# Agitation - Power Required

An agitated baffle vessel is being used to prepare a uniform solution of viscosity 2 cP, running the agitator at 100 rpm, so as to obtain a Reynolds number of 50,000. If the contents of the vessel are replaced by a solution of viscosity 4 cP, and the agitator rpm is increased to 200, by how much will the power requirement change?

Theory:

For agitated vessel, the following dimensional relationship is applicable.

NP = y (NRe, NFr, S1, S2,..., Sn)

Where NP = Power Number = P/(n3Da5r)

NRe = Reynolds number = nDa2r/m

NFr = Froude Number = n2Da/g

And S1, S2,..., Sn are shape factors.

n = rotation per unit time of agitator

Da = dia of impeller

r = density of fluid

m = viscosity of fluid

If the shape factors are remaining constant, then

NP = y (NRe, NFr)

For baffled vessel, NP is a function of only NRe provided the shape factors are remaining at constant value.

i.e.,

NP = y (NRe)

For various impeller configurations, and system geometry, NP vs. NRe chart is available to estimate the power required.

From the charts available, it could be seen that, for NRe > 10000, NP is independent of NRe, and remains at a constant value.

i.e., NP = constant, (for NRe > 10000), and viscosity is not a factor.

Calculations:

For the given problem,

For the case 1:

m1 = 2 cP

n1 = 100

NRe,1 = 50000

NRe,1 = n1Da2r/m1 = 50000

Therefore, Da2r = 50000 x 2 / 100 = 1000

For the case 2:

m2 = 4 cP

n2 = 200

NRe,2 = (Da2r) n2/m1 = 1000 x 200/4 = 50000

Here NRe is more than 10000. Therefore, NP = constant = K

NP = P/(n3Da5r) = K

i.e., P = Kn3Da5r

since Da and r are same for the two cases,

we can group KDa5r as a constant, say M.

i.e., P = Mn3

The ratio of power required for case 2 to 1 is,

P2/P1 = n23/n13

= 2003/1003 = 8.

The power required for the second case will be 8 times that of the first case. In other words, the power requirement will rise by, 100 x (8-1)/1 = 700%.
{"ft_lang_label":"__label__en","ft_lang_prob":0.86803347,"math_prob":0.99911034,"size":1859,"snap":"2019-26-2019-30","text_gpt3_token_len":605,"char_repetition_ratio":0.10566038,"word_repetition_ratio":0.011019284,"special_character_ratio":0.3415815,"punctuation_ratio":0.17142858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99906236,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-19T19:18:12Z\",\"WARC-Record-ID\":\"<urn:uuid:982ee3b6-b158-4a03-b113-664d04cfba14>\",\"Content-Length\":\"8054\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3dbf4b8f-1b9b-4aa3-bf4b-e29a42c91092>\",\"WARC-Concurrent-To\":\"<urn:uuid:b129df32-9217-4bbd-8f77-d4257352a302>\",\"WARC-IP-Address\":\"118.67.248.103\",\"WARC-Target-URI\":\"http://msubbu.in/problems/MechanicalOperations/Agitation.html\",\"WARC-Payload-Digest\":\"sha1:HJZO5LAJE2NUAIC76YUJJTRFQIHVYIZT\",\"WARC-Block-Digest\":\"sha1:3GOK7YM57AXVKB5MOOARFTNCRCVPMUY7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526337.45_warc_CC-MAIN-20190719182214-20190719204214-00049.warc.gz\"}"}