Chapter 1, Problem 1.1P

Convert the following quantities from English to SI units:

a. 98 Btu/(hr-ft-F)

b. 0.24 Btu/(lbm-F)

C. 0.04 Ibm/(ft-hr)

d. 1050 Btu/lbm

e. 1.0 ton (cooling)

f. 14.7 lbf/in.²

To determine

To Convert: $98\text{Btu}/\left(\text{hr-ft-F}\right)$ from English to SI units.

Answer to Problem 1.1P

Thequantity in SI units is, $169.6\text{W}/\text{m-k}$ .

Explanation of Solution

Given:

The given quantity is,

  $98\text{Btu}/\left(\text{hr-ft-F}\right)$

Calculation:

Consider the given quantity.

  $98\text{Btu}/\left(\text{hr-ft-F}\right)$

Convert the quantity to SI units.

  $\begin{array}{l}98\text{Btu}/\text{hr}=0.2931\text{W}\\ 1\text{ft}=0.3048\text{m}\\ 1\text{F}=255.93\text{K}\end{array}$

The quantity in SI units is converted as,

  $\begin{array}{c}98\text{Btu}/\left(\text{hr-ft-F}\right)=98\times 1.7307\text{W}/\text{m-k}\\ =169.6\text{W}/\text{m-k}\end{array}$

To determine

To convert: $0.24\text{Btu}/\left(\text{lbm-F}\right)$ from English to SI units.

Answer to Problem 1.1P

The quantity in SI units is $1\text{kJ}/\text{kg-}°\text{C}$ .

Explanation of Solution

Given:

The given quantity is,

  $0.24\text{Btu}/\left(\text{lbm-F}\right)$

Calculation:

Consider the given quantity.

  $0.24\text{Btu}/\left(\text{lbm-F}\right)$

Convert the quantity to SI units.

  $1\text{Btu}/\text{lbm-F}=4.1868\text{kJ}/\text{kg-°C}$

The quantity in SI units is converted as,

  $\begin{array}{c}0.24\text{Btu}/\left(\text{lbm-F}\right)=0.24\times 4.1868\text{kJ}/\text{kg-°C}\\ =1.004\text{kJ}/\text{kg-}°\text{C}\\ \approx 1\text{kJ}/\text{kg-}°\text{C}\end{array}$

To determine

To convert: $0.04\text{lbm}/\left(\text{ft-hr}\right)$ from English to SI units.

Answer to Problem 1.1P

The quantity in SI units is $16.5\times {10}^{-6}\text{kg}/\text{m-s}$ .

Explanation of Solution

Given:

The given quantity is,

  $0.04\text{lbm}/\left(\text{ft-hr}\right)$

Calculation:

Consider the given quantity.

  $0.04\text{lbm}/\left(\text{ft-hr}\right)$

Convert the quantity to SI units.

  $\begin{array}{l}1\text{lbm}=0.4536\text{kg}\\ 1\text{ft}=0.3048\text{m}\end{array}$

The quantity in SI units is converted as,

  $\begin{array}{c}0.04\text{lbm}/\left(\text{ft-hr}\right)=0.04\times 0.4536\text{kg}\times \frac{1}{0.3048\text{m}\times \text{3600}\text{s}}\\ =16.5\times {10}^{-6}\text{kg}/\text{m-s}\end{array}$

To determine

To convert: $1050\text{Btu}/\text{lbm}$ from English to SI units.

Answer to Problem 1.1P

The quantity in SI units is $2442.3\text{J}/\text{g}$ .

Explanation of Solution

Given:

The given quantity is,

  $1050\text{Btu}/\text{lbm}$

Calculation:

Consider the given quantity.

  $1050\text{Btu}/\text{lbm}$

Convert the quantity to SI units.

  $1\text{Btu}/\text{lbm}=2.326\text{J}/\text{g}$

The quantity in SI units is converted as,

  $\begin{array}{c}1050\text{Btu}/\text{lbm}=1050\times 2.326\text{J}/\text{g}\\ =2442.3\text{J}/\text{g}\end{array}$

To determine

To convert: $1.0\text{ton}\left(\text{cooling}\right)$ from English to SI units.

Answer to Problem 1.1P

The quantity in SI units is $3.517\text{kW}$ .

Explanation of Solution

Given:

The given quantity is,

  $1.0\text{ton}\left(\text{cooling}\right)$

Calculation:

Consider the given quantity.

  $1.0\text{ton}\left(\text{cooling}\right)$

Convert the quantity to SI units.

  $\begin{array}{l}1.0\text{ton}\left(\text{cooling}\right)=200\text{Btu}/\text{min}\\ 1\text{Btu}/\text{min}=0.017585\text{kW}\end{array}$

The quantity in SI units is converted as,

  $\begin{array}{c}1.0\text{ton}\left(\text{cooling}\right)=200\text{Btu}/\text{min}\times 0.017585\text{kW}\\ =3.517\text{kW}\end{array}$

To determine

To convert: $14.7\text{lbf}/{\text{in}}^{\text{2}}$ from English to SI units.

Answer to Problem 1.1P

The quantity in SI units is $101356.5\text{N}/{\text{m}}^{\text{2}}$ .

Explanation of Solution

Given:

The given quantity is,

  $14.7\text{lbf}/{\text{in}}^{\text{2}}$

Calculation:

Consider the given quantity.

  $14.7\text{lbf}/{\text{in}}^{\text{2}}$

Convert the quantity to SI units.

  $\begin{array}{l}1\text{lbf}/{\text{in}}^{\text{2}}=1\text{psi}\\ 1\text{psi}=6895\text{N}/{\text{m}}^{\text{2}}\end{array}$

The quantity in SI units is converted as,

  $\begin{array}{c}14.7\text{lbf}/{\text{in}}^{\text{2}}=14.7\times 6895\text{N}/{\text{m}}^{\text{2}}\\ =101356.5\text{N}/{\text{m}}^{\text{2}}\end{array}$