Page 15 - BASIC CONCEPTUAL OF THERMOFLUID
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CHAPTER 1: CONCEPTUAL PRINCIPLE IN THERMOFLUIDS
Example 1.2
If 5.6 m³ of oil and weights 46000 N, determine:
i. Mass density, ρ in unit kg/m³
ii. Specific weight, ω
iii. Specific gravity of oil, S
Solution:
V = 5.6 m³
ii. = = 837.33 × 9.81 = 8214.21 2
W = 46000 N = 4689.1 kg
837.33
iii. = = 1000 = 0.837
4689.1 kg
i. = = = 837.33
5.6 m 3
Example 1.3
Determine the specific volume if it mass is 500 g and the volume is 400 cm³.
Solution:
m = 0.5 kg
1000 ³
= 400 cm³ × = 0.0004 ³
100 ³
3
0.0004 m 3
i. = = = 0.0008
0.5
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