A rectangular channel 6 m wide with a depth of flow of 3m has a mean velocity of 1.5 m/sec. The channel undergoes a smooth, grad
ual contraction to a width of 4.5 m. (a) Calculate the depth and velocity in the contracted section (Hint: y2 is between 2.6~2.9m). (b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction. In each case, identify any assumptions that you made.
1 answer:
Answer:
The answer for (1) 2.07 m/s (2) F= 63.8kN
Explanation:
Solution
By applying the energy equation from the approach section 1 to the contacted section 2 with negligible head losses and assuming a horizontal bottom
we have y₁ + V₁²/ 2g = y₂ + q₂²/2gy₂²
Where
q is =flow rate
y = the depth
g = the acceleration
Kindly find an attached copy of the final solution for this question.
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Answer:
The temperature T= 648.07k
Explanation:
T1=input temperature of the first heat engine =1400k
T=output temperature of the first heat engine and input temperature of the second heat engine= unknown
T3=output temperature of the second heat engine=300k
but carnot efficiency of heat engine =
where Th =temperature at which the heat enters the engine
Tl is the temperature of the environment
since both engines have the same thermal capacities <em> </em> therefore
We have now that
multiplying through by T
multiplying through by 300
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The temperature T= 648.07k