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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:

Conduction

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.

Convection

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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Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b

svetlana [45]

complete question:

attached

Answer:

2356.11 W/m^2

6100.11 W/m^2

Explanation:

Assumptions:

1. Steady-state conditions.

2. The cake is placed in a large surrounding.

3. Heat flux delivered to the cake is due to convection and radiation.

Case 1

Since convection feature is disabled the mode of heat transfer associated with this situation is through free convection and radiation.

q''(free) = [q''(free convection+q''(radiation) ]W/m^2

= h_free(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

= 3 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -

(24+273K)^4 ]

= 468 +1881.11

= 2356.11 W/m^2

Case 2

Since convection feature is enabled or activated the mode of heat transfer associated with this situation is through forced convection and radiation.

q''(free) = [q''(forced convection+q''(radiation) ]W/m^2

= h_forced(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

= 27 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -

(24+273K)^4 ]

= 4212 +1881.11

= 6100.11 W/m^2

1. The total heat flux is is 2.58 times higher when the convection feature is activated. Therefore the cake will bake faster during this condition.

2. The contribution of convection heat flux under natural(free) convection is very low as compared to the contribution during forced convection.

3. The heat transfer due to radiation is same in both the cases.

4. Only 19.9 % of the total heat flux is contributed by free convection in the first case.

5. In the second case 69 % of the total heat flux is contributed by forced convection.

5 0

3 years ago

The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k

Vesnalui [34]

<h3>CSMA/CD Protocol: </h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

Example:

$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

Efficiency of CSMA/CD:

In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

7 0

3 years ago

A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

Part 2

For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

And we can quantify the decrease using the relative change:

$\% Change = \frac{5s}{55 s} *100 = 9.09\%$ of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

8 0

3 years ago

Andreyy89

Answer:

Explanation:

5 + 2 = 7

4 0

2 years ago

There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter

weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

u_e = 0.5*e_o * E^2

- Plug in the values given:

u_e = 0.5*8.854 *10^-12 *145^2

u_e = 9.30777 * 10^-8 J/m^3

5 0

3 years ago