. Were you able to observe ???? = 0 in the circuit you constructed during lab? Why or why not? Hint: What value of resistance wo

Answer:

Explanation:

A liquid food containing 12% total solids is heated via steam injection at a pressure of 232.1 kPa (see Fig. E3.3). The product starts at a temperature of 50°C and has a flow rate of 100 kg/min, being elevated to a temperature of 120°C. The specific heat of the product varies with its composition as follows:

and the

specific heat of the product at 12% total solids is 3.936 kJ/(kg°C). The goal is to calculate the quantity and minimum quality of steam required to ensure that the leaving product has 10% total solids.

Given

Product total solids in () = 0.12

Product mass flow rate () = 100 kg/min

Product total solids out () = 0.1

Product temperature in () = 50°C

Product temperature out () = 120°C

Steam pressure = 232.1 kPa at () = 125°C

Product specific heat in () = 3.936 kJ/(kg°C)

The mass equation is:

Also

Therefore:

The energy balance equation is:

 kJ/(kg°C)

By substituting values into the energy equation:

From the properties of saturated steam at 232.1 kPa,

= 524.99 kJ/kg

= 2713.5 kJ/kg

% quality =

Any steam quality above 63.5% will result in higher total solids in the heated product.

Answer:

total expense for the new boiler = $229706.825

total expense for new boiler = $127512

Explanation:

provided information

initial power p1 = 80 kW

price C = $160000

cost index CI 1 = 187

cost index CI 2= 194

capacity factor f = 0.6

subsequent power p2 = 120 kW

present cost = $18000

to determine

total expense and cost for 40 kW

solution

we evaluate CN cost for the new boiler and CO cost for the existing boiler

where x represents the capacity of the new boiler

first we compute the current cost of the old boiler which is

current cost CO = C × .............1

substituting the value here

current cost = 160000 ×

updated current cost = $165989.304

and

employing power sizing strategy for 124 kW

CN/CO = ...............2

insert value and calculate CN

CN/CO =  

CN / 165989.304 =  

CN = 211706.825

therefore the new expense = $211706.825

hence

total expense for the new boiler amounts to

total expense = new expense + current cost

total exp = 211706.825 + 18000

boiler expense total = $229706.825

and

concerning a 40 kW unit, the calculated new cost will be

applying equation 2

CN/CO =

CN / 165989.304 =  

CN = 109512

thus the new cost is $109512

therefore

total expense for the new boiler is

total exp = new cost + current cost

total expense = 109512 + 18000

total cost for the new boiler = $127512

Answer:

Explanation:

Un dato importante: la división entera elimina la parte fraccionaria. Por lo tanto, el promedio de 8, 10, 5 y 4 se presenta como 6, no 6.75.

Observación: Las pruebas incluyen cuatro valores de entrada muy grandes cuyo producto causa desbordamiento. No necesitas hacer nada especial, solo ten en cuenta que la salida no representa el producto correcto (en realidad, el resultado de cuatro números positivos es negativo; sorprendente).

Envía lo anterior para evaluación. Tu programa no pasará las últimas pruebas (eso es normal) hasta que completes la segunda parte a continuación.

Parte 2: Además, se debe calcular e imprimir el producto y promedio usando aritmética de punto flotante.

Presenta cada número de punto flotante con tres dígitos después del punto decimal, lo cual puedes hacer así: System.out.printf("%.3f", tuValor);

Ejemplo: Si la entrada es 8, 10, 5, 4, la salida sería:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

double average_arith = (num1 + num2 + num3 + num4) / 4.0;

double product_arith = num1 * num2 * num3 * num4;

int result1 = (int) average_arith;

int result2 = (int) product_arith;

System.out.printf("%d %d\n", result2, result1);

System.out.printf("%.3f %.3f\n", product_arith, average_arith);

}

}

Expected output: 1600.000 6.750

Response:

a) 144.000 seconds

b) and c) Battery voltage and power graphs are in the attached image.

     where D:{0<t h="" />

d) 1620 J

Description:

a) The initial response is derived via a rule of three

b) Using the line equation from the starting point (0 seconds, 1.5 V)

where m denotes the slope.

where V represents voltage in volts and t signifies time in seconds

along with P and m.

d) By evaluating that count.