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FLUID MECHANICS 
Midterm Exam 
Name:................................................. Open book, closed notes 
 (Circle the right answer) (except for a 1-page formula sheet) 
Note: If needed, take the atmosperic pressure to be 100 kPa, the water density to be 1000 
kg/m3, the air density to be 1.25 kg/m3, and the mercury density to be 13,600 kg/m3. Also, take 
g = 9.81 m/s2. In the unlikely event that all the answers are wrong, ignore the answers and 
write down your answer next to the question. Each question is worth 17 points (119 total). 
20 kg/s 
 
1) Orange juice at 20°C (ρjuice=1020 kg/m3) is flowing through a U-type reducing bend 
that joins two streams, as shown in the figure. If the outlet diameter is 8 cm, determine 
the outlet velocity of the orange juice. 
55 kg/s 
 
(a) 10.7 m/s (b) 3.7 m/s (c) 6.8 m/s (d) 14.6 m/s (e) 75 m/s 
 
3 m
2 m 
4 m2) The average density of icebergs is about 917 kg/m3. Determine the 
percentage of the total volume of an iceberg submerged in seawater of density 
1042 kg/m3. 
a) 12% b) 50% c) 88% d) 92% e) 100% 
 
3) The flow of water from a reservoir is controlled by an L-shaped gate hinged at 
point A, as shown in the figure. The mass of the weight at B is 5100 kg. If the gate 
opens when the water height is 2 m, determine the width of the gate. 
 
(a) 0.45 m (b) 2.2 m (c) 5.1 m (d) 4.4 m (e) 2.9 m 
 
4) For a fruit juice, shear stresses are measured by a viscometer as a function of 
shear rate (the velocity gradient). When the shear stress in N/m2 is plotted 
against the shear rate in the range of 0 to 50 s-1, a straight line having a slope of 
0.4 is obtained. What is the viscosity of this fluid when the shear rate is 31.4 s-1? 
a) 78.5 Pa.s b) 12.6 Pa.s c) 31.8 Pa.s d) 3.97 Pa.s e) 0.4 Pa.s 
 
5) A water pipe is connected to a double-U tube manometer, as 
shown in the figure. For given specific gravities and fluid column 
heights, determine the gage pressure at the center of the pipe. 
 
a) 4.5 kPa b) 17 kPa c) 30 kPa d) 44 kPa e) 63 kPa 
36 kg/s
0.3 m 
1.2 m 
0.8 m 
0.7 m
 
6) Orange juice is flowing through a horizontal U-type 
reducing bend at a rate of 36 kg/s. The cross-sectional area, 
gage pressure, and velocity are 168 cm2, 57.5 kPa and 2.10 
m/s at the inlet, and 80 cm2, 50 kPa and 4.37 m/s at the 
outlet. Calculate the force applied to the reducing bend by the 
orange juice in the horizontal direction. 
 
(a) 1284 N (b) 1599 N (c) 1133 N (d) 866 N (e) 233 N 
13 m 
12 m h 
Pump
Sea level 
 
7) The water level in a tank is 12 m above the bottom. A hose is 
connected to the bottom of the tank, and the nozzle at the end of the 
hose is pointed straight up. The tank is at 13 m above the sea level, 
and the water surface is open to the atmosphere. In the line leading 
from the tank to the nozzle there is a pump, which increases the 
water pressure by 50 kPa. Determine the maximum height above the 
sea level to which the water stream could rise. 
 
(a) 17 m (b) 42 m (c) 30 m (d) 25 m (e) 75 m 
 
SOLUTIONS 
 
 
 
20 kg/s 
55 kg/s 
1) Orange juice at 20°C (ρjuice=1020 kg/m3) is flowing through a U-type 
reducing bend that joins two streams, as shown in the figure. If the outlet 
diameter is 8 cm, determine the outlet velocity of the orange juice. 
 
