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Chapter 4

Q. 4.P.7

Air, at a pressure of 10 MN/m² and a temperature of 290 K, flows from a reservoir through a mild steel pipe of 10 mm diameter and 30 m long into a second reservoir at a pressure P_2. Plot the mass rate of flow of the air as a function of the pressure P_2. Neglect any effects attributable to differences in level and assume an adiabatic expansion of the air. μ = 0.018 mN s/m², γ = 1.36.

Step-by-Step

Verified Solution

G/A is required as a function of P_2 \cdot  v_2 cannot be found directly since the downstream temperature T_2 is unknown and varies as a function of the flowrate. For adiabatic flow, v_2 may be calculated from equation 4.77 using specified values of G/A and substituted in equation 4.72 to obtain the value of P_2. In this way the required data may be calculated.

8\left(R / \rho u^2\right)(l / d)=\left[\frac{\gamma-1}{2 \gamma}+\frac{P_1}{v_1}\left(\frac{A}{G}\right)^2\right]\left[1-\left(\frac{v_1}{v_2}\right)^2\right]-\frac{\gamma+1}{\gamma} \ln \left(\frac{v_2}{v_1}\right)

(equation 4.77)

0.5(G / A)^2 v_1^2+[\gamma /(\gamma-1)] P_1 v_1=0.5(G / A)^2 v_2^2+[\gamma /(\gamma-1)] P_2 v_2

(equation 4.72)

or:                 \frac{0.5(G / A)^2\left(v_1^2-v_2^2\right)+[\gamma /(\gamma-1)] P_1 v_1}{[\gamma /(\gamma-1)] v_2}=P_2

When P_2=P_1=10  MN / m ^2,  G / A=0.
If G/A is 2000 kg/m²s, then:

R e=\left(0.01 \times 2000 / 0.018 \times 10^{-3}\right)=1.11 \times 10^6

When e / d=0.0002, R / \rho u^2=0.0028 from Fig. 3.7 and:

v_1=(22.4 / 29)(290 / 273)(0.1013 / 10)=0.0083  m ^3 / kg

Substituting in equation 4.77:

8(0.0028)(30 / 0.01)=\left[\frac{0.36}{2 \times 1.36}+\frac{10 \times 10^6}{0.0083}\left(\frac{1}{2000}\right)^2\right]

\times\left[1-\left(\frac{0.0083}{v_2}\right)^2\right]-\frac{2.36}{1.36} \ln \left(\frac{v_2}{0.0083}\right)

and: v_2=0.00942 m ^3 / kg.
Substituting for v_2 in equation 4.72 gives:

P_2=\left[0.5(2000)^2\left(0.0083^2-0.00942^2\right)+(1.36 / 0.36) 10 \times 10^{6 \times 0.0083}\right] /(1.36 / 0.36)

\times 0.00942

and: P_2=8.75  MN / m ^2.

In a similar way the following table may be produced.

G / A\left( kg / m ^2 s \right) v_2\left( m ^3 / kg \right) P_2\left( MN / m ^2\right)
0 0.0083 10.0
2000 0.00942 8.75
3000 0.012 6.76
3500 0.0165 5.01
4000 0.025 3.37
4238 0.039 2.04

These data are plotted in Fig. 4a. It is shown in Section 4.5.4, Volume 1, that the maximum velocity which can occur in a pipe under adiabatic flow conditions is the sonic velocity which is equal to \sqrt{\gamma P_2 v_2}.

From the above table \sqrt{\gamma P_2 v_2} at maximum flow is:

\sqrt{1.36 \times 2.04 \times 10^6 \times 0.039}=329  m / s

The temperature at this condition is given by P_2 v_2= R T / M, and:

T_2=\left(29 \times 0.039 \times 2.04 \times 10^6 / 8314\right)=227  K

The velocity of sound in air at 227 K = 334 m/s, which serves as a check on the calculated data.

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