Exam (elaborations) TEST BANK FOR Thermodynamics An Engineering Approach 7th Edition By Yunus A. Cengel and Michael A. Boles (Solution Manual)

Exam (elaborations) TEST BANK FOR Thermodynamics An Engineering Approach 7th Edition By Yunus A. Cengel and Michael A. Boles (Solution Manual) Solutions Manual for Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011 Chapter 1 INTRODUCTION AND BASIC CONCEPTS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-2 Thermodynamics 1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle. 1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill. 1-3C There is no truth to his claim. It violates the second law of thermodynamics. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-3 Mass, Force, and Units 1-4C The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English system. You should get into the habit of never writing the unit “lb”, but always use either “lbm” or “lbf” as appropriate since the two units have different dimensions. 1-5C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit. 1-6C There is no acceleration, thus the net force is zero in both cases. 1-7E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives 210.5 lbm 1 lbf 32.174 lbm ft/s 32.10 ft/s 210 lbf 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = ⎯⎯→ = = g W mg m W Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be lbf 35.8 = ⎟⎠ ⎞ ⎜⎝ ⎛ ⋅ = = 2 2 32.174 lbm ft/s W mg (210.5 lbm)(5.47 ft/s ) 1 lbf 1-8 The interior dimensions of a room are given. The mass and weight of the air in the room are to be determined. Assumptions The density of air is constant throughout the room. Properties The density of air is given to be ρ = 1.16 kg/m3. ROOM AIR 6X6X8 m3 Analysis The mass of the air in the room is m = ρV = (1.16 kg/m3 )(6×6×8 m3 ) = 334.1kg Thus, N 3277 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = 2 2 1 kg m/s 1 N W mg (334.1 kg)(9.81 m/s ) preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-4 1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 0.5% is to be determined. 0 z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 ×10−6 z) In our case, W = 0.995Ws = 0.995mgs = 0.995(m)(9.81) Substituting, 0.995(9.81) = (9.81− 3.32×10−6 z) ⎯⎯→ z = 14,774 m ≅ 14,770 m Sea level 1-10 The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be W = mg = (200 kg)(9.6 m/s2 ) = 1920N 1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units. Analysis Applying Newton's second law, the weight is determined in various units to be 0.240 Btu/lbm F 0.240 kcal/kg C 1.005 J/g C 1.005 kJ/kg K ° ⋅ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⋅ ° ⋅ ° = ⋅ ° ° ⋅ = ⎟⎠ ⎞ ⎜⎝ = ⋅ ° ⎛ ° ⋅ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ = ⋅ ° ⎛ ⋅ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⋅ ° ⋅ = ⋅ ° 4.1868 kJ/kg C (1.005 kJ/kg C) 1 Btu/lbm F 4.1868 kJ (1.005 kJ/kg C) 1 kcal 1000 g 1 kg 1 kJ (1.005 kJ/kg C) 1000 J 1 kJ/kg C (1.005 kJ/kg C) 1 kJ/kg K p p p p c c c c preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-5 1-12 A rock is thrown upward with a specified force. The acceleration of the rock is to be determined. Analysis The weight of the rock is 29.37 N 1 kg m/s 1 N (3 kg)(9.79 m/s ) 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ W = mg = Then the net force that acts on the rock is Fnet = Fup − Fdown = 200 − 29.37 =170.6 N Stone From the Newton's second law, the acceleration of the rock becomes 2 m/s 56.9 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = 1 N 1 kg m/s 3 kg 170.6 N 2 m a F preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-6 1-13 Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. "The weight of the rock is" W=m*g m=3 [kg] g=9.79 [m/s2] "The force balance on the rock yields the net force acting on the rock as" F_up=200 [N] F_net = F_up - F_down F_down=W "The acceleration of the rock is determined from Newton's second law." F_net=m*a "To Run the program, press F2 or select Solve from the Calculate menu." SOLUTION a=56.88 [m/s^2] F_down=29.37 [N] F_net=170.6 [N] F_up=200 [N] g=9.79 [m/s2] m=3 [kg] W=29.37 [N] m [kg] a [m/s2] 1 2 3 4 5 6 7 8 9 10 190.2 90.21 56.88 40.21 30.21 23.54 18.78 15.21 12.43 10.21 1 2 3 4 5 6 7 8 9 10 0 40 80 120 160 200 m [kg] a [m/s2] preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-7 1-14 During an analysis, a relation with inconsistent units is obtained. A correction is to be found, and the probable cause of the error is to be determined. Analysis The two terms on the right-hand side of the equation E = 25 kJ + 7 kJ/kg do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally homogeneous; that is, every term in the equation will have the same unit. Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage. 1-15 A resistance heater is used to heat water to desired temperature. The amount of electric energy used in kWh and kJ are to be determined. Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of electric energy used in 2 hours becomes Total energy = (Energy per unit time)(Time interval) = (4 kW)(2 h) = 8 kWh Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ, Total energy = (8 kWh)(3600 kJ/kWh) = 28,800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy. 1-16 A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is to be obtained for the filling time. Assumptions Gasoline is an incompressible substance and the flow rate is constant. Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unit of time is ‘seconds’. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have t [s] ↔ V [L], and V& [L/s} It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate. Therefore, the desired relation is = & V t V Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-8 1-17 A pool is to be filled with water using a hose. Based on unit considerations, a relation is to be obtained for the volume of the pool. Assumptions Water is an incompressible substance and the average flow velocity is constant. Analysis The pool volume depends on the filling time, the cross-sectional area which depends on hose diameter, and flow velocity. Also, we know that the unit of volume is m3. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have V [m3] is a function of t [s], D [m], and V [m/s} It is obvious that the only way to end up with the unit “m3” for volume is to multiply the quantities t and V with the square of D. Therefore, the desired relation is V = CD2Vt where the constant of proportionality is obtained for a round hose, namely, C =π/4 so that V = (πD2/4)Vt. Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach. 1-18 It is to be shown that the power needed to accelerate a car is proportional to the mass and the square of the velocity of the car, and inversely proportional to the time interval. Assumptions The car is initially at rest. Analysis The power needed for acceleration depends on the mass, velocity change, and time interval. Also, the unit of power W& is watt, W, which is equivalent to W = J/s = N⋅m/s = (kg⋅m/s2)m/s = kg⋅m2/s3 Therefore, the independent quantities should be arranged such that we end up with the unit kg⋅m2/s3 for power. Putting the given information into perspective, we have W& [ kg⋅m2/s3] is a function of m [kg], V [m/s], and t [s] It is obvious that the only way to end up with the unit “kg⋅m2/s3” for power is to multiply mass with the square of the velocity and divide by time. Therefore, the desired relation is W& is proportional to mV 2 / t or, W& = CmV 2 / t where C is the dimensionless constant of proportionality (whose value is ½ in this case). Discussion Note that this approach cannot determine the numerical value of the dimensionless numbers involved. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-9 Systems, Properties, State, and Processes 1-19C This system is a region of space or open system in that mass such as air and food can cross its control boundary. The system can also interact with the surroundings by exchanging heat and work across its control boundary. By tracking these interactions, we can determine the energy conversion characteristics of this system. 1-20C The system is taken as the air contained in the piston-cylinder device. This system is a closed or fixed mass system since no mass enters or leaves it. 1-21C Any portion of the atmosphere which contains the ozone layer will work as an open system to study this problem. Once a portion of the atmosphere is selected, we must solve the practical problem of determining the interactions that occur at the control surfaces which surround the system's control volume. 1-22C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 1-23C If we were to divide the system into smaller portions, the weight of each portion would also be smaller. Hence, the weight is an extensive property. 