Solucionario Potter

CHAPTER 4

The Integral Forms of the Fundamental Laws
4.1 v a) No net force may act on the system: ΣF = 0. b) The energy transferred to or from the system must be zero: Q - W = 0. v ˆ ˆ ˆ c) If V3n = V3 ⋅ n2 = 10i ⋅ (− j ) = 0 is the same for all volume elements then v D v v D v ΣF = V ∫ dm , or ΣF = (mV ). Since mass is constant for a system Dt v v Dt v v DV DV v v ΣF = m . Since = a, Σ F =ma. Dt Dt

4.2

v 1 Extensive properties: Mass, m; Momentum, mV ; kinetic energy, mV 2 ; 2 potential energy, mgh; enthalpy, H. v Associated intensive properties (divide by the mass): unity, 1; velocity, V; V2 /2; gh; H/m = h (specific enthalpy). Intensive properties: Temperature, T; time, t; pressure, p; density, ρ; viscosity, µ.
(B)

4.3 4.4

System (t ) = V 1 c.v.(t ) = V 1 System (t +∆t ) = V 1 + V c.v.(t + ∆t ) = V
1 2

1

2

4.5

System ( t ) = V 1 + V c.v.( t ) = V 1 + V System (t + ∆t ) = V
2 2

2

+V
2

3

1

2 pump

3

c.v.( t + ∆t ) = V 1 + V

50

4.6

a) The energy equation (the 1st law of Thermo). b) The conservation of mass. c) Newton’s 2nd law. d) The energy equation. e) The energy equation.

4.7

n ˆ

v

n ˆ

v

n ˆ
v

vω

v

n ˆ

n ˆ

4.8

n ˆ n ˆ
v v

n ˆ
v v

n ˆ
v

4.9

n ˆ
v

v

n ˆ
v

n ˆ

v

n ˆ
v

n ˆ
4.10

n ˆ

v

1 $ 1 $ $ j $ $ n1 = − i− j = −0.707(i + $ ) . n 2 = 0.866 $ − 0.5 $ . i j 2 2 v ˆ ˆ j ˆ V1n = V1 ⋅ n1 = 10i ⋅ [ −0.707( i + ˆ )] = −7.07 fps v $ V2 n = V2 ⋅ n2 = 10i$ ⋅ ( 0.866i$ − 0.5 $) = 8.66 fps j v ˆ ˆ ˆ V3n = V3 ⋅ n2 = 10i ⋅ (− j ) = 0

$n3 = −$ . j

4.11

v $ flux = ηρn ⋅ VA
$ j $ flux1 = ηρ[−0.707 (i + $ )] ⋅ 10i A / 0.707 = −10ηρA $ $ flux2 = ηρ( 0.866i − 0.5 $ ) ⋅ 10i A / 0.866 = 10ηρA j flux3 = ηρ( − $ ) ⋅ 10$ = 0 j iA
3

51

4.12

v $ $ (B ⋅ n) A = 15(0.5i + 0.866 $ ) ⋅ $(10 × 12) j j
= 15 × 0.866 × 120 = 1559 cm 3 Volume = 15 sin 60 o × 10 × 12 = 1559 cm 3

4.13

The control volume must be independent oftime. Since all space coordinates are integrated out on the left, only time remains; thus, we use an ordinary derivative to differentiate a function of time. But, on the right, we note that ρ and η may be functions of (x, y, z, t); hence, the partial derivative is used.

4.14
2 c.v. (0) = c.v. (∆t) = volume 1 system (∆t) is in volumes 1 and 2 1

1

4.15

2 3 1

system (∆t) = V1 + V 2 + V3 c.v. (∆ t) = V1 + V2

4.16
system boundary at (t + ∆t)

4.17

If fluid crosses the control surface only on areas A1 and A2 , v v v $ $ $ ρn ⋅ VdA = ∫ ρn ⋅ VdA + ∫ ρn ⋅ V dA = 0 ∫
c .s . A1 A2

For uniform flow all quantities are constant over each area: v v $ $ ρ 1n 1 ⋅ V 1 ∫ dA + ρ 2 n 2 ⋅ V 2 ∫ dA = 0
2 v v $ $ Let A 1 be the inlet so n 1 ⋅ V1 = −V1 and A 2 be the outlet so n 2 ⋅ V2= V 2 . Then

A1

A

−ρ 1V1 A 1 + ρ 2 V2 A 2 = 0
or

ρ 2 A 2V 2 = ρ 1A 1V 1

52

4.18

Use Eq. 4.4.2 with m V representing the mass in the volume: v dm V dm V $ 0= + ∫ ρn ⋅ V dA = + ρA 2 V 2 − ρA 1V 1 dt dt c. s . dm V & = + ρQ − m . dt Finally, dmV & = m − ρQ . dt Use Eq. 4.4.2 with m S representing the mass in the sponge: v dm S dm S $ 0= + ∫ ρn ⋅ VdA = + ρA 2 V2 + ρA 3V 3 − ρA1V1 dt dt
= dm S & + m 2 + ρA 3V 3 − ρQ 1 . dt

4.19

Finally, dm S & = ρQ 1 − m 2 − ρA 3 V 3 . dt 4.20 (D)
& m = ρ AV = p 200 AV = π × 0.042 × 70 = 0.837 kg/s . RT 0.287 × 293

4.21

A1V1 = A2V2.

π×

1.25 2 2.5 2 × 60 = π × V2. 144 144

∴V2 = 15 ft/sec.

1.25 2 1.25 2 & m = ρAV = 1.94π × 60 = 3.968 slug/sec. Q = AV = π × 60 = 2.045 ft 3 / sec. 144 144 4.22 A1 V1 = A2 V2 . π ×.0252 × 10 = (2π × .6 × .003)V2 . ∴V2 = 1.736 m/s. 2 m = ρAV = 1000π ×.025 × 10 = 19.63 kg/s. Q = AV=π × .0252 × 10 = 0.01963 m 3 / s. &

4.23 4.24

& min = ρA1 V1 + ρA2 V2 . 200 = 1000 π × .0252 × 25 + 1000 Q2 . ∴Q2 = 0.1509 m 3 / s.
p1 40 × 144 7 × 144 = = .006455 slug/ft3 . ρ 2 = = .000963 slug/ft3 . RT1 1716 × 520 1716 × 610 & m .2 & m = ρAV . ∴V1 = = . ∴V1 = 355 fps. 2 ρ 1 A 1 (π × 2...