Hydrostatics Question. The lock system between the Charles River and Boston harb

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Hydrostatics Question. The lock system between the Charles River and Boston harbor is designed to control the flow of water from the river due to an extreme change in the level of the water in the Boston Harbor between low and high tide. The tides in Boston Harbor change 15 feet each day from low to high tide. The level of the river is a constant 18 feet deep all day long and at low title the harbor is 14 feet deep. The lock consists of two swinging gates that are 22 feet wide and extend to the river/harbor seabed. No water can leak through the locks when they are closed. The two walls are 50 feet apart as shown in the drawing. When a ship wants to go from the harbor to the river one side of the locks is opened to allow the water in the middle chamber to equalize to the same height as the harbor. When the water in the chamber is at the same level then the ship enters the locks. The gate is closed behind him. A pump then begins to fill or drain the chamber in order to equalize the depth in the chamber to the depth in the river. Once the water level in the chamber is equal to the height of the water in the river, the gate on the river side is opened and the vessel can exit safely. When the vessel has exited the gate closes and the water in the chamber remains at the height until the next vessel goes through. Figure 3: Lock System between the Charles River and Boston Harbor Determine the maximum possible force on the gate adjacent to the harbor when it is closed. Does this occur at high or low tide and what level is the water in the middle chamber? (Hint: Ignore the dynamic forces due to filling the chamber and/or currents in the harbor) Where is this force acting (center of pressure)? You are asked to redesign the lock gates and are given three options sketched on the next page. Use your intuition and knowledge of hydrostatics to determine which would be the most feasible design. It is possible that more than one design will work.

Explanation / Answer

The water level in the chamber can be either 14 ft (Low-tide harbour inlet), 18 ft (river inlet) or 29 ft (High-tide harbour inlet), assuming the harbour and river to be infinite bodies of water. Thus, the maximum force on the harbour-side gate will occur at a level of 18 ft in the chamber and 29 ft on the harbour side.

P1: Pressure from the Harbour water, P2: Pressure from Chamber water, g: Gravity accleration

P1 = Density of Water * g * 8.8392 (in metres) / 2 = 43356.276 N/m2 (Division by two to consider the variation of Hydrostatic pressure with height)

F1 = P1 * Area = P1 * width * 8.8392 = 2569.81 kN

Similarly,

F2 = Desnity of Water * g * 5.4864 (metres) * Width * 5.4864 / 2 = 990.037 kN

Total Force on the Gate = F1 - F2 = 1579.773 kN ---- Ans (a)

 

Now for point of pressure,

F1*8.8392 + F2*5.4864 = TotalForce * PressurePointHeight

Thus,

PressurePointHeight = (F1*8.8392 + F2*5.4864) / TotalForce = 4.103 metres above base level. --- Ans (b)

 

Please upload images of choices for option (c) so that I can help you with that as well.

Also, please convert answers to British units.

Cheers!

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