Air flows through a Air (rhoair= 12Kg / m3 ) flows through a 6cm diameter inclin

Question

Air flows through a

Air (rhoair= 12Kg / m3 ) flows through a 6cm diameter inclined duct at the rate of 0.045m3/s. (See Figure 2)The duct diameter reduces to 4cm through a reducer. The pressure change is measured by a water manometer. The elevation difference between the center of the duct at the two points where the arms of the manometer attach to the duct is 0.20m. These points are marked by 1 and 2 in the figure. P water = l000Kg / m3 Find V, and V2. From the Bernoulli Equation, find the pressure difference between 1 and 2 Find the differential height, h, between the fluid levels of the arms of the manometer, using the result of part b and the calculation from the manometer.

Explanation / Answer

Given Flow rate of air, Q = 0.045 m^3/s

We know

Q = A1*V1 = A2*V2

a)

A1*V1 = 0.045

(pi/4)*D1^2*V1 = 0.045

(pi/4)*0.06^2*V1 = 0.045

V1 = 15.924 m/s

Also

A2*V2 = 0.045

(pi/4)*0.04^2*V2 = 0.045

V2 = 35.828 m/s

b) Applying Bernoulis Equation at Point 1 and 2

P1 + 1/2*rho(air)*V1^2 + rho(air)*g*Z1 = P2 + 1/2*rho(air)*V2^2 + rho(air)*g*Z2

Given

(Z1 - Z2) = 0.2 m

P1 - P2 = 1/2*rho(air)*(V2^2-V1^2) - rho(air)*g*(Z1-Z2)

P1-P2 = 1/2*1.2*(35.828^2 - 15.924^2) - 1.2*9.81*0.2

(P1-P2) = 615.69 Pa

c) From Manometer

We can write

P2 = P1 + rho(air)*g*(H+h) - rho(water)*g*h

Here

H = 0.2 m

g*h*{rho(water) - rho(air)} = (P1-P2) + rho(air)*g*H

9.81*h*998.8 = 618.0444

h = 63.08 mm

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