The shortest is sequoia with only two consonants.
If you consider y to be a vowel, the shortest is a tie between audiometry and aureomycin each with 4 non-y consonants.
Because of the vowel frequencies in English, the UNIX command is most efficient if invoked as:
grep -i u /usr/dict/words | grep -i o | grep -i a | grep -i i | grep -i e
To find words with the vowels in order (A E I O U), use the command:
grep 'a.*e.*i.*o.*u' /usr/dict/words
which will find abstemious, facetious, adventitious and sacreligious (facetious is probably the best known). Notice two of these have an extra i but the puzzle seldom requires a solution with exactly five vowels.
If you want the vowels in order and count y as a vowel, note that all four are adjectives to which you can add -ly. Thus, facetiously is the unique shortest-length solution with all six vowels, and accomplishes this with 5 non-vowel letters.