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zavuch27
2 months ago
9

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

what will be the percentage change in its volume? If the vat has a diameter of 2 m, how much will the water level rise due to this temperature increase?
Engineering
1 answer:
Mrrafil [318]2 months ago
8 0

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = -\frac{dp}{dV/V}................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = \frac{\pi }{4} *d^2*l(i)................3

Final volume v2 = \frac{\pi }{4} *d^2*l(f)................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

Substituting all values yields

0.0130 = \frac{\pi }{4} *2^2*dl

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

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Answer:

Here is the JAVA program:

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed to 4 assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two arrays origList[] and offsetAmount[] get assigned their values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // variable for storing the product results

for(i = 0; i <= origList.length - 1; i++){

/* iterates from 0 to the end of origList */

/* multiplies each origList entry with the corresponding offsetAmount entry, stores results in product */

product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }

System.out.println(product); }}

Output:

80 180 80 400

Explanation:

If you wish to print the product of origList alongside offsetAmount values vertically, this can be done in this manner:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

origList[i] *= offsetAmount[i];

System.out.println(origList[i]);}

}}

Output:

80

180

80

400

The program is shown with the output as a screenshot along with the example's input.

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