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Chapter 6: Problem 3

\(\mathrm{A}\) horizontal flow of water is described by the velocity field \(\vec{V}=(-A x+B t) \hat{i}+(A y+B t) \hat{j},\) where \(A=1 \mathrm{s}^{-1}\) and \(B=2 \mathrm{m} / \mathrm{s}^{2}, x\) and \(y\) are in meters, and \(t\) is in seconds. Find expressions for the local acceleration, the convective acceleration, and the total acceleration. Evaluate these at point (1,2) at \(t=5\) seconds. Evaluate \(\nabla p\) at the same point and time.

Short Answer

Expert verified
The local acceleration at (x=1, y=2) at time t=5 seconds is 2\hat{i} + 2\hat{j} m/s虏, the convective acceleration at the same point and time is -4\hat{i} - 7\hat{j} m/s虏, and the total acceleration is 11\hat{i} + 8\hat{j} m/s虏. The gradient of pressure can't be calculated without a function for the pressure.

Step by step solution

01

Calculating Local Acceleration

The local acceleration \( a_l \) is calculated by differentiating \( \vec{V} \) with respect to time \( t \). Thus: \( a_l = \frac{d \vec{V}}{d t} \). After differentiating we get: \( a_l=\frac{d(-A x+B t)}{d t} \hat{i}+\frac{d(A y+B t)}{d t} \hat{j} = B\hat{i} + B\hat{j} \).
02

Calculating Convective Acceleration

The convective acceleration \( a_c \) is calculated by taking the dot product of velocity \( \vec{V} \) and the gradient of velocity. Thus: \( a_c = \vec{V} \cdot \nabla \vec{V} \). After performing the operation we get: \( a_c=(-A x+B t)(-A)+(-A y+B t)(A) = -A^2 x + B A t \hat{i} - A^2 y + B A t \hat{j} \).
03

Calculating Total Acceleration

The total acceleration \( a_t \) is the sum of local acceleration \( a_l \) and convective acceleration \( a_c \). Thus: \( a_t = a_l + a_c = (B\hat{i} + B\hat{j}) + (-A^2 x + B A t \hat{i} - A^2 y + B A t \hat{j}) = (B- A^2 x + B A t)\hat{i} + (B - A^2y + B A t)\hat{j} \).
04

Evaluating Accelerations at given point and time

Substitute the given point (x=1, y=2) and time t=5 into the above derivations for accelerations. After substituting, we get: \( a_l = B\hat{i} + B\hat{j} = 2\hat{i} + 2\hat{j} \) m/s虏, \( a_c = -A^2 x + B A t \hat{i} - A^2 y + B A t \hat{j} = (1 - 5)\hat{i} + (-2 - 5)\hat{j} = -4\hat{i} - 7\hat{j} \) m/s虏, \( a_t = (B- A^2 x + B A t)\hat{i} + (B - A^2y + B A t)\hat{j} = ( 2 - 1 + 10)\hat{i} + (2 - 4 + 10)\hat{j} = 11\hat{i} + 8\hat{j} \) m/s虏.
05

Evaluating Pressure Gradient

In fluid dynamics, pressure is a product of the velocity field, and hence, we would need to know the function form of pressure to calculate its gradient. Without additional information regarding the pressure function, we can't calculate the gradient. Hence, \( \nabla p \) can't be calculated.

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Most popular questions from this chapter

In a laboratory experiment, water flows radially outward at moderate speed through the space between circular plane parallel disks. The perimeter of the disks is open to the atmosphere. The disks have diameter \(D=150 \mathrm{mm}\) and the spacing between the disks is \(h=0.8 \mathrm{mm} .\) The measured mass flow rate of water is \(\dot{m}=305 \mathrm{g} / \mathrm{s}\). Assuming frictionless flow in the space between the disks, estimate the theoretical static pressure between the disks at radius \(r=50 \mathrm{mm} .\) In the laboratory situation, where some friction is present, would the pressure measured at the same location be above or below the theoretical value? Why?

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