Hibbeler Fluid Mechanics 1st ed solution

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.



1–1. Represent each of the following quantities with combi­
nations of units in the correct SI form, using an appropriate
prefix: (a) GN # mm, (b) kg >mm, (c) N>ks2, (d) kN>ms.




Solution
a) GN # mm = (109)N(10-6)m = 103 N # m = kN # m Ans.
3 -6 9
b) kg>mm = (10 )g>(10 )m = 10 g>m = Gg>m Ans.

c) N>ks2 = N>(103 s)2 = 10-6 N>s2 = mN>s2 Ans.
3 -6 9
d) kN>ms = (10 )N>(10 )s = 10 N>s = GN>s Ans.




Ans:
a) kN # m
b) Gg>m
c) µN>s2
d) GN>s

1

,© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.



1–2. Evaluate each of the following to three signifi­
cant figures, and express each answer in SI units using
an appropriate prefix: (a) (425 mN)2, (b) (67 300 ms)2,
(c) 3 723 ( 106 ) 4 1>2 mm.




Solution
a) (425 mN)2 = 3 425 ( 10-3 ) N 4 2 = 0.181 N2 Ans.

b) (67 300 ms) = 3 67.3 ( 10
2 3
)( 10 ) s 4 = 4.53 ( 10 ) s
-3 2 3 2
Ans.

c) 3 723 ( 106 ) 4 1>2 mm = 3 723 ( 106 ) 4 1>2 ( 10-3 ) m = 26.9 m Ans.




Ans:
a) 0.181 N2
b) 4.53 ( 103 ) s2
c) 26.9 m

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,© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.



1–3. Evaluate each of the following to three signifi­
cant figures, and express each answer in SI units using
an appropriate prefix: (a) 749 mm>63 ms, (b) (34 mm)
(0.0763 Ms)>263 mg, (c) (4.78 mm)(263 Mg).




Solution
a) 749 mm>63 ms = 749 ( 10-6 ) m>63 ( 10-3 ) s = 11.88 ( 10-3 ) m>s
= 11.9 mm>s Ans.

b) (34 mm)(0.0763 Ms)>263 mg = 3 34 ( 10-3 ) m4 3 0.0763 ( 106 ) s4 > 3 263 ( 10-6 )( 103 ) g4
= 9.86 ( 106 ) m # s>kg = 9.86 Mm # s>kg Ans.

c) (4.78 mm)(263 Mg) = 3 4.78 ( 10 ) m 4 3 263 ( 10 ) g 4
-3 6

= 1.257 ( 106 ) g # m = 1.26 Mg # m Ans.




Ans:
a) 11.9 mm>s
b) 9.86 Mm # s>kg
c) 1.26 Mg # m

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, © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.



*1–4. Convert the following temperatures: (a) 20°C to
degrees Fahrenheit, (b) 500 K to degrees Celsius, (c) 125°F
to degrees Rankine, (d) 215°F to degrees Celsius.




Solution
5
a) TC = (T - 32)
9 F
5
20°C = (T - 32)
9 F
TF = 68.0°F Ans.

b) TK = TC + 273
500 K = TC + 273
TC = 227°C Ans.

c) TR = TF + 460
TR = 125°F + 460 = 585°R Ans.

5
d) TC = (T - 32)
9 F
5
TC = (215°F - 32) = 102°C Ans.
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