Water at a gauge pressure of 3.0 atm at street level flows into an office buildi
ID: 1985807 • Letter: W
Question
Water at a gauge pressure of 3.0 atm at street level flows into an office building at a speed of .40 m/s through a pipe 5.0 cm in diameter. The pipe tapers down to 2.2 cm in diameter by the top floor, 15 m above, where the faucet has been left open. Calculate the flow velocity and the gauge pressure in such a pipe on the top floor. Assume no branch pipes and ignore viscosity.Explanation / Answer
hi, this is a same question with just different numericals . I am posting both question and answer. Please rate me Lifesaver !!!! Water at a gauge pressure of 3.8 atm at street level flows into an office building at a speed of 0.55 m/s through a pipe 5.0 cm in diameter. The pipe tapers down to 2.6 cm in diameter by the top floor, 20 m above, when the faucet has been left open. Calculate the flow velocity and the gauge pressure in such a pipe on the top floor. Assume no branch pipes and ignore viscosity. a)What is flow velocity? I got the right answer 2.03 m/s b) Pressure? in atm Four lawn sprinkler heads are fed by a 1.9 cm diameter pipe. The water comes out of the heads at an angle of 31° to the horizontal and covers a radius of 8.4 m. (a) What is the velocity of the water coming out of each sprinkler head? (Assume zero air resistance.) m/s (b) If the output diameter of each head is 3.0 mm, how many liters of water do the four heads deliver per second? L/s (c) How fast is the water flowing inside the 1.9 cm diameter pipe? m/s A 7.0 cm diameter pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 33.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow? m3/s 1) From Bernoulli's Law we have p1 + rho*g*y1 + 1/2*rho*v1^2 = p2 + rho*g*y2 + 1/2*rho*v2^2 Letting y1 =0 we have So 3.8atm + 1/2*rho*v1^2 = p2 + rho*g*y2 +1/2*rho*v2^2 Now at the top floor the velocity changes but A*v = constant..So v(top) = A*v(bottom)/Atop = pi*0.025^2*0.55m/s/(pi*0.013^2) = 2.03m/s Now the pressure p2 = 3.8atm + 1/2*rho*v1^2 - rho*g*y2 -1/2*rho*v2^2 = 3.8x1.013x10^5Pa + 1/2*1000kg/m^3*(0.55m/s)^2 - 1000kg/m^3*9.8m/s^2*20m -1/2*1000kg/m^3*(2.03m/s)^2 = 1.87x10^5Pa 2) Now the horizontal component of water = v0*cos(31) ..and x = 8.4m = v0*cos(31) therefore v0 = 8.4m/cos(31) = 9.8m/s b) Mass flow rate = rho (density)*A*v = 1000kg/m^3*pi*(1.5x10^-3m)^2*9.8m/s = 6.93x10^-2m^3 = 6.93x10^-2m^3*(1000 L/1m^3) = 69.3 L/s c) From continuity A1*v1 = A2*v2 .. so v1 = A2/A1*v1 = pi*(1.50x10^-3)^2/(pi*0.95x10^-2)^2*9.8m… = 0.24m/s 3) From Bernoulli's eqn we have p1 + rho*g*y1 + 1/2*rho*v1^2 = p2 + rho*g*y2 + 1/2*rho*v2^2 assuming the pipe is level then y1 = y2..so the eqn reduces to p1 - p2 = 1/2*rho*(v2^2 - v1^2) but from continuity we have A1*v1 = A2*v2 and since A = pi*d^2/4 then v2 = v1*A1/A2 reduces to v1*d1^2/d2^2 So p1 - p2 = 1/2*rho*(v1^2 *(d1/d2)^4 - v1^2) = 1/2*rho*v1^2((d1/d2)^2 - 1) => v1 = sqrt(2*(p1 - p2)/(rho*((d1/d2)^4-1)) = sqrt(2*(33x10^3 - 24x10^3)/(1000*(7/4)^4 -1)) = v1 = 2.15m/s So vol flow rate = A*v = pi((3.5x10^-2m)^2)*2.15m/s = 8.27x10^-3m^3/s
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