Download 10.53: PROBLEM DEFINITION Situation: Water is draining out of a

10.53: PROBLEM DEFINITION
Situation:
Water is draining out of a tank through a galvanized iron pipe.
ks = 0.006 in = 5 × 10−4 ft, L = 10 ft, H = 4 ft.
The pipe is 1-in schedule 40 NPS, D = 1.049 in = 0.08742 ft.
Find:
Velocity in the pipe (ft/s).
Flow rate (cfs)
Assumptions:
Steady flow.
Component head loss is zero.
Turbulent flow. Also, α2 = 1.0.
Properties:
Water (70 ◦ F), Table A.5, ρ = 1.94 slug/ ft3 , γ = 62.3 lbf/ ft3 , ν = 1.06 × 10−5 ft2 / s.
PLAN Classify this problem as case 2 (V is unknown), then
1. Write the energy eqn., the Darcy-Weisbach eqn., etc. to produce a set of 4
equations with 4 unknowns.
2. Solve the set of equations using a computer program (we used TK Solver).
3. Find the flow rate with Q = V A.
SOLUTION
1. Governing equations:
• Energy equation (section 1 on water surface, section 2 at exit plane)
p1
V2
V2
p2
+ α1 1 + z1 + hp =
+ α2 2 + z2 + ht + hL
γ
2g
γ
2g
V22
0 + 0 + (H + L) + 0 = 0 +
+ 0 + hf
2g
• Darcy-Weisbach:
hf = f
78
L V2
D 2g
(1)
(2)
• Swamee-Jain:
• Reynolds number:
0.25
f=£
¡ ks
log10 3.7D +
Re =
5.74
Re0.9
¢¤2
VD
ν
(3)
(4)
2. Solution of Eqs. (1) to (4):
hf = 11.04 ft
Re = 113800
f = 0.0326
V = 13.8 ft/ s
3. Flow rate equation:
πD2
V
Q =
!
Ã4
π (0.08742 ft)2
(13.8 ft/ s) = 8. 283 1 × 10−2 ft3 / s
=
4
Q = 0.0828 cfs
REVIEW
• Notice that the turbulent flow assumption is valid because Re > 2300.
• An easy way to solve case 2 and case 3 problems is to acquire a computer
program that can solve coupled, non-linear equations.
• Notice that most of the elevation head (14 ft) is converted to head loss (11 ft) .
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