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adell
21 day ago
7

In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in

5.0 min. The piece is 400 mm long and 150 mm in diameter. Using a feed = 0.30 mm/rev and a depth of cut = 4.0 mm, what cutting speed must be used to meet this machining time requirement?
Engineering
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The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re
mote1985 [299]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, g = 9.81 m/s^2

mass flow rate, \dot{m} = 60.6 kg/s

flow density, \rho = 1000 kg/m^3

efficiency of the pump, \eta = 0.74

output gauge pressure, p_o = 344.75 kPa

input gauge pressure, p_i = 68.95 kPa

cross-sectional area of output pipe, A_o = 0.069 m^2

cross-sectional area of input pipe, A_i = 0.093 m^2

height of discharge, z_o = 1.22 m - 0.61 m = 0.61 m (evaluated at pump’s maximum height of 1.22 m)

input height, z_i = 0 m

hydraulic power of the pump,P =? kW

Initially, the volumetric flow (Q) needs to be determined

Q = \frac{\dot{m}}{\rho}

Q = \frac{60.6 kg/s}{1000 kg/m^3}

Q = 0.0606 m^3/s

Next, compute the velocity (v) for both input and output

v_o = \frac{Q}{A_o}

v_o = \frac{0.0606 m^3/s}{0.069 m^2}

v_o = 0.88 m/s

v_i = \frac{Q}{A_i}

v_i = \frac{0.0606 m^3/s}{0.093 m^2}

v_i = 0.65 m/s

Subsequently, the total head (H) can be calculated

H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}

H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}

H = 28.74m

Finally, the computation of pump power is done as follows

P = \frac{Q \, \rho \, g \, H}{\eta}

P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}

P = 23.09 kW

6 0
3 months ago
Water flows steadily through a horizontal nozzle, discharging to the atmosphere. At the nozzle inlet the diameter is D1; at the
mote1985 [299]
The inlet gauge pressure must be 61.627 Psi.
4 0
3 months ago
A spherical hot-air balloon is initially filled with air at 120 kPa and 20°C with an initial diameter of 5 m. Air enters this ba
mote1985 [299]

Answer:

The duration is 17.43 minutes.

Explanation:

Based on the provided information, the initial diameter is 5 m

the velocity is 3 m/s

and the final diameter is 17 m.

To find the solution, we will use the volume change equation expressed as

ΔV = \frac{4}{3} \pi * (rf)^3 - \frac{4}{3} \pi * (ri)^3.............1

where ΔV represents the change in volume, rf is the final radius, and ri is the initial radius.

Calculating ΔV yields

ΔV = \frac{4}{3} \pi * (8.5)^3 - \frac{4}{3} \pi * (2.5)^3

ΔV = 2507 m³.

Thus,

Q = velocity × Area

Q = 3 × π ×(0.5)² = 2.356 m³/s.

Next, the change in time can be expressed as

Δt = \frac{\Delta V}{Q}

Δt = \frac{2507}{2.356}

Δt = 1046 seconds.

Therefore, the total change in time amounts to 17.43 minutes.

8 0
3 months ago
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
8. When supplying heated air for a building, one often chooses to mix in some fresh outside air with air that has been heated fr
pantera1 [306]
Refer to the explanation for a detailed breakdown.
5 0
2 months ago
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