Chapter 3, Problem 3.1P

Perform the following estimations without using a calculator.

1. Estimate the mass of water (kg) in an Olympic-size swimming pool.
2. A drinking glass is being filled from a pitcher. Estimate die mass flow rate of the water (g/s).
3. Twelve male heavyweight boxers coincidentally get on the same elevator in Great Britain. Posted on the elevator wall is a sign that gives the maximum safe combined weight of the passengers, Wmax, in stones. (A stone is a unit of mass equal to 14 lb_m. It is commonly used in England as a measure of body weight, which, like the numerical equivalence between the lb_m and lb_f, is only valid at or near sea level.) If you were one of the boxers, estimate the lowest value of W_niaxfor which you would feel comfortable remaining on the elevator.
4. The Trans-Alaska Pipeline has an outside diameter of 4 ft and extends 800 miles from the North Slope of Alaska to the northernmost ice-free port in Valdez, Alaska. How many barrels of oil arc required to fill the pipeline?
5. Estimate the volume of your body (cm³) in two different ways. (Show your work.)
6. A solid block is dropped into water and very slowly sinks to the bottom. Estimate its specific gravity.

Interpretation Introduction

(a)

Interpretation:

The mass of water (kg) in an Olympic-size pool should be estimated.

Concept introduction:

The density is the ratio of mass and volume given as:

$\rho =\frac{m}{V}$

Here, m is mass and V is volume

Answer to Problem 3.1P

$2.5\times {10}^6\text{kg}$.

Explanation of Solution

Let,

The size of a pool is $\text{50}\text{m}\text{×}\text{25}\text{m}\text{×}\text{2}\text{m}$

The density of water is $1000kg/m^3$

Since, the formula for density is:

$\rho =\frac{m}{V}$

Thus,

$m=\rho \times V$

The mass of water is:

$\begin{array}{l}\text{=}\text{50}\text{m}\text{×}\text{25}\text{m}\text{×}\text{2}\text{m}\text{×}\text{1000}\text{kg}\text{/}{\text{m}}^{\text{3}}\\ \text{=}\text{2}\text{.5}\text{×}{\text{10}}^{\text{6}}\text{kg}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The mass flow rate of the water (g/s) should be estimated.

Concept introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

The relationship between US and SI units are,

$\begin{array}{l}1{\text{cm}}^{\text{3}}=0.0\text{0105669}\text{qt}\\ 1\text{oz}=0.0\text{3125}\text{qt}\end{array}$

Answer to Problem 3.1P

$100\text{g/s}$.

Explanation of Solution

We can assume that the pitcher is a ‘large jug’.

In US standard system the volume of a drinking water glass is 8 oz and we can assume that the glass can be filled within 2 seconds using the pitcher.

The answer can be calculated as:

$\begin{array}{l}=\left(\frac{\text{8}\text{oz}}{\text{2}\text{s}}\right)=\text{4}\text{oz}\text{/s}\\ \text{Convert}\text{units}\text{to}\text{SI}\text{system}\\ =\text{(4}\text{oz}\text{/s)}\left(\frac{\text{1}\text{qt}}{\text{32}\text{oz}}\right)\left(\frac{{\text{10}}^{\text{6}}{\text{cm}}^{\text{3}}}{\text{1056}\text{.68}\text{qt}}\right)\left(\frac{\text{1}\text{g}}{{\text{cm}}^{\text{3}}}\right)\approx 10\text{0}\text{g/s}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The lowest value of maximum weight (W_max) for which you would feel comfortable remaining on the elevator should be estimated.

Concept introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor. A stone is a unit of mass equal to 14 lb_m.

Answer to Problem 3.1P

220 stones.

Explanation of Solution

A boxer who weight 220 lb_m or above is considered as the boxer in heavy weight category.

The weight of a boxer can be considered as 220 lb_m.

So calculate the minimum weight,

$W_{\max }\ge \left(\frac{12\times 2\text{20}{\text{lb}}_{\text{m}}}{14{\text{lb}}_{\text{m}}}\right)1\text{stone}\approx 220\text{stones}$

Where $W_{\max }$ is maximum weight.

