ENGINEERING COLLEGE

for a rankine cycle with one stage of reheat between turbines, there are how many relevant pressures?

Answers

Answer 1

Answer:

The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.

For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:

Boiler pressure (P1): This is the pressure at which the water is heated in the boiler before entering the first turbine.
High-pressure turbine outlet pressure (P2): This is the pressure at the outlet of the first turbine before the steam is sent to the reheater.
Reheat pressure (P3): This is the pressure at which the steam is reheated before entering the second turbine.
Low-pressure turbine outlet pressure (P4): This is the pressure at the outlet of the second turbine, which is also the condenser pressure.

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Analyze that, “Convection is equal to the Conduction plus fluid flow.”

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

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.

Answers

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

$K_(H)=(H_(A)K_(A)+H_(A)K_(A)+H_(A)K_(A))/(H_(A)+H_(B)+H_(C))$

Put the value into the formula

$K_(H)=(8.0*25+2,0*142+34*40)/(8.0+2.0+34)$

$K_(H)=41.9\ m/d$

We need to calculate the vertical conductivity

Using formula of vertical conductivity

$K_(V)=(H_(A)+H_(B)+H_(C))/((H_(A))/(K_(A))+(H_(B))/(K_(B))+(H_(C))/(K_(C)))$

Put the value into the formula

$K_(V)=(8.0+2.0+34)/((8.0)/(25)+(2.0)/(142)+(34)/(40))$

$K_(V)=37.2\ m/d$

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

8.2.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new

Answers

Answer:

The output will be (3, 4) becomes (8, 10)

Explanation:

#include <stdio.h>

//If you send a pointer to a int, you are allowing the contents of that int to change.

void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){

*xNew = (xVal+1)*2;

*yNew = (yVal+1)*2;

}

int main(void) {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, &xValNew, &yValNew);

printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);

return 0;

}

A cube with 1 m on a side is located in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. Find the total charge contained in the cube if the charge is given by p_v = x^2 ye^-z mC/m^3

Answers

Answer:

4.61 mC

Explanation:

The cube has 1 m side in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. The charge density is given as:

$\rho_v=x^2ye^(-z) \ mC/m^3$

Charge density is the charge per unit length or area or volume. It is the amount of charge in a particular region.

The charge Q is given as:

$Q=\int\limits_v {\rho_v} \, dv \nQ=\int\limits_v {\rho_v} \, dv=\int\limits^2_(x=0)\int\limits^2_(y=0)\int\limits^2_(z=0) {x^2ye^(-z)} \, dxdydz\n$

$Q=\int\limits^2_(x=0) {x^2} \, dx \int\limits^2_(y=0) {y} \, dy \int\limits^2_(z=0) {e^(-z)} \, dz \n\nQ=((1)/(3) [x^3]^2_0)((1)/(2) [y^2]^2_0)(-1 [e^(-z)]^2_0)\n\nQ=(-1)/(6) ([x^3]^2_0)( [y^2]^2_0)( [e^(-z)]^2_0)\n\nQ=(-1)/(6)[2^3-0^3][2^2-0^2][e^(-2)-e^0]\n\nQ=(-1)/(6)(8)(4)(0.1353-1)=(-1)/(6)(8)(4)(-0.8647)\n\nQ=4.61\ mC$

Use Euler’s Method: ????????????????=???? ????????????????=−????????−????3+????cos(????) ????(0)=1.0 ????(0)=1.0 ????=0.4 ????=20.0 ℎ=0.01 ????=10000 ▪ Write the data (y1, y2) to a file named "LASTNAME_Prob1.dat" Example: If your name is John Doe – file name would be "DOE_Prob1.dat" ▪ Plot the result with lines using GNUPLOT (Hint: see lecture 08) ▪ Submit full code (copy and paste). Plot must be on a separate page. ▪ Run the code again and plot the result for: ????=0.1 ????=11.0

Answers

Answer:

Too many question marks

Explanation:

Assuming that the following three variables have already been declared, which variable will store a Boolean value after these statements are executed? choice = true;
again = "false";
result = 0;

a. choice
b. again
c. result
d. none of these are Boolean variables

Answers

Answer:

Explanation:

Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.

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