What is the discharge coefficient for the valve in Example 9.6?Example 9.6Even though the...

What is the discharge coefficient for the valve in Example 9.6?

Example 9.6

Even though the valve trains in engines operate transiently, the flow over the valves may be analyzed using steady-state head loss by invoking a quasi-steady assumption. The mass flow rate m is believed to be a function of the gas density ρ, the viscosity µ, the throat area of the valve A, and the pressure drop over the valve ΔP = P1 – P2.

SOLUTION First we write the units for each variable:

There are five variables and three repeated units (kg, m, s), so using the pi theorem we can reduce this problem to 5 – 3 = 2 nondimensional parameters, such as = f(µ*). First let’s find the nondimensional forms (pi groups) of the mass flow rate and viscosity. There are three different choices of parameters to eliminate kg from /it , but only one choice, ΔP, to eliminate s (we exclude µ, to be used for the Other nondimensional parameter). Thus   leaves us with units of  . Dividing by the square root of density will eliminate kg, and then dividing by A will eliminate, m, so that the final form of is

The viscosity µ, can be nondimensionalized using ρ, A, and ΔP:

 For some engine and valve configurations, the is constant for a wide range of Reynolds numbers. Assuming this holds true for our engine, if the mass flow rate of air is measured to be 0.86 kg/s when A = 0.004 m2, ρ = 1.2 kg/m3, and ΔP = 40.000 Pa, we can predict the mass flow rate for the same conditions and a pressure drop of 10,000 Pa in the actual engine. If , then . Since the areas and densities are constant between the two cases, the mass flow rate scaling is