91Ó°ÊÓ

The velocity field in the region shown is given by \(\vec{V}=(a j+b y k)\) where \(a=10 \mathrm{m} / \mathrm{s}\) and \(b=5 \mathrm{s}^{-1}\). For the \(1 \mathrm{m}\) \(\times 1 \mathrm{m}\) triangular control volume (depth \(w=1 \mathrm{m}\) perpendicular to the diagram), an element of area (1) may be represented by \(, d \vec{A}_{1}=w d z j-w d y k\) and an element of area (2) by \(d A_{2}=-w d y k\) (a) Find an expression for \(\vec{V} \cdot d A_{1}\) (b) Evaluate \(\int_{A_{1}} V \cdot d A_{1}\) (c) Find an expression for \(V \cdot d A_{2}\) (d) Find an expression for \(\vec{V}\left(\vec{V} \cdot d A_{2}\right)\) (c) Evaluate \(\int_{A_{2}} V\left(V \cdot d A_{2}\right)\)

Step by step solution

01

Expression for \(\vec{V} \cdot d \vec{A_{1}}\)

Substitute the given vectors into the dot product formula. It yields: \(\vec{V} \cdot d \vec{A_1} = (10j + 5y k) \cdot (w dz j - w dy k)\). Here, \(j \cdot j = 1\) and \(k \cdot k = 1\), so \(\vec{V} \cdot d \vec{A_1} = 10w dz - 5wy dy\)

02

Evaluation of \(\int_{A_{1}} V \cdot d A_{1}\)

Substitute the expression previously found into the integral and integrate over the given triangular control volume: \(\int_{A_1} (10w dz - 5wy dy) \) = \(\int_0^1 \int_0^{1-z} (10w - 5wy) dy dz\). As \(w = 1m\), this simplifies to \(\int_0^1 \int_0^{1-z} (10 - 5y) dy dz\) = \(\int_0^1 (10(1-z) - 5(1-z)^2 /2) dz = 10/2 - 5/6 = 5/3 m^2/s\)

03

Expression for \(V \cdot d A_{2}\)

Substitute the given vectors into the dot product formula: \(\vec{V} \cdot d \vec{A_2} = (10j + 5y k) \cdot (-w dy k)\). Here \(j \cdot k = 0\) and \(k \cdot k = 1\), so \(\vec{V} \cdot d \vec{A_2} = -5wy dy\)

04

Expression for \(\vec{V}\left(\vec{V} \cdot d A_{2}\right)\)

Substitute the previous result into the expression for \(\vec{V}\left(\vec{V} \cdot d A_{2}\right)\): \(\vec{V} * -5wy dy = -5wy (10j + 5y k) dy = -50wy^2 j dy - 25w y^3 k dy\)

05

Evaluation of \(\int_{A_{2}} V\left(V \cdot d A_{2}\right)\)

Substitute the expression previously found into the integral and integrate over the given control volume: \(\int_{A_2} -50wy^2 j dy - 25w y^3 k dy = -50w \int_0^1 y^2 dy - 25w \int_0^1 y^3 dy = -50/3w j - 25/4w k = -50/3 j m^3/s - 25/4 k m^3/s\)

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