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2 months ago

6

A long, horizontal, pressurized hot water pipe of 15cm diameter passes through a room where the air temperature is 24degree C. T

he pipe surface temperature is 130 degree C and the emissivity of the surface is 0.95. Neglecting radiation loss from the pipe, what is the rate of the heat transfer to the room air per meter of pipe length

1 answer:

Daniel [329]2 months ago

5 0

The heat transfer rate to the surrounding air per meter of pipe length is quantified as 521.99 W/m. Given the negligible radiation losses from the pipe, convection remains the sole method of heat transfer. The rate of heat transfer via convection can be defined as such, using the specified heat transfer coefficient of 10.45 for air and calculating the surface area of the pipe.

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A heating cable is embedded in a concrete slab for snow melting. The heating cable is heated electrically with joule heating to

Daniel [329]

Answer:

18.7 mm

Explanation:

The law of heat conduction as described by Fourier for a plate is:

q = -k/t * ΔT

In this equation

q: heat transfer per unit time and area

k: coefficient of thermal conductivity

t: plate's thickness

ΔT: difference in temperature

By rearranging the terms, we find:

t = -k/q * ΔT

t = -1.4/(-1200) * 16 = 0.0187 m = 18.7 mm (the negative value of q indicates heat is being released from the plate)

Thus, the concrete slab can have a maximum thickness of 18.7 mm.

6 0

1 month ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

Kisachek [356]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The maximum permitted number of users in the system equals:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) Each cell's side length equals:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

The least duration for moving across a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum duration for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

Approximately, the time is calculated as the mean of the minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The maximum allowed number of users in the system totals:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each cell's side is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

The shortest time to traverse a cell measures:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The longest time to traverse a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is calculated by averaging the minimum and maximum durations:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

1 month ago

3.24 Program: Drawing a half arrow (Java) This program outputs a downwards facing arrow composed of a rectangle and a right tria

alex41 [359]

Answer:

This is the JAVA code:

import java.util.Scanner; // to capture user input

public class DrawHalfArrow{ // class declaration

public static void main(String[] args) { // main function starts here

Scanner scnr = new Scanner(System.in); //input reading

int arrowBaseHeight = 0; // variable for arrow base height

int arrowBaseWidth = 0; // variable for arrow base width

int arrowHeadWidth = 0; // variable for arrow head width

// prompts user for height, base width, and head width

System.out.println("Enter arrow base height: ");

arrowBaseHeight = scnr.nextInt(); // reads integer input

System.out.println("Enter arrow base width: ");

arrowBaseWidth = scnr.nextInt();

/* loop for asking user for arrow head width until it exceeds base width */

while (arrowHeadWidth <= arrowBaseWidth) {

System.out.println("Enter arrow head width: ");

arrowHeadWidth = scnr.nextInt(); }

//begin nested loop for base

//outer loop for outputting based on arrow base height

for (int i = 0; i < arrowBaseHeight; i++) {

//inner loop produces the stars

for (int j = 0; j <arrowBaseWidth; j++) {

System.out.print("*"); } //prints stars

System.out.println(); }

//used for the width of the arrowhead

int k = arrowHeadWidth;

//outer loop for the head height

for (int i = 1; i <= arrowHeadWidth; i++)

{ for(int j = k; j > 0; j--) {//inner loop for stars

System.out.print("*"); } //print stars

k = k - 1;

System.out.println(); } } } //marks end of asterisks for a new line

Explanation:

The code prompts for arrow dimensions, including height, base width, and head width. The program verifies that the head width must be larger than the base width through a condition. This loop instigates until the user provides a valid head width.

The loop manages the output of the arrow base corresponding to the specified height. Thus point (1) is fulfilled.

A nested loop is utilized to present an arrow base matching the defined width. The inner loop creates stars, establishing the width for the base, while the outer loop iterates according to the specified height. This satisfies point (2).

A temporary variable, k, saves the initial arrow head width for necessary adjustments.

The final nested loop outputs an arrow head corresponding to the specified width with stars. Therefore, point (3) is achieved.

The value of k decreases by 1 for each execution of the nested loop, ensuring each subsequent line contains fewer stars.

The output screenshot is attached.

3 0

2 months ago

cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge wit

grin007 [323]

Answer:

final temperature = 26.5°

Explanation:

The water's initial volume is calculated as 1 x 1 x 1 = 1 $m^{3}$

The starting temperature of the water is 20° C

Water density = 1000 kg/ $m^{3}$

Volume of the copper block is determined as 0.46 x 0.46 x 0.46 = 0.097 $m^{3}$

The copper block's initial temperature is 100° C

Copper density = 8960 kg/ $m^{3}$

The water's final volume after accounting for the copper block is 1 - 0.097 = 0.903 $m^{3}$

Assumptions:

the tank is adiabatic, meaning there’s no heat exchange through its walls
the tank is completely filled, lacking space for any air to cool the water
the total thermal energy in the tank is comprised of both the water’s heat energy and that of the copper block.

The mass of the remaining water in the tank can be calculated using density x volume = 1000 x 0.903 = 903 kg

The specific heat capacity for water, c = 4186 J/K-kg

Total heat content of the water Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules

The copper's mass is calculated as density x volume = 8960 x 0.097 = 869.12 kg

Copper's specific heat capacity is 385 J/K-kg

The heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules

The overall heat in the system totals up to 75.59 + 33.46 = 109.05 Mega-joules

This heat will be evenly dispersed across the system

The heat energy of the water in the system is expressed as mcT

where T signifies the final temperature

= 903 x 4186 x T = 3779958T

For copper, the heat will be

mcT = 869.12 x 385 = 334611.2T

The combined heat from both will equal the total heat of the system, meaning

3779958T + 334611.2T = 109.05 x $10^{6}$

4114569.2T = 109.05 x $10^{6}$

Thus, the final temperature T = (109.05 x $10^{6}$ )/4114569.2 = 26.5°

3 0

2 months ago

A test machine that kicks soccer balls has a 5-lb simulated foot attached to the end of a 6-ft long pendulum arm of negligible m

choli [298]

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:

$mgh=\frac{1}{2} mv^{2} \\v=\sqrt{2gh} =\sqrt{2*32.2*6} =19.66ft/s$

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:

$e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{1}*cos30 )} \\0.8=\frac{v_{2}cos0-v_{2}*cos30 }{19.66*cos30-(0*cos30)} \\v_{2} =\frac{v_{2}-13.6 }{cos30}$

where O=θ

The aggregate momentum equation is:

$mv-mv_{1} =mv_{2} +mv_{2} cos(O+30)\\$ , where O = θ

$mv-mv_{1} =m(\frac{v_{2}-13.6 }{cos30} )+mv_{2} cos(O+30)$

When we incorporate gravitational acceleration:

$mgv-mgv_{1} =mg(\frac{v_{2}-13.6 }{cos30} )+mgv_{2} cos(O+30)\\5*19.66-1*0=5*(\frac{v_{2}-13.6 }{cos30} )+1*v_{2} cos(O+30)$

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:

$e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{2}*cos30) } \\0.8=\frac{v_{2}cosO-v_{2}cos30 }{19.66*cos30-(-10*cos30)} \\v_{2} =\frac{v_{2}cos30-20.55}{cos30}$

where O = θ

$mgv-mgv_{1} =mgv_{2} +mgv_{2} cos(O+30)\\5*19.66-1*0=5v_{2} +v_{2} cos(O+30)\\98.3=5v_{2}+v_{2} cosOcos30-v_{2} sinOsin30$

Solving for v₂ yields:

v₂ = 31.9 ft/s

6 0

2 months ago