Oil of specific gravity 0.761 flows from tank A to tank E as shown in Fig. 7-16. Lost head items may be assumed to be as follows:

$A \text { to } B=0.60 \frac{V_{12}^{2}}{2 g}$ .

$B \text { to } C=9.0 \frac{V_{12}^{2}}{2 g}$ .

$C \text { to } D=0.40 \frac{V_{6}^{2}}{2 g}$ .

$D \text { to } E=9.0 \frac{V_{6}^{2}}{2 g}$ .

Find (a) the flow Q in cfs, (b) the pressure at C in psi, and (c) the horsepower at C, datum E.

Question

Oil of specific gravity 0.761 flows from tank A to tank E as shown in Fig. 7-16. Lost head items may be assumed to be as follows:

[latex]A ext { to } B=0.60 \frac{V_{12}^{2}}{2 g}[/latex].

[latex]B ext { to } C=9.0 \frac{V_{12}^{2}}{2 g}[/latex].

[latex]C ext { to } D=0.40 \frac{V_{6}^{2}}{2 g}[/latex].

[latex]D ext { to } E=9.0 \frac{V_{6}^{2}}{2 g}[/latex].

Find (a) the flow Q in cfs, (b) the pressure at C in psi, and (c) the horsepower at C, datum E.

Accepted Answer

(a) Using A to E, datum E,

At A A to B B to C C to D D to E at E.

[latex](0+ ext { negl }+40.0)-\left[\left(0.60 \frac{V_{12}^{2}}{2 g}+9.0 \frac{V_{12}^{2}}{2 g} ight)+\left(0.40 \frac{V_{6}^{2}}{2 g}+9.0 \frac{V_{6}^{2}}{2} ight) ight]=(0+ neg g +0)[/latex].

[latex] ext { or } 40.0=9.6\left(V_{12}^{2} / 2 g ight)+9.4\left(V_{6}^{2} / 2 g ight) . \quad ext { Also, } V_{12}^{2}=\left(\frac{1}{2} ight)^{4} V_{6}^{2}=\frac{1}{16} V_{6}^{2} [/latex].

Substituting and solving,

[latex]V_{6}^{2} / 2 g=4 ft , \quad V_{6}=16.0 ft / sec , \quad ext { and } \quad Q=\frac{1}{4} \pi\left(\frac{1}{2} ight)^{2} imes 16.0=3.14 cfs[/latex].

(b) Using A to C, datum A,

[latex](0+ neg 1+0)-(0.60+9.0) \frac{V_{12}^{2}}{2 g}=\left(\frac{p_{C}}{\gamma}+\frac{V_{12}^{2}}{2 g}+2 ight) ext { and } \frac{V_{12}^{2}}{2 g}=\frac{1}{16} \frac{V_{6}^{2}}{2 g}=\frac{1}{16}(4)=\frac{1}{4} ft[/latex].

[latex] ext { Then } p_{C} / \gamma=-4.65 ft ext { of oil gage and } p_{C}=(0.761 imes 62.4)(-4.65) / 144=-1.53 ext { psi gage. }[/latex].

The Bernoulli equation could have been applied from C to E with equally satisfactory results.
The two equations obtained by the two choices would not be independent, simultaneous equations.

[latex] ext { (c) Horsepower at } C=\frac{\gamma Q H_{C}}{550}=\frac{(0.761 imes 62.4)(3.14)(-4.65+0.25+42.0)}{550}=10.19 ext {, datum } E [/latex].

Question 7.34: Oil of specific gravity 0.761 flows from tank A to tank E as...

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