1-24C If we were to divide this system in half, both the volume and the number of moles contained in each half would be one-half that of the original system. The molar specific volume of the original system is N V v = and the molar specific volume of one of the smaller systems is N N V/ V v = = / 2 2 which is the same as that of the original system. The molar specific volume is then an intensive property. 1-25C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight. 1-26C A process during which a system remains almost in equilibrium at all times is called a quasi-equilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium processes. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-10 1-27C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric. 1-28C The state of a simple compressible system is completely specified by two independent, intensive properties. 1-29C The pressure and temperature of the water are normally used to describe the state. Chemical composition, surface tension coefficient, and other properties may be required in some cases. As the water cools, its pressure remains fixed. This cooling process is then an isobaric process. 1- 30C When analyzing the acceleration of gases as they flow through a nozzle, the proper choice for the system is the volume within the nozzle, bounded by the entire inner surface of the nozzle and the inlet and outlet cross-sections. This is a control volume since mass crosses the boundary. 1-31C A process is said to be steady-flow if it involves no changes with time anywhere within the system or at the system boundaries. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-11 1-32 The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km. Properties The density data are given in tabular form as 25 0 0.2 0.4 0.6 0.8 1 1.2 1.4 z, km ρ, kg/m3 r, km z, km ρ, kg/m3 6377 0 1.225 6378 1 1.112 6379 2 1.007 6380 3 0.9093 6381 4 0.8194 6382 5 0.7364 6383 6 0.6601 6385 8 0.5258 6387 10 0.4135 6392 15 0.1948 6397 20 0.08891 6402 25 0.04008 Analysis Using EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are: ρ(z) = a + bz + cz2 = 1.20252 – 0.z + 0.z2 for the unit of kg/m3, (or, ρ(z) = (1.20252 – 0.z + 0.z2)×109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give ρ = 0.60 kg/m3. (b) The mass of atmosphere can be evaluated by integration to be 4 [ (2 ) / 2 ( 2 ) / 3 ( 2 ) / 4 / 5] ( )4 ( ) 4 ( )( 2 ) 4 5 0 2 3 0 0 2 0 0 2 0 2 0 2 0 2 0 2 0 2 0 ar h r a br h a br cr h b cr h ch m dV a bz cz r z dz a bz cz r r z z dz h z h z V = + + + + + + + + = ∫ = ∫ + + + = ∫ + + + + = = π ρ π π where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = - 0., and c = 0. are the constants in the density function. Substituting and multiplying by the factor 109 for the density unity kg/km3, the mass of the atmosphere is determined to be m = 5.092×1018 kg Discussion Performing the analysis with excel would yield exactly the same results. EES Solution for final result: a=1.; b=-0.10167 c=0.; r=6377; h=25 m=4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9 preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-12 Temperature 1-33C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading, even if they are not in contact. 1-34C They are Celsius (°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system. 1-35C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may deviate. 1-36 A temperature is given in °C. It is to be expressed in K. Analysis The Kelvin scale is related to Celsius scale by T(K] = T(°C) + 273 Thus, T(K] = 37°C + 273 = 310 K 1-37E The temperature of air given in °C unit is to be converted to °F and R unit. Analysis Using the conversion relations between the various temperature scales, 762R 302 F = ° + = + = ° = ° + = + = ° (R) ( F) ( F) 1.8 ( C) 32 (1.8)(150) 32 T T T T 1-38 A temperature change is given in °C. It is to be expressed in K. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus, ΔT(K] = ΔT(°C) = 45 K preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-13 1-39E The flash point temperature of engine oil given in °F unit is to be converted to K and R units. Analysis Using the conversion relations between the various temperature scales, 457K 823R = = = = ° + = + = 1.8 823 1.8 (K) (R) (R) ( F) T T T T 1-40E The temperature of ambient air given in °C unit is to be converted to °F, K and R units. Analysis Using the conversion relations between the various temperature scales, 419.67 R 233.15 K 40 C = − + = = − + = = − ° = − + = − ° 40 459.67 40 273.15 40 C ( 40)(1.8) 32 T T T 1-41E The change in water temperature given in °F unit is to be converted