The maximum volume flow rate of water is determined as 0.029 m³/s.
<h3>Power of the pump</h3>
The power of the pump is watt is calculated as follows;
1 hp = 745.69 W
7 hp = ?
= 7 x 745.69 W
= 5,219.83 W
<h3>Mass flow rate of water</h3>
η = mgh/P
mgh = ηP
m = ηP/gh
m = (0.82 x 5,219.83)/(9.8 x 15)
m = 29.12 kg/s
<h3>Maximum volume rate</h3>
V = m/ρ
where;
- ρ is density of water = 1000 kg/m³
V = (29.12)/(1000)
V = 0.029 m³/s
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Answer:
Explanation:
volume of 20.9 N
= 20.9 / 11.5 m³
= 1.8174 m³
In one hour 1.8174 m³ flows
in one second volume flowing = 1.8174 / 60 x 60
= 5 x 10⁻⁴ m³
Rate of volume flow = 5 x 10⁻⁴ m³ / s .
Maximum shear stress in the pole is 0.
<u>Explanation:</u>
Given-
Outer diameter = 127 mm
Outer radius,
= 127/2 = 63.5 mm
Inner diameter = 115 mm
Inner radius,
= 115/2 = 57.5 mm
Force, q = 0
Maximum shear stress, τmax = ?
τmax 
If force, q is 0 then τmax is also equal to 0.
Therefore, maximum shear stress in the pole is 0.
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)

Now we can use the equation that defines the percentage of increase, in this case for speed

Now we use the equation obtained in the previous step, and replace values

the percent increase in the velocity of air is 25.65%