1 At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power

Answer:

The velocity at exit U_2 is 578.359 m/s

The exit diameter d_e is 1.4924 cm

Explanation:

Provided data includes:

Length of the nozzle L = 25 cm

Inlet diameter d_i = 5 cm

At the nozzle entrance (state 1): Temperature T_1 = 325 °C, Pressure P_1 = 700 kPa, Velocity U_1 = 30 m/s, Enthalpy H_1 = 3112.5 kJ/kg, Volume V_1 = 388.61 cm³/gAt the nozzle exit (state 2): Temperature T_2 = 250 °C, Pressure P_2 = 350 kPa, Velocity U_2, Enthalpy H_2 = 2945.7 kJ/kg, Volume V_2 = 667.75 cm³/g To determine:a. Exit Velocity U_2b. Exit Diameter d_e

a.The Energy Equation can be represented by:ΔH + ΔU² / 2 + gΔz = Q + WAssuming Q = W = Δz = 0Substituting the values yields:

(H_2 - H_1) + (U²_2 - U²_1)  / 2 = 0From which we can derive U_2 = sqrt((2* (H_1 - H_2 )) + U²_1) with the calculations leading to U_2 = sqrt ( 2 * 10^3 * (3112.5 -2945.7) + 900) yielding U_2 = 578.359 m/s

b.

Using mass balance approach, we have U_1 * A_1 / V_1 = U_2 * A_2 / V_2

This leads to U_1 * d_i² / V_1 = U_2 * d_e² / V_2, thus d_e = d_i * sqrt((U_1 / U_2) * (V_2 / V_1)). Hence, d_e = 5 * sqrt((30 / 578.359) * (667.75 / 388.61)) computes to d_e = 1.4924 cm

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 = ................1

And ............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

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % =  × 100

dV % =  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = ................3

Final volume v2 = ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  

dV =

Substituting all values yields

0.0130 =

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

where D refers to diameter, L refers to length and f refers to feed while V represents speed

Tc =

Tc =

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 ×

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

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