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Nata [24]
3 years ago
9

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

ials to be dried. The base is maintained at a temperature of 370K, while the dome of the dryer is maintained at 1000 K. If both surfaces behave as blackbody, determine the drying rate per unit length experienced by the wet materials.
Engineering
1 answer:
Anestetic [448]3 years ago
6 0

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}

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Whitepunk [10]

Answer:

I don't really know but i have some info for you...

Explanation:

The cold forging manufacturing process increases the strength of a metal through strain hardening at a room temperature. On the contrary the hot forging manufacturing process keeps materials from strain hardening at high temperature, which results in optimum yield strength, low hardness and high ductility.

7 0
3 years ago
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A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s
grandymaker [24]

Answer:

the endurance strength  S_e = 421.24  MPa

Explanation:

From the given information; The objective is to estimate the endurance strength, Se, in MPa .

To do that; let's for see the expression that shows the relationship between the ultimate tensile strength and Brinell hardness number .

It is expressed as:

200 \leq H_B \leq 450

S_{ut} = 3.41 H_B

where;

H_B = Brinell hardness number

S_{ut} =  Ultimate tensile strength

From ;

S_{ut} = 3.41 H_B; replace 290 for H_B ; we have

S_{ut} = 3.41 (290)

S_{ut} = 988.9 MPa

We can see that the derived value for the ultimate tensile strength when the Brinell harness number = 290 is less than 1400 MPa ( i.e it is 988.9 MPa)

So; we can say

S_{ut} < 1400

The Endurance limit can be represented by the formula:

S_e ' = 0.5 S_{ut}

S_e ' = 0.5 (988.9)

S_e ' = 494.45 MPa

Using Table 6.2 for parameter for Marin Surface modification factor. The value for a and b are derived; which are :

a = 1.58

b =  -0.085

The value of the surface factor can be calculate by using the equation

k_a = aS^b_{ut}

K_a = 1.58 (988.9)^{-0.085

K_a = 0.8792

The formula that is used to determine the value of  k_b for the rotating shaft of size factor d = 10 mm is as follows:

k_b = 1.24d^{-0.107}

k_b = 1.24(10)^{-0.107}

k_b = 0.969

Finally; the the endurance strength, Se, in MPa if the rod is used in rotating bending is determined by using the expression;

S_e =k_ak_b S' _e

S_e= 0.8792×0.969×494.45

S_e = 421.24  MPa

Thus; the endurance strength  S_e = 421.24  MPa

8 0
3 years ago
If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M
Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

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- The signal bandwidth W = 20 MHz

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what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

                            C = W*log_2 ( 1 + P/N)

- Plug the values in:

                            C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

                            C = (20*10^6)*log_2 (25001)

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A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

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Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

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NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

= 0.3200 \times 40 g \times 21.30 mL \times  1L/1000mL

= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

Now

Moles of H_2SO_4 needed  is

= \frac{0.006816}{2}

= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

= \frac{moles}{Volume}

= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

=5.112 \times 10^{-5 M}

We simply applied the above formulas

4 0
3 years ago
What are the three elementary parts of a vibrating system?
zhenek [66]

Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring -  the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and  zero elasticity.

7 0
3 years ago
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