Chapter 5, Problem 5.30P

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To determine

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 $\stackrel{•}{m}=\frac{P_o{D}^2}{\stackrel{.}{m}{T}_o{}^{\frac{1}{2}}{R}^{\frac{1}{2}}}f\left(\frac{C_p}{R}\right)$

Explanation of Solution

Given information:

$\begin{array}{l}Massflowrate=\stackrel{•}{m}\hfill \\ Tankpressure=P_o\hfill \\ Temperature=T_o\hfill \\ Gasconstant=R\hfill \\ Specificheat=C_p\hfill \\ Nozzlediameter=D\hfill \end{array}$

ṁ is a function of P₀, To,R,c_p and D.

ṁ = f (ṁ, P₀, T₀, R, C_p, D)

or g(ṁ, P₀, T₀, R, C_p, 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 C_P = [L² T^-2 θ^- 1]

Dimensions of T_O = [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: -

$\begin{array}{l}{\pi }_1=P^{a_1}{T}_o{}^{b_1}{R}^{c_1}{D}^{d_1}\stackrel{.}{m}\\ \left[M^0{L}^0{T}^0{\theta }^0\right]={\left[M{L}^{-1}{T}^{-2}\right]}^{a_1}{\left[\theta \right]}^{b_1}{\left[L^2{T}^{-2}{\theta }^{-1}\right]}^{c_1}{\left[L\right]}^{d_1}\left[M{T}^{-1}\right]\\ \left[M^0{L}^0{T}^0{\theta }^0\right]=\left[M^{a_1}{}^{+1}{L}^{-a_1}{{}^{+2c_1}}^{+d_1}{T}^{-}{{}^{2a_1}}^{-2c_1}{}^{-1}{\theta }^{b_1}{}^{-c_1}\right]\end{array}$

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 :-

$\begin{array}{l}{\pi }_1=P_o{}^{-1}{T}_o{}^{\frac{1}{2}}{R}^{\frac{1}{2}}{D}^{-2}\stackrel{.}{m}\\ {\pi }_1=\frac{T_o{}^{\frac{1}{2}}{R}^{\frac{1}{2}}\stackrel{.}{m}}{P_o{D}^2}\end{array}$

Second π-term :-

$\begin{array}{l}{\pi }_2=P^{a_2}{T}_o{}^{b_2}{R}^{c_2}{D}^{d_2}{C}_p\\ \left[M^0{L}^0{T}^0{\theta }^0\right]={\left[M{L}^{-1}{T}^{-2}\right]}^{a_2}{\left[\theta \right]}^{b_2}{\left[L^2{T}^{-2}{\theta }^{-1}\right]}^{c_{{}_2}}{\left[L\right]}^{d_2}\left[L^2{T}^{-2}{\theta }^{-1}\right]\\ \left[M^0{L}^0{T}^0{\theta }^0\right]=\left[M^{a_2}{L}^{-a_2}{{}^{+2c_2}}^{+d_2+2}{T}^{-}{{}^{2a_2}}^{-2c_2}{}^{-2}{\theta }^{b_2}{}^{-c_2-1}\right]\end{array}$

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 :-

$\begin{array}{l}{\pi }_2=P^0{T}_o{}^0{R}^{-1}{D}^0{C}_p\\ {\pi }_2=\frac{C_p}{R}\end{array}$

Thus π₁ = f(π₂)

$\begin{array}{l}{\pi }_1=f\left({\pi }_2\right)\\ \frac{T_o{}^{\frac{1}{2}}{R}^{\frac{1}{2}}\stackrel{.}{m}}{P_o{D}^2}=f\left(\frac{C_p}{R}\right)\\ \stackrel{.}{m}=\frac{P_o{D}^2}{\stackrel{.}{m}{T}_o{}^{\frac{1}{2}}{R}^{\frac{1}{2}}}f\left(\frac{C_p}{R}\right)\end{array}$

Conclusion:

The required dimensionless function is

$\stackrel{.}{m}=\frac{P_o{D}^2}{\stackrel{.}{m}{T}_o{}^{\frac{1}{2}}{R}^{\frac{1}{2}}}f\left(\frac{C_p}{R}\right)$.