The 10 ft by 4 ft plate of Problem 13.44 is held in water (50°F) moving at 4 ft/sec parallel to its length. Assuming laminar conditions in the boundary layer at the leading edge of the plate, (a) locate where the boundary layer flow changes from laminar to turbulent, (b) estimate the thickness of

Question

The 10 ft by 4 ft plate of Problem 13.44 is held in water (50°F) moving at 4 ft/sec parallel to its length. Assuming laminar conditions in the boundary layer at the leading edge of the plate, (a) locate where the boundary layer flow changes from laminar to turbulent, (b) estimate the thickness of the boundary layer at this point, and (c) compute the friction drag of the plate.

Accepted Answer

[latex] ext { (a) Reynolds number Re }=V L / v=(4)(10) /\left(1.410 imes 10^{-5} ight)=2,840,000 [/latex].

This value of Reynolds number indicates that the boundary layer flow is in the transition range. Assuming the critical Reynolds number is 500,000, the location of the end of the laminar flow conditions can be estimated by using

[latex]\frac{x_{c}}{L}=\frac{ ext { critical } \operatorname{Re}}{ ext { full plate } \operatorname{Re}} \quad ext { or } \quad x_{c}=(10)\left(\frac{500,000}{2,840,000} ight)=1.76 ft[/latex].

(b) The thickness of the boundary layer at this point is estimated to be

[latex]\delta_{c}=\frac{5.20 x_{c}}{\sqrt{\operatorname{Re}_{c}}}=\frac{(5.20)(1.76)}{\sqrt{500,000}}=0.0129 ft[/latex].

(c) The friction drag can be computed by adding to the drag from the laminar boundary layer up to [latex]x_{c}[/latex] (see Figure 13-17) the drag from the turbulent boundary layer, B to C. The latter value of drag is obtained by computing the drag for turbulent flow along the entire plate (A to C) less the drag of the fictitious turbulent layer from A to B.
(1) Laminar drag, A to B, on one side.

[latex] ext { drag force }=C_{D} ho A \frac{V^{2}}{2}=\frac{1.328}{\sqrt{\operatorname{Re}_{c}}} ho A \frac{V^{2}}{2}=\frac{1.328}{\sqrt{500,000}}(1.94)(4 imes 1.76) \frac{4^{2}}{2}=0.205 lb [/latex].

(2) Turbulent drag, A to C, if conditions were turbulent for the entire length of the plate.

[latex]\operatorname{drag} ext { force }=C_{D} ho A \frac{V^{2}}{2} ext { (one side) }[/latex]

[latex]=\frac{0.074}{\operatorname{Re}^{0.20}} ho A \frac{V^{2}}{2}=\frac{0.074}{(2,840,000)^{0.20}}(1.94)(4 imes 10) \frac{4^{2}}{2}=2.352 lb[/latex]

(3) Fictitious turbulent drag, A to B.

[latex]\operatorname{drag} ext { force }=C_{D} ho A \frac{V^{2}}{2} ext { (one side) }[/latex]

[latex]=\frac{0.074}{\operatorname{Re_c}^{0.20}} ho A \frac{V^{2}}{2}=\frac{0.074}{(500,000)^{0.20}}(1.94)(4 imes 1.76) \frac{4^{2}}{2}=3.94 lb[/latex]

Total drag force (two sides) = (2)[0.205 + (2.352 - 0.586)] = 3.94 Ib.

Had the Reynolds number for the entire plate been over [latex]10^{7}[/latex], equation (13) at the beginning of this chapter should be used in part (2) above.
A weighted value of [latex]C_{D}^{\prime}[/latex] might be obtained for the entire plate by equating the above total drag force to the drag force expression, as follows.

[latex]=\frac{0.455}{\left(\log _{10} \operatorname{Re} ight)^{2.58}} \quad ext { for } 10^{6}<\operatorname{Re}<10^{9}[/latex] (13).

[latex] ext { total drag force }=2 C_{D}^{\prime} ho A \frac{V^{2}}{2}, \quad 3.94=2 C_{D}^{\prime}(1.94)(4 imes 10) \frac{4^{2}}{2}, \quad C_{D}^{\prime}=0.00317[/latex].

Question 13.46: The 10 ft by 4 ft plate of Problem 13.44 is held in water (5...

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