Answer to Question #238575 in Molecular Physics | Thermodynamics for Angelo

Question #238575

Two gaseous streams enter a combining tube and leave as a single mixture. These data apply at the entrance section:


For one gas : A1 = 55 cm2 U1 = 400fps    v1 = 10 ft3/lbm

For another gas: A2 = 50 cm2  m2 = 12.22lbm/s    р = 00.12lbm/ft3

At exit section: U3 = 350 fps      v3 = 8ft3/lbm

Find (a) U2 at section 2 (b) mass flow rate at the exit


1
Expert's answer
2021-09-17T15:32:43-0400

"A_1 = 55 \\;cm^2 = 0.0592 \\;ft^2 \\\\\n\nU_1 = 400 \\;fps \\\\\n\nv_1 = 10 \\;ft^3\/lb \\\\\n\nP_1 = 0.1 \\;\\frac{lb}{ft^3} \\\\\n\nA_2 = 50 \\;cm^2 = 0.0538 \\;ft^2 \\\\\n\nm_2 = 12.22 lb\/s \\\\\n\nP_2 = 0.12 \\;lb\/ft^3 \\\\\n\nU_3 = 350 \\;fps \\\\\n\nv_3 = 8 \\;ft^3\/lb \\\\\n\nP_3 = \\frac{1}{8} \\; lb\/ft^3"

(a) To find speed at section 2

Mass flow rate at section 2:

"m_2 = P_2A_2U_2 \\\\\n\nU_2 = \\frac{m_2}{P_2A_2} \\\\\n\n= \\frac{12.22 \\;lb\/ft^3}{0.12 \\; lg\/ft^3 \\times 0.0538 \\;ft^2} \\\\\n\n= 1892.21 \\;fps"

(b) To find mass flow rate at the exit

"m_1=P_1A_1U_1 \\\\\n\nm_1 = 0.1 \\times 0.0592 \\times 400 = 2.368 \\; lb\/s"

Conservation of mass

"m_3 = m_1+m_2 \\\\\n\n= 2.368 + 12.22 \\\\\n\n= 14.588 \\;lb\/s"


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Comments

CJTS
30.09.21, 08:26

Thank you very much❤️

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