to °C, K and R units. Analysis Using the conversion relations between the various temperature scales, 10R 5.6K 5.6 C Δ = ° = Δ = = Δ = = ° 10 F 10 /1.8 10 /1.8 T T T 1-42E A temperature range given in °F unit is to be converted to °C unit and the temperature difference in °F is to be expressed in K, °C, and R. Analysis The lower and upper limits of comfort range in °C are = 18.3°C − = ° − ° = 1.8 65 32 1.8 ( C) ( F) 32 T T = 23.9°C − = ° − ° = 1.8 75 32 1.8 ( C) ( F) 32 T T A temperature change of 10°F in various units are 5.6K 5.6 C 10R Δ = Δ ° = = = ° Δ ° Δ ° = Δ = Δ ° = (K) ( C) 1.8 10 1.8 ( F) ( C) (R) ( F) T T T T T T preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-14 Pressure, Manometer, and Barometer 1-43C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure. 1-44C The blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to the body. For a constant volume of blood to be discharged by the heart, the blood pressure must increase to overcome the increased resistance to flow. 1-45C No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled. 1-46C If the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube will be the same. 1-47C Pascal’s principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example of Pascal’s principle is the operation of the hydraulic car jack. 1-48E The pressure given in psia unit is to be converted to kPa. Analysis Using the psia to kPa units conversion factor, kPa 1034 = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = 1 psia P (150 psia) 6.895 kPa 1-49 The pressure in a tank is given. The tank's pressure in various units are to be determined. Analysis Using appropriate conversion factors, we obtain (a) 2 kN/m 1500 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 1 kPa (1500 kPa) 1 kN/m 2 P (b) 2 s kg/m 1,500,000 ⋅ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 1 kN 1000 kg m/s 1 kPa (1500 kPa) 1 kN/m 2 2 P (c) 2 s kg/km 000 1,500,000, ⋅ = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 1 km 1000 m 1 kN 1000 kg m/s 1 kPa (1500 kPa) 1 kN/m 2 2 P preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-15 1-50E The pressure in a tank in SI unit is given. The tank's pressure in various English units are to be determined. Analysis Using appropriate conversion factors, we obtain (a) 2 lbf/ft 31,330 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 1 kPa (1500 kPa) 20.886 lbf/ft 2 P (b) psia 217.6 = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 2 2 2 2 1 lbf/in 1 psia 144 in 1 ft 1 kPa P (1500 kPa) 20.886 lbf/ft 1-51E The pressure given in mm Hg unit is to be converted to psia. Analysis Using the mm Hg to kPa and kPa to psia units conversion factors, psia 29.0 = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = 6.895 kPa 1 psia 1 mm Hg P (1500 mm Hg) 0.1333 kPa 1-52 The pressure given in mm Hg unit is to be converted to kPa. Analysis Using the mm Hg to kPa units conversion factor, kPa 166.6 = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = 1 mm Hg P (1250 mm Hg) 0.1333 kPa preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-16 1-53 The pressure in a pressurized water tank is measured by a multi-fluid manometer. The gage pressure of air in the tank is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air-water interface. Properties The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m3, respectively. Analysis Starting with the pressure at point 1 at the air-water interface, and moving along the tube by adding (as we go down) or subtracting (as we go up) th e ρgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives P1 + ρ water gh1 + ρ oil gh2 − ρ mercurygh3 = Patm PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course Solving for P1, P1 = Patm −ρ water gh1 − ρ oil gh2 + ρ mercurygh3 or, P1 − Patm = g(ρ mercuryh3 − ρ waterh1 − ρ oilh2 ) Noting that P1,gage = P1 - Patm and substituting, = 56.9 kPa ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ − = − 2 2 3 2 3 3 1,gage 1000 N/m 1 kPa 1 kg m/s (850 kg/m )(0.3 m)] 1 N P (9.81 m/s )[(13,600 kg/m )(0.46 m) (1000 kg/m )(0.2 m) Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly. 1-54 The barometric reading at a location is given in height of mercury column. The atmospheric pressure is to be determined. Properties The density of mercury is given to be 13,600 kg/m3. Analysis The atmospheric pressure is determined directly from = 100.1 kPa ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = 2 2 3 2 atm 1000 N/m 1 kPa 1 kg m/s (13,600 kg/m )(9.81 m/s )(0.750 m) 1 N P ρgh preparation. If you are a student using this Manual, you are using it without permission. 