Ask question

Ask question

All categories

-Dominant- [34]

3 years ago

9

Assess the capabilities of a hydroelectric power plant from the following field data: Estimated water flow rate, 40 m3/s River i

nlet at 1 atm, 10 oC Discharge at 1 atm, 10.2 oC, 200 m below the intake?

1 answer:

ryzh [129]3 years ago

6 0

Answer:

This hydroelectric power plant has the capability of producing around 78.4 MW of electric energy.

Explanation:

First, we calculate the pressure the water exerts or gains when it travels 200 m below the intake. W have the data:

Density of water = ρ = 1000 kg/m^3

Acceleration due to gravity = g = 9.8 m/s²

Height lost = h = 200 m

Volume flow rate = 40 m^3/s

The pressure difference, is now given as:

ΔP = ρgh

ΔP = (1000 kg/m^3)(9.8 m/s²)(200 m)

ΔP = 1960000 Pa = 1.96 MPa

Now, we calculate the power capacity of this flow:

Power = ΔP(Volume flow rate)

Power = (1960000 N/m²)(40 m^3/s)

Power = 78400000 Watt

<u>Power = 78.4 MW</u>

Thus, this plant can produce at max 78.4 MW. The actual power produced will be slightly less than this due of the losses and efficiency of turbine system used.

You might be interested in

In the engineering design and prototyping process, what is the advantage of drawings and symbols over written descriptions?

MrMuchimi

The advantages that can be associated to

drawings and symbols over written descriptions in engineering design and prototyping process are;

Communicate design ideas as well as technical information to engineers.

Symbols and drawings can be universal which means it is easy to interpret any where by professionals.

An engineering drawing serves as complex dimensional object and symbol use by engineer to communicate.

Drawings and symbols makes it easier to communicate design ideas and technical information to engineers and and how the process will go.

Therefore, drawings and symbols is universal to all engineer unlike written one.

Learn more at:

brainly.com/question/20925313?referrer=searchResults

4 0

3 years ago

What is a radio wave made up of? Molecules? Electrons? Other?

mafiozo [28]

Radio waves are radiated by charged particles when they are accelerated. They are produced artificially by time-varying electric currents, consisting of electrons flowing back and forth in a specially-shaped metal conductor called an antenna. ... Radio waves are received by another antenna attached to a radio receiver.

4 0

3 years ago

Read 2 more answers

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago

What is the biggest expectation when engineers test out designs?

forsale [732]

The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.

6 0

4 years ago

Read 2 more answers

Please help!!!! WHICH 2 SKILLS N ABILITIES R ESSENTIAL FOR A SHIP CAPTAIN?? A. Good vision B. Public speaking skills C. Leadersh

Brilliant_brown [7]

A and C are the correct answers

6 0

3 years ago

Read 2 more answers