Problem 8.90

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Problem 8.90 [Difficulty: 3]
Given: Data on flow from reservoir
Find: Head from pump; head loss
Solution:
Basic equations
p3
ρ g⋅ α
V3
2
2 g⋅⋅+ z3+
⎛⎜⎜⎝
⎞
⎠
p4
ρ g⋅ α
V4
2
2 g⋅⋅+ z4+
⎛⎜⎜⎝
⎞
⎠−
hlT
g
= HlT= for flow from 3 to 4
p3
ρ g⋅ α
V3
2
2 g⋅⋅+ z3+
⎛⎜⎜⎝
⎞
⎠
p2
ρ g⋅ α
V2
2
2 g⋅⋅+ z2+
⎛⎜⎜⎝
⎞
⎠−
∆hpump
g
= Hpump= for flow from 2 to 3
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 = V3 = V4 (constant area pipe)
Then for the pump Hpump
p3 p2−
ρ g⋅=
Hpump 450 150−( ) 103×
N
m2
⋅ m
3
1000 kg⋅×
kg m⋅
s2 N⋅
× s
2
9.81 m⋅×= Hpump 30.6 m=
In terms of energy/mass hpump g Hpump⋅= hpump 9.81
m
s2
⋅ 30.6× m⋅ N s
2⋅
kg m⋅×= hpump 300
N m⋅
kg
⋅=
For the head loss from 3 to 4 HlT
p3 p4−
ρ g⋅ z3+ z4−=
HlT 450 0−( ) 103×
N
m2
⋅ m
3
1000 kg⋅×
kg m⋅
s2 N⋅
× s
2
9.81 m⋅× 0 35−( ) m⋅+= HlT 10.9 m=
In terms of energy/mass hlT g HlT⋅= hlT 9.81
m
s2
⋅ 10.9× m⋅ N s
2⋅
kg m⋅×= hlT 107
N m⋅
kg
⋅=