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IrinaK [193]
3 years ago
5

Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be

ing charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas. Consider a packed bed of 75-mm-diameter aluminum spheres (p = 2,700 kg/m^3; c = 950 J/kg*K; k = 240 W/m*K) and a charging process for which gas enters the storage unit at a temperature of 300 degrees C. The initial temperature of the spheres is Ti = 25 degrees C and the convection heat transfer coefficient is h = 75 W/m^2*K.
a. How long does it take a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere?
b. Is there any advantage to using copper (p = 8,900 kg/m^3; c = 380 J/kg*K; k = 390W/m*K) instead of aluminum?
c. Consider the same packed bed operating conditions, but with Pyrex (p = 2,200 kg/m^3; c = 840 J/kg*K; k = 1.4 W/m*K) used instead of aluminum. How long does it take a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere?
Engineering
1 answer:
hammer [34]3 years ago
3 0

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit  = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

<u>B) The physical properties are copper</u>

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

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name the three exposure techniques in photolithography. what are the alternatives to photolithography in ic processing?
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<h3>What is Photolithography?</h3>

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6 0
1 year ago
Your boss needs to recommend either the in-situ soil or the borrow material. Although you do not have much background on soil be
Andrej [43]

Answer: see attached file for the answer

7 0
3 years ago
The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
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Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

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Answer:

a. Using a straight forward scheme, the RC4 algorithm stores 2064 bits in the interval state.

b. 1700 bits

Explanation:

See RC4 Algorithm attached with details

3 0
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Answer:

Take a 3 to 8 decoder with active low outputs

Assuming you are familiar with the functioning of decoders,

The three inputs of decoder of course are the first, second and the carry bit which you feed to the subtractor.

Next we examine the truth table of the full subtractor i formatted in the picture.

Then write the minterms for the difference output and borrow output from the given truth table picture I have mentioned before!!

Explanation:

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