Question

A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose with a speed of 7.0 feet per second. If you then put your thumb over the end of the hose so the opening is only a circle 0.25" across, with what speed will the water stream out now?

Answer 1

Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
$v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \: \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142$
$we \: can \: only \: assume \:that \\ flow \: (q) \:stays \: same \: since \: it \\ isnt \: impeded \: by \: anything \\ thus \:it \: (q)\: stays \: the \: same \: \\ so \: 4q \: can \: be \: removed \: from \: \\ the \: equation$
$then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to > > \\ (v1 \times {d1}^{2} \pi) = (v2 \times {d2}^{2}\pi)$
$both \: \pi \: will \: cancel \: each \: other \: out \: \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \: =$

$(v1 \times {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times {0.052}^{2}) = (v2 \times {0.021}^{2}) \\ divide \: both \: sides \: by \: {0.021}^{2} \\ to \: solve \: for \: v2 > >$
$v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2}) \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second$
new velocity coming out of the hose then is
44 ft/sec