Chapter 6.3, Problem 35PPS

To determine

To order: given group of rates from least to greatest.

Answer to Problem 35PPS

  $18\text{Ib/min}<\text{500}\text{kg/h}<\text{5}\text{oz/s}\text{.}$

Explanation of Solution

Given information:

Given group

500 kg/h, 5 oz/s, 18Ib/min.

Calculation:

Use the conversions,

  $\begin{array}{l}1\text{oz=0}\text{.0283}\text{kg,}\text{1}\text{Ib=}\text{0}\text{.454}\text{kg}\\ \text{1}\text{hr=3600}\text{sec,}\text{1}\text{min=60}\text{sec}\end{array}$

Convert all the units in kg/hr.

  $\begin{array}{c}5\text{oz/s=}\frac{5\times 0.0283\text{kg}}{1\text{hr}}\times 3600\\ =509.4\text{kg/hr}\end{array}$

And

  $\begin{array}{c}18\text{Ib/min=}\frac{18\times 0.454\text{kg}}{1\text{hr}}\times 60\\ =490.32\text{kg/hr}\end{array}$

Now order is $490.32$ kg/hr, 500 kg/h, $509.4$ kg/hr.

Hence, $18\text{Ib/min}<\text{500}\text{kg/h}<\text{5}\text{oz/s}\text{.}$