Question

(6) A reducing elbow is used to deflect water flow at a rate of 30 kg/s in a horizontal pipe upward by an angle of 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 150 cm² at the inlet and 25 cm² at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03.

          (6)
A reducing elbow is used to deflect water flow at a rate of 30 kg/s in a horizontal pipe upward by an angle of 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 150 cm² at the inlet and 25 cm² at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03.
        
Show more…

Added by Jeremy K.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
(6) A reducing elbow is used to deflect water flow at a rate of 30 kg/s in a horizontal pipe upward by an angle of 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 150 cm² at the inlet and 25 cm² at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03.
Close icon
Play audio
Feedback
Powered by NumerAI
Danielle Fairburn David Collins
Jennifer Stoner verified

Sri K and 95 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
1-a-reducing-elbow-is-used-to-deflect-water-flow-at-a-rate-of-14-kgs-in-a-horizontal-pipe-upward-3o-while-accelerating-it-the-elbow-discharges-water-into-the-atmosphere-the-cross-sectional-a-13077

1. A reducing elbow is used to deflect water flow at a rate of 14 kg/s in a horizontal pipe upward 30' while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 113 cm2 at the inlet and 7 cm2 at the outlet. The elevation difference between the centers of the outlet and the inlet is 30 cm. The weight of the elbow and the water in it is considered to be negligible. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place.

Adi S.

a-90-elbow-in-a-horizontal-pipe-is-used-to-direct-water-flow-upward-at-a-rate-of-40-kgs-the-diameter-of-the-entire-elbow-is-10-cm-the-elbow-discharges-water-into-the-atmosphere-and-thus-the-66949

A 90° elbow in a horizontal pipe is used to direct water flow upward at a rate of 40 kg/s. The diameter of the entire elbow is 10 cm. The elbow discharges water into the atmosphere, and thus the pressure at the exit is the local atmospheric pressure. The elevation difference between the centers of the exit and the inlet of the elbow is 50 cm. The weight of the elbow and the water in it is considered to be negligible. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03 at both the inlet and the outlet.

Penny R.

water-flows-steadily-through-the-reducing-elbow-shown-the-elbow-is-smooth-and-short-and-the-flow-accelerates-so-the-effect-of-friction-is-small-the-volume-flow-rate-is-q-127-l-s-the-elbow-is-85968

Water flows steadily through the reducing elbow shown. The elbow is smooth and short, and the flow accelerates, so the effect of friction is small. The volume flow rate is Q =1.27 L/ s. The elbow is in a horizontal plane. Estimate the gage pressure at section (1). Calculate the x-component of the force exerted by the reducing elbow on the supply pipe. Show transcribed image text

Sri K.


*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,727 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,131 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,560 solutions

*

Transcript

-
00:01 So here in this question we are given an elbow, let's say this is the elbow.
00:06 Here, the volume here in this part is v1 and which is here the volume, this is v2.
00:13 The pressure in this first part, or we can say that first, this is p1 and area is a1.
00:20 Here the pressure is p2 and area is a2.
00:24 And this distance is 40 centimeters.
00:31 This is making an angle of 45 degree with the horizontal so from here we are given the water mass flow rate that is equals to 30 kilogram per second we are given the area of section 1 that is a 1 which will be equals to 150 centimeter square that will be equals to 0 .015 square we are given the area of section 2 that is a 2 that will be equals to 25 centimeter square that will comes out to be 0 .0025 meter square so from here mass of the water elbow, m -not will comes out to be 50 kilogram and the momentum flux factor that is beta will come out to be 1 .03.
01:11 So from here we can say that m -0 will be rho a1 v1 that will be equals to r -a2 v2.
01:18 So from here m -not is 30 that will be equals to density of water is thousand multiplied by the area that is 0 .015 multiplied by the v1 that will be equals to thousand multiplied by the 0 .0025 multiplied by the a2.
01:34 So individually equating all this term.
01:38 So from the first part we get the value of v1 as 2 meter per second and we get the value of v2 as 12 meter per second.
01:46 So here we are applying the bernoulli's theorem on section 1 and 2.
01:53 So according to the bernoulli's theorem, p1 divided by the row g plus v1 raised to the power 2 divided by the 2g will be equals to p2 divided by the row g plus v2 raised to the power 2 divided by the 2g plus z 2 so from here p 2 is equal to p a so from here we can say that p a divided by the row multiply by the g will be equals to 10 .3 meter so from here we get the value of p a which comes out to be 11043 pascal so from here we can say that p1 divided by the thousand multiply by the 9 .8 plus 4 divided beta to multiply by the 9 .8, that will be close to 10 .3 plus 144 divided by the 2 multiply by the 9 .8 plus 0 .40.
02:40 So from here, after simplifying this term, we get the value of p1 as 174971 .16 pascal.
02:47 And from here, this is the value of p1.
02:50 So from here we can say that vix is 2 meter per second.
02:54 V1 .y is 0...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade


Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever