The exit mass flow rate as a function of pressure, temperature, gas constant, specific heat and nozzle diameter.
Answer to Problem 5.30P
Required dimensionless function
Explanation of Solution
Given information:
ṁ is a function of P0, To,R,cp and D.
ṁ = f (ṁ, P0, T0, R, Cp, D)
or g(ṁ, P0, T0, R, Cp, D) = 0
Dimensions of ṁ = [MT-1]
Dimensions of P = [ML-1 T-2]
Dimensions of R = [L2 T-2 θ- 1]
Dimensions of D = [L]
Dimensions of CP = [L2 T-2 θ- 1]
Dimensions of TO = [θ]
Total number of variables m = 6
Number of fundamental variables n = 4 [MLTθ]
So, Number of (terms = m-n = 2(π1,π2)
Calculation:
Taking (P0, T0, R, D) as repeating variables: -
comparing powers of [MLTθ]
a1+1 = 0
-a1+2c1+d1 = 0
-2a1-2c1-1 = 0
b1-c1 = 0
solving above equations
a1 = -1
b1 = ½
c1 = ½
d1 = -2
so first π-term :-
Second π-term :-
comparing powers of [MLTθ]
a2 = 0
-a2+2c2+d2+2 = 0
-2a2-2c2-2 = 0
b2-c2-1 = 0
solving above equations
a2 = 0
b2 = 0
c2 = -1
d2 = 0
so second π-term :-
Thus π1 = f(π2)
Conclusion:
The required dimensionless function is
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