An oil issues from the pipe in the Figure below at Q = 35 ft

Name :_________________ Quiz: No. 9 Time: 15 minutes Student ID# : ____________ Course: ME 5160, Fall 2021

--- The exam is closed book and closed notes.

An oil issues from the pipe in the Figure below at Q = 35 ft

/h. (a) Assume laminar flow and neglect minor losses to find the kinematic viscosity (ν) of the oil in ft

/s. (b) Verify the laminar flow assumption. (c) For the same flow rate and using the viscosity you found in part a, find the pipe diameter D for which the flow begins to not be laminar, i.e. Re

=2000.

Hint: Calculate Velocity at pipe and derive ℎ

using Energy equation.

Then use below ℎ

formula to get 𝜈 ℎ

= 𝑓

𝐷 𝑉²

2𝑔

; 𝑓 =

𝑅𝑒_𝐷(for laminar);

𝑅𝑒

=

𝜈

Energy equation:

( 𝑝 𝜌𝑔 + 𝑉

2𝑔 + 𝑧)

1

= ( 𝑝 𝜌𝑔 + 𝑉

2𝑔 + 𝑧)

2

+ ℎ

(2)

Name :_________________ Quiz: No. 9 Time: 15 minutes Student ID# : ____________ Course: ME 5160, Fall 2021

---

Solution:

KNOWN: Q, H, L, D FIND: (a) ν, (b) Re, (c) D

ASSUMPTIONS: the pipe flow is laminar and α ≈ 1; no minor losses ANALYSIS:

(a)

𝑉 = 𝑄

𝐴 = 𝑄 𝜋 4 𝐷

=

( 35 3600 ) 𝜋

4 ( 0.5 12 )

2

= 7.13 𝑓𝑡 𝑠

The energy equation between (1) at the free surface and (2) at the pipe exit:

( 𝑝 𝜌𝑔 + 𝑉

2𝑔 + 𝑧)

1

= ( 𝑝 𝜌𝑔 + 𝑉

2𝑔 + 𝑧)

2

+ ℎ

𝑝

𝜌𝑔 + 0

2𝑔 + 𝑧

= 𝑝

𝜌𝑔 + 𝑉

2𝑔 + 𝑧

+ ℎ

ℎ

= 𝐻 − 𝑉

2𝑔 = (10) − (7.13)

2(32.2) = 9.21 𝑓𝑡

Assuming laminar pipe flow:

ℎ

= 𝑓 𝐿 𝐷

𝑉

2𝑔 ; 𝑓

= 64 𝑅𝑒

Replace and simplify:

ℎ

= 64 𝑉𝐷 𝜈

𝐿 𝐷

𝑉

2𝑔 = 32𝜈𝐿𝑉 𝑔𝐷

(0.5)

(1)

(0.5)

(1.5) (1.5)

(0.5)

Name :_________________ Quiz: No. 9 Time: 15 minutes Student ID# : ____________ Course: ME 5160, Fall 2021

---

Solve for ν:

𝜈 = ℎ

𝑔𝐷

32𝐿𝑉 ⁼

(9.21)(32.2) ( 0.5 12 )

2

32(6)(7.13) = 0.000376 = 3.76𝐸 − 4 𝑓𝑡

𝑠

(b) Check Re:

𝑅𝑒

= 𝑉𝐷

𝜈 = (7.13) ( 0.5 12 )

(3.76𝐸 − 4) ≈ 790 𝑂𝐾, 𝑙𝑎𝑚𝑖𝑛𝑎𝑟

(c) Find Re

in terms of Q:

𝑉 = 𝑄 𝜋

4 𝐷

; 𝑅𝑒

= 𝑉𝐷

𝜈 → 𝑅𝑒

= 𝑄 𝜋 4 𝐷

𝐷

𝜈 = 4𝑄 𝜋𝜈𝐷

𝑅𝑒

= 2000 = 4𝑄

𝜋𝜈𝐷 → 𝐷 = 4 ( 35 3600 )

(2000)𝜋(3.76𝐸 − 4) = 0.01646 𝑓𝑡 = 0.197 𝑖𝑛

(0.5)

(0.5) (0.5)

(1.5)

(0.5) (0.5)

(0.5)