Chapter 1, Problem 74AP

Assume it takes 7.00 minutes to fill a 30.0-gal gasoline tank. (a) Calculate the rate at which the tank is filled in gallons per second. (b) Calculate the rate at which the tank is filled in cubic meters per second. (c) Determine the time interval, in hours, required to fill a 1.00-m³ volume at the same rate. (1 U.S. gal = 231 in.³)

To determine

The rate at which the tank is filled in gallons per second.

Answer to Problem 74AP

The rate at which the tank is filled in gallons per second is $7.14\times {10}^{-2}\text{gal/s}$.

Explanation of Solution

Given Info: The fill the tank $30\text{gal}$ gasoline, it requires $7\min$.

Explanation:

Calculating the rate at which the tank will fill is,

$\begin{array}{c}rate=\frac{\left(30\text{gal}\right)}{\left(7\min \right)}\frac{\left(1\min \right)}{\left(60\sec \right)}\\ =7.14\times {10}^{-2}\text{gal/s}\end{array}$

Thus, the rate at which the tank will fill is $7.14\times {10}^{-2}\text{gal/s}$.

Conclusion:

Therefore, the rate at which the tank will fill is $7.14\times {10}^{-2}\text{gal/s}$.

To determine

The rate at which the tank is filled in cubic meter per second.

Answer to Problem 74AP

The rate at which the tank is filled in cubic meter per second is $2.70\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}$.

Explanation of Solution

Given Info: The fill the tank $30\text{gal}$ gasoline, it requires $7\min$.

Explanation:

From unit conversion,

$1{\text{m}}^{\text{3}}={10}^3\text{L}$

$1\min =60\text{s}$

$1\text{gal}=3.786\text{L}$

Calculating the rate at which the tank will fill is,

$\begin{array}{c}rate=\frac{\left(30\text{gal}\right)}{\left(7\min \right)}\frac{\left(1\min \right)}{\left(60\sec \right)}\left(\frac{3.786\text{L}}{1\text{gal}}\right)\left(\frac{1{\text{m}}^{\text{3}}}{{10}^3\text{L}}\right)\\ =2.7\times {10}^{-4}{\text{m}}^3\text{/s}\end{array}$

Thus, the rate at which the tank will fill is $2.70\times {10}^{-4}{\text{m}}^3\text{/s}$.

Conclusion:

Therefore, the rate at which the tank will fill is $2.70\times {10}^{-4}{\text{m}}^3\text{/s}$.

To determine

The time interval required to fill the tank of $1{\text{m}}^{\text{3}}$ volume at the same rate

Answer to Problem 74AP

The time interval required to fill the tank at rate $1{\text{m}}^{\text{3}}$

Explanation of Solution

Given Info: The rate at which the tank will fill is $2.70\times {10}^{-4}{\text{m}}^3\text{/s}$. The volume of the tank is $1{\text{m}}^{\text{3}}$.

Explanation:

From unit conversion,

$1\text{h}=3600\text{s}$

Formula to calculate the time interval is,

$t=\frac{V}{R}$

Here,

$V$ is the volume

$R$ is rate of filling

$t$ is time

Substitute $2.7\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}$ for $R$, $1{\text{m}}^{\text{3}}$ for $V$, to calculate $t$,

$\begin{array}{c}t=\left[\frac{\left(1{\text{m}}^{\text{3}}\right)}{\left(2.7\times {10}^{-4}{\text{m}}^3\text{/s}\right)}\right]\left(\frac{1\text{h}}{3600\text{s}}\right)\\ =1.03\text{h}\end{array}$

Thus, the required time interval to fill the tank is $1.03\text{h}$.

Conclusion:

Therefore, the required time interval to fill the tank is $1.03\text{h}$.