AME 30331-01 (Fall 13) – Fluid Mechanics
Midterm 1
October 7, 2013 11:35 AM – 12:25 PM
Name/ndID: _____________________________________ Format: - This exam consists of two problems (10 points each + 2 bonus points in problem 2). Rules: - The exam is open book, open notes. assumptions must be be justified to get the maximum number number of points points for each each question. question. - All assumptions - Clearly indicate the final answer/result by underlining it or drawing a box around it. - Place your name/ndID in the top, right-hand corner of each page. - Use only black/blue ink. Honor code: “As a member of the Notre Dame community, I will not participate in or tolerate academic dishonesty.”
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Problem 1
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Problem 2
Total
AME 30331-01 (Fall 13) – Fluid Mechanics
Midterm 1
October 7, 2013 11:35 AM – 12:25 PM
PROBLEM 1 (10 pts) A rectangular gate (height: h , width normal to the page: 1) blocks the end of a freshwater channel. The gate is located at a depth h below the free surface. a)
(2 pts) The water density varies with depth as: ( z )
z 0 1 . 2h
Derive an expression for the gage pressure distribution p( z ) in the channel. The pressure distribution can be obtained from the equation of fluid statics along the vertical direction: dp g dz (note the positive sign of the term on the right-hand side due to the downward orientation of the z-axis). Using the density distribution given in the problem:
z z 0 g 1 dp 0 g 1 dz dz 2h 2h
dp Integration yields:
z 2 p( z ) 0 g z C , where 4h
C is a constant of integration.
At the free-surface, the gage pressure is equal to zero (since the absolute pressure is atmospheric): p( z 0) 0 C 0 . Therefore, the gage pressure distribution is:
z 2 p( z ) 0 g z 4h b) (4 pts) What is the resultant hydrostatic force exerted by the water on the gate? At a depth z , the water exerts a force dF on a surface element dA of the gate. By definition, the magnitude of that force is: dF p( z)dA . Here, the surface element is a rectangle of height dz and width (normal to the page) equal to 1. Therefore:
z 2 dF 0 g z dz 4h The resultant hydrostatic force is obtained by integrating dF over the entire gate surface:
AME 30331-01 (Fall 13) – Fluid Mechanics
Midterm 1
October 7, 2013 11:35 AM – 12:25 PM
2h
z 2 z2 z3 4h2 8h3 h2 h3 3 7 0 g 0 gh2 F 0 g z dz 0 g z h 4h 2 12 2 12h h 2 12h 2 12 h 2h
Therefore:
F
c)
25 0 gh2 12
(4 pts) Determine the coordinate z c of the center of pressure (to simplify the integrations, consider to calculate all moments with respect to point O).
At depth z , the pressure force dF generates a moment d M about point O:
z 2 d M zdF 0 gz z dz 4 h The resultant moment exerted on the gate about point O is: 2h
2h z 2 z3 z 4 8h3 16h4 h3 h4 z3 2 g z z dz g z dz g g M 0 0 0 0 3 16h z h zh 4 h 4 h 3 16 h 3 16 h h 2h
Therefore: M
157 0 gh3 48
This moment can also be expressed as: M zc F
z c
25 0 gh2 12
Equating those 2 expressions yields:
z c
250 gh2 12
157 0 gh3
Therefore:
z c
157h 100
48
AME 30331-01 (Fall 13) – Fluid Mechanics
Midterm 1
October 7, 2013 11:35 AM – 12:25 PM
PROBLEM 2 (10 pts + 2 bonus pts) The aerodynamic performance of a semi-trailer truck of width b is tested in a wind tunnel (see Figure below). The wind tunnel is four times larger than the truck width ( H 4b ). Since the problem is twodimensional, all dimensions normal to the page will be assumed equal to 1. The wind tunnel is operated to generate a steady flow of air (constant density
)
with a uniform
velocity U at the entrance. The presence of the truck in the air flow generates a wake downstream, characterized by low flow velocity. As a result, the velocity profile at the exit section of the wind tunnel can be modeled as a non-uniform profile with magnitude varying between U behind the wake and 1.2U outside the wake (see Figure below).
a)
(3 pts) Using a control volume approach, determine the velocity measured at the exit section of the wind tunnel, behind the wake (i.e., determine the coefficient ). State all your assumptions and describe all the steps of the calculation. Hint: you might want to split the control surface at the outlet into 2 sections, each characterized by a uniform velocity profile.
