This question was previously asked in

TNTRB 2017 ME Official Question Paper

- 190 W
- 276 W
- 407 W
- 655 W

Option 2 : 276 W

Free

TNTRB 2017 ME Official Question Paper

641

150 Questions
190 Marks
180 Mins

__Concept:__

Power requirement to overcome the pressure drop, P = ρ × Q × g × hf = ΔP × Q

where, Q = discharge through pipe, h_{f} = friction head

The discharge through pipe is givena s, Q = A × V

__Calculation__

__Given__

Velocity, V = 1.3 m/s, Diameter, D = 15 cm = 0.15 m, Pressure drop, ΔP = 12 kPa

Pressure drop head = Friction loss head

\(\frac{{{P_1} - {P_2}}}{{\rho g}} = {h_f}~or~{\rm{\Delta }}P = \rho g{h_f}\)

**Power requirement:**

\(P = {\rm{\Delta }}P × A × V = 12 × {10^3} × \frac{\pi }{4} × {0.15^2} × 1.3\)

**P = 276 W**

India’s **#1 Learning** Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Trusted by 2,18,55,675+ Students

Testbook Edu Solutions Pvt. Ltd.

1st & 2nd Floor, Zion Building,

Plot No. 273, Sector 10, Kharghar,

Navi Mumbai - 410210

[email protected]
Plot No. 273, Sector 10, Kharghar,

Navi Mumbai - 410210

Toll Free:1800 833 0800

Office Hours: 10 AM to 7 PM (all 7 days)