Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 8 (Q.No. 7)
7.
A vessel of 4 m3 contains an oil which weighs 30 kN. The specific weight of the oil is
Discussion:
5 comments Page 1 of 1.
Tejajay said:
7 years ago
The specific weight (w) = p*g = (30/4) * (9.81).
Where, density(p) = m/v = (30/4).
[since,1 kg = 9.81 N] (9.81 is canceled).
w = 30/4 = 7.5 KN/m3.
Where, density(p) = m/v = (30/4).
[since,1 kg = 9.81 N] (9.81 is canceled).
w = 30/4 = 7.5 KN/m3.
Biswaranjan sahoo said:
5 years ago
Specific weight of oil (w) = oil of weights(4) /volume of the vessel(30).
= 30/4,
= 7.5 KN/M^3.
= 30/4,
= 7.5 KN/M^3.
Sid said:
7 years ago
Specific weight= mass /volume.
We have mass and volume,
Specific weight=30/4 =7.5.
We have mass and volume,
Specific weight=30/4 =7.5.
(1)
GANESH KULKARNI said:
1 decade ago
Sp.weight of oil = weight of the oil/volume of the vessel = 30/4.
= 7.5KN/m3.
= 7.5KN/m3.
(1)
Daniel said:
8 years ago
Good @Ganesh.
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