Answer:
A) 5 MPa , 55 MPa
B) maximum stress = 55 MPa, maximum shear stress = 25 MPa
Explanation:
using the given Data
free surface of a solid body
α
= 50 MPa, α
= 10 MPa , t
= -15 MPa
attached below is the detailed solution to the question
Answer:
Given that

LHS of above given equation have dimension
.
Now find the dimension of RHS
Dimension of P =
.
Dimension of d=
.
Dimension of μ =
.
Dimension of L=
.
So
![\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5Cdfrac%7B%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%5E2%7D%7B%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%7D)
![\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5BM%5E0L%5E%7B1%7DT%5E%7B-1%7D%5D)
It means that both sides have same dimensions.
Answer:
Sprockets.
Explanation:
A chain drive is an efficient technique used for the transmission of mechanical power from one point to another. For example, it is used for transmitting power to the wheels of a bicycle, motorcycle, motor vehicle and other machineries such as chain saw etc.
Chain drive ratio is the ratio between the rotational speeds of the input and output sprockets of a roller chain drive system. This ultimately implies that, chain drive ratio is the ratio of the number of teeth on the driving sprocket (T1) divided to the number of teeth on the driven sprocket (T2).
Also, the chain drive ratio can be calculated by dividing the number of teeth on the large sprocket by the number of teeth on the small sprocket.
Additionally, the rotational speed of a sprocket is measured in revolutions per minute (RPM).
One of the issues with the roller chain is that, as the roller chain moves round the sprocket link by link, it affects its speed (surge) due to the change in acceleration and deceleration i.e the rise and fall of its pitch line.
Answer:
P2 = 3.9 MPa
Explanation:
Given that
T₁ = 290 K
P₁ = 95 KPa
Power P = 5.5 KW
mass flow rate = 0.01 kg/s
solution
with the help of table A5
here air specific heat and adiabatic exponent is
Cp = 1.004 kJ/kg K
and k = 1.4
so
work rate will be
W = m × Cp × (T2 - T1) ..........................1
here T2 = W ÷ ( m × Cp) + T1
so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290
T2 = 838 k
so final pressure will be here
P2 = P1 ×
..............2
P2 = 95 × 
P2 = 3.9 MPa