Question

Suppose the container is fabricated from steel having a thickness of .500 inch and an average density of 7.85 g/cm3. (You may assume that the container is cylindrical with hemispherical ends.) What is the mass of the filled container assuming that liquid propane has a density of .585 g/cm3. The total length is 10 ft, the height is 4 ft, and each hemispherical end is 2 ft.

Answer 1

Answer on Question #45648, Chemistry, Other

Suppose the container is fabricated from steel having a thickness of .500 inch and an average density of $7.85\mathrm{g/cm}^3$. (You may assume that the container is cylindrical with hemispherical ends.) What is the mass of the filled container assuming that liquid propane has a density of $.585\mathrm{g/cm}^3$. The total length is 10 ft, the height is 4 ft, and each hemispherical end is 2 ft.

Solution:

Given:

$$\begin{array}{l} d = 0.500 \text{ inch} = 0.5 \cdot 0.0254 = 0.0127 \text{ m} \\ \rho_{\text{steel}} = 7.85 \mathrm{g/cm}^3 = 7850 \mathrm{kg/m}^3 \\ \rho_{\text{prop}} = 0.585 \mathrm{g/cm}^3 = 585 \mathrm{kg/m}^3 \\ r = 2 \mathrm{ft} = 2 \cdot 0.3048 = 0.6096 \mathrm{m} \\ m = ? \\ \end{array}$$

The volume of container is

$$V = \text{hemisphere volumes} + \text{cylinder portion}$$

The length of the cylinder portion = $10 \mathrm{ft} - 2 \cdot 2 \mathrm{ft} = 6 \mathrm{ft}$.

Thus,

$$\begin{array}{l} V = \frac{4}{3} \pi r^3 + \pi r^2 \cdot 6 = \frac{4}{3} \cdot \pi \cdot 2^3 + \pi \cdot 2^2 \cdot 6 = \frac{104}{3} \pi = 108.91 \mathrm{ft}^3 \\ 1 \text{ cubic foot} = 0.028317 \text{ cubic meters} \\ \end{array}$$

Thus,

$$V = 108.91 \mathrm{ft}^3 = 108.91 \cdot 0.028317 = 3.084 \mathrm{m}^3$$

The mass of liquid propane is

$$m_{\text{prop}} = V \rho_{\text{prop}} = 3.084 \cdot 585 = 1804.14 \mathrm{kg}$$

The mass of empty container is

$$m_{\text{steel}} = S d \rho_{\text{steel}}$$

where $S$ is surface area of the container

$$\begin{array}{l} S = \text{surface area of a cylinder} + \text{surface area of sphere} \\ S = 2 \pi r h + 4 \pi r^2 = 2 \pi r (h + 2 r) = 2 \pi \cdot 2 \cdot (6 + 2 \cdot 2) = 40 \pi = 125.664 \mathrm{ft}^2 \\ 1 \text{ square foot} = 0.0929 \text{ square meters} \\ \end{array}$$

So,

$$125.664 \cdot 0.0929 \quad 11.674 \, \text{m}^2$$

Thus,

$$m_{\text{steel}} \quad 11.674 \cdot 0.0127 \cdot 7850 \quad 1163.84 \, \text{kg}$$

The total mass of the filled container is

$$m \quad m_{\text{steel}} + m_{\text{prop}} \quad 1163.84 + 1804.14 \quad 2967.98 \, \text{kg} \approx 2968 \, \text{kg}$$

To pounds conversion

$$\text{lb} \quad \text{kg} \cdot 2.2046$$ $$m \quad 2967.98 \, \text{kg} \quad 2967.98 \cdot 2.2046 \, \text{lb} \quad 6543.21 \, \text{lb}$$

Answer: $m$ 2967.98 kg 6543.21 lb.

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