Answer:
düz devam sağa dön bı tane Kabe var
Answer: Question 1: Efficiency is 0.6944
Question 2: speed of similar pump is 2067rpm
Explanation:
Question 1:
Flow rate of pump 1 (Q1) = 300gpm
Flow rate of pump 2 (Q2) = 400gpm
Head of pump (H)= 55ft
Speed of pump1 (v1)= 1500rpm
Speed of pump2(v2) = ?
Diameter of impeller in pump 1= 15.5in = 0.3937m
Diameter of impeller in pump 2= 15in = 0.381
B.H.P= 6.0
Assuming cold water, S.G = 1.0
eff= (H x Q x S.G)/ 3960 x B.H.P
= (55x 300x 1)/3960x 6
= 0.6944
Question 2:
Q = A x V. (1)
A1 x v1 = A2 x V2. (2)
Since A1 = A2 = A ( since they are geometrically similar
A = Q1/V1 = Q2/V2. (3)
V1(m/s) = r x 2π x N(rpm)/60
= (0.3937x 2 x π x 1500)/2x 60
= 30.925m/s
Using equation (3)
V2 = (400 x 30.925)/300
= 41.2335m/s
To rpm:
N(rpm) = (60 x V(m/s))/2 x π x r
= (60 x 41.2335)/ 2× π × 0.1905
= 2067rpm.
Answer:
L = 5,955 m
Explanation:
For this exercise we must use the relation
R = ρ L / A
where R is the resistance that indicates that it is 1 Ω, the resistivity is taken from the tables ρ = 2.82 10⁻⁸ Ω m, L is the length of the wire and A is the cross section.
As it indicates to us in volume of aluminum to use we divide the two terms by the length
R / L = ρ L / (A L)
the volume of a body is its area times its length, therefore
R / L = ρ L / V
R = ρ L² / V
we clear the length of the wire
L = √ R V /ρ
we reduce the volume to SI units
v = 1 cm³ (1m / 10² cm)³ = 1 10⁻⁶ m
let's calculate
L = √ (1 1 10⁻⁶ / 2.82 10⁻⁸)
L = √ (0.3546 10²)
L = 5,955 m
