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20 days ago

A converging–diverging nozzle is designed to produce a Mach number of 2.5 with air. (a) What operating pressure ratio (prec/pt i

nlet) will cause this nozzle to operate at the first, second, and third critical points? (b) If the inlet stagnation pressure is 150 psia, what receiver pressures represent operation at these critical points? (c) Suppose that the receiver pressure were fixed at 15 psia. What inlet pressures are necessary to cause operation at the critical points?

Engineering

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2.31 LAB: Simple statistics Part 1 Given 4 integers, output their product and their average, using integer arithmetic. Ex: If th

iogann1982 [368]

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

5 0

3 months ago

Read 2 more answers

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val

grin007 [323]

Answer: a) U2 = 164.737kJ/kg

b) Q2 = 22.6kJ

Explanation: see attachment below

6 0

2 months ago

(a) For BCC iron, calculate the diameter of the minimum space available in an octahedral site at the center of the (010) plane,

choli [298]

Response:

a) The diameter available is 0.0384 nm

b)This space is less than the size of a carbon atom, which has a radius of 0.077 nm, indicating that the carbon atom won't occupy these sites.

Clarification:

For BCC iron

According to the information in Appendix B, the lattice parameter (a) is determined to be 0.2866 nm

BCC iron encompasses 4 atomic radii, thus the body diagonal length = $a(3)^\frac{1}{2}$

which represents the atomic radius of BCC iron

4r = $a(3)^\frac{1}{2}$

Substituting the value of (a) from Appendix B, set as 0.2866 nm

4r = $0.2866 nm (3)^\frac{1}{2}$

leading to r = 0.4964 nm / 4 = 0.1241 nm

Refer back to Appendix C, where the atomic radius of BCC iron is stated as 0.1241 nm, assuming the atomic sizes for iron remain consistent.

Thus, the radius ratio = 0.62

According to Figure 3.2, the space necessary for an interstitial at the BCC site exists between atoms located at the FCC site, containing two atoms, each equal to a radius of 0.2482 nm

The diameter of the minimum space available

$d_{a} = a - r_{a}$

$r_{a} = atomic radii$ = 0.2482 nm

With a = 0.2666 nm

therefore

d $_{a}$ = 0.2866 nm - 0.2482 nm = 0.0384 nm

When comparing with the diameter of a carbon atom

This space is smaller than that of a carbon atom which has a radius of 0.077 nm, confirming that the carbon atom will not be able to occupy these positions.

7 0

3 months ago