρ water g h =.7 cm of water! := p( z) = 61 Pa Air will flow out of a window at the top of the skyscraper. This effect is sometimes called the chimney effect. The pressure difference between inside and outside p is ( 300m) = 61 Pa Problem
If an altimeter can measure the difference in elevation of 0.5 m at sea level when the temperature is 0 C, the what pressure difference must the altimeter be capable of measuring? What is the ratio of this pressure differe divided by atmospheric pressure? Note that you can use either constant temperature or constant temperature gradient equations. Solution 9 M gas := 1000 kg mol R gas := 8.31447 T := 73.15K + 0K joule molek z := 0.5m elevation above sea level p atm := 101kPa M gas g( z) p ( z) := p atm exp R gas T p ( z) = 1.009941 10 5 Pa p atm p ( z) = 5.89 Pa p atm p ( z) p ratio := p p ratio = 5.8 10 5 atm This means that the altimeter must have 5-6 digits of precision and thus the pressure sensor and associated circuitry must be very well designed. The pressure ratio is p ratio = 5.8 10 5 Proble The barometric pressure reading on your GPS altimeter is 99,15 Pa, your elevation is 35 m, and the temperature is 0 C. You take a hike and the pressure reading changes to 98,085 Pa and the temperature remains almost constant. What is your new elevation? Solution 3 p 1 := 9915Pa p := 98085Pa z 1 := 35m T := 73.15K + 0K 9 M gas := 1000 kg mol
R gas := 8.31447 joule molek ( ) M gas g z z 1 p = p 1 exp R gas T ln z := z 1 p R p gas T 1 M gas g z = 415 m Problem 4 An air and water manometer is used to measure the pressure drop through a 0-cm-diameter sand filter. The top of the manometer is pressurized with air at 10 kpa (gage pressure). The water source for the filter is a large tank. A) What is the pressure in kpa at point A? B) What is the pressure in kpa at point B (5 cm above the base)? C) What is the change in pressure (in kpa) through the filter due to head loss (this is the change in pressure NOT due to the elevation change)? D) Calculate the elevation of the water in the supply tank relative to the bottom of the filter column. E) Calculate the magnitude and direction of the force acting on the horizontal surface at the base of the column. 10 kpa 70 cm 60 cm 5 cm ρ := 1000 kg p g := 10kPa A B 5 cm 65 cm Solution 4 A) Pressure at point A p A := p g 5cmρg p A = 9.51 kpa B) Pressure at point B p B := p g + 0cmρg p B = 11.961 kpa The pressure difference between the two points p B p A =.45 kpa The pressure difference due to elevation is simply the elevation difference times the specific weight. p elevation := 60cmρg p elevation = 5.884 kpa
C) The pressure difference that isn't due to elevation is the change in pressure due to head loss ( p B p A ) p elevation = 3.43 kpa D) The water surface elevation in the supply tank can be calculated from the pressure at point A and the elev of point A. p A supplytank := + 65cm supplytank 16 cm ρg = water surface elevation above base of filter E) The magnitude of the force acting on the filter bottom is the pressure at the centroid times the area. p bottom := p B + 5cmρg p bottom = 1.45 kpa F bottom := p bottom A filter πd filter d filter := 0cm A filter := A 4 filter = 314 cm F bottom = 391 N normal to the surface Problem 5 An entrepreneurial friend proposes making forms for concrete steps using the design at the right. The concrete will be added through the top step and a mechanical vibrator will be used to make sure the concrete completely fills the form. You suspect that concrete may try to lift the form and so you are going to calculate how much force must be applied to prevent the form from lifting off of the floor. Calculate the location of the resulting force. b Open top w w:= 1m b := 0cm h := 15cm ρ concrete := 400 kg h Open bottom Solution 5 The pressure in the concrete increases with depth. The horizontal treads will have a vertical force acting up. F find the pressure on the treads. Use atmospheric pressure as the pressure datum since that is the pressure actin down on the outside of the form. p toptread := ρ concrete gh p toptread = 3.53 kpa p bottomtread := ρ concrete gh A tread := wb A tread = 0. m ( ) A tread p bottomtread = 7.061 kpa F r := p toptread + p bottomtread F r =.118 10 3 N
