1.4 Perform the following unit conversions:

1

1.4 Perform the following unit conversions:

(a) 3 33 in. 61 ft 1 in. 12 L 1 ft 0353.0

L 1 ←

(b) Btu 616.0 kJ 1.0551 Btu 1 J 10 kJ 1 J 506 3

 ←

(c) s lbfft

596.99

Btu 1 lbfft 78.177 s 3600 h 1 kW 1 Btu/h 4133 kW .1350  

 ←

(d) min lb 50 min 1 s 60 kg 0.4536 lb 1 g 10 kg 1 s g 783 3

 ←

(e) 2 32 in.lbf 09.44 kPa 1 Pa 10 Pa 894.86 lbf/in. 1 kPa 043 ←

(f) s ft 54.0 s 3600 h 1 m 1 ft .28083 h m 55 333

 ←

(g) s ft 57.45 s 3600 h 1 m 1 ft .28083 km 1 m 01 h km 05 3

 ←

(h) ton1 lbf 2000 ton1

N .44824

lbf 1

N 8968 ←

(Fundamentals of Engineering Thermodynamics, 9e Michael Moran, Howard Shapiro, Daisie Boettner, Margaret Bailey) (Solution Manual, For Complete File, Download link at the end of this File) 1 / 4

1

1.5 Perform the following unit conversions:

(a) L 2 m 01 L 1 cm 10 m 1 in. .0610240 cm 1 in. 122 33- 3 23 3 3

 ←

(b) kJ 0551.1 lbfft 37.567 kJ 1 lbfft 78.177 

 ←

(c) kW 57.74 hp 341.1 kW 1 hp 001  ←

(d) s kg 126.0 lb 2.2046 kg 1 s 3600 h 1 h lb

0001 ←

(e) bar 027.2 N/m 10 bar 1 Pa 1 N/m 1 lbf/in. 1 Pa 894.86 in.lbf

9.3922

25 2 22

 ←

(f) s m 18.1 s 60 min 1 ft 1 m .0283170 min ft 5002 3 3 33

 ←

(g) h km 7.120 mile/h 1 km/h .60931 h mile

57 ←

(h) N 8896 lbf 1

N .44824

ton1 lbf 0002 ton1 ←

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1 1.6 Which of the following food items weighs approximately one newton?

• a grain of rice
• a small strawberry
• a medium-sized apple
• a large watermelon
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1 1.7 A fully-loaded shipping container has a mass of 30,000 kg. If local acceleration of gravity is 9.81 m/s 2 , determine the container’s weight, in kN.

KNOWN: A fully-loaded shipping container has a specified mass. The local acceleration of gravity is known.

FIND: Determine the container’s weight.

SCHEMATIC AND GIVEN DATA:

ENGINEERING MODEL :

• Local gravitational acceleration is constant at 9.81 m/s

2 .

ANALYSIS: The force due to gravitational acceleration is computed using Eq. 1.1, where Fgrav is the container weight and acceleration is local gravitational acceleration (g).

Fgrav = mg

Substituting values and solving give

Fgrav =N 10 kN 1 m/skg 1 N 1 s m 81.9kg) 000,30( 322        = 294 kN Shipping Container m= 30,000 kg g= 9.81 m/s 2 F grav= ?

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