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In a triangular notch there is an error of 4% in observing the head, the error in the computation of discharge
- a)10%
- b)4%
- c)6%
- d)2.5%
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
In a triangular notch there is an error of 4% in observing the head, t...
Concept:
Discharge through a triangular notch,
Q = 8/15 Cd √2 g tan θ/2 × H5/2
Where Q = discharge, Cd = Co-efficient of discharge, H = Head above the bottom of the notch, and θ = enclosed angle of the vee.
Differentiating equation w.r.t. H,
we have
dQ / dH = 8/15 Cd √2 g tanθ/2 × 5 /2 H3 / 2
dQ = 8/15 Cd √ 2 g tanθ/2 × 5 /2 H3/2 × d H
∴ dQ/Q = 5/2 dH/H = 2.5 dH/H
Calculation:
Given:
dH/H X 100 = 4%
The percentage error in computing discharge
= 2.5 dH/H = 2.5 x 4 = 10%
Discharge through a triangular notch,
Q = 8/15 Cd √2 g tan θ/2 × H5/2
Where Q = discharge, Cd = Co-efficient of discharge, H = Head above the bottom of the notch, and θ = enclosed angle of the vee.
Differentiating equation w.r.t. H,
we have
dQ / dH = 8/15 Cd √2 g tanθ/2 × 5 /2 H3 / 2
dQ = 8/15 Cd √ 2 g tanθ/2 × 5 /2 H3/2 × d H
∴ dQ/Q = 5/2 dH/H = 2.5 dH/H
Calculation:
Given:
dH/H X 100 = 4%
The percentage error in computing discharge
= 2.5 dH/H = 2.5 x 4 = 10%
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Community Answer
In a triangular notch there is an error of 4% in observing the head, t...
Concept:
Discharge through a triangular notch,
Q = 8/15 Cd √2 g tan θ/2 × H5/2
Where Q = discharge, Cd = Co-efficient of discharge, H = Head above the bottom of the notch, and θ = enclosed angle of the vee.
Differentiating equation w.r.t. H,
we have
dQ / dH = 8/15 Cd √2 g tanθ/2 × 5 /2 H3 / 2
dQ = 8/15 Cd √ 2 g tanθ/2 × 5 /2 H3/2 × d H
∴ dQ/Q = 5/2 dH/H = 2.5 dH/H
Calculation:
Given:
dH/H X 100 = 4%
The percentage error in computing discharge
= 2.5 dH/H = 2.5 x 4 = 10%
Discharge through a triangular notch,
Q = 8/15 Cd √2 g tan θ/2 × H5/2
Where Q = discharge, Cd = Co-efficient of discharge, H = Head above the bottom of the notch, and θ = enclosed angle of the vee.
Differentiating equation w.r.t. H,
we have
dQ / dH = 8/15 Cd √2 g tanθ/2 × 5 /2 H3 / 2
dQ = 8/15 Cd √ 2 g tanθ/2 × 5 /2 H3/2 × d H
∴ dQ/Q = 5/2 dH/H = 2.5 dH/H
Calculation:
Given:
dH/H X 100 = 4%
The percentage error in computing discharge
= 2.5 dH/H = 2.5 x 4 = 10%
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In a triangular notch there is an error of 4% in observing the head, the error in the computation of dischargea)10%b)4%c)6%d)2.5%Correct answer is option 'A'. Can you explain this answer?
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