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Problem 4.2
A mass of 3 kg falls freely a distance of 5 m before contacting a spring attached to the ground. If the spring stiffness is 400 N/m, what is the maximum spring compression?

Given: Data on mass and spring Find: Maximum spring compression Solution The given data is M 3 kg h 5m k 400 N m

Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring(t spring has gravitional potential energy and the spring elastic potential energy) Total mechanical energy at initial state E1 Mgh

Total mechanical energy at instant of maximum compression xE2

M g ( x)

1 2 kx 2

Note: The datum for zero potential is the top of the uncompressed spring But E1 E2 M g ( x) 1 2 kx 2 0

so

Mgh

Solving for x

x

2

2Mg x k

2Mgh k

x

Mg kMg k

2

2Mgh k

x

3 kg

9.81

m s
2

m 400 N m s
2

3 kg

9.81

m 400 N

2

2

3 kg 9.81

m s
2

5m

m 400 N

x

0.934 m

Note that ignoring the loss of potential of the mass due to spring compression x gives 2Mgh k

x

x

0.858 m

Note that the deflection if the mass is dropped from immediately above the spring is 2Mg k

x

x

0.147 m Problem 4.4

Problem 4.5

Problem 4.7

Problem 4.8

Given: Data on cooling of a can of soda in a refrigerator Find: How long it takes to warm up in a room Solution The First Law of Thermodynamics for the can (either warming or cooling) is dT dt dT dt k Mc T is the

Mc

k T

Tamb

or

A T

Tamb

where

A

where M is the can mass, c temperature, and Tamb is theambient temperature dT T Tamb T ( t) Tamb

Separating variables

A dt

Integrating

Tinit

Tamb e

At

where Tinit is the initial temperature. The available data from the coolling can now be used to ob a value for constant A Given data for cooling Tinit ( 25 273) K Tinit 298 K

Tamb

(5

273) K

Tamb

278 K

T

( 10

273) K

T

283 K

when

t

10 hr

HenceA

1

ln

Tinit T

Tamb Tamb
4 1

1 3 hr

1 hr 3600 s

ln

298 283

278 278

A

1.284

10

s

Then, for the warming up process

Tinit

( 10

273) K

Tinit

283 K

Tamb

( 20

273) K

Tamb

293 K

Tend

( 15

273) K

Tend

288 K

with

Tend

Tamb

Tinit

Tamb e

A

Hence the time τ is

Tinit Tamb 1 ln A Tend Tamb

s 1.28410
4

ln

283 288

293 293

5.398 10 s

3

1.5 hr

Problem 4.10

Given: Data on velocity field and control volume geometry Find: Several surface integrals Solution
dA1 dA2 V wdzˆ j wdzˆ j ˆ azˆ bk j ˆ wdyk dA1 dA2 V dzˆ j dzˆ j ˆ 10 zˆ 5k j ˆ dyk

(a)

V dA1

ˆ 10 zˆ 5k j
1

dzˆ j
1

ˆ dyk

10 zdz 5dy
1 0 1

(b)
A1

V dA1
0

10 zdz
0

5dy

5z 25y 0

0

(c)

V dA2

ˆ j 10 zˆ 5k

dzˆ j

10 zdz

(d)

V V dA2

ˆ 10 zˆ 5k 10 zdz j
1

(e)
A2

V V dA2
0

ˆ 10 zˆ 5k 10 zdz j

100 3 ˆ z j 3 0

1

ˆ 25 z 2 k

1 0

33.3 ˆ j

ˆ 25k

Problem 4.11

Given: Data on velocity field and control volume geometry Find: Volume flow rate and momentum flux through shaded area Solution
dA dxdzˆ j ˆ dxdyk

V

ˆ axibyˆ j

V

ˆ xi

yˆ j

(a)

Volume flow rate

Q

V dA
A 3 1 A

ˆ xi
1

yˆ j 3 ydz

dxdzˆ j
1

ˆ dxdyk 3 2 2 z dz 6 z 3z 2
1 0

ydzdx
0 0 0

0

Q

3

m3 s

(b)

Momentum flux

V V dA
A A 3 1

ˆ xi

yˆ j

ydxdz
1

ˆ xy i dzdx
0 0 0

3 y 2 dz
1 0

x2 2

3

2z
0

z2

4 z 13 3

1 0

9 2

4 3

3.167

Problem 4.12

Problem4.13

Problem 4.18

Problem 4.19

Given: Data on flow through device Find: Velocity V3; plot V3 against time; find when V3 is zero; total mean flow Solution Governing equation: For incompressible flow (Eq. 4.13) and uniform flow

V dA

VA

0

Applying to the device (assuming V3 is out)

V1 A1

V2 A2

V3 A3

0

V3

V1 A1 V2 A2 A3

10 e

t 2 m

s

0.1 m

2...