The control volume is the volume of air contained within the test section of the wind tunnel:
AME 30331-01 (Fall 13) – Fluid Mechanics
Midterm 1
October 7, 2013 11:35 AM – 12:25 PM
Since the question is about a velocity parameter, the equation to be considered is the continuity equation:
d V CS V dA 0 t CV Assumptions: - Steady flow - Incompressible flow - 1 inlet, 2 outlets - Uniform velocity and density over each inlet/outlet Reduced continuity equation:
V dA V dA 1
1
1
2
2
2
V3 dA3 0 , 3
which can be rewritten:
AV 1 1 A2V2 A3V 3 0 Substituting the velocities and dimensions provided in the problem:
4bU 3b(1.1U ) b( U ) 0 Therefore:
0.7
b) (5 pts) The drop in air pressure between the entrance and the exit of the wind tunnel is measured as Pentrance
Pexit P . Neglecting the shear stress exerted by the air flow on the walls of the wind
tunnel, determine the drag force, D1 , exerted by the air flow on the truck. Express your result as a function of
, U , b
and P .
The general form of the linear momentum balance is:
Vd V CS V V dA F on CV t CV Using the assumptions listed in the previous question and considering only the forces exerted along the x-direction, this equation reduces to:
V 1
1 x
V1 dA1 V2 x V2 dA2 V3 x V3 dA3 Fx on CV 2
3
Substituting the velocities and dimensions provided in the problem: U (U 4b) (1.1U )(1.1U 3b) (0.7U )(0.7Ub) P(4b)
R x ,
AME 30331-01 (Fall 13) – Fluid Mechanics
Midterm 1
October 7, 2013 11:35 AM – 12:25 PM
where R x is the resultant force exerted by the truck on the CV. Therefore:
4 U 2b 3(1.12 ) U 2b (0.72 ) U 2 b P(4b) R x and
R x U 2b 4 3 1.12 0.7 2 4bP The drag force is the force exerted by the air in the CV on the truck: D1
R x 4bP 0.12 U 2b
c) (2 pts) In order to improve the aerodynamics of the truck, a device called TrailerTail is installed at the rear of the semi-trailer. The device consists of three panels that deploy to modify the aerodynamic profile of the truck (see Figure A below). Wind tunnel measurements indicate that when the device is used, the flow in the wake region is “normalized” and exhibits a velocity magnitude much closer to that of the incoming flow (see Figure B below).
The flow configuration with the TrailerTail deployed can be approximated as that depicted below.
AME 30331-01 (Fall 13) – Fluid Mechanics
Midterm 1
Assuming that the pressure drop is still Pentrance
October 7, 2013 11:35 AM – 12:25 PM
Pexit P , determine the drag force
air on the truck equipped with TrailerTail. Express your result as a function of P , an integral of the form
b 2
b 2
D2 exerted , b
by the
and U and
...dy (you are not required to evaluate this integral).
Using the same equation as that developed for part b):
V 1
1 x
V1 dA1 V2 x V2 dA2 V3 x V3 dA3 Fx on CV 2
3
Given the new velocity profile at the outlet: 2
y 2 4 bU 3bU (1.2) 4.8U 2 dy 4bP R x y b 2 b 2
2
b 2
2
Since the drag force is the opposite of R x : 2
D2
d)
y 2 4bP 0.32 bU 4.8U 2 dy y b 2 b b 2
2
(bonus question: 2 pts) Determine the percent reduction in drag achieved by this technology 3 4 when 1.2 kg/m , b 2.4 m , U 29 m/s and P 10 Pa .
Performing the integration in part c):
D2
b 2
y5 2 2 4bP 0.32 bU 2 4.8U dy 4bP 0.32 bU 2 4.8U 4 y 0 b 4 5b 0 2
b2
y 4
2
Therefore:
4.82 4bP bU 0.32 D2 4bP 0.32 bU 2 4.8U 160b4 80 2
2
b5
2
Finally: D2
4bP 0.608 bU 2
The percent reduction in drag is given by: D1 D2 D1
4bP 0.12 U 2b 4bP 0.608 bU 2 4bP 0.12 U 2b
Numerically: D1 D2 D1
1.23%
0.488 bU 2 4bP 0.12 U 2b
