Problem set #8
1) A large artery in a person’s body can be approximated by a tube of diameter 9 mm and length 0.35 m. Also
assume that blood has a viscosity of approximately 4 x 10 -3 N·s/m2, a specific gravity of 1.0, and that the
pressure at the beginning of the artery is equivalent to 120 mm Hg. If the flow were steady (it is not) with V =
0.2 m/s, determine the pressure at the end of the artery if it is oriented (a) vertically up (flow up) or (b)
horizontal.
(a)
Vertical
P2 = P1  gL  ef = 15.99  (9.81)(0.35)  0.1107 = 12.45 kPa
(b)
Horizontal
P2 = P1  ef = 15.99  0.1107 = 15.88 kPa
2) A liquid with SG = 0.96, μ = 9.2 x 10-4 N·s/m2, and vapor
pressure pv = 12000 N/m2 (abs) is drawn into the syringe
as is indicated in the given figure. What is the maximum
flowrate if cavitation is not to occur in the syringe?
Entrance loss coefficient = 0.5, exit loss coefficient = 1.
Q = VA = 1.81 ×

(0.25×10-3)2 = 8.9×10-8 m3/s
4
Check if laminar flow
Re =
VD = 9601.81 0.25103

9.2104
= 473 < 2000, laminar flow
3) As shown in the given figure, the velocity profile for laminar flow in a pipe is quite different from that for
turbulent flow. With laminar flow the velocity profile is parabolic; with turbulent flow at Re = 10,000 the velocity
profile can be approximated by the power-law profile shown. (a) For laminar flow, determine at what radial
location you would place a Pitot tube if it is to measure the average velocity in the pipe. (b) Repeat part (a) for
turbulent flow with Re = 10,000.
(a) For laminar flow: πR2Vave = 2πVc 
R
0



r 2
R 
r
1 
  R 
= 0.5  r = 0.707R
(b) For turbulent flow:
0.2

r
1  = 50   r  = 0.750 or r = 0.750R
R  66  R 






2
rdr

4) Water flows at a rate of 0.040 m3/s in a 0.12-m-diameter pipe that contains a sudden contraction to a 0.06m-diameter pipe. (a) Determine the pressure drop across the contraction section. (b) How much of this
pressure difference is due to losses and (c) how much is due to kinetic energy changes?
Solution
P1  P1 = 40.0 kPa + 93.8 kPa = 133.8 kPa
Pressure drop due to losses = 40.0 kPa
Pressure drop due to kinetic energy changes = 93.8 kPa
5) Air flows through the fine mesh gauze shown in the given figure with an average velocity of 1.50 m/s in the
pipe. Determine the loss coefficient for the gauze. Air density is 1.23 kg/m3.
Gauze
V
Water
8 mm
K=
2 78.48
1.231.52
= 56.7
(1)
6) The pressure at section (2) shown in the given Figure is not to
fall below 60 psi when the flowrate from the tank varies from 0 to
1.0 cfs and the branch line is shut off. Determine the minimum
height, h, of the water tank under the assumption that (a) minor
losses are negligible, (b) minor losses are not negligible with
Kentrance = 0.5, Kelbow = 0.7, and Ktee = 0.4. Water viscosity =
1.21×10-5 ft2/s.
10 ft
All pipe is 6-in diameter
smooth plastic with
flanged fittings
h
Branch line
(2)
6 ft
a) Neglect minor losses
600 ft
with 15
o
90 elbows
h = 141.79 + 0.0126h  h = 143.6 ft
2
b) Include minor losses: ef = 4f
LV
D 2
V2
+ K
2
h = 143.88 + 0.0126h  h = 145.7 ft
7) A fan is to produce a constant air speed of 40 m/s
throughout the pipe loop shown in the given Figure. The 3-mdiameter pipes are smooth, and each of the four elbows has a
loss coefficient of 0.30. Determine the power that the fan adds
to the air. Air: viscosity = 1.79×10-5 kg/ms, density = 1.23
kg/m3.
Solution
Ws = Qhp = (1.23)(9.81)(π×1.52)(40)(111) = 3.79×105 = 379 kW
900 ft
Main
line
8) Water flows from the nozzle attached to
the spray tank shown in the given Figure.
Determine the flowrate if the loss coefficient
for the nozzle (based on upstream
conditions) is 0.75 and the Fanning friction
factor for the rough hose is 0.03.
Solution
If 0.11 is the Darcy friction factor then
v(m) = 3.08524, Q(m3/s) = 0.000545207
If 0.11 is the Fanning friction factor then
v(m) = 2.00734, Q(m3/s) = 0.000354726
9) Water flows through two sections of the vertical pipe shown
in the Figure. The bellows connection cannot support any force
in the vertical direction. The 0.40-ft-diameter pipe weighs 0.2
lb/ft, and the friction factor is assumed to be 0.02. At what
velocity will the force, F, required to hold the pipe be zero?
Solution
Apply the momentum equation to the control
volume from the bellows (1) to the exit (2)
P1A1  WH2O  Wpipe =
m (V2  V1) = 0 since V2 = V1
P1 = (WH2O + Wpipe)/A1 = [LA1 + L(Wpipe/L)]/A1
P1 = L + (0.2L)/(π×0.22) = L + 1.59L
P1
g
V12 P2
+ z1 +
=
2g  g
V22
+ z2 +
2g
+ hL
P2 = 0, z1 = 0, z2 = L, , V1 = V2 = V
L V 2
P1 = L + f
= L + 1.59L
D 2
Therefore f
V 2
= 1.59
2D
V=
2D(1.59)
=
f
2(0.4)(1.59)
(1.94)(0.02)
= 5.73 ft/s
8.100) The flowrate between tank A and tank B shown in the Figure is to be increased by 35% (i.e., from Q to
1.35Q) by the addition of a second pipe (indicated by the dotted lines) running from node C to tank B. If the
elevation of the free surface in tank A is 30 ft above that in tank B, determine the diameter, D, of this new
pipe. Neglect minor losses and assume that the friction factor for each pipe is 0.02.
Solution