A 20-mm thick draw batch furnace front is subjected to uniform heat flux on the inside surface, while the outside surface is sub
1 answer:
Answer:
hello your question is incomplete attached below is the complete question
<em>answer</em> :
To ( inside temperature ) = 598 K
TL ( outside temperature ) = 594 k
Explanation:
a) Determine the surface temperature To and TL based on the known conditions provided in the drawing
To ( inside temperature ) = 598 K
TL ( outside temperature ) = 594 k
attached below is the detailed solution
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Answer:
Explanation:
Given the data in the question;
L = 46 in
Ga = 5 × 10³ ksi
Gs = 11 × 10³ ksi
Outside diameter da = 5 in
ds = 4 in
Tb = 3 kip.in
Now,
Ja = polar moment of Inertia of Aluminum;
Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 )
Js = polar moment of inertia of steel
Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )
Ta is torque transmitted by Aluminum
Ts is torque transmitted by steel
{composite member }
T = Ta + Ts ------ let this be equation m1
Now, we use the relation;
T/J = G∅/L
JG∅ = TL
∅ = TL/GJ
so, for aluminum rod ∅ = TaLa/GaJa
for steel rod ∅ = TsLs/GsJs
but we know that, ∅a = ∅s = ∅
so
[TaLa/GaJa] = [TsLs/GsJs]
also, we know that, La = Ls = L
∴ [Ta/GaJa] = [Ts/GsJs]
we solve for Ta
TaGsJs = TsGaJa
Ta = TsGaJa / GsJs
we substitute
Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]
Ta = 0.66Ts
now, we substitute 0.66Ts for Ta and 3 for T in equation 1
T = Ta + Ts
3 = 0.66Ts + Ts
3 = 1.66Ts
Ts = 3 / 1.66
Ts = 1.8072 ≈ 1.81 kip-in
so
∅ = TsLs / GsJs
we substitute
∅ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )
∅ = 83.26 / 276460.1535
∅ = 0.000301
∅ = 3.01 × 10⁻⁴ rad
so
∅ = ∅
= 3.01 × 10⁻⁴ rad
Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad
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
Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean :
Standard deviation :
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula, , the z-value corresponds to 2.39 will be :-
z-value corresponds to 2.60 will be :-
Using the standard normal table for z, we have
P-value =
Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
Answer:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.
