Question #233489

liquid flows through an equal percentage valve at a rate of 4m^3/h and valve is 25% open. When the valve opens to 50%, the flow rate increases to 8m^3/h. Assume that the pressure drop across the valve and the density of the liquid remains constant whe the valve opens at 70%, find the flow rate.

Expert's answer

In this case

F(ω)F(ω) is Laplace transform of f(t)f(t). We choose f(t)=(1−e−t)f(t)=(1−e−t) and it is easy to show that

F(s)=L[1−e−t]=1s−1s+1F(s)=L[1−e−t]=1s−1s+1

For equal percentage valve f=f_{o}e^{bx}

Given, f = 2 at x = 0.1

f = 3 at x = 0.2

x = % of opening

f = flow rate

Now at x = 0.5 , what is f= 10.3 m^{3} h

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