What is the answer???

A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055

Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

$K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}$

So the mean conductivity

$K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )$

$T_m=\dfrac{160+273+40+273}{2}$

$T_m=373 K$

$K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )$

$K_m=0.112 \frac{W}{mK}$

So heat conduction through cylinder

$Q=kA\dfrac{\Delta T}{L}$

$Q=0.112\times \pi \times 0.15^2\times 120$

Q=0.95 W/m

4 0

3 years ago

Will Give Brainliest

Tom [10]

Answer:

A hole within a hole

Explanation:

8 0

3 years ago

Write using about 10-15 lines for each of the six materials (metals, ceramics, glasses, polymers, composites, and semiconductors

Svetradugi [14.3K]

Answer:

See Answer below- Explanation is the entire answer

Explanation:

Metals:

Properties: Ductile, good heat conductivity, good electrical conductivity, high strength;

Drawbacks: Relatively high weight, reactive with oxygen to create oxides- corrosion is presented;

Examples: steel, aluminum alloys, brass, copper, titanium

Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

Ceramics:

Properties: Brittle, poor heat conductors, poor electrical conductors, high wear resistance, corrosion resistance;

Drawbacks: Deforms by fracturing, shock resistance is low, no conductivity of electricity;

Examples: concrete, tungsten carbide, diamond

Applications: bricks for constructions, clay pots to keep heat, cutting tools for metals;

Glasses:

Properties: amorphous, transparent, high weight

Drawbacks: poor conductors of heat and electricity; brittle; low shock resistance;

Examples: Silica, lead glass, glaze;

Applications: windows, protection screens;

Polymers:

Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

Drawbacks: low strength, low operating temperatures;

Examples: polyethylene, nylon, ABS-plastic, rubber;

Applications: toys, tires, insulation covers for the wires.

Composites:

Properties: high strength to weight ratio, can get combination of properties from the used materials, rarely conductive, good shock resistance;

Drawbacks: high cost, hard to recycle, expensive;

Examples: steel-reinforced concrete, carbon fiber, fiber glass, Nomex, sandwich roof panels;

Applications: buildings, bullet proof vests, body of the Formula 1 cars, rockets, roof panels.

Semiconductors:

Properties: brittle, change conductive behavior under certain scenario, poor heat conductors;

Drawbacks: hard to manufacture, expensive;

Examples: Silicon-based semiconductors, Germanium-based semiconductors, Ga-based semiconductors;

Applications: chips, LED, diodes, transistors, op-amps, microprocessors.

8 0

3 years ago

An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.

saul85 [17]

Answer:

a. $L_o$ = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, $L_o$ = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

$L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)$

For the school, $K_{LL}$ = 2

Therefore, we have;

$L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf$

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, $W_d$ = b × $W_{D + L}$ =

∴ $W_d$ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, $W_d$ = 812.8 ft./lb

d. For the uniformly distributed load, we have;

$V_{max}$ = 812.8 × 26/2 = 10566.4 lbs

$M_{max}$ = 812.8 × 26²/8 = 68,681.6 ft-lbs

$v_{max}$ = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

$P_L$ = 1000 lb

$W_{D}$ = 20 × 16 = 320 lb/ft.

$V_{max}$ = 1,000 + 320 × 26/2 = 5,160 lbs

$M_{max}$ = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

$v_{max}$ = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0

3 years ago

import java.util.Scanner; public class FindSpecialValue { public static void main (String [] args) { Scanner scnr = new Scanner(

Hitman42 [59]

Answer:

Java program explained below

Explanation:

FindSpecialNumber.java

import java.util.Scanner;

public class FindSpecialNumber {

public static void main(String[] args) {

//Declaring variable

int number;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//getting the input entered by the user

System.out.print("Enter a number :");

number = sc.nextInt();

/* Based on user entered number

* check whether it is special number or not

*/

if (number == -99 || number == 0 || number == 44) {

System.out.println("Special Number");

} else {

System.out.println("Not Special Number");

}

_______________

Output#1:

Enter a number :-99

Special Number

Output#2:

Enter a number :49

Not Special Number

7 0

3 years ago