Question

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.

Answer 1

The absolute air pressure,
`P_A` on top of reservoir is 3858.84 kPa

Step-by-step explanation:

Given

`P_A +(\rho v^2_A)/(2) + \rho gz_A =P_B +(\rho v^2_B)/(2) + \rho gz_B+ (1)/(2) \rho \cdot K L\cdot{u^2}`

Since

`z_A = z_B`

`P_A +(\rho v^2_A)/(2) =P_B +(\rho v^2_B)/(2) + (1)/(2) \rho \cdot K L\cdot{u^2}`

Also,

`v_A = v_B`, hence

`P_A =P_B + (1)/(2) \rho \cdot K L\cdot{u^2}`

KL = 0.5 + 2 +0.2 = 2.7

Pressure loss of pipe =
`\Delta P = (4fL\rho v^2)/(2D)`

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,
`\Delta P = (4fL\rho v^2)/(2D)` = 3733071.965 Pa

`P_A =95000 + (1)/(2)999.7 \cdot 2.7\cdot{4.775^2} +3733071.965` = 3858839.04 Pa

The absolute air pressure,
`P_A` on top of reservoir = 3858.84 kPa