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MET330 20xx0xE Applied Fluid Mechanics Instructor: Instructors name Week 2 Review Assignment - 1 I pledge to support the Honor System of ECPI. I will refrain from any form of...

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MET330 20xx0xE Applied Fluid Mechanics


Instructor: Instructors name


Week 2 Review Assignment - 1



I pledge to support the Honor System of ECPI. I will refrain from any form of academic dishonesty or deception, such as cheating or plagiarism. I am aware that as a member of the academic community, it is my responsibility to turn in all suspected violators of the honor code.
I understand that any failure on my part to support the Honor System will be turned over to a Judicial Review Board for determination. I will report to the Judicial Review Board hearing if summoned.




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Date:

Review Assignment Problems
NOTE: Handwritten work is not acceptable. All work must be typed and submitted on a Word Document.
a. Identify and list all variables from the problem.
. Write the equation(s) you will use that are found in the textbook. Do not use any other equations.
c. Substitute your variables into the equation. YOU MUST INCLUDE UNITS IN ALL PLACES OF YOUR EQUATIONS.
d. You must show all of your work, even “simple” calculations. During your calculations, please use 3 decimal places or 3 significant decimal places. This does not apply to financial cost calculations.
e. Clearly identify your answer with co
ect units.

· CORRECT METHOD:
· INCORRECT METHOD: p2 = (W * a)/g + p1 = a^2 - Ap – μ = 0
6.30. Oil for a hydraulic system (sg = 0.90) is flowing at 2.35 × 10−3 m3/s. Calculate the weight flow rate and mass flow rate.
Known values:
Governing equations:
Calculations:
6.39. When 2000 L/min of water flows through a circular section with an inside diameter of 300mm that later reduces to a 150mm diameter, calculate the average velocity of flow in each section.
Known values:
Governing equations:
Calculations:
6.53. From the list of standard hydraulic steel tubing in Appendix G.2, select the smallest size that would ca
y 2.80 L/min of oil with a maximum velocity of 0.30 m/s.
Known values:
Governing equations:
Calculations:
6.61. Water at 10C is flowing from point A to point B through the fa
icated section shown at the rate of 0.37 m3/s. If the pressure at A is 66.2 kPa, calculate the pressure at B.
Figure for 6.61
Known values:
Governing equations:
Calculations:
6.88. What depth of fluid above the outlet nozzle is required to deliver 200 gal/min of water from the tank shown? The nozzle has a 3-in diameter.
Figure for 6.88
Known values:
Governing equations:
Calculations:
7.4. A long DN 150 Schedule 40 steel pipe discharges 0.085 m3/s of water from a reservoir into the atmosphere as shown. Calculate the energy loss in the pipe.
Figure for 7.4
Known values:
Governing equations:
Calculations:
7.11. A submersible deep-well pump delivers 745 gal/h of water through a 1-in Schedule 40 pipe when operating in the system shown. An energy loss of 10.5 lb-ft/lb occurs in the piping system.
(a) Calculate the power delivered by the pump to the water.
(b) If the pump draws 1 hp, calculate its efficiency.
Known values:Figure for 7.11
Governing equations:
Calculations:
7.23. Calculate the power delivered to the hydraulic motor shown if the pressure at A is 6.8 MPa and the pressure at B is 3.4 MPa. The motor inlet is a steel hydraulic tube with 25mm OD × 1.50mm wall and the outlet is a tube with 50mm OD × 2.00mm wall. The fluid is oil (sg = 0.90) and the velocity of flow is 1.50 m/s at point B.Figure for 7.23
Known values:
Governing equations:
Calculations:
7.25. Calculate the power delivered by the oil to the fluid motor shown if the volume flow rate is 0.25 m3/s. There is an energy loss of 1.40 N∙m/N in the piping system. If the motor has an efficiency of 75 percent, calculate the power output.
Figure for 7.25
.25
Known values:
Governing equations:
Calculations:
    Page 1
Answered 1 days After Jan 30, 2023

Solution

Dr Shweta answered on Jan 31 2023
38 Votes
Ans 1.
Known values: sg = 0.90, flow rate = 2.35 × 10−3 m3/s.
Governing equations: ρoil = Sg * ρw
Calculations: ρoil = 0.9 * 1000 Kg/m3
Mass flow rate of oil is
m* = ρoil * V = 900 * 2.35 * 10-3 = 2.115 Kg/s
Weight flow rate of oil = W = m*g = 20115 *9.8 = 20.75N/s.
Ans 2.
Known values: flow rate 2000 L/min, inside diameter of 300mm, later reduces to 150mm diameter.
Governing equations: A1. V1=A2.V2=2m3/min
Calculations:
3.14*0.3*0.3/4*V1=2/60
V1=0.47m/s
similarly
V2=1.88m/s
Ans 3.
Known values: 2.80 L/min of oil = 4.66 * 10-5m3/s, maximum velocity of 0.30 m/s.
Governing equations: Q=VA
Calculations: flow area=
A = 4.666 * 10-5m3/s/0.3 = 1.55 * 10-4m2
A = π/4Di2
π/4Di2 = 1.55 * 10-4m2
Di2 = 4 * 1.55 * 10-4m2/ 3.14
Di = 0.014 m or 14 mm
From the standard table:
D0 = 20 mm
t = 1.2 mm
Di =17.6 mm
Flow area (A) = 2.433 * 10-4m2
Ans 4. Known values: T = 10 ͦ C, rate = 0.37 m3/s, pressure at A = 66.2 kPa, density of water at 10 ͦ C =999.77 kg/m3, dB = 0.6 m, dA = 0.3 m, hB = 4.5 m
Governing equations:...
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