1-17 1-55 The gage pressure in a liquid at a certain depth is given. The gage pressure in the same liquid at a different depth is to be determined. Assumptions The variation of the density of the liquid with depth is negligible. Analysis The gage pressure at two different depths of a liquid can be expressed as P1 = ρgh1 and P2 = ρgh2 h2 2 h1 1 Taking their ratio, 1 2 1 2 1 2 h h gh gh P P = = ρ ρ Solving for P2 and substituting gives = = (42 kPa) = 126 kPa 3 m 9 m 1 1 2 2 P h h P Discussion Note that the gage pressure in a given fluid is proportional to depth. 1-56 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined. Assumptions The liquid and water are incompressible. Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be 1000 kg/m3. Then density of the liquid is obtained by multiplying its specific gravity by the density of water, SG (0.85)(1000 kg/m3 ) 850 kg/m3 2 ρ = ×ρ H O = = Analysis (a) Knowing the absolute pressure, the atmospheric pressure can be determined from Patm h P = 96.0 kPa ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − = − 2 3 2 atm 1000 N/m 1 kPa (145 kPa) (1000 kg/m )(9.81 m/s )(5 m) P P ρgh (b) The absolute pressure at a depth of 5 m in the other liquid is = 137.7 kPa ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = + = + 2 3 2 atm 1000 N/m 1 kPa (96.0 kPa) (850 kg/m )(9.81 m/s )(5 m) P P ρgh Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-18 1-57E It is to be shown that 1 kgf/cm2 = 14.223 psi . Analysis Noting that 1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in = 2.54 cm, we have 2.20463 lbf 1 N 1 kgf 9.80665 N (9.80665 N ) 0.22481 lbf = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = = and psi 14.223 = = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = = 2 2 2 2 2 14.223 lbf/in 1 in 2.54 cm 1 kgf/cm 2.20463 lbf/cm (2.20463 lbf/cm ) 1-58E The pressure in chamber 3 of the two-piston cylinder shown in the figure is to be determined. Analysis The area upon which pressure 1 acts is 2 2 2 1 1 7.069 in 4 (3 in) 4 =π =π = D A F1 F3 F2 and the area upon which pressure 2 acts is 2 2 2 2 2 1.767 in 4 (1.5 in) 4 =π =π = D A The area upon which pressure 3 acts is given by 2 A3 = A1 − A2 = 7.069 −1.767 = 5.302 in The force produced by pressure 1 on the piston is then (7.069 in ) 1060 lbf 1 psia (150 psia) 1 lbf/in 2 2 1 1 1 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ F = P A = while that produced by pressure 2 is (250 psia)(1.767 in2 ) 441.8 lbf F1 = P2 A2 = = According to the vertical force balance on the piston free body diagram F3 = F1 − F2 =1060 − 441.8 = 618.3 lbf Pressure 3 is then = = = 117 psia 2 3 3 3 5.302 in 618.3 lbf A F P preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-19 1-59 The pressure in chamber 1 of the two-piston cylinder shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives 2 2 3 1 2 1 1 2 3 1 P A P (A A ) P A F F F + − = + = F1 F3 F2 which when solved for P1 gives ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = + − 1 2 3 1 2 1 2 1 A A P A A P P since the areas of the piston faces are given by the above equation becomes A =πD2 / 4 = 908 kPa ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎠ ⎞ ⎜⎝ + − ⎛ ⎟⎠ ⎞ ⎜⎝ = ⎛ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = 2 2 2 1 2 3 2 1 2 1 2 10 (700 kPa) 1 4 10 (2000 kPa) 4 1 D D P D D P P 1-60 The mass of a woman is given. The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined. Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes. 2 One foot carries the entire weight of a person during walking, and the shoe is sized for walking conditions (rather than standing). 3 The weight of the shoes is negligible. Analysis The mass of the woman is given to be 70 kg. For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be 2 m 1.37 = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = = 2 2 2 1000 N/m 1 kPa 1 kg m/s 1 N 0.5 kPa (70 kg)(9.81 m/s ) P mg P A W Discussion This is a very large area for a shoe, and such shoes would be impractical to use. Therefore, some sinking of the snow should be allowed to have shoes of reasonable size. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-20 1-61 The vacuum pressure reading of a tank is given. