The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.
<h3>What is power?</h3>
Power is the energy transferred per unit time.
Torque is find out by
P = 2πNT/60
10000 = 2π x 2000 x T / 60
T =47.74 N.m
The gear ratio Ne / Ns =4/1
Ns =2000/4 = 500
Ts =Ps x 60/(2π x 500)
Ts =190.96 N.m
Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))
τ max =T/J x D/2
where d₁ = 30mm = 0.03 m
d₀ = 30 +(2x 4) = 38mm =0.038 m
Substitute the values into the equation, we get
τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)
τ max = 28.98 MPa.
Thus, the maximum shear stress in the tube is 28.98 MPa.
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Answer:
See explanation
Explanation:
Solution:-
- The shell and tube heat exchanger are designated by the order of tube and shell passes.
- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.
- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.
- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.
- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:
U ∝ v^( 0.8 ) .... ( turbulence )
- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.
Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).
Answer:
Efficiency based on Otto cycle.
Effotto = 47.47%
Explanation:
Efficiency based on Otto cycle.
effotto = 1 – (V2 / V1)^γ-1
effotto = 1 – (1 / 5)^1.4 - 1
effotto = 47.47%
Answer:
Explanation:
First we compute the characteristic length and the Biot number to see if the lumped parameter
analysis is applicable.
Since the Biot number is less than 0.1, we can use the lumped parameter analysis. In such an
analysis, the time to reach a certain temperature is given by the following
From the data in the problem we can compute the parameter, b, and then compute the time for
the ratio (T – T)/(Ti
– T)