Respuesta :
Answer:
P=3.31 hp (2.47 kW).
Explanation:
Solution
Curve A in Fig1. applies under the conditions of this problem.
S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt
The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.
32.2
Fig. 32.2 Dimension of turbine agitator
The Reynolds number is calculated. The quantities for substitution are, in consistent units,
D a =2â ft
n= 90/ 60 =1.5 r/s
Îź = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s
Ď = 93.5 lb/ft3 g= 32.17 ft/s2
NRc = (( D a) 2 n Ď)/ Îź = 2 2 Ă1.5Ă93.5 8.06Ă 10 â3 =69,600
From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P Ă (n) 3 Ă ( D a )5 Ă Ď g c
The power P= 5.8Ă93.5Ă (1.5) 3 Ă (2) 5 / 32.17 =1821â ftâlb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).