How much profit does the fossil fuel industry make

What must engineers keep in mind so that their solutions will be appropriate?

vekshin1

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

7 0

3 years ago

Write an expression that evaluates to true if the value of variable lastName is greater than the string Dexter.

gregori [183]

Answer:

lastName.compareTo("Dexter")>0

Explanation:

The expression that evaluates to true if the value of variable lastName is greater than the string Dexter is; lastName.compareTo("Dexter")>0.

4 0

3 years ago

A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m² of water vapor in equilibrium at 800 kPa. Heat is transferre

docker41 [41]

Answer:

Initial temperature = 170. 414 °C

Total mass = 94.478 Kg

Final volumen = 33.1181 m^3

Diagram = see picture.

Explanation:

We can consider this system as a close system, because there is not information about any output or input of water, so the mass in the system is constant.

The information tells us that the system is in equilibrium with two phases: liquid and steam. When a system is a two phases region (equilibrium) the temperature and pressure keep constant until the change is completed (either condensation or evaporation). Since we know that we are in a two-phase region and we know the pressure of the system, we can check the thermodynamics tables to know the temperature, because there is a unique temperature in which with this pressure (800 kPa) the system can be in two-phases region (reach the equilibrium condition).

For water in equilibrium at 800 kPa the temperature of saturation is 170.414 °C which is the initial temperature of the system.

to calculate the total mass of the system, we need to estimate the mass of steam and liquid water and add them. To get these values we use the specific volume for both, liquid and steam for the initial condition. We can get them from the thermodynamics tables.

For the condition of 800 kPa and 170.414 °C using the thermodynamics tables we get:

Vg (Specific Volume of Saturated Steam) = 0.240328 m^3/kg

Vf (Specific Volume of Saturated Liquid) = 0.00111479 m^3/kg

if you divide the volume of liquid and steam provided in the statement by the specific volume of saturated liquid and steam, we can obtain the value of mass of vapor and liquid in the system.

Steam mass = *0.9 m^3 / 0.240328 m^3/kg = 3.74488 Kg

Liquid mass = 0.1 m^3 /0.00111479 m^3/kg = 89.70299 Kg

Total mass of the system = 3.74488 Kg + 89.70299 Kg = 93,4478 Kg

If we keep the pressure constant increasing the temperature the system will experience a phase-change (see the diagram) going from two-phase region to superheated steam. When we check for properties for the condition of P= 800 kPa and T= 350°C we see that is in the region of superheated steam, so we don’t have liquid water in this condition.

If we want to get the final volume of the water (steam) in the system, we need to get the specific volume for this condition from the thermodynamics tables.

Specific Volume of Superheated Steam at 800 kPa and 350°C = 0.354411 m^3/kg

We already know that this a close system so the mass in it keeps constant during the process.

If we multiply the mass of the system by the specific volume in the final condition, we can get the final volume for the system.

Final volume = 93.4478 Kg * 0.354411 m^3/kg = 33.1189 m^3

You can the P-v diagram for this system in the picture.

For the initial condition you can calculate the quality of the steam (measure of the proportion of steam on the mixture) to see how far the point is from for the condition on all the mix is steam. Is a value between 0 and 1, where 0 is saturated liquid and 1 is saturated steam.

Quality of steam = mass of steam / total mass of the system

Quality of steam = 3.74488 Kg /93.4478 Kg = 0,040 this value is usually present as a percentage so is 4%.

Since this a low value we can say that we are very close the saturated liquid point in the diagram.

6 0

4 years ago

For this homework you are writing a main function, with 3 subfunctions. 2. The "main" function is to be named arm_prop a. Inputs

ra1l [238]

Answer:

MATLAB CODE:

clc;

clear all;

x=[1 2 3 4]; %Input vector of variable size

[gm,ra,hm]=arm_prop(x) %Main function calling , output values are hm, ra, hm and imput is x

%% Main function

function [gm,ra,hm]=arm_prop(x)

[gm]=geo_mean(x); %Calling Sub function 1

[ra]=rms_avg(x); %Calling Sub function 2

[hm]=har_mean(x); %Calling Sub function 3

%Sub function 2 =Rms average

function [gm]=geo_mean(x)

n=numel(x); %calculating number of elemnts in vector x

temp=1; %Temporary value for multiplying each and every value of x

for i=1:1:n % For loop run for number of elements in x

temp=temp*x(i); %Multliplyng each and every element of x

end

gm=temp^(1/n); % Calculating geometric mean

end

%Sub function 2 =Rms average

function [ra]=rms_avg(x)

temp=0; %Temporary value for adding square of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+x(i)^2; %Adding square of each and every element of x

end

ra=(temp/N)^(1/2); % Calculating Rms average

end

%Sub function 3 =Harmonic mean

function [hm]=har_mean(x)

temp=0; %Temporary value for adding reciprocal of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+(1/x(i)); %Adding sreciprocal of each and every element of x

end

hm=N/temp; % Halculating Harmonic mean

end

Results:

gm =

2.2134

ra =

2.7386

hm =

1.9200

Here in one main function. 3 sub functions are written. Every sub function returns one output. As all 3 sub functions are inside main function, main function returns 3 output values gm, ra, hm.

Explanation:

3 0

4 years ago

A platform and a weight totaling 900 N is held in position by a vertical water jet of 5 cm diameter. Determine the velocity of t

BlackZzzverrR [31]

Answer:

v = 21.409 m/s

Explanation:

Given data:

Total weight of the platform and weight together, W = 900 N

Diameter of the water jet = 5 cm = 0.05 m

Now,

the force exerted by the water jet is balancing the platform

also,

Force exerted by the water jet = ρAv²

where,

ρ is the density of the water = 1000 kg/m³

A is the area of the outlet of the jet = $\frac{\pi}{4}d^2=\frac{\pi}{4}(0.05)^2$

or

A = 0.00196 m²

v is the velocity of the jet

thus,

W = ρAv²

or

900 = 1000 × 0.00196 × v²

or

v = 21.409 m/s

Hence, the velocity of the jet is 21.409 m/s

7 0

3 years ago