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4 years ago

11

What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weigh

t of the gasoline in it when it is full?

1 answer:

Natasha_Volkova [10]4 years ago

8 0

Answer:

1088.89 Pa

Explanation:

According to the Newton's second law of motion:-

$Force=Mass\times Acceleration$

Mass = 50.0 kg

Acceleration = g = 9.81 m/s²

So,

$Force=50.0\times 9.8\ kgm/s^2$

Force = 490 N

Area of the base = $length\times breath$ = $0.500\times 0.900$ m² = 0.45 m²

<u>Pressure = Force/Area = $\frac{490\ N}{0.45\ m^2}$ = 1088.89 Pa</u>

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2 years ago

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The given data is as follows.

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$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

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Formula for equivalent diameter is as follows.

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= $4 \times \frac{1}{3} ft$

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(b) Formula for velocity floe is as follows.

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V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

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(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

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4 years ago