Concept explainers
Find the average density of the gas exiting the engine.
Answer to Problem 45P
The average density of the gas exiting the engine in kgm3, slugsm3, and lbmft3 are 0.155 kgm3, 3×10−4 slugsm3 and 0.0097 lbmft3 respectively.
Explanation of Solution
Given data:
Consider that the inlet area of the engine as 5 ft2.
Ain=5 ft2 (1)
Consider that the outlet area of the engine as 4 ft2.
Aout=4 ft2 (2)
Formula used:
Consider the following expression.
(air mass flow rate)in+(fuel mass flow rate)in=(combustion mass flow rate)out
In this problem, the fuel mass flow rate is neglected since it is negligible value when compared air mass flow rate.
Consider that the mass flow rate of air enters is equal to the mass flow rate of exists.
(mass flow rate)in=(mass flow rate)out (3)
Write the expression for the mass flow rate.
mass flow rate=(density)(volume flow rate) (4)
Calculation:
Convert the unit of the value in equation (1) from ft2 to m2.
Ain=5×(1(3.28084)2) m2 [∴1 m =3.28084 ft]=0.464 m2
Convert the unit of the value in equation (2) from ft2 to m2.
Aout=4×(1(3.28084)2) m2 [∴1 m =3.28084 ft]=0.372 m2
Substitute equation (4) in (3),
[(density)(volume flow rate)]in=[(density)(volume flow rate)]outρin(volume flow rate)in=ρout(volume flow rate)out(ρin)(Vin)(Ain)=(ρout)(Vout)(Aout) [∴volume flow rate=(velocity)(area)]
Substitute 0.45 kgm3 for ρin, and 180 ms for Vin, 650 ms for Vout, 0.464 m2 for Ain, and 0.372 m2 for Aout,
(0.45 kgm3)(180 ms)(0.464 m2)=(ρout)(650 ms)(0.372 m2)
ρout=0.155 kgm3 (5)
Convert the unit of the value in equation (5) from kgm3 to slugsm3.
ρout=0.155 (114.5939) slugsm3 [∴1 slug=14.5939 kg]=3×10−4 slugsm3
Convert the unit of the value in equation (5) from kgm3 to lbmft3.
ρout=0.155 kgm3(2.20462 lbm1 kg)1 m3(3.28084)3 ft3 [∴1 kg=2.20462 lbm, and 1 m =3.28084 ft]=0.0097 lbmft3
Conclusion:
Thus, the average density of the gas exiting the engine in kgm3, slugsm3, and lbmft3 are 0.155 kgm3, 3×10−4 slugsm3 and 0.0097 lbmft3 respectively.
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