Answer:d
Explanation:
(d) chain drive belt drive gear drive
Answer:
D
Explanation:
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Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ
= displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ
------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
P > 142.5 N (→)
the motion sliding
Explanation:
Given
W = 959 N
μs = 0.3
If we apply
∑ Fy = 0 (+↑)
Ay + By = W
If Ay = By
2*By = W
By = W / 2
By = 950 N / 2
By = 475 N (↑)
Then we can get F (the force of friction) as follows
F = μs*N = μs*By
F = 0.3*475 N
F = 142.5 N (←)
we can apply
P - F > 0
P > 142.5 N (→)
the motion sliding
Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 -
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 -
= 1100/3
= 733.33 K

Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;

Hence, from;
, we have
5912500 = 90 × A × 341.67

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².