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gregori [183]
3 years ago
10

Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the w

ire mesh on top of the grill is covered with steaks initially at 274 K. The distance between the coal bricks and the steaks is 0.20 m. Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.

Engineering
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

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Pls help me it’s due today
hichkok12 [17]

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

6 0
3 years ago
A gear train has two gears. The driver gear has 8 teeth and a diametral pitch of 6 teeth/inch. the follower gear has 24 teeth. W
Sliva [168]

Answer:

18 teeth/inch

Explanation:

Given that: i. driver gear has 8 teeth and diametral pitch of 6 teeth/inch.

                  ii. follower gear has 24 teeth.

Let the followers diametral pitch be represented by x.

Then,

8 teeth ⇒  6 teeth/inch

24 teeth ⇒ x teeth/inch

So that;

x = \frac{24 x 6}{8}

   = \frac{144}{8}

   = 18 teeth/inch

The diametral is 18 teeth/inch

3 0
3 years ago
What is the least count of screw gauge?<br> (a) 0.01 cm<br> (b) 0.001 cm<br> (c) 0.1 cm<br> (d) 1 mm
Nonamiya [84]
Its 0.001

0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m
3 0
3 years ago
Consider laminar, fully developed flow in a channel of constant surface temperature Ts. For a given mass flow rate and channel l
Pachacha [2.7K]

Answer:

Please see attachment.

Explanation:

8 0
3 years ago
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K
Mashcka [7]

Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

\Delta S=\dfrac{Q}{T}

Now by putting the values

20.3=\dfrac{Q}{400}

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU  = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU  = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

8 0
3 years ago
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