An overhead 25m-long, uninsulated industrial steam pipe of 100-mm diameter, is routed through a building whose walls and air are

Answer:

a) v₂ = 26.6 ft/s

b) v₂ = 31.9 ft/s

Explanation:

a) Applying the principle of energy conservation for the soccer arm from the point of impact to its lowest point:

The ball's velocity in the tangential direction is given by:

v₂*sinθ = v₁*sin30

With the values:

if v₁ = 0

θ = 0º

The restitution coefficient is:

where O=θ

The aggregate momentum equation is:

, where O = θ

When we incorporate gravitational acceleration:

Isolating v₂ results in:

v₂ = 26.6 ft/s

b) To find the ball's velocity, we utilize:

v₂ * sinθ = v₁ * sin30

if v₁ = 10 ft/s

then v₂ * sinθ = 10 * sin30

thus sinθ = 5/v₂

The restitution coefficient remains:

where O = θ

Solving for v₂ yields:

v₂ = 31.9 ft/s

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the fin in question, L = 0.15 m

- The fin's surface temperature, Ts = 250°C

- The velocity of free stream air, U = 80 km/h

- The air temperature, Ta = 27°C

- The flow is parallel over both sides of the fin, assuming turbulent flow conditions throughout.

Find:-

What is the heat removal rate per unit width of the fin?

Solution:-

- Steady state conditions are assumed, along with negligible radiation and turbulent flow conditions.

- From Table A-4, we gather air properties (T = 412 K, P = 1 atm ):

    Dynamic viscosity, v = 27.85 * 10^-6 m²/s  

    Thermal conductivity, k = 0.0346 W / m.K

    Prandtl number Pr = 0.69

- Compute the Nusselt Number (Nu) corresponding to - turbulent conditions - using the relevant relationship as follows:

Where,    Re_L: Average Reynolds number for the entire length of fin:

- The convective heat transfer coefficient (h) can now be derived from:

- The heat loss rate q' per unit width can be established using the convection heat transfer formula and should be multiplied by (x2) since the airflow is present on both sides of the fin:

- Ultimately, the heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The thermal loss per unit width (q') attributed to radiation:

Where, a signifies the Stefan-Boltzmann constant = 5.67*10^-8

- It is observed that radiation losses are not insignificant, accounting for 20% of thermal loss by convection. As the emissivity (e) of the fin is unspecified, this value is dismissed from the calculations as it pertains to the provided information.

Answer:

Combustion

Explanation:

This relates to an internal-combustion engine.

Answer:

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Explanation:

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Response:

Refer to the explanation

Clarification:

Code:

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void displayNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

displayNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

displayNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

displayNumPattern(num1, num2);

}

}

See attached example output