Niagara Falls is capable of producing a total power output of 2.575 MW at steady-state during the night when flow over the falls
is diverted through a hydroelectric turbine. This is done at night so as not to detract from tourism during the daytime. Changes in temperature and pressure of the flow through the turbine are negligible. The intake to the turbine is located at the top of Niagara River and exits at the bottom of the falls, 22 m below. The velocity of the water in the penstock upstream of the turbine is, on average, 0.6 m/s and exits the turbine, operating adiabatically, at 0.9 m/s. - What is the mass flow rate of water through the turbine at night, in kg/s?
2 answers:
Answer:
Explanation:
This is the question based on First Law of Thermodynamics with Steady Flow Energy Equation for Turbine.
The total energy entering the turbine control volume is equal to the total energy leaving the turbine control volume.
The calculation is attached
Answer:
m= 11953.67 kg/s
Explanation:
Given : power output P = 2.575 MW
Head available H = 22 m
Inlet velocity V1 = 0.6 m/s
Outlet velocity V2 = 0.9 m/s
Solution:
Here the hydraulic energy available at the inlet of turbine is given by,
Eh= mgH
Where m is mass flow rate in kg/s
g is acceleration due to gravity
H is head available.
Part of this energy is converted in to electrical energy of 2575000 W and remaining is at outlet of Turbine in the form of kinetic energy i.e. (see attachment)
KE =\frac{mv^{2}}{2}
So we can write energy balance equation as, (see attachment)
mgH =\frac{mv^{2}}{2} + 2575000
m(gH -\frac{v^{2}}{2}) = 2575000
m(9.81*22 -\frac{0.9^{2}}{2}) = 2575000
m= 2575000/215.415
m= 11953.67 kg/s
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Answer:
The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.
Explanation:
The detailed steps and appropriate calculation is as shown in the attached file.
Answer:
t = 30.1 sec
Explanation:
If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.
If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.
We are only interested in the component of the velocity across the paper, that can be calculated as follows:
vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.
We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:
vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s
Applying the definition of average velocity, we can solve for t:
t = =
⇒ t = 30.1 sec