Respuesta :
Answer:
a. The mass flow rate through the compressor = 0.07 [tex]\frac{lb}{sec}[/tex]
b. The value of area at inlet = 0.1337 [tex]ft^{2}[/tex]
Explanation:
Pressure at inlet [tex]P_{1}[/tex] = 15 psia = 103.4 kpa
Temperature at inlet [tex]T_{1}[/tex] = 70 °F = 294.3 K
Pressure at outlet [tex]P_{2}[/tex] = 200 psia = 1378 kpa
Temperature at outlet [tex]T_{2}[/tex] = 600 °F = 588.7 K
outlet area [tex]A_{2}[/tex] = 0.01 [tex]ft^{2}[/tex] = 0.00093 [tex]m^{2}[/tex]
Outlet velocity [tex]V_{2}[/tex] = 100 [tex]\frac{ft}{s}[/tex] = 30.48 [tex]\frac{m}{s}[/tex]
Inlet velocity [tex]V_{1}[/tex] = 50 [tex]\frac{ft}{s}[/tex] = 15.24 [tex]\frac{m}{s}[/tex]
Gas constant for helium R =2.077 [tex]\frac{KJ}{kg k}[/tex]
(a). Mass flow rate is given by the formula m = [tex]\rho_{2}[/tex] × [tex]A_{2}[/tex] × [tex]V_{2}[/tex] -------- (1)
Where [tex]\rho_{2}[/tex] is density at outlet. it is calculated by
⇒ [tex]\rho_{2}[/tex] = [tex]\frac{P_{2} }{R T_{2} }[/tex]
⇒ [tex]\rho_{2}[/tex] = [tex]\frac{1378}{2.077}[/tex] × [tex]\frac{1}{588.7}[/tex]
⇒ [tex]\rho_{2}[/tex] = 1.127 [tex]\frac{kg}{m^{3} }[/tex]
Put the value of [tex]\rho_{2}[/tex] in equation ( 1 ) we get
Mass flow rate m = 1.127 × 0.00093 × 30.48
⇒ m = 0.032 [tex]\frac{kg}{sec}[/tex]
⇒ m = 0.07 [tex]\frac{lb}{sec}[/tex]
This is the mass flow rate through the compressor.
(b). Mass flow rate is constant through the compressor so it is also given by
⇒ m = [tex]\rho_{1}[/tex] × [tex]A_{1}[/tex] ×[tex]V_{1}[/tex] ----------- ( 2 )
⇒ [tex]\rho_{1}[/tex] = [tex]\frac{P_{1} }{R T_{1}}[/tex]
Put all the values in above formula we get
⇒ [tex]\rho_{1}[/tex] = [tex]\frac{103.4}{2.077}[/tex] × [tex]\frac{1}{294.3}[/tex]
⇒ [tex]\rho_{1}[/tex] = 0.169 [tex]\frac{kg}{m^{3} }[/tex]
Put this value of [tex]\rho_{1}[/tex] in equation ( 2 ) we get,
⇒ m = 0.169 × [tex]A_{1}[/tex] × 15.24
⇒ 0.032 = 0.169 × [tex]A_{1}[/tex] × 15.24
⇒ [tex]A_{1}[/tex] = 0.01242 [tex]m^{2}[/tex] = 0.1337 [tex]ft^{2}[/tex]
This is the value of area at inlet.
