D. Keffer - ChE 240: Heat Transfer and Fluid Flow
Homework Assignment Number Four Solution
Assigned: Wednesday, February 3, 1999
Due: Wednesday, February 10 1999 BEFORE LECTURE STARTS.
Problem 1. Geankoplis, problem 2.8-4, page 110
We have water.
m3
q = 0.050
s
o
Expanding bend. α = 120 , D1 = 0.0762m , D 2 = 0.2112m , p1 = 68,940Pa( gage )
Neglect energy losses.
q 0.050
m
=
= 10 .96
π 2
s
A1
D1
4
q
0.050
m
=
= 1.43
v2 =
π 2
A2
s
D2
4
D v ρ 0.0762 ⋅10.96 ⋅1000
NRe1 = 1 1 =
= 8351520
µ
0.001
D v ρ 0.2112 ⋅1.43 ⋅1000
NRe 2 = 2 2 =
= 302016
µ
0.001
v1 =
turbulent.
Mechanical energy balance:
0=
∆p ∆v 2
+
+ g∆z + Ŵs +
ρ
2α
0=
p2 − 68940 − 101325 1.432 − 10.962
+
+ 0+ 0+ 0
1000
2
∑ F̂
p2 = 229303 Pa(abs ) = 127978 Pa(gage )
From momentum balance on page 74:
& v 2 cos α 2 − m
& v1 cos α1 + p2 A 2 cos α 2 − p1A1 cos α1
Rx = m
& v 2 sin α 2 − m
& v1 sin α1 + p2 A 2 sin α 2 − p1A1 sin α1 + mt g
Ry = m
& = ρv1A 1 = ρv 2 A 2 = 50 kg
m
s
We must realize that:
except
α1 = 0o and α 2 = 120o .
Then we know everything to plug into the above equations
mt , the total mass inside the pipe, which we cannot calculate without knowing the volume inside the pipe.
D. Keffer - ChE 240: Heat Transfer and Fluid Flow
R x = 50 ⋅1.43 ⋅(− 0.5 )− 50 ⋅10.96 ⋅1 +
π
π
229303 ⋅ D22 ⋅(− 0.5 )− (68940 + 101325 ) D12 ⋅1 = − 5376.8N
4
4
π
R y = 50 ⋅1.43 ⋅(0.866 )− 0 + 229303 ⋅ D22 ⋅0.866 − 0 + mt (9.8)
4
R y = 7018 + mt (9.8 )N
Problem 2. Geankoplis, problem 2.8-7, page 110
v1 = 30.5
m
, D1 = 0.01
s
Vane is U-shaped so
α1 = 0o and α 2 = 180o
Ignore pressure terms for a free jet because pressure is the same everywhere.
& v 2 cos α 2 − m
& v1 cos α1
Rx = m
& v 2 sin α 2 − m
& v1 sin α1 + mt g
Ry = m
We ignore the
& = ρv1A 1 = 1000 ⋅30 .5 ⋅π 0.012 = 1000 ⋅30.5 ⋅7.85 ⋅10 − 5 = 2.40 kg
m
4
s
Assume that the cross-sectional area of flow is constant so
& = ρv1A 1 = ρv 2 A 2
m
and
v1 = v 2 , then
R x = 2.40 ⋅30.5 ⋅(− 1)− 2.40 ⋅30.5 ⋅(1)= − 146.4N
R y = 2.40 ⋅30.5 ⋅(0 )− 2.40 ⋅30.5 ⋅(0 )+ mt g = 0 if we neglect gravity.
Problem 3. Geankoplis, problem 2.9-2, page 110
constant density, laminar flow, steady state, horizontal between 2 flat parallel plates. Separated by a distance of
2y o .
vx
2
2
pL )y o   y  
1 − 
y 

2µL

  o 

(p −
(y) = 0
for
− yo < y < yo
This derivation follows the derivation done in class for the fluid flowing down a plate.
The control volume is a rectangular cube of dimensions
V = LW∆y where W is some arbitrary width in the z-direction, L is the length in the flow direction, x, and
delta y is the incremental width in the direction perpendicular to the parallel plates.
Then the shell momentum balance becomes:
D. Keffer - ChE 240: Heat Transfer and Fluid Flow
0 = LWτyx | y + ∆y − LWτyx | y +
∆yW v y (ρv )| z =L − ∆yW v y (ρv )|z =0
p∆yW | z =L − p∆yW | z =0
The two convective terms in the middle cancel. Divide by
0=
τyx | y + ∆y − τyx | y
∆y
dτyx
dy
=−
+
LW∆y
p | z =L − p |z =0
L
∆p
L
Integrate, using the Boundary Condition given in the problem statement.
τyx ( y )
∫dτyx
τyx ( y = 0 )
τyx = −
y
y
∆p
∆p
=− ∫
dy = −
∫dy
L
L
y =0
y =0
∆p
y
L
(LINEAR PROFILE OF STRESS)
Remember Newton’s Law of Viscosity:
dv x
dy
dv
∆p
− µ x =−
y
dy
L
τ yx = − µ
µ
v x( y )
y
∆p
∫dv x = L ∫ydy
y =0
v x ( y =0 )
v x( y o )
y
∆p o
µ ∫dv x =
∫ydy
L
v
y
x( y )
2
y
∆p 
y2 
o


µ(v x ( y o ) − v x ( y ))=
−

L 
2
2


∆p 2 
y2 


(0 − v x ( y ))=
y o 1−
2


2µL
 yo 
D. Keffer - ChE 240: Heat Transfer and Fluid Flow
(
p 0 − pL ) 2 
y2 
y2 
∆p 2 



 PARABOLIC PROFILE
v x (y) = −
y o 1−
y o 1−
=−
2
2




2µL
2µL
 yo 
 yo 
Problem 4. Geankoplis, problem 2.10-1, page 111
Use Hagen-Poiseuille eqn to measure viscosity;
v=
∆pD 2
Hagen-Poiseuille Equation,
32µL
µ=
∆pD 2
32 v L
∆p = ρgh = 996 ⋅9.8 ⋅0.131 = 1289 .1Pa
q
q
5.33 ⋅10 − 7
m
v= =
=
= 0.13745
A π D 2 π 0.002222 2
s
4
4
1289 .1 ⋅0.002222 2
µ=
= 0.009129 kg / m / s
32 ⋅0.13745 ⋅0.1585