Answer :
Answer:
The answer is VN =37.416 m/s
Explanation:
Recall that:
Pressure (atmospheric) = 100 kPa
So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles
Now,
Pabs =Patm + Pgauge = 800 KN/m²
Thus
PT/9.81 + VT²/2g =PN/9.81 + VN²/2g
Here
Acceleration due to gravity = 9.81 m/s
800/9.81 + 0
= 100/9.81 + VN²/19.62
Here,
9.81 * 2= 19.62
Thus,
VN²/19.62 = 700/9.81
So,
VN² =1400
VN =37.416 m/s
Note: (800 - 100) = 700
Answer:
[tex]V2 = 37.417ms^{-1}[/tex]
Explanation:
Given the following data;
Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.
Nozzle outlets = 100kPa = 100000pa.
Density of water = 1000kg/m³.
We would apply, the Bernoulli equation between the inlet and outlet;
[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]
Where, V1 is approximately equal to zero(0).
Z[tex]z_{1} = z_{2}[/tex]
Therefore, to find the maximum velocity, V2;
[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]
[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]
[tex]V2 = \sqrt{2(800-100)}[/tex]
[tex]V2 = \sqrt{2(700)}[/tex]
[tex]V2 = \sqrt{1400}[/tex]
[tex]V2 = 37.417ms^{-1}[/tex]
Hence, the maximum velocity, V2 is 37.417m/s
