Explanation Given: Stratosphere begins at z = 11 km . Calculation: Refer Figure 2-11, “Approximate temperature distribution in the U.S standard temperature” Absolute temperature as 15 ° C . Calculate the constant ( T 0 ) using the relation. T = T 0 − C z Here, a constant is C , the absolute temperature is T , and the height is z . Substitute T = 15 ° C and z = 0 . 15 = T 0 − C ( 0 ) T 0 = 15 ° C Refer Figure 2-11, “Approximate temperature distribution in the U.S standard temperature” from textbook. The temperature of stratosphere is constant value of − 56.5 ° C in the region 11 km < z < 20 .1 km . Thus, the temperature at an elevation of z = 15 km is − 56.5 ° C . Calculate the constant C using the relation. T C = T 0 − C z (I) Substitute T C = − 56.5 ° C , T 0 = 15 ° C , and z = 11 km . − 56.5 = 15 − C ( 11 × 10 3 m ) C ( 11 × 10 3 m ) = 71.5 C = 6.5 × 10 − 3 °C / m Substitute C = 6.5 × 10 − 3 °C / m for C and T 0 = 15 ° C in Equation (I). T C = 15 − ( 6.5 × 10 − 3 ) z (II) Calculate the absolute temperature for 0 km using the relation. T = 273 + T C Substitute Equation (II) for T C . T = 273 + 15 − ( 6.5 × 10 − 3 ) z = 288 − ( 6.5 × 10 − 3 ) z (III) Calculate the unit weight of air ( γ ) using the relation. γ = ρ g Calculate pressure ( p ) using the relation. p = − γ z Substitute γ = ρ g in the above equation. p = − ( ρ g ) z Differentiate on both sides. d p = − ( ρ g ) d z ρ = − d p g d z (IV) Write the expression for the Ideal gas law. p = ρ R T p = ( − d p g d z ) R T d p p = − g R T d z (V) Refer Appendix A, “Physical Properties of Gases at Standard Atmospheric Pressure 101.3 kPa (SI Units)” Gas constant ( R ) as 286.9 J / kg .K . Calculate the value of p for 11 km by method of integration of Equation (5). ∫ p 0 p d p p = ∫ z 0 z − g R T d z Here, the height at ground level is z 0 and the pressure when height 0 is p 0 . Substitute z 0 = 0 , p 0 = 101 × 10 3 Pa , g = 9.81 m / s 2 , R = 286.9 J / kg .K and Equation (III) for T in the above equation. ∫ 101 × 10 3 p d p p = − 9.81 286.9 ∫ 0 z d z 288 − ( 6.5 × 10 − 3 ) z ( ln p ) 101 × 10 3 p = − 0.03419 ∫ 0 z d z ( 288 − 6

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 24P

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The temperature and pressure at an elevation of 15 km.

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