Answer:
Monopsony
Explanation:
Kinda like a monopoly, except instead of one producer it is one buyer.
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:
1) a. Customers requiring AC electric power transmission for powering remote devices which may include a subsea transmission system where power is distributed to subsea devices
b. Customers that utilize commutator-type motors
2) The primary voltage is 150 volts
Explanation:
The parameters given are;
Number of primary winding,
= 50 turns
Number of secondary winding,
= 150 turns
Voltage in secondary winding,
= 450 volts
Voltage in primary winding = 
The relation between
,
,
and
is as follows;

Which gives;

From which we have;

The primary voltage = 150 volts.
Answer:
The answer is 2.715 In
Explanation:
An estimated 60% of annual precipitation in a watershed (drainage area = 20000 acres) is evaporated. If the average annual river flow at the outlet of the basin has been observed to be 2.5 cfs, determine the annual precipitation (inches) in the basin
The annual precipitation in inches in the basin is 2.715 inches.
The solution and steps is explained in the attachment.
I hope i have been able to help.