Interpretation Introduction

(d)

Interpretation:

The number of barrels of oil required to fill the pipeline should be calculated.

Concept introduction:

The pipeline has an outside diameter of 4 ft and extends 800 miles.

The volume of a pipe (V) is given by,

$V=\frac{\pi D^{2}L}{4}$

Where, D = diameter of the pipe and L = length of the pipe

Answer to Problem 3.1P

$10^7\text{barrels}$.

Explanation of Solution

Let,

There are 42 gals in 1 barrel.

1 mile = 5880 ft

So the number of barrels in the pipe line can be calculated as,

$\begin{array}{l}{N}_{\text{Barrels}}=\frac{\pi D^{2}L}{4}\\ =\left(\frac{3.14}{4}\right){\left(\text{4}\text{ft}\right)}^{\text{2}}\left(\text{800}\text{miles}\right)\left(\frac{\text{5880}\text{ft}}{\text{1}\text{mile}}\right)\left(\frac{\text{7}\text{.48}\text{gal}}{\text{1}{\text{ft}}^{\text{3}}}\right)\left(\frac{\text{1}\text{barrel}}{\text{42}\text{gal}}\right)\\ \approx \frac{\left(3\right)\left(4\right)\left(5\right)\left(800\right)\left(5000\right)\left(7\right)}{\left(4\right)\left(4\right)\left(10\right)}\\ \approx 1\times 10^7\text{barrels}\end{array}$

Interpretation Introduction

(e)

Interpretation:

The volume of your body should be estimated in two different ways.

Concept introduction:

The density is the ratio of mass and volume given as:

$\rho =\frac{m}{V}$

Here, m is mass and V is volume

Answer to Problem 3.1P

The answer is $1\times 10^{5}c{m}^3$.

Explanation of Solution

Given Information:

The volume of the person is $\text{6ft}\text{×}\text{1}\text{ft}\text{×}\text{0}\text{.5}\text{ft}$ and the weight of the person is $\text{150}{\text{lb}}_{\text{m}}$.

Calculation:

Method 1:

Assume that the volume of the person can be considered as,

Height = 6 ft, width = 1 ft, thickness (distance between heal and toe) = 0.5 ft

Unit conversion,

1 ft³= 28 317 cm³

The volume of the person,

$\text{6ft}\text{×}\text{1}\text{ft}\text{×}\text{0}\text{.5}\text{ft}\left(\frac{\text{28,317}{\text{cm}}^{\text{3}}}{\text{1}{\text{ft}}^{\text{3}}}\right)\approx 1\times 10^{\text{5}}{\text{cm}}^{\text{3}}$

Method 2:

Assume,

The weight of the person to be 150 lb_m

The density of the person to be 62.4 lb_m / ft³

Volume = mass / density

The volume of the person,

$150{\text{lb}}_{\text{m}}\left(\frac{\text{1}{\text{ft}}^{\text{3}}}{\text{62}\text{.4}{\text{lb}}_{\text{m}}}\right)\frac{\text{28,317}{\text{cm}}^{\text{3}}}{\text{1}{\text{ft}}^{\text{3}}}\approx \frac{150\times 3\times 10^4}{60}\approx \text{1}\text{×}{\text{10}}^{\text{5}}{\text{cm}}^{\text{3}}$

Interpretation Introduction

(f)

Interpretation:

The specific gravity of a solid block in water should be calculated.

Concept introduction:

The specific gravity of a body with respect to water is calculated by taking ratio of density of that body to the density of water. It is also related to molar mass as follows:

$S=\frac{{\rho }_A}{{\rho }_B}=\frac{M_A}{M_B}$

Here, ${\rho }_A$ is density of body A and ${\rho }_B$ is density of water.

Answer to Problem 3.1P

$1.05$

Explanation of Solution

Since, the solid block sinks into the water very slowly, the density of block should be close to the density of water. So, assume density of the block to be 1050 kg / m³.

The Specific gravity is calculated as follows:

$S=\frac{{\rho }_A}{{\rho }_B}$

Putting the values,

$S=\frac{\text{1050}\text{kg}\text{/}{\text{m}}^{\text{3}}}{\text{1000}\text{kg/}{\text{m}}^{\text{3}}}=1.05$