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Anonymous
Anonymous asked in Science & MathematicsPhysics · 9 years ago

A long horizontal hose of diameter 4.6 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.8 cm. Water squirts from the nozzle at velocity 26.0 m/sec. Assume that the water has no viscosity or other form of energy dissipation,

I figured out the first two need help with the last question!

1) What is the velocity of the water in the hose ?

m/s 3.98

2) What is the pressure differential between the water in the hose and water in the nozzle ?

Pa 330079

3) How long will it take to fill a tub of volume 70.0 liters with the hose ?

sec

Relevance
• donpat
Lv 7
9 years ago

Get the flow areas:

---------------------------------------------------------------------------

AFH = ( pi / 4 ) ( DH^2 )

AFH = ( pi / 4 ) ( 4.6^2 ) = 16.6190 sq. cm.

AFNZ = ( pi / 4 ) ( DNZ^2 )

AFNZ = ( pi / 4 ) ( 1.8^2 ) = 2.5447 sq. cm.

Get the water volumetric flow rate :

------------------------------------------------------------------

QF = ( AFH ) ( VH ) = ( AFNZ ) ( VFNZ )

QF = ( 26.0 ) ( 2.5447 ) ( 1 / 100^2 ) ( 1000 ) = 6.616 L per sec. <-------------------------

VH = QF / AFH

VH = ( 6.616 ) ( 1/1000 ) ( 100^2 ) / (16.6190 ) = 3.98 m. per sec. <-----------------------------

Apply Bernoulli Equation to get PH - PNZ :

---------------------------------------------------------------

PH - PNX = ( rho ) ( g ) ( VNZ^2 - VH^2 ) ( 1/2g )

PH - PNZ = ( 1000) ( 9.807 ) ( 26.0^2 - 3.98^2 )( 1/ 2 ) ( 1/ 9.807 )

PH - PNZ = 330,080 Pa <------------------------------------------------

Get time to fill the tub :

---------------------------------------------------------------------

tF = VT / QF

tF = 70.0 / 6.616 = 10.6 sec. <-----------------------------------------

Note :

------------------------------------------------------------------------

QF = QFH = QFNZ for incompressible flow