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:
T₂ =93.77 °C
Explanation:
Initial temperature ,T₁ =27°C= 273 +27 = 300 K
We know that
Absolute pressure = Gauge pressure + Atmospheric pressure
Initial pressure ,P₁ = 300+1=301 kPa
Final pressure ,P₂= 367+1 = 368 kPa
Lets take temperature=T₂
We know that ,If the volume of the gas is constant ,then we can say that
Now by putting the values in the above equation we get
The temperature in °C
T₂ = 366.77 - 273 °C
T₂ =93.77 °C
Answer:
The mechanical advantage is 3 to 1
Explanation:
A frictionless pulley with three support ropes carries equal tension on each of the ropes thus;
Tension in each pulley rope = T
Total tension in the 3 ropes = 3 × T = 3·T
Direction of the tension forces on each rope = Unidirectional
Total force provided by the 3 ropes = 3·T
Therefore, a force, T, applied at the end of the rope will result in a lifting force of 3·T
Hence, the mechanical advantage = 3·T to T which is presented as follows;
The mechanical advantage = 3 to 1.