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3 years ago

Explain the difference in the heat transfer modes of conduction and convection.

Engineering

1 answer:

LiRa [457]3 years ago

4 0

Explanation:

Conduction is the heat transfer through a stationary matter by the physical contact.

For example, the heat transferred between electric burner of stove and bottom of pan is transferred by the process of conduction.

Convection is heat transfer by macroscopic movement of the fluid. The particles of the fluid carry the current within themselves.

For example, the water is the pot is warmed overall by heat transferred by the process of convection.

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In a lab, scientists grew several generations of offspring of a plant using the method shown. What conclusion can you make about

exis [7]

Answer: c) they have low genetic variability among them.

When a plant is grown for several generations of offspring of a plant, then there are some common things which are to be noted which are found similar in the offspring and in the parent of the offspring. The flowers and fruits and the time or season they come in are absolutely the same.

6 0

1 year ago

What would the Select lines need to be to send data for the fifth bit in an 8-bit system (S0 being the MSB and S2 being the LSB)

Maurinko [17]

Answer:

A. S0 = 1, S1 = 0, S2 = 0

lines need to send data for the fifth bit in an 8 bit system

5 0

3 years ago

An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t

IRINA_888 [86]

Answer:

a) 28 stations

b) Rp = 21.43

E = 0.5

Explanation:

Given:

Average downtime per occurrence = 5.0 min

Probability that leads to downtime, d= 0.01

Total work time, Tc = 39.2 min

a) For the optimum number of stations on the line that will maximize production rate.

Maximizing Rp =minimizing Tp

Tp = Tc + Ftd

$= \frac{39.2}{n} + (n * 0.01 * 5.0)$

$= \frac{39.2}{n} + (n * 0.05)$

At minimum pt. = 0, we have:

dTp/dn = 0

$= \frac{-39.2}{n^2} + 0.05 = 0$

Solving for n²:

$n^2 = \frac{39.2}{0.05} = 784$

$n = \sqrt{784} = 28$

The optimum number of stations on the line that will maximize production rate is 28 stations.

b) $Tp = \frac{39.2}{28} + (28 * 0.01 * 5)$

Tp = 1.4 +1.4 = 2.8

The production rate, Rp =

$\frac{60min}{2.8} = 21.43$

The proportion uptime,

$E = \frac{1.4}{2.8} = 0.5$

3 0

3 years ago

Ai r is compressed by a 30-kW compressor from P1 to P2. The air t emperature i s maintained constant at 25oC during thi s proces

RUDIKE [14]

Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

We'll assume the following

1. It is a steady-flow process;

2. The changes in the kinetic energy and the potential energy are negligible;

3. Lastly, the air is an ideal gas

Energy balance will be required to calculate heat loss;

mh1 + W = mh2 + Q where W = Q.

Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;

Rate of Entropy Change = -Q/T

Where Q = 30Kw

T = Temperature of air = 25°C = 298K

Rate = -30/298

Rate = -0.100671140939597 KW/K

Rate = -0.10067kW/K

Hence, the rate of entropy change of the air is -0.10067kW/K

3 0

3 years ago

How to find the voltage(B Aab) in series parallel circuit?

Sindrei [870]

Answer:

Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

(1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

(V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

4 0

3 years ago