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be ρ = 13,590 kg/m3. Analysis The atmospheric (or barometric) pressure can be expressed as P 30 kPa abs 100.0 kPa 1000 N/m 1 kPa 1 kg m/s 1 N (13,590 kg/m )(9.807 m/s )(0.750 m) 2 2 3 2 atm = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = P = ρ gh Then the absolute pressure in the tank becomes Patm = 750 mmHg Pabs = Patm − Pvac = 100.0 − 30 = 70.0 kPa 1-62E The vacuum pressure given in kPa unit is to be converted to various units. Analysis Using the definition of vacuum pressure, = − = − = 18 kPa = 98 80 not applicable for pressures below atmospheric pressure abs atm vac gage P P P P Then using the conversion factors, = 18 kN/m2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 1 kPa (18 kPa) 1 kN/m 2 Pabs = 2.61lbf/in2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 6.895 kPa (18 kPa) 1 lbf/in 2 Pabs psi 2.61 = ⎟⎠ ⎞ ⎜⎝ = ⎛ 6.895 kPa (18 kPa) 1 psi Pabs Hg mm 135 = ⎟⎠ ⎞ ⎜⎝ = ⎛ 0.1333 kPa (18 kPa) 1 mm Hg Pabs preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-21 1-63 A mountain hiker records the barometric reading before and after a hiking trip. The vertical distance climbed is to be determined. 630 mbar h = ? Assumptions The variation of air density and the gravitational acceleration with altitude is negligible. Properties The density of air is given to be ρ = 1.20 kg/m3. Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain 740 mbar (0.740 0.630) bar 100,000 N/m 1 bar 1 kg m/s 1 N (1.20 kg/m )(9.81 m/s )( ) ( ) / 2 2 3 2 air bottom top air bottom top − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = − = − h gh P P W A P P ρ It yields h = 934 m which is also the distance climbed. 1-64 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible. Properties The density of air is given to be ρ = 1.18 kg/m3. The density of mercury is 13,600 kg/m3. 675 mmHg Analysis Atmospheric pressures at the top and at the bottom of the building are h 695 mmHg 92.72 kPa 1000 N/m 1 kPa 1kg m/s 1N (13,600 kg/m )(9.81 m/s )(0.695 m) ( ) 90.06 kPa 1000 N/m 1 kPa 1 kg m/s 1 N (13,600 kg/m )(9.81 m/s )(0.675 m) ( ) 2 2 3 2 bottom bottom 2 2 3 2 top top = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = P gh P ρgh ρ Taking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtain (92.72 90.06) kPa 1000 N/m 1 kPa 1 kg m/s 1 N (1.18 kg/m )(9.81 m/s )( ) ( ) / 2 2 3 2 air bottom top air bottom top − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = − = − h gh P P W A P P ρ It yields h = 231 m which is also the height of the building. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-22 1-65 Problem 1-64 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. P_bottom=695 [mmHg] P_top=675 [mmHg] g=9.81 [m/s^2] "local acceleration of gravity at sea level" rho=1.18 [kg/m^3] DELTAP_abs=(P_bottom-P_top)*CONVERT(mmHg, kPa) "[kPa]" "Delta P reading from the barometers, converted from mmHg to kPa." DELTAP_h =rho*g*h*Convert(Pa, kPa) "Delta P due to the air fluid column height, h, between the top and bottom of the building." DELTAP_abs=DELTAP_h SOLUTION DELTAP_abs=2.666 [kPa] DELTAP_h=2.666 [kPa] g=9.81 [m/s^2] h=230.3 [m] P_bottom=695 [mmHg] P_top=675 [mmHg] rho=1.18 [kg/m^3] 1-66 A man is standing in water vertically while being completely submerged. The difference between the pressures acting on the head and on the toes is to be determined. Assumptions Water is an incompressible substance, and thus the density does not change with depth. htoe Properties We take the density of water to be ρ =1000 kg/m3. hhead Analysis The pressures at the head and toes of the person can be expressed as Phead = Patm + ρghhead and Ptoe = Patm + ρghtoe where h is the vertical distance of the location in water from the free surface. The pressure difference between the toes and the head is determined by subtracting the first relation above from the second, Ptoe − Phead = ρghtoe − ρghhead = ρg(htoe − hhead ) Substituting, kPa 17.2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ − = 2 2 3 2 toe head 1000N/m 1kPa 1kg m/s 1N P P (1000 kg/m )(9.81 m/s )(1.75 m - 0) Discussion This problem can also be solved by noting that the atmospheric pressure (1 atm = 101.325 kPa) is equivalent to 10.3-m of water height, and finding the pressure that corresponds to a water height of 1.75 m. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-23 1-67 A gas contained in a vertical piston-cylinder device is pressurized by a spring and by the weight of the piston. The pressure of the gas is to be determined. Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield W = mg P Patm Fspring PA = Patm A+W + Fspring Thus, = 147 kPa ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × + = + + = + −4 2 2 2 spring atm 1000 N/m 1 kPa 35 10 m (95 kPa) (3.2 kg)(9.81 m/s ) 150 N A mg F P P 1-68 Problem 1-67 is reconsidered. The effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder is to be investigated. The pressure against the spring force is to be plotted, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. g=9.81 [m/s^2] P_atm= 95 [kPa] m_piston=3.2 [kg] {F_spring=150 [N]} A=35*CONVERT(cm^2, m^2) W_piston=m_piston*g F_atm=P_atm*A*CONVERT(kPa, N/m^2) "From the free body diagram of the piston, the balancing vertical forces yield:" F_gas= F_atm+F_spring+W_piston P_gas=F_gas/A*CONVERT(N/m^2, kPa) Fspring [N] Pgas [kPa] 0 50 100 150 200 250 300 350 400 450 500 104 118.3 132.5 146.8 161.1 175.4 189.7 204 218.3 232.5 246.8 500 100 120 140 160 180 200 220 240 260 Fspring [N] Pgas [kPa] preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-24 1-69 Both a gage and a manometer are attached to a gas to measure its pressure. For a specified reading of gage pressure, the difference between the fluid levels of the two arms of the manometer is to be determined for mercury and water. Properties The densities of water and mercury are given to be ρwater = 1000 kg/m3 and be ρHg = 13,600 kg/m3. Analysis The gage pressure is related to the vertical distance h between the two fluid levels by g P P gh h ρ ρ gage gage = ⎯⎯→ = (a) For mercury, 0.60 m 1 kN 1000 kg/m s 1 kPa 1 kN/m (13,600 kg/m )(9.81 m/s ) 80 kPa 2 2 3 2 gage = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = g P h ρ Hg (b) For water, 8.16 m 1 kN 1000 kg/m s 1 kPa 1 kN/m (1000 kg/m )(9.81 m/s ) 80 kPa 2 2 3 2 H O gage 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = g P h ρ preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-25 1-70 Problem 1-69 is reconsidered. The effect of the manometer fluid density in the range of 800 to 13,000 kg/m3 on the differential fluid height of the manometer is to be investigated. Differential fluid height against the density is to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Let's modify this problem to also calculate the absolute pressure in the tank by supplying the atmospheric pressure. Use the relationship between the pressure gage reading and the manometer fluid column height. " Function fluid_density(Fluid$) "This function is needed since if-then-else logic can only be used in functions or procedures. The underscore displays whatever follows as subscripts in the Formatted Equations Window." If fluid$='Mercury' then fluid_density=13600 else fluid_density=1000 end {Input from the diagram window. If the diagram window is hidden, then all of the input must come from the equations window. Also note that brackets can also denote comments - but these comments do not appear in the formatted equations window.} {Fluid$='Mercury' P_atm = 101.325 [kPa] DELTAP=80 [kPa] "Note how DELTAP is displayed on the Formatted Equations Window."} g=9.807 [m/s^2] "local acceleration of gravity at sea level" rho=Fluid_density(Fluid$) "Get the fluid density, either Hg or H2O, from the function" "To plot fluid height against density place {} around the above equation. Then set up the parametric table and solve." DELTAP = RHO*g*h/1000 "Instead of dividiing by 1000 Pa/kPa we could have multiplied by the EES function, CONVERT(Pa,kPa)" h_mm=h*convert(m, mm) "The fluid height in mm is found using the built-in CONVERT function." P_abs= P_atm + DELTAP "To make the graph, hide the diagram window and remove the {}brackets from Fluid$ and from P_atm. Select New Parametric Table from the Tables menu. Choose P_abs, DELTAP and h to be in the table. Choose Alter Values from the Tables menu. Set values of h to range from 0 to 1 in steps of 0.2. Choose Solve Table (or press F3) from the Calculate menu. Choose New Plot Window from the Plot menu. Choose to plot P_abs vs h and then choose Overlay Plot from the Plot menu and plot DELTAP on the same scale." 