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3 years ago

How does heat conduction differ from convection?

Engineering

2 answers:

Helga [31]3 years ago

7 0

Explanation:

Conduction:

Heat transfer in the conduction occurs due to movement of molecule or we can say that due to movement of electrons in the two end of same the body. Generally, phenomenon of conduction happens in the case of solid . In conduction heat transfer takes places due to direct contact of two bodies.

Convection:

In convection heat transfer of fluid takes place due to density difference .In simple words we can say that heat transfer occur due to motion of fluid.

Kaylis [27]3 years ago

4 0

Answer:

Conduction occurs in solid whereas convection occurs in fluids.

Explanation:

<u>Conduction</u>

1. Conduction is a process of transfer of heat energy by direct contact.

2. It occurs in solids through molecular collisions of matter.

3. Conduction is due to temperature difference in the body.

4. It is a slow process.

5. Conduction process needs matter for heat transfer but does not require bulk motion of matter.

6. The rate of heat transfer in conduction is governed by Fourier's Law of Conduction.

<u>Convection</u>

1. Convection is a process of heat energy transfer by actual motion of matter.

2. It occurs in fluids by actual flow of matter.

3.Convection is due to density difference.

4. Convection is also a slow process.

5. It does not require any bulk transport of matter.

6. The rate of heat transfer is governed by Newton's Law of Cooling.

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A force is specified by the vector F= 160i + 80j + 120k N. Calculate the angles made by F with the positive x-, y-, and z-axis.

Rashid [163]

Answer:

1) Angle with x-axis = 42.03 degrees

2) Angle with y-axis =68.2 degrees

3) Angle with z-axis = 56.14 degrees

Explanation:

given any vector $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$

and any x axis the angle between them is given by

$\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}$

Angle between the vector and y axis is given by

$\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}$

Similarly angle between z axis and the vector is given by

$\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}$

5 0

4 years ago

Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a

guapka [62]

Answer:

$K_v=12.34$

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo = Discharge / area of the orifice

Vo = $\frac{0.01}{\frac{\pi}{4}0.045^2}$

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, $V_v$ = Discharge / area of the orifice

$V_v$ = $\frac{0.01}{\frac{\pi}{4}0.0675^2}$

$V_v$ = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

25 m = $K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}$

where,

$K_v$ is the loss coefficient for the gate valve

on substituting the values, we get

25 m = $10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}$

$K_v\frac{2.79^2}{2\times9.81}$ = 4.898

$K_v=12.34$

3 0

4 years ago

declare integer product declare integer number product = 0 do while product < 100 display ""Type your number"" input number p

Brilliant_brown [7]

Full Question

1. Correct the following code and

2. Convert the do while loop the following code to a while loop

declare integer product

declare integer number

product = 0

do while product < 100

display ""Type your number""

input number

product = number * 10

loop

display product

End While

Answer:

1. Code Correction

The errors in the code segment are:

a. The use of do while on line 4

You either use do or while product < 100

b. The use of double "" as open and end quotes for the string literal on line 5

c. The use of "loop" statement on line 7

The correction of the code segment is as follows:

declare integer product

declare integer number

product = 0

while product < 100

display "Type your number"

input number

product = number * 10

display product

End While

2. The same code segment using a do-while statement

declare integer product

declare integer number

product = 0

display "Type your number"

input number

product = number * 10

display product

while product < 100

4 0

4 years ago

HEY Y'ALL!!! WILL GIVE BRAINLEST AND THANKS+A LOT OF POINTS!!!!! So in class our teacher wants to help us think out of the box m

Elden [556K]

Answer:

2) phone

3) snake

Explanation:

NO EXPLANATION

5 0

3 years ago

Read 2 more answers

An insulated piston–cylinder device initially contains 1 m3 of air at 120 kPa and 17°C. Air is now heated for 15 min by a 200-W

irakobra [83]

Answer:

∆S1 = 0.5166kJ/K

∆S2 = 0.51826kJ/K

Explanation:

Check attachment for solution

4 0

3 years ago