8.5 The friction factor for a 25 mm diameter 11.5 m long pipe is 0.004. The conditions of air at entry are P₁ = 2.0 bar, T₁ = 301 K, M₁ = 0.25 Determine the mass flow rate, and the pressure, temperature and the Mach number at exit. (m = 0.098 kg/s, p₂ = 1.002 bar, T₂ = 290.8 K, M₂ = 0.49).

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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Need solutions for 8.5 and 8.6
8.5 The friction factor for a 25 mm diameter 11.5 m long pipe is 0.004. The conditions of air at entry
are
P₁ = 2.0 bar, T₁ = 301 K, M₁ = 0.25
Determine the mass flow rate, and the pressure, temperature and the Mach number at exit.
1.002 bar, T₂ = 290.8 K, M₂ = 0.49).
(m = 0.098 kg/s, P2
-
8.6 Repeat Example 8.4 for a gas having y = 1.3 and R = 0.469 kJ/kg-K. The values calculated are
given in Table 8.6.
Table 8.6 Comparison of isothermal and adiabatic flows
Lmax
L
metre
metre
Process
Isothermal
Adiabatic
87 AL
5.79
M=0.5
P2
bar
2.143
M = 0.7
6.535
2.107
M=0.5 M=0.7
T₂
K
312
6.618
Mt = 0.877
301.58
P₁
bar
1.71
1.425
T₁
K
312
281.58
8.139
M* = 1
con coefficient of friction of 0.003. Air enters the pipe
Transcribed Image Text:8.5 The friction factor for a 25 mm diameter 11.5 m long pipe is 0.004. The conditions of air at entry are P₁ = 2.0 bar, T₁ = 301 K, M₁ = 0.25 Determine the mass flow rate, and the pressure, temperature and the Mach number at exit. 1.002 bar, T₂ = 290.8 K, M₂ = 0.49). (m = 0.098 kg/s, P2 - 8.6 Repeat Example 8.4 for a gas having y = 1.3 and R = 0.469 kJ/kg-K. The values calculated are given in Table 8.6. Table 8.6 Comparison of isothermal and adiabatic flows Lmax L metre metre Process Isothermal Adiabatic 87 AL 5.79 M=0.5 P2 bar 2.143 M = 0.7 6.535 2.107 M=0.5 M=0.7 T₂ K 312 6.618 Mt = 0.877 301.58 P₁ bar 1.71 1.425 T₁ K 312 281.58 8.139 M* = 1 con coefficient of friction of 0.003. Air enters the pipe
Example 8.4. Air enters a long circular duct (d = 12.5 cm, f = 0.0045) at a Mach number
0.5, pressure 3.0 bar and temperature 312 K. If the flow is isothermal throughout the duct
determine (a) the length of the duct required to change the Mach number to 0.7, (b) pressure and
temperature of air at M = 0.7, (c) the length of the duct required to attain limiting Mach number,
and (d) state of air at the limiting Mach number.
Compare these values with those obtained in adiabatic flow.
Solution. (a)
4f Lmax
D
4f Lmax
D
M₂
4fL
D
M₁ = 0.5
=
=
1-YM²
YM²
1-1.4 x 0.25
1.4 x 0.25
1-1.4 x 0.49
1.4 x 0.49
-+ In YM²
+ ln 1.4 x 0.25 = 0.807
+ In 1.4 x 0.49 = 0.0808
= 0.807-0.0808 = 0.7262
SE
CC
ne
T
Transcribed Image Text:Example 8.4. Air enters a long circular duct (d = 12.5 cm, f = 0.0045) at a Mach number 0.5, pressure 3.0 bar and temperature 312 K. If the flow is isothermal throughout the duct determine (a) the length of the duct required to change the Mach number to 0.7, (b) pressure and temperature of air at M = 0.7, (c) the length of the duct required to attain limiting Mach number, and (d) state of air at the limiting Mach number. Compare these values with those obtained in adiabatic flow. Solution. (a) 4f Lmax D 4f Lmax D M₂ 4fL D M₁ = 0.5 = = 1-YM² YM² 1-1.4 x 0.25 1.4 x 0.25 1-1.4 x 0.49 1.4 x 0.49 -+ In YM² + ln 1.4 x 0.25 = 0.807 + In 1.4 x 0.49 = 0.0808 = 0.807-0.0808 = 0.7262 SE CC ne T
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