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A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plate located at a distance of 0.01 cm from it, the fluid separating them being water with viscosity of 0.001 Ns/m2. The power required to maintain velocity will be
- a)0.05 W
- b)0.07 W
- c)0.09 W
- d)0.11 W
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plat...
Concept:
The tangential shear stress between two adjoining is proportional to velocity gradient in direction of perpendicular to the layer. This is known as law of viscosity.
Shear force acting on the moving plate is given by F = τ × A
Power required to maintain velocity U is given by P = F × U
The tangential shear stress between two adjoining is proportional to velocity gradient in direction of perpendicular to the layer. This is known as law of viscosity.
Shear force acting on the moving plate is given by F = τ × A
Power required to maintain velocity U is given by P = F × U
Calculation:
Given μ = 0.001 Ns/m2, U = 0.3 m/s, Y = 0.01 cm = 0.01 × 10-2 m, A = 0.1 m.;
Velocity gradient or rate of shear strain is given by
τ = 0.001 × 3000 = 3 N/m2
F = 3 × 0.1 = 0.3 N,
P = 0.3 × 0.3 = 0.09 W
Given μ = 0.001 Ns/m2, U = 0.3 m/s, Y = 0.01 cm = 0.01 × 10-2 m, A = 0.1 m.;
Velocity gradient or rate of shear strain is given by
τ = 0.001 × 3000 = 3 N/m2
F = 3 × 0.1 = 0.3 N,
P = 0.3 × 0.3 = 0.09 W
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Community Answer
A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plat...
To calculate the power required to maintain velocity between two plates separated by a fluid, we can use the formula:
Power = Force × Velocity
Let's break down the problem step by step:
1. Calculation of Force:
- The force required to maintain velocity is the drag force acting on the plate.
- The drag force can be calculated using the formula: Drag force = viscosity × area × velocity / distance
- Given values:
- Viscosity (η) = 0.001 Ns/m2
- Area (A) = 0.1 m2
- Velocity (V) = 30 cm/s = 0.3 m/s
- Distance (d) = 0.01 cm = 0.0001 m
- Plugging in the values:
Drag force = 0.001 Ns/m2 × 0.1 m2 × 0.3 m/s / 0.0001 m
= 0.001 N × 0.3 / 0.0001
= 3 N
2. Calculation of Power:
- Power = Force × Velocity
- Given values:
- Force = 3 N
- Velocity = 0.3 m/s
- Plugging in the values:
Power = 3 N × 0.3 m/s
= 0.9 W
Therefore, the power required to maintain velocity between the two plates is 0.9 W. None of the given options match this answer.
It seems that there might be an error in the given options, as none of them correspond to the calculated power.
Power = Force × Velocity
Let's break down the problem step by step:
1. Calculation of Force:
- The force required to maintain velocity is the drag force acting on the plate.
- The drag force can be calculated using the formula: Drag force = viscosity × area × velocity / distance
- Given values:
- Viscosity (η) = 0.001 Ns/m2
- Area (A) = 0.1 m2
- Velocity (V) = 30 cm/s = 0.3 m/s
- Distance (d) = 0.01 cm = 0.0001 m
- Plugging in the values:
Drag force = 0.001 Ns/m2 × 0.1 m2 × 0.3 m/s / 0.0001 m
= 0.001 N × 0.3 / 0.0001
= 3 N
2. Calculation of Power:
- Power = Force × Velocity
- Given values:
- Force = 3 N
- Velocity = 0.3 m/s
- Plugging in the values:
Power = 3 N × 0.3 m/s
= 0.9 W
Therefore, the power required to maintain velocity between the two plates is 0.9 W. None of the given options match this answer.
It seems that there might be an error in the given options, as none of them correspond to the calculated power.
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A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plate located at a distance of 0.01 cm from it, the fluid separating them being water with viscosity of 0.001 Ns/m2. The power required to maintain velocity will bea)0.05 Wb)0.07 Wc)0.09 Wd)0.11 WCorrect answer is option 'C'. Can you explain this answer?
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