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2 months ago

estimate the area for a landfill for 12000 p producing waste for 10 y. assume that the national average is

Engineering

1 answer:

alex41 [359]2 months ago

4 0

Answer:

1.015 ha.

Explanation:

To calculate the landfill area required for 12,000 people producing waste over 10 years, follow these steps:[STEP ONE: Calculate the average solid waste generated per person per year (kg p^-1 ^y(kg/py)).

According to the problem, the average solid waste produced is 2.78 kg per person daily (kg/pd), hence converting to kg/py involves:

2.78 × 365 days = 1014.7 kg/py.

STEP TWO: Determine yearly volume of refuse per person.

Thus, volume = 1014.7 kg/py ÷ 500 kg/m^3 = 2.03 m^3 per person per year.

STEP THREE: Calculate total solid waste volume over 10 years for 12,000 individuals.

Total waste volume over 10 years = 10 × 12,000 × 2.03 = 243,600 m^3.

STEP FOUR: Find the required area for the landfill.

Note: The total height for the landfill should be 20 + 4 = 24m.

Thus, the area for the landfill = 243,600 m^3 / 24m = 10,150 m^2.

If 10,000 m^2 equals 1 ha, then 10,150 m^2 ÷ 10,000 m^2 = 1.015 ha.

(f). Ensure to expand the landfill area for enhancements.

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Steam flows at steady state through a converging, insulated nozzle, 25 cm long and with an inlet diameter of 5 cm. At the nozzle

iogann1982 [368]

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

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

Here, A = π*d² / 4

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

6 0

2 months ago

A hydrogen-filled balloon to be used in high altitude atmosphere studies will eventually be 100 ft in diameter. At 150,000 ft, t

mote1985 [299]

Answer:

The calculated result is 11.7 ft

Explanation:

You can apply the combined gas law, which incorporates Boyle's law, Charles's law, and Gay-Lussac's Law, because hydrogen demonstrates ideal gas behavior under these specific conditions.

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

where the subscripts indicate "p" for pressure, "V" for volume, and "T" for temperature (in Kelvin) at varying moments. Let's denote $t_1$ as the balloon at 150,000 ft so

$p_1 = 0.14 \ lb/in^2$

$V_1 = \frac{4}{3} \pi R_1^3 = 523598.77 \ ft^3$

and $T_1 = -67^\circ F = 218.15\ K$ .

Then $t_2$ represents the point at which the balloon is on the ground.

$p_2 = 14.7 \ lb/in^2$ and $T_2 = 68^\circ F = 293.15\ K$ .

Based on the first equation

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}$ , we find

$V_2 = 6701.07 ft^3$ and consequently the radius turns out to be

$R_2 = \sqrt[3]{\frac{3 V_2}{4 \pi}} = 11.7 \ ft$ .

5 0

2 months ago

In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini

mote1985 [299]

Answer:

provided below

Explanation:

8 0

2 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

2 months ago

Read 2 more answers