Verizon 12:44 AM @ * 70%@iD. + session. masteringphysics.com C Practice Problem

Question

Verizon 12:44 AM @ * 70%@iD. + session. masteringphysics.com C Practice Problem 13.9 Let's begin by looking at a flow problem in which the diameter of a pipe changes along the flow path. We will need to use both the continuity equation and Bernoulli's equation for this problem Water enters a house through a pipe with an inside diameter of 2.0 cm at a gauge pressure of 4.0 × 10 Pa (about 4 atm, or 60 lb/in.2). The cold-water pipe leading to the second-floor bathroom 5.0 m above is 1.0 em in diameter. Find the flow speed and gauge pressure in the bathroom when the flow speed at the inlet pipe is 2.0 m/s. How much time would be required to fill a 100 L bathtub with cold water? SOLUTION SET UP Let point 1 be at the inlet pipe and point 2 at the bathroom. The speed t of the water at the bathroom is obtained from the continuity equation. Alth = A2U2. We take yl = 0 at the inlet and ½ = 5.0 m at the bathroom. We are given P1 an Bernoullis equation. To find the time required to fill the bathtub, we use the volume flow-rate relationship V/4t = Au. The bathtub's volume is 100 L = 100 × 10-3 m3 d vy: we can find P2 from SOLVE From the continuity equation, the flow speed t in the bathroom is v2 T2V1 = T(0.50 cm)" (2.0 m/s) = 8.0 m/s From Bernoulls equation, the gauge pressure P2 in the bathroom is 4.0 × 105 Pa ½ (1.0 x 103 kg/)(64 /s2-4.0 /s*) -(1.0 x 103 kg/) (9.50 m/s)(5.0 m) 3.2 × 10s Pa 3.2 atm 47 lb/in.2 = ( gaige pressure ) = The volume flow rate is A2v2 (0.50 × 10-2 m)2(8.0 m/s) 6.3 x 10-4 m3/s 0.63 L/s V/t = The time needed to fill the tub is 100 L/(0.63 L/s) 160 s. REFLECTNote that when the water is turned off, the second term on the right of the pressure equation vanishes, and the pressure rises to 3.5 × 10 Pa. In fact, when the fluid is not moving, Bemoullis equation reduces to the pressure relationship that we derived for a fluid at rest Part A - Practice Problem:

Explanation / Answer

let's begin with finding velocity for faucet

v2= ( pi )(1x 10^-2)^2 (2m/s) / ( pi ) (0.5x 10^-2)^2= 8 m/s

usng bernoulli

P1+ 1000 (9.8) ( 0) + 1/2 ( 1000) ( 8^2) = 4x 10^5 + 1000 (9.8) ( 2.8) + 1/2 (1000) (2^2)

P1= 500 ( 4 - 64) + 4x 10^5 + 27440 = 427440 -500 ( 60) =397440 Pa

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