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Semmy [17]
3 years ago
6

Air exits a turbine at 200 kPa and 1508C with a volumetric flow rate of 7000 liters/s. Modeling air as an ideal gas, determine t

he mass flow rate, in kg/s.
Physics
1 answer:
liraira [26]3 years ago
8 0

Answer:

mass flow rate = 11.464 kg/sec

Explanation:

given data

P = 200 kPa = 2 × 10^{5}  Pa

temperature = 150°C  = 423 k

volumetric flow rate = 7000 liters/s

solution

first we get here molar flow rate that is express as

molar flow rate  = \frac{Pv}{RT}   .................1

put here value

molar flow rate  = \frac{2 \times 10^5 \times 7}{8.314\times 423}  

molar flow rate  = 398.06 mol/sec

and

now we get mass flow rate

mass flow rate = molar flow rate × average  mol weight ..............2

mass flow rate = 398.06 × 28.8

mass flow rate = 11464.3 g/sec

mass flow rate = 11.464 kg/sec

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To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

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Where,

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\sum F = m_b a

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In the case of mass A,

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T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

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T_BR_0-T_AR_0=I\alpha

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Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

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The intensity of an electromagnetic wave is given by

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P is the power

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P=182 kW = 1.82\cdot 10^5 W is the power

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