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Marina CMI [18]
2 years ago
6

The water behind Hoover Dam in Nevada is 221 m higher than the Colorado River below it. At what rate must water pass through the

hydraulic turbines of this dam to produce 50 MW of power if the turbines are 100 percent efficient?
Engineering
1 answer:
dexar [7]2 years ago
3 0

Answer:

23.06262m^3/s

Explanation:

The volume flow rate of the water is determined from the needed power output and the elevation difference:

Where, height (h) =221m, power(w)=50MW=50*10^6w

Density of water (ρ)=1000kg/m^3

Efficiency of turbine(η)=100%=1

V=W/ρηgh

=50*10^6m^3/(1)*(1000)*(9.81)*(221)s=23.06262m^3/s

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An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [
crimeas [40]

Answer:

Modulus of resilience will be 3216942.308j/m^3

Explanation:

We have given yield strength \sigma _y=818MPa

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience =\frac{\sigma _y^2}{2E}, here \sigma _y is yield strength and E is elastic modulus

Modulus of resilience =\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3  

5 0
3 years ago
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0
3 years ago
Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "<strong>{0}</strong>".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "<em>{0}</em>".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0
3 years ago
Engineering Questions
valkas [14]
5 is the correct one to choose for this
6 0
2 years ago
Read 2 more answers
A growing trend in urban design is the concept of a rooftop garden. If every building in a city were to install a rooftop garden
vlabodo [156]

Answer:The Urban heat island temperature will be REDUCED.

Two Impacts of Rooftop gardens

1) provision of shade against Sunlight.

2) It helps to purify the air around the building.

Explanation: Rooftop gardens are gardens made on top of the roofs of buildings, it is a Green initiative aimed at helping to improve the overall Environment.

Rooftop gardens have several significant benefits which includes

Reduction of the surrounding temperatures and the Urban heat Island temperatures.

Rooftop gardens helps to shade the roof from the direct impacts of harsh weather conditions.

Generally, plants are known as air purifiers as they remove the excess Carbondioxide around the environment through photosynthesis, and they also help to release water vapor which will help to improve the humidity of the environment.

5 0
2 years ago
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