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The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation?
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The space between two large flat and parallel walls 25 mm appart is fi...
Given Information:
- Distance between two large flat and parallel walls = 25 mm
- Absolute velocity of liquid = 0.7 N/m^2
- Plate dimensions = 25 x 25 cm
- Plate velocity = 0.15 m/s
- Plate distance from one wall = 6 mm
To determine the force exerted by the liquid on the plate, we need to calculate the shear stress on the plate due to the liquid flow. This can be done using the following formula:
Shear stress = viscosity x (velocity gradient)
where viscosity is the dynamic viscosity of the liquid and velocity gradient is the rate of change of velocity with respect to distance.
Calculation of Velocity Gradient:
- Velocity of plate = 0.15 m/s
- Distance of plate from one wall = 6 mm = 0.006 m
- Distance of plate from other wall = 25 mm - 6 mm = 19 mm = 0.019 m
- Velocity gradient = (0.15 m/s) / (0.019 m - 0.006 m) = 9.375 m/s^2
Calculation of Shear Stress:
- Dynamic viscosity of the liquid is not given, so we cannot directly calculate the shear stress. However, we can use the concept of wall shear stress to estimate the shear stress on the plate.
- Wall shear stress is the shear stress exerted by a fluid on a solid surface in contact with it. It is given by the formula:
Wall shear stress = (viscosity x velocity) / distance from wall
- Assuming that the liquid flow is parallel to the walls and the plate, we can estimate the wall shear stress at the plate location as follows:
- Wall shear stress = (viscosity x 0.15 m/s) / 0.006 m
- We can assume that the wall shear stress is constant across the distance between the walls, since the flow is parallel to the walls and the plate. Therefore, the shear stress on the plate can be estimated as follows:
- Shear stress on plate = Wall shear stress x (distance between walls / plate width)
- Shear stress on plate = (viscosity x 0.15 m/s) / 0.006 m x (0.025 m / 0.019 m)
- Shear stress on plate = 0.785 x viscosity N/m^2
Therefore, the force exerted by the liquid on the plate can be calculated as follows:
- Force = Shear stress x Plate area
- Force = (0.785 x viscosity N/m^2) x (0.25 m^2) = 0.19625 x viscosity N
Conclusion:
- The force exerted by the liquid on the plate is 0.19625 times the dynamic viscosity of the liquid in Newtons.
- Distance between two large flat and parallel walls = 25 mm
- Absolute velocity of liquid = 0.7 N/m^2
- Plate dimensions = 25 x 25 cm
- Plate velocity = 0.15 m/s
- Plate distance from one wall = 6 mm
To determine the force exerted by the liquid on the plate, we need to calculate the shear stress on the plate due to the liquid flow. This can be done using the following formula:
Shear stress = viscosity x (velocity gradient)
where viscosity is the dynamic viscosity of the liquid and velocity gradient is the rate of change of velocity with respect to distance.
Calculation of Velocity Gradient:
- Velocity of plate = 0.15 m/s
- Distance of plate from one wall = 6 mm = 0.006 m
- Distance of plate from other wall = 25 mm - 6 mm = 19 mm = 0.019 m
- Velocity gradient = (0.15 m/s) / (0.019 m - 0.006 m) = 9.375 m/s^2
Calculation of Shear Stress:
- Dynamic viscosity of the liquid is not given, so we cannot directly calculate the shear stress. However, we can use the concept of wall shear stress to estimate the shear stress on the plate.
- Wall shear stress is the shear stress exerted by a fluid on a solid surface in contact with it. It is given by the formula:
Wall shear stress = (viscosity x velocity) / distance from wall
- Assuming that the liquid flow is parallel to the walls and the plate, we can estimate the wall shear stress at the plate location as follows:
- Wall shear stress = (viscosity x 0.15 m/s) / 0.006 m
- We can assume that the wall shear stress is constant across the distance between the walls, since the flow is parallel to the walls and the plate. Therefore, the shear stress on the plate can be estimated as follows:
- Shear stress on plate = Wall shear stress x (distance between walls / plate width)
- Shear stress on plate = (viscosity x 0.15 m/s) / 0.006 m x (0.025 m / 0.019 m)
- Shear stress on plate = 0.785 x viscosity N/m^2
Therefore, the force exerted by the liquid on the plate can be calculated as follows:
- Force = Shear stress x Plate area
- Force = (0.785 x viscosity N/m^2) x (0.25 m^2) = 0.19625 x viscosity N
Conclusion:
- The force exerted by the liquid on the plate is 0.19625 times the dynamic viscosity of the liquid in Newtons.
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The space between two large flat and parallel walls 25 mm appart is fi...
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The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation?
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The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation?.
The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The space between two large flat and parallel walls 25 mm appart is filled with a liquid of absolute velocity .7 na/m^2 , within this space a thin plate 25x25cm is towed with a velocity of .15 m/s at 6 mm from one wall and parallel to the walls . Assuming the liner velocity between the plate and walls determine the force exerted by the liquid on the plate. Please give me the explanation?.
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