Two water reservoirs A and B are connected to each other through a 39-m-long, 2-cm-diameter cast iron pipe with a sharp-edged entrance. The pipe also involves a swing check valve and a fully open gate valve. The water level in both reservoirs is the same, but reservoir A is pressurized by compressed air while reservoir B is open to the atmosphere at 95 kPa. If the initial flow rate through the pipe is 1.5 L/s, determine the absolute air pressure on top of reservoir A. Take the water temperature to be 10°C. The density and dynamic viscosity of water at 10°C are rho = 999.7 kg/m3 and μ = 1.307 × 10–3 kg/m·s. The loss coefficient is KL = 0.5 for a sharp-edged entrance, KL = 2 for a swing check valve, KL = 0.2 for the fully open gate valve, and KL = 1 for the exit. The roughness of a cast iron pipe is ε = 0.00026 m.

Question

contestada

Respuesta :

oeerivona oeerivona · Answer 1 · 2020-04-09T07:21:02+02:00

Answer:

The absolute air pressure, [tex]P_A[/tex] on top of reservoir is 3858.84 kPa

Explanation:

Given

[tex]P_A +\frac{\rho v^2_A}{2} + \rho gz_A =P_B +\frac{\rho v^2_B}{2} + \rho gz_B+ \frac{1}{2} \rho \cdot K L\cdot{u^2}[/tex]

Since

[tex]z_A = z_B[/tex]

[tex]P_A +\frac{\rho v^2_A}{2} =P_B +\frac{\rho v^2_B}{2} + \frac{1}{2} \rho \cdot K L\cdot{u^2}[/tex]

Also,

[tex]v_A = v_B[/tex], hence

[tex]P_A =P_B + \frac{1}{2} \rho \cdot K L\cdot{u^2}[/tex]

KL = 0.5 + 2 +0.2 = 2.7

Pressure loss of pipe = [tex]\Delta P = \frac{4fL\rho v^2}{2D}[/tex]

f from Moody chart at μ = 1.307 × 10⁻³ kg/m·s and

Roughness ε = 0.00026 m whereby, relative roughness is ε/d = (0.00026 m)/ 0.02 m = 0.013

f= 0.042 and

Q = v·A = 1.5 L/s = 0.0015 m³/s

v = Q/A where A = π·(0.02²)/4 = 3.1416×10⁻⁴ m²

v = (0.0015 m³/s)÷(3.1416×10⁻⁴ m²) =4.775 m/s

Pressure loss of pipe, [tex]\Delta P = \frac{4fL\rho v^2}{2D}[/tex] = 3733071.965 Pa

[tex]P_A =95000 + \frac{1}{2}999.7 \cdot 2.7\cdot{4.775^2} +3733071.965[/tex] = 3858839.04 Pa

The absolute air pressure, [tex]P_A[/tex] on top of reservoir = 3858.84 kPa