An inventor claims that he wants to build a dam to produce hydroelectric power. He correctly realizes that civilization uses a lot more electricity during the day than at night, so he thinks he has stumbled upon a great untapped energy supply. His plan is to install pumps at the bottom of the dam so that he can pump some of the water that flows out from the generators back up into the reservoir using the excess electricity generated at night. He reasons that if he did that, the water would just flow right back down through the generators the next day producing power for free. What is wrong with his plan?

Answers

Answer 1

Answer:

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.

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The sports car has a weight of 4500-lb and a center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are us=0.5 and uk=0.3, respectively. Neglect the mass of the wheels.

Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold climates because it is light, fashionable, and a very effective thermal insulator. So, it is no surprise that such clothing has largely replaced thick and heavy old-fashioned coats. Consider a jacket made up of six layers of 0.1 mm thick synthetic fabric (k = 0.026W/m.K) with 1.2 mm thick air space (k = 0.026 W/m.K) between the fabric layers. Assuming the inner surface temperature of the jacket to be 25˚C and the surface area to be 1.25 m2 , determine the heat loss through the jacket when the temperature of the outdoors is -5˚C and the heat transfer co-efficient of outer surface is 25 W/m2 .K. What would be the thickness of a wool fabric (k = 0.035W/m.K) if the person has to achieve the same level of thermal comfort wearing a thick wool coat instead of a jacket. (30 points)

Answers

Answer:

Q=127.66W

L=9.2mm

Explanation:

Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.

To solve any problem related to heat transfer, the general equation is used

Q = delta / R

Where

Q = heat

Delta = the temperature difference

R = is the thermal resistance by conduction, convection and radiation

to solve this problem we propose the previous equation

Q = delta / R

later we find R

$R=[tex]r=(6L1)/(AK1) +(5L2)/(AK2)+(1)/(Ah)$

$R=(6(0.0001))/((1.25)(0.026)) +(5(0.012))/((1.25)(0.026))+(1)/((25)(1.25)) =0.235 K/w$

Q=(25-(-5))/0.235=127.66W

part b

we use the same ecuation with Q=127.66

Q = delta / R

Δ $R=(L)/(KA) +(1)/(hA) \nR=(L)/((0.035)(1.25)) +(1)/((25)(1.25))\n R=22.85L+0.032\nQ=(T1-T2)/R\n\n127.66=(25-(-5))/(22.85L+0.032)\nsolving for L\nL=9.2mm$

Make a copy of the pthreads_skeleton.cpp program and name it pthreads_p2.cpp Modify the main function to implement a loop that reads 10 integers from the console (user input) and stores these numbers in a one-dimensional (1D) array (this code will go right after the comment that says ""Add code to perform any needed initialization or to process user input""). You should use a global array for this.

Answers

Answer:

The solution code is as follows:

#include <iostream>
using namespace std;
int main()
{
int myArray [10] = {};
int i;
for( i = 0; i < 10; i++ )
{
cout <<"Enter an integer: ";
cin>> myArray[i];
}
}

Explanation:

Firstly, we initialize a 10-elements array, myArray (Line 7) with no values.

Next, we create a for-loop (Line 10). Within the loop, we prompt user to enter an integer and assign the input value to the current element of myArray (Line 12-13).

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.

Answers

Answer:

Time period = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration $9.4 * 10^(-5) kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4* 10^(-5) * 62 = 582.8* 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8* 10{-5}}{1.156} = 5.04* 10^(-3) m^3/s$

Time period $= (210)/(5.04* 10^(-3)) = 41654.08 s$

A horizontal curve on a single-lane highway has its PC at station 1+346.200 and its PI at station 1+568.70. The curve has a superelevation of 6.0% and is designed for 120 km/h. The limiting value for coefficient of side friction at 120 km/h is 0.09. What is the station of the PT? Remember that 1 metric station = 1000 m.

Answers

Answer:

The solution is given in the attachments.

Aluminum alloys are made by adding other elements to aluminum to im prove its properties, such as hardness or tensile strength. The following table shows the composition of ve commonly used alloys, which are known by their alloy numbers (2024, 6061, and so on) [Kutz, 1999]. Obtain a matrix algorithm to compute the amounts of raw materials needed to pro duce a given amount of each alloy. Use MATLAB to determine how much raw material of each type is needed to produce 1000 tons of each alloy. Composition of aluminum alloys Alloy %Cu %Mg %Mn %Si %Zn 2024 6061 7005 0 7075 356.0 0 0.6 0 0 0 0 4.5 5.6 0 0.6 0 0 2.5 0.3 0

Answers

Answer:

Cu= 60

Mg= 67

Mn= 6

Si= 76

Zn= 101

Explanation:

Solution steps :

1) Creating matrix A holds the composition of each raw material.

2) Submission of composition of each raw material.

3) Multiplying each amount by the total amount needed to be produced.

note

find the attached code

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?