Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change
If everyone died then the too people were not on the plane they were near by when the plane crashed and were not onbard.
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.