0 2200 4400 6600 8800 11000 ρ [kg/m^3] hmm [mm] Manometer Fluid Height vs Manometer Fluid Density ρ [kg/m3] hmm [mm] 800 2156 3511 4867 6222 7578 8933 10289 11644 13000 10197 3784 2323 1676 1311 1076 913.1 792.8 700.5 627.5 preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-26 1-71 The air pressure in a tank is measured by an oil manometer. For a given oil-level difference between the two columns, the absolute pressure in the tank is to be determined. Patm = 98 kPa AIR 0.36 m Properties The density of oil is given to be ρ = 850 kg/m3. Analysis The absolute pressure in the tank is determined from = 101.0 kPa ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = + = + 2 3 2 atm 1000N/m 1kPa (98 kPa) (850 kg/m )(9.81m/s )(0.36 m) P P ρgh 1-72 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. AIR P 15 mm Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level. (b) The absolute pressure in the duct is determined from = 102 kPa ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = + = + 2 2 3 2 atm 1000 N/m 1 kPa 1 kg m/s 1 N (100 kPa) (13,600 kg/m )(9.81 m/s )(0.015 m) P P ρgh 1-73 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. 45 mm AIR P Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level. (b) The absolute pressure in the duct is determined from = 106 kPa ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = + = + 2 2 3 2 atm 1000 N/m 1 kPa 1 kg m/s 1 N (100 kPa) (13,600 kg/m )(9.81 m/s )(0.045 m) P P ρgh preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-27 1-74E The systolic and diastolic pressures of a healthy person are given in mmHg. These pressures are to be expressed in kPa, psi, and meter water column. Assumptions Both mercury and water are incompressible substances. Properties We take the densities of water and mercury to be 1000 kg/m3 and 13,600 kg/m3, respectively. Analysis Using the relation P = ρgh for gage pressure, the high and low pressures are expressed as 10.7 kPa 16.0 kPa 1000N/m 1kPa 1kg m/s (13,600 kg/m )(9.81 m/s )(0.08 m) 1N 1000N/m 1kPa 1kg m/s (13,600 kg/m )(9.81 m/s )(0.12 m) 1N 2 2 3 2 low low 2 2 3 2 high high = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = P gh P gh ρ ρ Noting that 1 psi = 6.895 kPa, 2.32 psi 6.895kPa (16.0 Pa) 1 psi high = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ P = and 1.55 psi 6.895kPa (10.7 Pa) 1 psi low = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ P = For a given pressure, the relation P = ρgh can be expressed for mercury and water as P = ρ water ghwater and P = ρ mercuryghmercury . Setting these two relations equal to each other and solving for water height gives h mercury water mercury P water ghwater mercuryghmercury hwater h ρ ρ = ρ = ρ → = Therefore, 1.09m 1.63m = = = = = = (0.08 m) 1000 kg/m 13,600 kg/m (0.12 m) 1000 kg/m 13,600 kg/m 3 3 mercury, low water mercury water, low 3 3 mercury, high water mercury water, high h h h h ρ ρ ρ ρ Discussion Note that measuring blood pressure with a “water” monometer would involve differential fluid heights higher than the person, and thus it is impractical. This problem shows why mercury is a suitable fluid for blood pressure measurement devices. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-28 1-75 A vertical tube open to the atmosphere is connected to the vein in the arm of a person. The height that the blood will rise in the tube is to be determined. Assumptions 1 The density of blood is constant. 2 The gage pressure of blood is 120 mmHg. Properties The density of blood is given to be ρ = 1050 kg/m3. Blood h Analysis For a given gage pressure, the relation P = ρgh can be expressed for mercury and blood as P = ρ blood ghblood and P = ρ mercuryghmercury . Setting these two relations equal to each other we get P = ρ blood ghblood = ρ mercuryghmercury Solving for blood height and substituting gives = = (0.12 m) = 1.55m 1050 kg/m 13,600 kg/m 3 3 mercury blood mercury hblood h ρ ρ Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein. This explains why IV tubes must be placed high to force a fluid into the vein of a patient. 1-76 A diver is moving at a specified depth from the water surface. The pressure exerted on the surface of the diver by water is to be determined. Assumptions The variation of the density of water with depth is negligible. Properties The specific gravity of seawater is given to be SG = 1.03. We take the density of water to be 1000 kg/m3. Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water which is taken to be 1000 kg/m3: Patm Sea h P SG (1.03)(1000 kg/m3 ) 1030 kg/m3 2 ρ = × ρ H O = = The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location: = 404 kPa ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = + = + 2 3 2 atm 1000 N/m (101 kPa) (1030 kg/m )(9.807 m/s )(30 m) 1 kPa P P ρgh preparation. If you are a student using this Manual, you are using it without permi