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Midterm f09 solution
1. ENSC283 INTRODUCTION TO FLUID MECHANICS
27 February 2009
Midterm Examination M. Bahrami
This is a 2-1/2 hours, closed-book and notes examination.
You are permitted to use one 8.5 in.× 11 in. crib sheet (double-sided) and the
property tables.
There are 5 questions to be answered. Read the questions very carefully.
Clearly state all assumptions.
It is your responsibility to write clearly and legibly.
Problem 1: (20 marks)
A stream of water of diameter ݀ ൌ 0.1 ݉ flows steadily from a tank of diameter
ܦ ൌ 1 ݉ as shown in the figure. Determine the flow rate, ܳ, needed from the inflow pipe
if the water depth remains constant, (݄ ൌ 2 ݉).
Problem 1: (Solution)
݄ ൌ 2 ݉
݀ ൌ 0.1 ݉
ܳ
ܦ ൌ 1 ݉
For steady, incompressible flow, the Bernoulli equation applied between points 1 and 2
ଵ
1
2
ଶ ߛݖଵ ൌ ଶ
ߩܸଵ
1
2
ଶ ߛݖଶ
ߩܸଶ
ENSC283 Mid-term W2009 1
2. ݄ ൌ 2 ݉
݀ ൌ 0.1 ݉
ܳ
ܦ ൌ 1 ݉
1
2
With the assumptions that ଵ ൌ ଶ ൌ 0, ݖଵ ൌ ݄ and ݖଶ ൌ 0,
ଵ
ଶ ܸଵ
ଶ ݄݃ ൌ ଵ
ଶ (I)
ଶ ܸଶ
Although the water level remains constant, there is an average velocity, ܸଵ, across section
1 because of the flow from the tank. For steady incompressible flow, conservation of
mass requires ܳଵ ൌ ܳଶ, where ܳ ൌ ܣܸ. Thus, ܣଵܸଵ ൌ ܣଶܸଶ or
ܦଶܸଵ ൌ ݀ଶܸଶ
Hence,
ܸଵ ൌ ቀௗ
ଶ
ܸଶ (II)
ቁ
Combining equations (I) and (II)
ܸଶ ൌ ඩ
2݄݃
1 െ ቀ݀ܦ
ସ ൌ ඪ
ቁ
2 ቀ9.81 ቂ݉
ݏଶቃቁ ሺ2 ሾ݉ሿሻ
1 െ ൬0.1ሾ݉ሿ
1 ሾ݉ሿ ൰
݉
ݏ
ସ ൌ 6.26 ቂ
ቃ
Thus,
ENSC283 Mid-term W2009 2
3. ܳ ൌ ܣଵܸଵ ൌ
ߨሺ0.1 ሾ݉ሿሻଶ
4
݉
ݏ
ൈ 6.26 ቂ
ቃ ൌ 0.0492 ቈ
݉ଷ
ݏ
Problem 2: (20 marks)
A uniform block of steel (SG = 7.85) will float at a mercury-water interface as shown in
the figure. What is the ratio of the distance ܽ and ܾ.
Problem 2: (Solution)
water steel
block
mercury: SG = 13.36
a
b
Let ܮ be the block length into the paper, ܹ is the block width and let ߛ be the water
specific weight. Then the vertical force balance on the block is
7.85ߛሺܽ ܾሻܮܹ ൌ 1ߛܽܮܹ 13.56ߛܾܮܹ
Canceling ܹ, ܮ and ߛ from both sides of this equation and rearranging,
7.85ܽ 7.85ܾ ൌ ܽ 13.56ܾ ՜
ܽ
ܾ
ൌ
13.56 െ 7.85
7.85 െ 1
ൌ 0.834
ENSC283 Mid-term W2009 3
4. Problem 3: (20 marks)
Kerosene at 20Ԩ flows through the pump shown in figure at 0.065 ݉ଷ/ݏ. Head losses
between 1 and 2 are 2.4 ݉, and the pump delivers 6 ܹ݇ to the flow. What should the
mercury-manometer reading ݄ be? (ߛ௦ ൌ 8016.2 ܰ/݉ଷ and ߛ௨௬ ൌ
133210 ܰ/݉ଷ)
1.5݉
ܸଵ
ܦଵ ൌ 7.5 ܿ݉
mercury
݄ ൌ?
