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2 years ago

What colour is best for radiative heat transfer? a. Black b. Brown c. Blue d. White

Engineering

1 answer:

GarryVolchara [31]2 years ago

8 0

Answer:

The correct answer is option 'a': Black

Explanation:

As we know that for an object which is black in color it absorbs all the electromagnetic radiation's that are incident on it. Thus if we need to transfer energy to an object by radiation the most suitable color for the process is black.

In contrast to black color white color is an excellent reflector, reflecting all the incident radiation that may be incident on it hence is the least suitable material for radiative heat transfer.

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Expalin the application of diesel cycle in detail.

mars1129 [50]

Explanation:

Diesel cycle:

All diesel engine work on diesel cycle .In diesel cycle there are four process .These processes are as follows

1. Adiabatic reversible compression

2.Heat addition at constant pressure

3.Adiabatic reversible expansion

4.Constant volume heat rejection

In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.

Applications of diesel cycle:

Generally diesel cycle used for heavy vehicle or equipment because heavy vehicle or equipment is required high initial torque.So this cycle have lots of applications such as in industrial machining,in trucks,power plant,in mining ,in defense or military,large motors ,compressor and pump etc.

5 0

2 years ago

Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather m

Alchen [17]

Even though the content of many alcohol blends doesn't affect engine driveability, using gasoline with alcohol in warm weather may cause: decrease in fuel economy.

Mark brainliest
s

3 0

1 year ago

Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Repor

REY [17]

Answer:

$T_2=315.69k$

Explanation:

Initial Temperature $T_1=500K$

Initial Pressure $P_1=1000kPa$

Final Pressure $P_2=200kPa$

Generally the gas equation is mathematically given by

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}$

Where

n for $CO=1.4$

Therefore

$\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}$

$T_2=315.69k$

7 0

2 years ago

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

Klio2033 [76]

Answer:

A) ΔS_refrigerant = 0.70754 Kj/K

B) ΔS_space = -0.68441 Kj/K

C) ΔS_total = 0.02313 Kj/K

Explanation:

A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C

Converting to degree kelvin yields;

T = -18.77 + 273 = 255.23 K

Formula for entropy change of refrigerant is given as;

ΔS_refrigerant = Q_in/T_refrigerant

We are given Q = 180 KJ

Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K

B) Formula for entropy change of cooled space is given as;

ΔS_space = Q_out/T_s pace

T_space = -10°C = 273 - 10 = 263K

Thus, ΔS_space = -180/263 = -0.68441 Kj/K

C) the total entropy change would be;

ΔS_total = ΔS_refrigerant + ΔS_space

Thus,

ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K

8 0

2 years ago

A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked mater

Anestetic [448]

Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

$A_1 \ and \ A_2$

Thus, using the summation rule, the view factor $F_{11$ and $F_{12$ is as follows:

$F_{11}+F_{12}=1$

Let assume the surface (1) is flat, the $F_{11} = 0$

Now:

$0+F_{12}=1$

$F_{12}=1$

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder $A_2$ to the flat base surface $A_1$ ; we have:

$A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}$

Suppose, we replace DL for $A_1$ and

$A_2$ = $\dfrac{\pi D}{2}$

Then:

$F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64$

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

$Q_{21} = Q_{evaporation}$

But, before that; let's find the radian heat exchange occurring among the dome and the flat base surface:

$Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)$

where;

$\sigma = Stefan \ Boltzmann's \ constant$

$T_1 = base \ temperature$

$T_2 = temperature \ of \ the \ dome$

∴

$Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m$

Recall the energy balance formula;

$Q_{21} = Q_{evaporation}$

where;

$Q_{evaporation} = mh_{fg}$

here;

$h_{fg}$ = enthalpy of vaporization

m = the water mass flow rate

∴

$83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}$

6 0

2 years ago