CIVL 3120_2021F_Tutorial 05 QUIZ_Questions and Solutions

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Dec 6, 2023

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CIVL 3120 Hydraulics Tutorial 05: External flow QUIZ _SOLUTIONS Question: PSA (Submarine power) Determine the power required for a submarine to cruise horizontally and steadily at (35 + Last one digit of your Student ID number) km/h in seawater whose density is 1020 kg/m 3 . The submarine is ellipsoid in shape with a diameter of 4.5 m and a length of 26 m. Solution: A submarine is treated as an ellipsoid at a specified length and diameter. The powers required for this submarine to cruise horizontally in seawater is to be determined. Assumptions 1 The submarine can be treated as an ellipsoid. 2 The flow is turbulent. 3 The drag of the towing rope is negligible. 4 The motion of submarine is steady and horizontal. Assume Last one digit of your Student ID number = 0 Submarine velocity = (35 + Last one digit of your Student ID number) km/h = 35 +0 = 35 km/h Properties The drag coefficient for an ellipsoid with L/D = 26/4.5 = 5.78 is C D = 0.1 in turbulent flow (Table 11-2). The density of sea water is given to be 1020 kg/m 3 . Analysis Noting that 1 m/s = 3.6 km/h, the velocity of the submarine is equivalent to V = 35/3.6 = 9.72 m/s. The frontal area of an ellipsoid is A = π D 2 /4. Then the drag force acting on the submarine becomes 2 3 2 2 2 (1020 kg/m )(9.72 m/s) 1 kN (0.1)[ (4.5 m) / 4] 76.63 kN 2 2 1000 kg m/s D D V F C A ρ π = = = Noting that power is force times velocity, the power needed to overcome this drag force is drag 1 kW (76.63 kN)(9.72 m/s) 1 kN m/s D W F V = = = 745 kW ___________ 35km/h km/h Submarine
CIVL 3120 Hydraulics Question: PSB (Sun Ball Drag) Consider a person is driving a car with a sun ball on the car antenna. The frontal area of the ball is 2.0 × 10 −3 m 2 . A wind tunnel test with a similar sun ball resulted the drag coefficient to be 0.81 at nearly all air speeds. Estimate how much additional energy the person requires per year by having this ball on the car antenna. Use the following additional information: the person drives about 17,000 km per year at an average speed of 20.5 m/s. The overall car efficiency is 0.31 and the density air is 𝜌𝜌 air = 1.204 kg/m 3 . Solution: We are to estimate how much additional energy would be required per year by driving with sun ball on the car antenna. Properties ρ air = 1.204 kg/m 3 Analysis The additional drag force due to the sun ball is 2 1 air 2 = ρ D D F V C A = 1 2 1.204 𝑘𝑘𝑘𝑘 𝑚𝑚 3 ∗ � 20.5 𝑚𝑚 𝑠𝑠 2 0.81 2.0 10 −3 𝑚𝑚 2 = 0.41 𝑁𝑁 where A is the frontal area of the sphere. The work required to overcome this additional drag is force times distance. So, letting L be the total distance driven in a year, 2 1 drag air 2 Work = = ρ D D F L V C AL = 0.41 𝑁𝑁 ∗ 17000000 𝑚𝑚 = 6967000 𝐽𝐽 The energy required to perform this work is much greater than this due to overall efficiency of the car engine, transmission, etc. Thus, 2 1 drag air 2 required overall overall Work = = ρ η η D V C AL E = 6967000 𝐽𝐽 0 . 31 = 22475000 𝐽𝐽 = 22475 𝑘𝑘𝐽𝐽 _________
CIVL 3120 Hydraulics Question PSB Extra Part (Not part of Quiz): Estimate how many additional liters of fuel the person requires per year by having this ball on the car antenna. Use the following additional information: the fuel density, 𝜌𝜌 fuel =0.802 kg/L, and the energy value of the fuel is 44,020 kJ/kg. Solutions: The required energy is also equal to the energy (heating) value of the fuel EV times the mass of fuel required. In terms of required fuel volume , volume = mass/density. Thus, 2 1 fuel required required air 2 fuel required fuel fuel fuel overall / HV = = HV = ρ ρ ρ ρ η D m E V C AL V = (22475 𝑘𝑘𝐽𝐽 /44020 𝑘𝑘𝐽𝐽 𝑘𝑘𝑘𝑘 )/ (0.802 𝑘𝑘𝑘𝑘 𝐿𝐿 ) = 0.637 𝐿𝐿 ____________
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