Chapter 1, Problem 1.1P

(a)

Interpretation Introduction

Interpretation:

The conversion factor for foot-pounds force to kilowatt-hours is to be determined.

Concept Introduction:

Conversion factors are used to change the magnitude of a quantity from one unit system to another. These conversion factors are defined for base units which can be further used to derive it for other unit systems.

Answer to Problem 1.1P

The conversion factor for foot-pounds force to kilowatt-hours is $1{\text{ ft-lb}}_f=3.76614\times {10}^{-7}\text{ kWh}$ .

Explanation of Solution

From Appendix 1, the conversion factor for foot-pounds force $\left({\text{ft-lb}}_f\right)$ to kilowatt-hours $\left(\text{kWh}\right)$ is calculated as:

  $\begin{array}{c}1{\text{ ft-lb}}_f=\frac{1}{0.73756}\text{ J}\\ =\frac{1}{0.73756\times 1055.06}\text{ Btu}\\ =\frac{1}{0.73756\times 1055.06}\times \left(2.9307\times {10}^{-4}\right)\text{ kWh}\\ =3.76614\times {10}^{-7}\text{ kWh}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The conversion factor for gallons to liters is to be determined.

Concept Introduction:

Conversion factors are used to change the magnitude of a quantity from one unit system to another. These conversion factors are defined for base units which can be further used to derive it for other unit systems.

Answer to Problem 1.1P

The conversion factor for gallons to liters is $1\text{ gal}=3.7854\text{ L}$ .

Explanation of Solution

From appendix 1, the conversion factor for gallons $\left(\text{gal}\right)$ to liters $\left(\text{L}\right)$ is calculated as:

  $\begin{array}{c}1\text{ gal}=231\text{ in}{\text{.}}^3\\ =231\times {\left(2.54\right)}^3{\text{ cm}}^3\\ =\frac{231\times {\left(2.54\right)}^3}{1000}\text{ L}\\ =3.7854\text{ L}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The conversion factor for Btu per pound mole to joules per kilogram mole is to be determined.

Concept Introduction:

Conversion factors are used to change the magnitude of a quantity from one unit system to another. These conversion factors are defined for base units which can be further used to derive it for other unit systems.

Answer to Problem 1.1P

The conversion factor for Btu per pound mole to joules per kilogram mole is $1\frac{\text{Btu}}{\text{lb-mol}}=2326\frac{\text{J}}{\text{kg-mol}}$ .

Explanation of Solution

From Appendix 1, the conversion factor for Btu per pound mole $\left(\text{Btu}/\text{lb-mol}\right)$ to joules per kilogram mole $\left(\text{J}/\text{kg-mol}\right)$ is calculated as:

  $\begin{array}{c}1\frac{\text{Btu}}{\text{lb-mol}}=1055.06\frac{\text{J}}{\text{lb-mol}}\\ =\frac{1055.06}{0.45359237}\frac{\text{J}}{\text{kg-mol}}\\ =2326\frac{\text{J}}{\text{kg-mol}}\end{array}$