Problem 3: (Solution)
First establish two velocities, i.e. ܸଵ and ܸଶ,
ܸଵ ൌ
ܳ
ܣଵ
ൌ
pump
0.065 ݉ଷ
ݏ ൨
ܦଶ ൌ 15 ܿ݉
ቀߨ4ቁ ሺ7.5 ൈ 0.01ሾ݉ሿሻଶ
ܸଶ
ൌ 14.7 ቂ
݉
ݏ
ቃ
ܸଶ
ܸଵ
ൌ
ܣଵ
ܣଶ
ൌ ൬
ܦଵ
ܦଶ
൰
ଶ
՜ ܸଶ ൌ
1
4
݉
ݏ
ܸଵ ൌ 3.675 ቂ
ቃ
To apply a manometer analysis to determine the pressure difference between points 1 and
2,
ଶ െ ଵ ൌ ሺߛ െ ߛሻ݄ െ ߛΔݖ ൌ ቀ133210 ቂ ே
యቃ െ 8016.2 ቂ ே
యቃቁ ݄ െ ቀ8016.2 ቂ ே
యቃቁ ൈ
1.5ሾ݉ሿ ൌ 125193.8݄ െ 12024.3 ቂ ே
మቃ
Now apply the steady flow energy equation between points 1 and 2
ENSC283 Mid-term W2009 4
5. ଵ
ߛ
ܸଵ
ଶ
2݃
ݖଵ ൌ
ଶ
ߛ
ܸଶ
ଶ
2݃
ݖଶ ݄ െ ݄
where
݄ ൌ
ܲ
ߛܳ
ൌ
6000ሾܹሿ
݉ଷቃ ൈ 0.068 ݉ଷ
8016.2 ቂ ܰ
ݏ ൨
ൌ 11 ሾ݉ሿ
Thus,
ଵ
8016.2 ቂ ܰ
݉ଷቃ
ቀ14.7 ቂ݉ݏ
ቃቁ
ଶ
2ሺ9.81ሻ
0
ൌ
ଶ
8016.2 ቂ ܰ
݉ଷቃ
ቀ3.675 ቂ݉ݏ
ቃቁ
ଶ
2ሺ9.81ሻ
1.5 ሾ݉ሿ 2.4ሾ݉ሿ െ 11ሾ݉ሿ
՜ ଶ െ ଵ ൌ 139685 ቂ ே
మቃ ൌ 125193.8݄ െ 12024.3 ቂ ே
మቃ
or
݄ ൌ 1.211 ሾ݉ሿ
Problem 4: (20 marks)
Circular-arc gate ABC pivots about point O. For the position shown, determine (a) the
hydrostatic force on the gate (per meter of width into the paper); and (b) its line of action.
water
Problem 4: (Solution)
ܴ
O
ܴ
ߙ
ߙ
݈ଵ
ܮ
ܪ
݈ଶ ൌ
ENSC283 Mid-term W2009 5
2ܮ
3
ܣ ൌ
ܮܪ
2
݈ଵ ൌ
2ܴݏ݅݊ߙ
3ߙ
ܣ ൌ ߙܴଶ
ܥܩ
ܥܩ
ܪ
Hint:
A
B
C
6. The horizontal hydrostatic force is based on the vertical projection:
ܨு ൌ ߛ௪݄ீܣ௩ ൌ
ߛ௪ܪଶ
2
and the point of action for the horizontal force is 2H/3 below point C.
The vertical force is upward and equal to the weight of the missing water in the segment
ABC shown hatched in the following figure.
