Air is cooled as it flows through a 30-cm-diameter duct. The inlet conditions are Ma1 = 1.2, T01 = 350 K, and P01 = 240 kPa and the exit Mach number is Ma2 = 2.0. Disregarding frictional effects, determine the rate of cooling of air.

Given, Ma1 = 1.2, T01 = 350 K, P01 = 240 kPa Ma2 = 2.0 diameter of duct (d) = 30 cm

Answered: Air is cooled as it flows through a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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Air is cooled as it flows through a 30-cm-diameter duct. The inlet conditions are Ma₁ = 1.2, T₀₁ = 350 K, and P₀₁ = 240 kPa and the exit Mach number is Ma₂ = 2.0. Disregarding frictional effects, determine the rate of cooling of air.

Expert Solution

Step 1

Given,

Ma₁ = 1.2,

T₀₁ = 350 K,

P₀₁ = 240 kPa

Ma₂ = 2.0

diameter of duct (d) = 30 cm

Step 2

Taking the properties of air,

k = 1.4

c_p = 1.005 kJ/kg

R = 0.287 kJ/kg.K

Stagnation Properties:

Stagnation Temperture:

$T_{1} = T_{01} {(1 + \frac{k - 1}{2} M a_{1}^{2})}^{- 1} T_{1} = 350 {(1 + \frac{1.4 - 1}{2} 1 . 2^{2})}^{- 1} T_{1} = 271.7 K$

Stagnation Pressure:

$P_{1} = P_{01} {(1 + \frac{k - 1}{2} M a_{1}^{2})}^{\frac{- k}{(k - 1)}} P_{1} = 240 {(1 + \frac{1.4 - 1}{2} 1 . 2^{2})}^{\frac{- 1.4}{(1.4 - 1)}} P_{1} = 98.97 kPa$

Stagnation Density:

$ρ_{1} = \frac{P_{1}}{R T_{1}} ρ_{1} = \frac{98.97}{0.287 \times 271.7} ρ_{1} = 1.269 k g / m^{3}$

Step 3

The inlet velocity and the mass flow rate is:

$c_{1} = \sqrt{k R T_{1}} c_{1} = \sqrt{1.4 \times 0.287 \times 271.7 \times 1000} c_{1} = 330.4 m / s V_{1} = M a_{1} c_{1} V_{1} = 1.2 \times 330.4 V_{1} = 396.5 m / s$

$\dot{m_{a i r}} = ρ_{1} A_{c 1} V_{1} \dot{m_{a i r}} = 1.269 \times (π \times \frac{{0.3}^{2}}{4}) \times 396.5 \dot{m_{a i r}} = 35.57 kg / s$

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