(a) 10.7 m/s (b) 3.7 m/s (c) 6.8 m/s (d) 14.6 m/s (e) 75 m/s 
 
Answer 14.6 m/s 
 
Solution Solved by EES Software. Solutions can be verified by copying 
-and-pasting the following lines on a blank EES screen. 
 
M1=55 "kg/s" 
M2=20 "kg/s" 
M=m1+m2 
D2 = 0.08 "m" 
Rho = 1020 "kg/m3" 
Vel2=m/(rho*pi*D2^2/4) 
 
"Some Wrong Solutions with Common Mistakes:" 
W1_Vel2=m1/(rho*pi*D2^2/4) "Using the wrong mass flow rate" 
W2_Vel2=m/(rho*pi*D2^2) "Finding the area wrong" 
W3_Vel2=(m1-m2)/(rho*pi*D2^2/4) "Using the wrong mass flow rate" 
 
 
 
2) The average density of icebergs is about 917 kg/m3. Determine the percentage of the total volume of an 
iceberg submerged in seawater of density 1042 kg/m3. 
a) 12% b) 50% c) 88% d) 92% e) 100% 
 
Answer 88% 
 
"V_submerget/V_total = rho_iceberg/rho_fluid" 
rho_ice=917 "kg/m3" 
rho_fluid=1042 "kg/m3" 
Submerged%=rho_ice/rho_fluid*100 
 
"Some Wrong Solutions with Common Mistakes:" 
W1_s%= rho_ice/1000*100 "Using water instead of seawater" 
 
3 m
2 m 
4 m 
 
3) The flow of water from a reservoir is controlled by an L-shaped gate hinged at 
point A, as shown in the figure. The mass of the weight at B is 5100 kg. If the gate 
opens when the water height is 2 m, determine the width of the gate. 
 
(a) 0.45 m (b) 2.2 m (c) 5.1 m (d) 4.4 m (e) 2.9 m 
 
Answer 4.4 m 
 
rho=1000 "kg/m3" 
g=9.81 "m/s2" 
P_ave=rho*g*1 “Pa” 
M=5100 “kg” 
Weight=m*g 
F_R=P_ave*2*Width 
F_R*(1+2*2/3)=Weight*4 “N” 
 
 
4) For a fruit juice, shear stresses are measured by a viscometer as a function of shear rate (the 
velocity gradient). When the shear stress in N/m2 is plotted against the shear rate in the range of 0 
to 50 s-1, a straight line having a slope of 0.4 is obtained. What is the viscosity of this fluid when the 
shear rate is 31.4 s-1? 
a) 78.5 Pa.s b) 12.6 Pa.s c) 31.8 Pa.s d) 3.97 Pa.s e) 0.4 Pa.s 
 
Answer 0.4 Pa.s (the slope is the viscosity) 
 
 
5) A water pipe is connected to a double-U tube manometer, as shown in the figure For given specific gravities 
and fluid column heights, determine the gage pressure at the center of the pipe. 
0.3 m 
1.2 m 
0.8 m 
0.7 m
 
a) 4.5 kPa b) 17 kPa c) 30 kPa 
d) 44 kPa e) 63 kPa 
 
Answer 44 kPa 
 
g=9.81 "m/s2" 
rho_water=1000 "kg/m3" 
SG_oil=0.8 
SG_Hg=13.6 
SG_w=1 
 
h_oil1=0.8 "m" 
h_oil2=1.2 "m" 
h_Hg=0.3 "m" 
h_w=0.7 "m" 
Pg=(SG_oil*h_oil1+SG_Hg*h_Hg-SG_oil*h_oil2+SG_w*h_W)*g*rho_water/1000 "kPa" 
 
"Some Wrong Solutions with Common Mistakes:" 
W1_Pg=(SG_oil*h_oil1+SG_Hg*h_Hg+SG_oil*h_oil2+SG_w*h_W)*g*rho_water/1000 
W2_Pg=(SG_oil*h_oil1+SG_Hg*h_Hg-SG_oil*h_oil2-SG_w*h_W)*g*rho_water/1000 
W3_Pg=(SG_oil*h_oil1-SG_Hg*h_Hg+SG_oil*h_oil2+SG_w*h_W)*g*rho_water/1000 
W4_Pg=(SG_oil*h_oil1+SG_Hg*h_Hg-SG_oil*h_oil2+SG_w*h_W)*rho_water/1000 
 