ܴ
ߙ
ߙ
ܥ
ܪ/2
ܪ/2
ܣ
ܤ
݈
The segment ABC area can be calculated from
ܴ
ܣ௦ ൌ ߙܴଶ െ
ܱ
ܴܪܿݏߙ
2
This area gives the volume of water in segment ABC per unit width into the paper. Hence
the water weight is,
ܨ ൌ ߛ௪ܣ௦ ൌ ߛ௪ ൬ߙܴଶ െ
ܴܪܿݏߙ
2
൰
To calculate the point of action, center of weight for segment ABC should be calculated.
To do so, momentum balance around point O should be applied,
െ
ܪܴܿݏߙ
2
ൈ
2ܴܿݏߙ
3
ߙܴଶ ൈ
2ܴݏ݅݊ߙ
3ߙ
ൌ ൬ߙܴଶ െ
ܴܪܿݏߙ
2
൰ ݈
՜ ݈ ൌ
2ܴ
3
ቆ
2ܴݏ݅݊ߙ െ ܪܿݏଶߙ
2ߙܴ െ ܪܿݏߙ
ቇ
The net force is thus
ଶ ܨு
ܨ ൌ ටܨ
ଶ
ENSC283 Mid-term W2009 6
7. Per meter of width. This force is acting upward to the right at an angle of β where
ߚ ൌ ݐܽ݊ିଵ ቆ
ܪଶ
2ߙܴଶ െ ܪܴܿݏߙ
ቇ
Problem 5: (20 mark)
A rigid tank of volume ܸ ൌ 1݉ଷ is initially filled with air at 20Ԩ and
ൌ 100 ݇ܽ. At time ݐ ൌ 0, a vacuum pump is turned on and evacuates air at a
constant volume flow rate of ܳ ൌ 80 ܮ/݉݅݊ (regardless of pressure). Assume an ideal
gas and isothermal process.
(a) Set up a differential equation for this flow.
(b) Solve this equation for t as a function of ሺܸ, ܳ, , ሻ.
(c) Compute the time in minutes to pump the tank down to ൌ 20 ݇ܽ.
Problem 5: (Solution)
ݐ ൌ 0
ൌ 100 ݇ܽ
ܶ ൌ 20Ԩ
ܳ ൌ 80 ܮ/݉݅݊
The control volume encloses the tank, as shown is selected.
ݐ ൌ 0
CV
ൌ 100 ݇ܽ
ܶ ൌ 20Ԩ
Continuity equation for the control volume becomes
݀
݀ݐ
ܳ ൌ 80 ܮ/݉݅݊
൬න ߩ݀ݒ൰ ݉ሶ௨௧ െ ݉ሶ ൌ 0
Since no mass enters the control volume Σ mሶ୧୬ ൌ 0. Assuming that the gas density is
uniform throughout the tank ρdv ൌ ρv. Hence
ENSC283 Mid-term W2009 7
8. ݒ
݀ߩ
݀ݐ
ߩܳ ൌ 0
or
ௗఘ
ఘ ൌ െ ொ
௩ ݀ݐ (part a)
This differential equation can be easily solved yielding
݈݊ ൬
ߩ
ߩ
൰ ൌ െ
ܳݐ
ݒ
where ρ is the gas initial density. For an isothermal ideal gas, ρ/ρ ൌ p/p, therefore
ݐ ൌ െ ௩
ொ ݈݊ ቀ
బ
ቁ (part b)
For given values of Q ൌ 80L/min ൌ 0.08 mଷ/min, the time to pump a 1 mଷ tank down
from 100 to 20 kpa is
ݐ ൌ െ
1ሾ݉ଷሿ
0.08 ሾ ݉ଷ
݉݅݊ሿ
݈݊ ൬
20
100
൰ ൌ 20.1 ሾ݉݅݊ሿ
ENSC283 Mid-term W2009 8