Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 7 (Q.No. 1)
1.
The specific gravity of an oil whose specific weight is 7.85 kN/m3, is
Discussion:
14 comments Page 2 of 2.
Saranya said:
8 years ago
Thank you for all your explanation.
Subhajit Pamja said:
8 years ago
Specific Wt=wt /unit volume
= w/v= mg/v.
We can say m/v *g and that is = rho(p)* g
Given, oil specific wt (pg)=7.85 KN/m^3.
And we have specific gravity= sp.wt of the liquid/sp.wt of pure water at 4°c.
So, specific Wt. Of water =9.81kN/m^3.
And the and is 7.8/9.81=0.79=>0.8(unitless).
= w/v= mg/v.
We can say m/v *g and that is = rho(p)* g
Given, oil specific wt (pg)=7.85 KN/m^3.
And we have specific gravity= sp.wt of the liquid/sp.wt of pure water at 4°c.
So, specific Wt. Of water =9.81kN/m^3.
And the and is 7.8/9.81=0.79=>0.8(unitless).
Ganesh said:
8 years ago
Thanks for the explanation.
Ashok said:
7 years ago
Specific weight=ρ*g=7.85*100.
=>ρ=(7.85*1000)/9.81=800.
Sp.gravity=density of oil/density of water,
=800/1000=0.8.
=>ρ=(7.85*1000)/9.81=800.
Sp.gravity=density of oil/density of water,
=800/1000=0.8.
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