Answer:
The power transferred to the fluid is 1.8852 kW
Explanation:
Power = pressure drop × area × velocity
pressure drop = 20 kPa
area = πd^2/4 = 3.142×0.2^2/4 = 0.03142 m^2
velocity = 3 m/s
Power = 20 × 0.03142 × 3 = 1.8852 kJ/s = 1.8852 kW
The correct answer would be d
The objects will not move towards or away from each other
Hope this helps
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).
Answer:
Explanation:
,K=1.35
Clearance is 8%.
Heat added=15 KJ
We know that compression ratio 
r=13.5
We know that efficiency of otto cycle
W is the work out put and Q is the heat addition.
W=8.4 KJ
We know that Work =Mean effective pressure x swept volume.
Here swept volume 
Noe by putting the values
Work =Mean effective pressure x swept volume.
Answer: you’re bussy *in french accent*
Explanation:
it is very wide as you can see *also in french*