( ) F r := 3 ρ concrete gh A tread F r =.118 10 3 N An alternative solution would be to determine the mass of concrete that is "missing" above each of the treads Volume concrete := wb h + wb h Volume concrete := 3wbh Weight concrete := Volume concrete ρ concrete g Weight concrete := 3w bhρ concrete g Weight concrete =.118 10 3 N Missing concrete weight! To calculate the location of the resulting force take moments about the left edge of the form. b 0 p 3b = bottomtreada tread + p toptreada tread F r x r b and 3b are the distances to the centroids of the treads. x r := b ρ concreteghbw + ( ) 3b ρ concrete 3 ρ concrete gh bw ghbw 5b x r := x 6 r = 16.7 cm A counteracting force applied near the back of the first tread would balance the resulting hydrostatic force The resulting force is F r =.1 kn acting up at a location x r = 16.7 cm measured from the front edge of the first tread. Problem 6 I installed a radiant floor heating system in my friend's house. The house stayed warm all winter even though circulator pump that is supposed to pump the water through the system never turned on. Determine the pressu at point A in the hot water supply and at point B in the cold water return assuming that the pressure at the bot of the water heater where the cold water enters and that as soon as the water enters the water heater it is heate Also calculate the pressure difference. Note that points A and B are at the same elevation a distance h above location where the cold water enters the water heater. Explain why water flows even without the pump?
A T hot := 70 C T cold := 30 C ρ 70 := 977.8 kg ρ 30 := 995.7 kg p bottom := 700kPa h Hot water B 1000 990 Pipe loop that runs between joists in the basement ceiling to heat the first floor Centrifugal circulator pump. Note that water can flow through this pump even when it isn t running. Cold water return h := m kg/m3 ρ water 980 970 30 40 50 60 70 T Temperature in C Solution 6 p A := p bottom ρ 70 gh p A = 680.8 kpa p B := p bottom ρ 30 gh p B = 680.471 kpa ( ) gh p := ρ 30 ρ 70 ( ) p A p B = 351 Pa The pressure at A is higher than the pressure at B. This difference in pressure causes the water in the pipe loo flow in the direction of the arrows and thus hot water was continuously carried up into the basement ceiling. Problem 7 A bubbler system is used to measure the depth of water in a river. During a flood the water level increases from h 1 to h. Derive an equation for the minimum volume of air at atmospheric pressure that must be pumped into the bubbler by a peristatlic pump at point a to maintain the air water interface at the end of the bubbler tube during the increase in water depth. Use symbols for any additional parameters that you need to know in order to solve the problem.
a h := hm hh 1 1 := 1m Solution 7 PV nr T = We will need to use the ideal gas law and the relationship between depth and pressure. P = ρgh Vtube = volume of tube Find moles of air in the tube at the two different hydrostatic pressures. Difference must be supplied by pumping in air at atmospheric pressure Solve ideal gas law for n and substitute the equation for hydrostatic pressure to obtain the air pressure in the tube given submergence h. n 1 = ( ρ g h 1 + P atm ) V tube RT Note that we must use absolute pressures in the ideal gas law. n = ( ρ g h + P atm ) V tube RT ( ) ρg n n 1 = h h 1 V tube RT The volume of air in the tube is V tube T = nr P The volume of air that must be supplied is ( ) R V pumped = n n 1 T P atm The air is supplied by a peristaltic pump that is pumping air at atmospheric pressure. Thus the pumped volume is based on the volume of the required moles of air at atmospheric pressure. ( ) ρg V pumped = h h 1 V tube P atm Let L tube := 10m d tube := 3mm πd tube V tube := L tube 4 ρ 1 10 3 kg = P atm := 101kPa
V pumped ( h h 1 ) ρg V tube := V P pumped = 6.9 ml atm Problem 8 A rod slides inside a bronze bearing with a small gap lubricated with oil. Estimate the force required to slide the rod through the bearing. Assume the velocity distribution in the gap is linear. (The rod is not rotating!) The bearing supports the shaft over a length of rod diameter d := 5mm rod velocity V:= 0 cm s oil viscosity µ := 0.05N s m gap between rod and bearingt := 10µm L bearing := cm Length of bearing Solution 8 τ := A F := µ V t πdl bearing := Area of contact between shaft and bearing τa VπdL bearing F := µ t The force required to slide the rod isf = 0.314 N