1. Solucionario Cengel Mecanica De Fluidos

Preview text LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS. LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA.

VISITANOS PARA DESARGALOS GRATIS. Solutions Manual for Thermodynamics: An Engineering Approach Seventh Edition Yunus A.

Descargar gratis fisiopatologia mcphee. Cengel, Michael A. Boles 2011 Chapter 1 INTRODUCTION AND BASIC CONCEPTS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The Companies, Inc. And protected copyright and other state and federal laws. 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 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 any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or any means, electronic or otherwise, without the prior written permission of PROPRIETARY MATERIAL. 2011 The Companies, Inc.

Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Mass, Force, and Units The mentioned here must be 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 but always use either or as appropriate since the two units have different dimensions. In this unit, the word light refers to the speed of light. The unit is then the product of a velocity and time.

Hence, this product forms a distance dimension and unit. There is no acceleration, thus the net force is zero in both cases. The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying second law to the weight force gives W mg m 210 lbf W g 32.10 2 32.174 lbm 2 1 lbf 210.5 lbm Mass is invariant and the man will have the same mass on the moon.

Then, his weight on the moon will be 1 lbf W mg (210.5 lbm)(5.47 2 35.8 lbf 2 32.174 lbm 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 1.16 Analysis The mass of the air in the room is 3 3 m (1.16 )(6 6 8 m ) 334.1 kg ROOM AIR 6X6X8 m3 Thus, 1N W mg (334.1 kg)(9.81 2 1 kg 2 3277 N PROPRIETARY MATERIAL.

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Solucionario Cengel Mecanica De Fluidos

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 is to be determined. Z Analysis The weight of a body at the elevation z can be expressed as W mg m(9.807 3.32 z ) In our case, W 0.995W s 0.995mg s 0.995(m)(9.81) Substituting, 0.995(9.81) (9.81 3.32 10 0 z) z 14,774 m 14,770 m Sea level The mass of an object is given. Its weight is to be determined. Analysis Applying second law, the weight is determined to be W mg (200 kg)(9.6 2 ) 1920 N The specific heat of air given in a specified unit is to be expressed in various units. Analysis Applying second law, the weight is determined in various units to be 1 K 1.005 K c p (1.005 1 J 1 kg 1.005 c p (1.005 1 kJ g 1 kcal c p (1.005 0.240 4.1868 kJ 1 0.240 c p (1.005 4.1868 PROPRIETARY MATERIAL. 2011 The Companies, Inc.

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Problem 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. Weight of the rock force balance on the rock yields the net force acting on the rock Fnet Fup Fdown acceleration of the rock is determined from second Run the program, press F2 or select Solve from the Calculate SOLUTION 200 160 2 a 190.2 90.21 56.88 40.21 30.21 23.54 18.78 15.21 12.43 10.21 a m 1 2 3 4 5 6 7 8 9 10 120 80 40 0 1 2 3 4 5 6 m 7 8 9 10 PROPRIETARY MATERIAL. 2011 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 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 side of the equation E 25 kJ 7 do not have the same units, and therefore they cannot be added to obtain the total energy.

Multiplying the last term mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally that is, every term in the equation will have the same unit. Discussion Obviously this error was caused forgetting to multiply the last term mass at an earlier stage.

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 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 s) 3600 kJ, Total energy (8 kWh)(3600 28,800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy. 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 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 V and It is obvious that the only way to end up with the unit for time is to divide the tank volume the discharge rate.

Therefore, the desired relation is V Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities. PROPRIETARY MATERIAL. 2011 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Systems, Properties, State, and Processes 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 exchanging heat and work across its control boundary. Tracking these interactions, we can determine the energy conversion characteristics of this system.

The system is taken as the air contained in the device. This system is a closed or fixed mass system since no mass enters or leaves it. 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 control volume. Intensive properties do not depend on the size (extent) of the system but extensive properties do. 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. If we were to divide this system in half, both the volume and the number of moles contained in each half would be that of the original system.

The molar specific volume of the original system is v V N and the molar specific volume of one of the smaller systems is v 2 V N N which is the same as that of the original system. The molar specific volume is then an intensive property.

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 increasing weight. A process during which a system remains almost in equilibrium at all times is called a process. Many engineering processes can be approximated as being The work output of a device is maximum and the work input to a device is minimum when processes are used instead of processes. PROPRIETARY MATERIAL. 2011 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.

If you are a student using this Manual, you are using it without permission. A process during which the temperature remains constant is called a process during which the pressure remains constant is called and a process during which the volume remains constant is called isochoric. The state of a simple compressible system is completely specified two independent, intensive properties.

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. 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 the entire inner surface of the nozzle and the inlet and outlet This is a control volume since mass crosses the boundary. A process is said to be if it involves no changes with time anywhere within the system or at the system boundaries. PROPRIETARY MATERIAL. 2011 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.

If you are a student using this Manual, you are using it without permission. Temperature 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.

They are Celsius and kelvin (K) in the SI, and fahrenheit and rankine (R) in the English system. 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. A temperature is given in It is to be expressed in K. Analysis The Kelvin scale is related to Celsius scale 273 Thus, 273 310 K The temperature of air given in unit is to be converted to and R unit.

Analysis Using the conversion relations between the various temperature scales, T 1.8T 32 (1.8)(150) 32 T (R ) T 460 302 460 762 R A temperature change is given in It is to be expressed in K. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus, 45 K PROPRIETARY MATERIAL. 2011 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

The flash point temperature of engine oil given in unit is to be converted to K and R units. Analysis Using the conversion relations between the various temperature scales, T (R ) T 460 363 460 823 R T (K ) T (R ) 823 457 K 1.8 1.8 The temperature of ambient air given in unit is to be converted to K and R units. Analysis Using the conversion relations between the various temperature scales, T 32 T 273.15 233.15 K T 459.67 419.67 R The change in water temperature given in unit is to be converted to K and R units. Analysis Using the conversion relations between the various temperature scales, 10 1.8 10 1.8 5.6 K 10 R A temperature range given in unit is to be converted to unit and the temperature difference in is to be expressed in K, and R. Analysis The lower and upper limits of comfort range in are T T 32 65 32 18.3 1. 8 T T 32 75 32 23.9 1.

8 A temperature change of in various units are (R ) 10 R 10 1.8 1.8 (K ) 5.6 K PROPRIETARY MATERIAL. 2011 The Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The pressure in a tank in SI unit is given.

The pressure in various English units are to be determined. Analysis Using appropriate conversion factors, we obtain (a) 20.886 2 P (1500 kPa 1 kPa 31,330 2 (b) 20.886 2 P (1500 kPa 1 kPa 1 ft 2 144 in 2 1 psia 217.6 psia 1 2 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, kPa 1 psia P (1500 mm 29.0 psia 1 mm Hg 6.895 kPa 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 kPa P (1250 mm 1 mm Hg PROPRIETARY MATERIAL. 2011 The Companies, Inc.

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The pressure in a pressurized water tank is measured a 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 interface. Properties The densities of mercury, water, and oil are given to be 13,600, and 850 respectively.