I need a definite answer. A 6.8 cm diameter pipe gradually narrows to 4.0 cm. Wh
ID: 1984175 • Letter: I
Question
I need a definite answer.A 6.8 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?
Explanation / Answer
Volume flowrate-constant // V-dot=m3/s=A*v=cross-section area * speed V-dot = A * V = constant = (1/4)pD2v // V1* D12 = V2 D22 // V2 = V1*[D1/D2]2 = 2.89V1 ----- (1) P1 + 0.5rho V12 = P2 + 0.5 rho V22 // (P1-P2) = 0.5 rho[V22 - V12] -- (2) here whether P1, P2 are absolute pressure or gauge pressure, difference is same, >>water density = 1000 =rho (35-24)1000 = 0.5*1000[V22 - V12] // 11(2) = V22 - V12 // [V22 - V12] = 22 >>> put in v2 here from (1) // [2.89*2.89 - 1]V12 = 22 V12 = 2.992342324 // V1=1.729.. m/s // V-dot = vol flow rate =A1V1 = [piD12/4] v1 // [3.14*0.068*0.068/4]*1.729838814 = V-dot = 0.00628 m3/s
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