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 =2⋅ft
n= 90/ 60 =1.5 r/s
μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s
ρ = 93.5 lb/ft3 g= 32.17 ft/s2
NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600
From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c
The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).
