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Lena [83]
3 years ago
15

The water behind Hoover Dam is 206m higher than the Colorado river below it. At what rate must water pass through the hydraulic

turbines of this dam to produce 100 MW of power if the turbines are 100 percent efficient?
Engineering
2 answers:
Alja [10]3 years ago
4 0

Answer:

49.484 m³ / s

Explanation:

Volume flow rate = Power in W / (efficiency × density × height × acceleration due to gravity)

Volume flow rate = 100 × 10⁶ / ( 1 × 1000 kg/m³ × 206 m × 9.81 m/s²)

V = 49.484 m³ / s

Leokris [45]3 years ago
3 0

Answer:

m' = 4948.38 kg/s

Explanation:

For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion

GIVEN THAT

Power = 100 MW

rate of Potential energy = (m')*g*h

100*10^6 = (m')*9.81*206

m' = 4948.38 kg/s

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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

6 0
3 years ago
A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage
faltersainse [42]

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

4 0
3 years ago
An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava
N76 [4]

Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.

Explanation:

4 0
3 years ago
Read 2 more answers
On aircraft equipped with fuel pumps, when is the auxiliary electric driven pump used?.
pochemuha
In an airplane equipped with fuel pumps, the auxiliary electric fuel pump is used in the event the engine-driven fuel pump fails.. hope this helped !
6 0
2 years ago
A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th
shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

<u>a) Calculate Natural frequency </u>

Wn = √k/m = \sqrt{12000 /  5*10^{-3}  }

                   = 1549.19 rad/s  ≈ 246.56 Hz

<u>b) Bandwidth of instrument </u>

W / Wn = \sqrt{1-2(0.4)^2}

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

5 0
3 years ago
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