Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace

Answers

Answer 1
Answer:

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

(T_(0)-T_(x)  )/(T_(1)-T_(x)  ) = C_(1) e^{(-0.4888^(2)*Fo )}

= 0.4167 = 1.0396e^(-0.4888*Fo)

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo = ((k)/(pc_(p) ) ) ( (t)/((L/2)^2) )

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s


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Webster is giving a speech on the benefits of moving toward the use of windmill energy instead of having to rely on fossil fuels. Most likely, he will select which method of arrangement?

Answers

Answer:

the netuon arrangment

Explanation:

Which of the following results from the fact that skip signals refracted from the ionosphere are elliptically polarized?A. Digital modes are unusable

B. Either vertically or horizontally polarized antennas may be used for transmission or reception

C. FM voice is unusable

D. Both the transmitting and receiving antennas must be of the same polarization

Answers

Answer:

B

Explanation:

The fact that skip signals refracted from the ionosphere are elliptically polarized is a result of either vertically or horizontally polarized antennas may be used for transmission or reception.

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provideA. expansion joints
B. isolation joints
C. control joints
D. construction joints

Answers

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provide Control joints. The correct answer would be C.

Control joints are used to prevent random cracks from forming in large concrete slabs caused by shrinkage. These joints are placed at strategic locations in the slab to allow for the concrete to expand and contract without cracking. Expansion joints, on the other hand, are used to separate concrete from other structures or materials, and isolation joints are used to separate different sections of concrete.

Construction joints are used to connect two different pours of concrete. Therefore, the best option for preventing random cracks caused by shrinkage would be to use control joints.

Learn more about Control joints:

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A 15. μF capacitor is connected to a 50. V battery and becomes fully charged. The battery is removed and a slab of dielectric that completely fills the space between the plates is inserted. If the dielectric has a dielectric constant of 5.0, what is the voltage across the capacitor's plates after the slab is inserted?

Answers

Answer:

voltage across the capacitor's is 75 μF

Explanation:

given data

capacitor = 15 μF

voltage = 50V

dielectric constant k = 5

to find out

voltage across the capacitor's

solution

we will find here voltage across the capacitor's  by this formula

voltage across the capacitor's  = kC_(o)

here k is 5 and C_(o) = 15

put these value we get

voltage across the capacitor's  = kC_(o)

voltage across the capacitor's  = 5 ( 15 )  = 75 μF

so voltage across the capacitor's is 75 μF

A test bottle containing just seeded dilution water has its DO level drop by 0.8 mg/L in a 5-day test. A 300 mL BOD bottle filled with 30 mL of wastewater and the rest with seeded dilution water experiences a drop of 7.3 mg/L in the same period (5-day). Calculate the BOD5 of the wastewater.

Answers

Answer:

BOD_5 = =65.8 mg/l

Explanation:

dilution water DO level = 0.8 m/l

BOD level drop to 7.3 mg/l

we know that BOD at 5th day can be clculated by using following relationBOD_5 = ((D_1 -D_2) -(B_1-B_2)(1-P))/(P)

D_1 -D_2 - DO drop in BOD bottle

B_1-B_2 - dilution water drop

P= 30/300 = 0.1

BOD_5 = ((7.3) -(0.8)(1-0.1))/(0.1)

BOD_5 = =65.8 mg/l

In a TDM communication example, 15 voice signals are badlimited to 5kHz and transmitted simultaneously using PAM. What is a preliminary estimate for the required system bandwidth?(a) 10 kHz
(b) 75 kHz
(c) 80 kHz
(d) 160 kHz
(e) None of the above.

Answers

Answer:

Option D

160 kHz

Explanation:

Since we must use at least one synchronization bit, total message signal is 15+1=16

The minimum sampling frequency, fs=2fm=2(5)=10 kHz

Bandwith, BW required is given by

BW=Nfs=16(10)=160 kHz