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Tasya [4]
3 years ago
5

How long should the shafts remain in the furnace to achieve a desired centerline temperature of 800K? 2) Determine the temperatu

re gradient in the shaft at the outer surface of the shaft at the initial time when the steels enter the furnace? Note, the r coordinate extends along the direction from the center to outside.

Engineering
1 answer:
Phantasy [73]3 years ago
6 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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Compute the electrical resistivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length i
snow_tiger [21]
No no no no no no no no no no
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3 years ago
Two gases—neon and air—are expanded from P1 to P2 in a closed-system polytropic process with n = 1.2. _____ produces more work w
Makovka662 [10]

Answer:

Note that Air requires lesser work

Explanation:

Calculate  for general work done

SInce Gas constant 'R' for:  Neon = 0.4119KJ/kg.k , and Air = 0.287 kJ/kg·K

Calculate for work done of NEON

Calculate for work done of Air

See solution attached.

7 0
3 years ago
A socket can be driven using any of the following except for (A) a socket ratchet
AfilCa [17]

Answer:

a socket can be driven by a flexible head socket wrench, I believe

4 0
4 years ago
A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for
Svet_ta [14]

Answer:

Time taken by the 1\mu m diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = 1\times 10^{- 6} m

Diameter of the smaller droplet, R' = 0.5\times 10^{- 6} m

Now,

Volume of the droplet, V = \frac{4}{3}\pi R^{3}

Volume of the smaller droplet, V' = \frac{4}{3}\pi R'^{3}

Volume of the droplet ∝ Time taken for complete evaporation

Thus

\frac{V}{V'} = \frac{t}{t'}

where

t' = taken taken by smaller droplet

\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}

\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}

t' = 60\times 10^{- 9} s = 60 ns

5 0
4 years ago
The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%
Aleonysh [2.5K]

Answer:

4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

4 0
3 years ago
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