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A cylindrical tank with a base area of 2000 cm2 is being filled with water from a pipe with a cross-section area of 6.16 cm2. If the water flows through the pipe at 10 cm/s, calculate the depth of wat

Mathematics

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Answer

a

18.48 cm\text{18.48 cm}

Step-by-step solution

6 steps

Here's the solution to the problem:

Assumption: The speed of water flowing through the pipe is 10 cm/s10 \text{ cm/s}, as indicated in your scratch work. The question text is incomplete regarding this value.

Part a· 5 steps

Calculate the depth of water in the tank at 8:30 a.m.

  1. Calculate the volume of water flowing into the tank per second.

    Given the cross-section area of the pipe Apipe=6.16 cm2A_{pipe} = 6.16 \text{ cm}^2 and the speed of water v=10 cm/sv = 10 \text{ cm/s}.
    The inflow rate QinQ_{in} is:

    Qin=Apipe×vQ_{in} = A_{pipe} \times v Qin=6.16 cm2×10 cm/sQ_{in} = 6.16 \text{ cm}^2 \times 10 \text{ cm/s} Qin=61.6 cm3/sQ_{in} = 61.6 \text{ cm}^3/\text{s}

  2. Calculate the total time water flows.

    Water starts flowing at 6:00 a.m. and we need to calculate the depth at 8:30 a.m.
    Time duration t=8:30 a.m.−6:00 a.m.=2 hours 30 minutest = 8:30 \text{ a.m.} - 6:00 \text{ a.m.} = 2 \text{ hours } 30 \text{ minutes}.
    Convert this time to seconds:

    t=(2×60 minutes+30 minutes)×60 seconds/minutet = (2 \times 60 \text{ minutes} + 30 \text{ minutes}) \times 60 \text{ seconds/minute} t=(120 minutes+30 minutes)×60 seconds/minutet = (120 \text{ minutes} + 30 \text{ minutes}) \times 60 \text{ seconds/minute} t=150 minutes×60 seconds/minutet = 150 \text{ minutes} \times 60 \text{ seconds/minute} t=9000 st = 9000 \text{ s}

  3. Calculate the total volume of water in the tank.

    Vwater=Qin×tV_{water} = Q_{in} \times t Vwater=61.6 cm3/s×9000 sV_{water} = 61.6 \text{ cm}^3/\text{s} \times 9000 \text{ s} Vwater=554400 cm3V_{water} = 554400 \text{ cm}^3

  4. Convert the base area of the tank to cm2\text{cm}^2.

    Given the base area of the tank Atank=3 m2A_{tank} = 3 \text{ m}^2.

    Atank=3 m2×(100 cm/m)2A_{tank} = 3 \text{ m}^2 \times (100 \text{ cm/m})^2 Atank=3×10000 cm2A_{tank} = 3 \times 10000 \text{ cm}^2 Atank=30000 cm2A_{tank} = 30000 \text{ cm}^2

  5. Calculate the depth of water in the tank.

    The volume of water in the tank is Vwater=Atank×hV_{water} = A_{tank} \times h, where hh is the depth.

    h=VwaterAtankh = \frac{V_{water}}{A_{tank}} h=554400 cm330000 cm2h = \frac{554400 \text{ cm}^3}{30000 \text{ cm}^2} h=18.48 cmh = 18.48 \text{ cm}

    The depth of water in the tank at 8:30 a.m. is 18.48 cm\boxed{\text{18.48 cm}}.

Part b· 1 step

If the tank is 1.2M high and has a hole at the bottom through which water leaks at the rate of 11.6cm³ per second. Determine the time the tank will be filled.

  1. Calculate the net inflow rate.

    Inflow rate $Q_{in} = 61.6 cm^

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