The horse power transmitted through a pipe is maximum when the ratio of loss of head due to friction and total head supplied is
A. $$\frac{1}{3}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{2}{3}$$
Answer: Option A
Solution (By Examveda Team)
When fluid flows through a pipe, energy is lost due to friction, and this loss is called the head loss.The power transmitted by the fluid depends on the head available after accounting for this frictional loss.
Let the total head supplied be H, and let the head loss due to friction be hf.
Then, the head available for power transmission = H – hf
Horsepower transmitted ∝ (H – hf) × flow rate, and flow rate itself ∝ √(H – hf)
Hence, horsepower transmitted is proportional to (H – hf)1.5
To maximize power, differentiate the expression and solve: the power transmitted is maximum when hf = H / 3
That means, head loss due to friction is one-third of the total head supplied.
Therefore, the correct answer is: Option A — 1/3
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Comments (5)
A. 22.5 m/sec.
B. 33 m/sec.
C. 40 m/sec.
D. 90 m/sec.
A. the weight of the body
B. more than the weight of the body
C. less than the weight of the body
D. weight of the fluid displaced by the body
The difference of pressure between the inside and outside of a liquid drop is
A. $${\text{p}} = {\text{T}} \times {\text{r}}$$
B. $${\text{p}} = \frac{{\text{T}}}{{\text{r}}}$$
C. $${\text{p}} = \frac{{\text{T}}}{{2{\text{r}}}}$$
D. $${\text{p}} = \frac{{2{\text{T}}}}{{\text{r}}}$$
A. cannot be subjected to shear forces
B. always expands until it fills any container
C. has the same shear stress.at a point regardless of its motion
D. cannot remain at rest under action of any shear force
Power : force x vel : F * q/a : q * g * h( h : H - h friction) x Q / A
dP/dV : 0, we get h/h friction: 1/3
head loss due to friction (hf)= fLV²/2gD
Total head supplied (H)
When the power transmitted through pipe is maximum then
H - (3fLV²/2gD )=0
(H)-(3hf)=0
H=3hf
So, hf/H=1/3
Please tell the theory behind this answer
Pls explain
How?