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Calculate the volume, capacity, mass of fuel, and total weight for a rectangular tank with given dimensions and fuel density.

Mathematics

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5 parts
a

80 m3\text{80 m}^3

b

80000 L\text{80000 L}

c

704 kg\text{704 kg}

d

7406.24 N\text{7406.24 N}

e

2.30 MPa\text{2.30 MPa}

Step-by-step solution

12 steps

Here are the calculations for the tank and its supports:

Part a· 2 steps

The volume of the tank.

  1. Identify the dimensions of the tank.

    Length (LL) = 8 m8 \text{ m}
    Width (WW) = 5 m5 \text{ m}
    Height (HH) = 2 m2 \text{ m}

  2. Calculate the volume using the formula for a rectangular prism.

    V=L×W×HV = L \times W \times H V=8 m×5 m×2 mV = 8 \text{ m} \times 5 \text{ m} \times 2 \text{ m} V=80 m3V = 80 \text{ m}^3

    The volume of the tank is 80 m3\boxed{\text{80 m}^3}.

Part b

80000 L\text{80000 L}

The capacity of the tank.

  1. Convert the volume from cubic meters to liters.

    1 m3=1000 L1 \text{ m}^3 = 1000 \text{ L} Capacity=80 m3×1000 L1 m3Capacity = 80 \text{ m}^3 \times \frac{1000 \text{ L}}{1 \text{ m}^3} Capacity=80000 LCapacity = 80000 \text{ L}

    The capacity of the tank is 80000 L\boxed{\text{80000 L}}.

Part c

704 kg\text{704 kg}

Mass of the fuel in the tank when full.

  1. Identify the density of petrol and the volume of the tank.

    Density of petrol (ρ\rho) = 8.8 kg/m38.8 \text{ kg/m}^3
    Volume of tank (VV) = 80 m380 \text{ m}^3 (from part a)

  2. Calculate the mass of the fuel using the density formula.

    mfuel=ρ×Vm_{fuel} = \rho \times V mfuel=8.8 kg/m3×80 m3m_{fuel} = 8.8 \text{ kg/m}^3 \times 80 \text{ m}^3 mfuel=704 kgm_{fuel} = 704 \text{ kg}

    The mass of the fuel in the tank when full is 704 kg\boxed{\text{704 kg}}.

Part d

7406.24 N\text{7406.24 N}

The total weight needed to be supported by a constructed steel support.

  1. Identify the weight of the tank and the mass of the fuel.

    Weight of the tank (WtankW_{tank}) = 500 N500 \text{ N}
    Mass of fuel (mfuelm_{fuel}) = 704 kg704 \text{ kg} (from part c)
    Assume acceleration due to gravity (gg) = 9.81 m/s29.81 \text{ m/s}^2.

  2. Calculate the weight of the fuel.

    Wfuel=mfuel×gW_{fuel} = m_{fuel} \times g Wfuel=704 kg×9.81 m/s2W_{fuel} = 704 \text{ kg} \times 9.81 \text{ m/s}^2 Wfuel=6906.24 NW_{fuel} = 6906.24 \text{ N}
  3. Calculate the total weight to be supported.

    Wtotal=Wtank+WfuelW_{total} = W_{tank} + W_{fuel} Wtotal=500 N+6906.24 NW_{total} = 500 \text{ N} + 6906.24 \text{ N} Wtotal=7406.24 NW_{total} = 7406.24 \text{ N}

    The total weight needed to be supported is 7406.24 N\boxed{\text{7406.24 N}}.

Part e

2.30 MPa\text{2.30 MPa}

The tensile stress of the steel support when carrying the tank full of fuel.

  1. Identify the total weight, number of supports, and dimensions of each support.

    Total weight (WtotalW_{total}) = 7406.24 N7406.24 \text{ N} (from part d)
    Number of supports = 4
    Diameter of each support (dd) = 32 mm=0.032 m32 \text{ mm} = 0.032 \text{ m}

  2. Calculate the force on each support.

    Fone=WtotalNumber of supportsF_{one} = \frac{W_{total}}{\text{Number of supports}} Fone=7406.24 N4F_{one} = \frac{7406.24 \text{ N}}{4} Fone=1851.56 NF_{one} = 1851.56 \text{ N}
  3. Calculate the cross-sectional area of one support.

    Radius (rr) = d2=0.032 m2=0.016 m\frac{d}{2} = \frac{0.032 \text{ m}}{2} = 0.016 \text{ m}

    Aone=πr2A_{one} = \pi r^2 Aone=π×(0.016 m)2A_{one} = \pi \times (0.016 \text{ m})^2 Aone=π×0.000256 m2A_{one} = \pi \times 0.000256 \text{ m}^2 Aone≈0.000804247 m2A_{one} \approx 0.000804247 \text{ m}^2
  4. Calculate the tensile stress on each support.

    σ=FoneAone\sigma = \frac{F_{one}}{A_{one}} σ=1851.56 N0.000804247 m2\sigma = \frac{1851.56 \text{ N}}{0.000804247 \text{ m}^2} σ≈2302200 Pa\sigma \approx 2302200 \text{ Pa} σ≈2.30×106 Pa\sigma \approx 2.30 \times 10^6 \text{ Pa} σ≈2.30 MPa\sigma \approx 2.30 \text{ MPa}

    The tensile stress of the steel support is approximately 2.30 MPa\boxed{\text{2.30 MPa}}.

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