Solucionario Cengel 6Ta Edicion

5-1

Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Conservation of Mass 5-1C Mass, energy, momentum, and electric charge are conserved, and volume and entropy are not conserved during a process. 5-2C Mass flow rate is the amount of mass flowing through a cross-section per unit time whereas the volume flow rate is the amount of volume flowing through a cross-section per unit time. 5-3CThe amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteady-flow process. 5-4C Flow through a control volume is steady when it involves no changes with time at any specified position. 5-5C No, a flow with the same volume flow rate at the inlet and the exit is not necessarily steady (unless the density is constant). Tobe steady, the mass flow rate through the device must remain constant.

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill 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.

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5-6E A garden hose is used to fill a water bucket. The volume and mass flow rates of water,the filling time, and the discharge velocity are to be determined. Assumptions 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by splashing. Properties We take the density of water to be 62.4 lbm/ft3 (Table A-3E). Analysis (a) The volume and mass flow rates of water are

V& = AV = (πD 2 / 4)V = [π (1 / 12 ft) 2 / 4](8 ft/s) = 0.04363 ft 3/s
& m = ρV& = (62.4 lbm/ft 3 )(0.04363 ft 3 /s) = 2.72 lbm/s
(b) The time it takes to fill a 20-gallon bucket is

Δt =

⎞ ⎛ 1 ft 3 20 gal V ⎟ = 61.3 s ⎜ = & 0.04363 ft 3 /s ⎜ 7.4804 gal ⎟ V ⎠ ⎝

(c) The average discharge velocity of water at the nozzle exit is
Ve =

V&
Ae

=

V&
2 πDe / 4

=

0.04363 ft 3 /s [π (0.5 / 12 ft) 2 / 4]

= 32 ft/s

Discussion Note that for agiven flow rate, the average velocity is inversely proportional to the square of the velocity. Therefore, when the diameter is reduced by half, the velocity quadruples.

5-7 Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 2.21 kg/m3 at the inlet, and0.762 kg/m3 at the exit. Analysis (a) The mass flow rate of air is determined from the inlet conditions to be V1 = 40 m/s A1 = 90 cm2 AIR V2 = 180 m/s

& m = ρ1 A1V1 = (2.21 kg/m3 )(0.009 m 2 )(40 m/s) = 0.796 kg/s
& & & (b) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit area of the nozzle is determined to be
& m = ρ 2 A2V2 ⎯ ⎯→ A2 = & m 0.796 kg/s = = 0.0058 m 2 =58 cm 2 ρ 2V2 (0.762 kg/ m 3 )(180 m/s)

PROPRIETARY MATERIAL. © 2008 The McGraw-Hill 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.

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5-8E Steam flows in a pipe. The minimum diameter of the pipe for a given steam velocity is to be determined. AssumptionsFlow through the pipe is steady. Properties The specific volume of steam at the given state is (Table A-6E)
P = 200 psia ⎫ 3 ⎬ v 1 = 3.0586 ft /lbm T = 600°F ⎭

Steam, 200 psia 600°F, 59 ft/s

Analysis The cross sectional area of the pipe is
& m= 1

v

AcV ⎯ ⎯→ A =

& mv (200 lbm/s)(3.0586 ft 3/lbm) = = 10.37 ft 2 V 59 ft/s

Solving for the pipe diameter gives
A=

πD 2
4

⎯ ⎯→D =

4A

π

=

4(10.37 ft 2 )

π

= 3.63 ft

Therefore, the diameter of the pipe must be at least 3.63 ft to ensure that the velocity does not exceed 59 ft/s.

5-9 A water pump increases water pressure. The diameters of the inlet and exit openings are given. The velocity of the water at the inlet and outlet are to be determined. Assumptions 1 Flow through the pump is steady. 2...