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Question

Air flows at the rate of 1.5 m3/s through a horizontal pipe with a gradually reducing cross-section as shown in the figure. The two cross-sections of the pipe have diameters of 400 mm and 200 mm. Take the air density as 1.2kg/m3 and assume inviscid incompressible flow. The change in pressure (p2p1) (in kPa) between sections 1 and 2 is

A
-1.28
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B
2.56
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C
-2.13
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D
1.28
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Solution

The correct option is A -1.28
p1ρg+V212g+Z1=p2ρg+V222g+Z2
[where, Z = constant]
A1=π4×0.42=0.1256 m2
A2=π4×0.22=0.0314 m2
p2p1ρg=V21V222g
p2p11.2=12[Q2A21Q2A22]
p2p1=0.6×(1.5)2[1A211A22]
p2p1=1.35[10.0157710.000985]
=1.35[63.41-1015.22]
=-1284.94 Pa=-1.28 kPa

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