 
36 kg/s 
2 
1 6) Orange juice is flowing through a horizontal U-type reducing bend at a 
rate of 36 kg/s. The cross-sectional area, gage pressure, and velocity are 
168 cm2, 57.5 kPa and 2.10 m/s at the inlet, and 80 cm2, 50 kPa and 4.37 
m/s at the outlet. Calculate the force applied to the reducing bend by the 
orange juice in the horizontal direction). 
 
(a) 1284 N (b) 1599 N (c) 1133 N (d) 866 N (e) 233 N 
 
Answer 1599 N 
 
Solution Solved by EES Software. Solutions can be verified by copying 
-and-pasting the following lines on a blank EES screen. 
 
"From Momentum equation for steady uniform flow,F=(mV)_exit – (mV)_inlet. We take the approach 
direction as positive direction. Then the inlet jet velocity is positive, and the exit jet velocity is negative. 
 Then, -F=-P1A1-P2A2-mV1-mV2. 
 
M=36 "kg/s" 
P1=57500 "Pa" 
A1=0.0168 "m2" 
V1=2.1 "m/s" 
P2=50000 "Pa" 
A2=0.0080 "m2" 
V2=4.37 "m/s" 
 
F=-P1*A1-P2*A2-m*V1-m*V2 "N" 
 
"Some Wrong Solutions with Common Mistakes:" 
W1_F=-m*V1-m*V2 "ignoring pressures" 
W2_F=-P1*A1-P2*A2-m*V1+m*V2 "Wrong sign for outlet velocity" 
W3_F=P1*A1+P2*A2-m*V1-m*V2 "Wrong signs for pressure" 
W4_F=(-P1*A1-P2*A2)/1000-m*V1-m*V2 "Using kPa for pressure" 
 
 
 
13 m 
12 m h 
Pump
Sea level 
7) The water level in a tank is 12 m above the ground. A hose is 
connected to the bottom of the tank, and the nozzle at the end of the 
hose is pointed straight up. Thetank is at 13 m above the sea level, 
and the water surface is open to the atmosphere. In the line leading 
from the tank to the nozzle is a pump, which increases the water 
pressure by 50 kPa. Determine the maximum height above the sea 
level to which the water stream could rise. 
 
(a) 17 m (b) 42 m (c) 30 m (d) 25 m (e) 75 m 
 
Answer (b) 30 m 
 
Solution Solved by EES Software. Solutions can be verified by copying 
-and-pasting the following lines on a blank EES screen. 
 
"From Energy equation, P1/rho*g+Vel1^2/2g+z1+h_pump=P2/rho*g+Vel2^2/2g+z2+h_turbine+hL. 
Taking 1 and 2 at surfaces of tank and top of water jet and hL = 0, it simplifies to z1+h_pump=z2 
or h_pump=h-z1. Also, DeltaP_pump=rho*g*h_pump. Therefore, h=z1+DeltaP_pump/(rho*g). 
 
g=9.81 "m/s2" 
rho=1000 "kg/m3" 
z1=12+13 "m" 
DeltaP_pump=50000 "N/m2" 
h=z1+DeltaP_pump/(rho*g) 
 
"Some Wrong Solutions with Common Mistakes:" 
W1_h=z1+DeltaP_pump/(rho*g)/1000 "Using kPa - wrong conversion" 
W2_h=z1+DeltaP_pump/(rho) "Not using g" 
W3_h=12+DeltaP_pump/(rho*g) "ignoring sea level" 
 
	Midterm Exam
	SOLUTIONS
	F_R*\(1+2*2/3\)=Weight*4 “N”