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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

(a) To derive: The equation d p d z = − M g R t p and solve it for the “isothermal” case where T is constant to obtain the barometric pressure equation p ( z ) = p ( z 0 ) exp [ − M g ( z − z 0 ) R T ] . The drop in air pressure p between the top and bottom of a cylindrical volume element of height Δ z and cross section area A equals the weight of the air enclosed, per unit area: p ( z + Δ z ) − p ( z ) = − ρ ( z ) ( A Δ z ) g A = − ρ ( z ) g Δ z This can be written as, p ( z + Δ z ) − p ( z ) Δ z = − ρ ( z ) g . Let Δ z approaches 0, then lim Δ z → 0 p ( z + Δ z ) − p ( z ) Δ z = − lim Δ z → 0 ( ρ ( z ) g ) d p d z = − ρ g Given that, the density ρ = M p R T . Thus, the above differential equation becomes d p d z = − M p R T g . Here, R is the universal gas constant, M is the molar mass of the air, and g is gravitational constant. Suppose we assume that the absolute temperature T is the constant, then the term M g R T is a constant and take it as A . Thus, the differential equation becomes d p d z = − A p Separate the variables and rewrite the differential equation as follows. 1 p d p = − A d z Integrate both sides of the equation. ∫ 1 p d p = − ∫ A d z ln p = − A z + C p = e − A z + C p = k e − A z ( where k = e C ) For the initial height z 0 , the pressure is b . That is, p ( z 0 ) = b . Apply the initial condition p ( z 0 ) = b in the solution p = k e − A z to find the constant k . p ( z 0 ) = k e − A z 0 b = k e A z 0 k = b e A z 0 Substitute k = b e A z 0 in p = k e − A z . p ( z ) = b e A z 0 e − A z = p ( z 0 ) e − A ( z − z 0 ) ( ∵ p ( z 0 ) = b ) = p ( z 0 ) e − M g ( z − z 0 ) R T ( Substitute back, A = M g R T ) = p ( z 0 ) exp ( − M g ( z − z 0 ) R T ) Thus, the barometric pressure equation is p ( z ) = p ( z 0 ) exp [ − M g ( z − z 0 ) R T ] . (b) To derive: The solution p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } when the temperature also varies with altitude T = T ( z ) . From part (a), the differential equation is d p d z = − M p R T g . Here, R is the universal gas constant, M is the molar mass of the air, and g is gravitational constant. Suppose we assume that the temperature varies with altitude T = T ( z ) . Then, the differential equation becomes d p d z = − M g R p T ( z ) Separate the variables and rewrite the differential equation as follows. 1 p ( z ) d p = − M g R d z T ( z ) The value of z starts at z 0 because the initial condition is p ( z 0 ) = b . Also, the end value is unknown. Therefore, the limits of integration would be [ z 0 , z ] . Integrate both sides of the equation from ζ = z 0 to ζ = z where ζ be the variable of integration. ∫ z 0 z 1 p ( ζ ) d p = − ∫ z 0 z M g R d ζ T ( ζ ) [ ln p ( ζ ) ] z 0 z = − M g R ∫ z 0 z d ζ T ( ζ ) l n p ( z ) − l n p ( z 0 ) = − M g R ∫ z 0 z d ζ T ( ζ ) ln ( p ( z ) p ( z 0 ) ) = − M g R ∫ z 0 z d ζ T ( ζ ) Simplify further, p ( z ) p ( z 0 ) = exp { − M g R ∫ z 0 z d ζ T ( ζ ) } p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } Thus, the required solution p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } is derived. (c) The height of the building when the absolute temperature varies as T ( z ) = 288 − 0.0065 ( z − z 0 ) . The height of the building is 168.1 m _ . Given: The barometric pressure at the top of building is p ( z ) = 99 , 000 Pa . The barometric pressure at the base is p ( z 0 ) = 101 , 000 Pa . The absolute temperature varies as T ( z ) = 288 − 0.0065 ( z − z 0 ) . The universal gas constant R is, 8.31 N-m mol-K . The molar mass of the air M is 0.029 kg mol . The gravitational constant g is 9.8 m sec 2 Calculation: Given that, the absolute temperature varies as T ( z ) = 288 − 0.0065 ( z − z 0 ) . From part (b), if the temperature also varies with altitude T = T ( z ) , then the solution p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } . Substitute the known values in the solution as follows. 99 , 000 = 101 , 000 exp { − ( 0.029 ) ( 9.8 ) 8.31 ∫ z 0 z d ζ 288 − 0.0065 ( ζ − z 0 ) } 99 , 000 101 , 000 = exp { − 0.2842 8.31 ∫ z 0 z d u − 0.0065 u } ( Take u = 288 − 0.0065 ( ζ − z 0 ) , then d u = − 0.0065 d ζ ) ln 99 101 = 0.2842 0.054015 [ ln u ] z 0 z 0.054015 0.2842 ln 99 101 = [ ln ( 288 − 0.0065 ( ζ − z 0 ) ) ] z 0 z ( Substitute back, u = 288 − 0.0065 ( ζ − z 0 ) ) Simplify further, ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 ( z − z 0 ) ) − ln ( 288 − 0.0065 ( z 0 − z 0 ) ) ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 ( z − z 0 ) ) − ln ( 288 − 0.0065 ( 0 ) ) ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 h ) − ln ( 288 ) ( ∵ h = z − z 0 ) ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 h 288 ) Raise e to the power of both sides. ( 99 101 ) 0.054015 0.2842 = 288 − 0.0065 h 288 288 ( 99 101 ) 0.054015 0.2842 = 288 − 0.0065 h 288 ( 99 101 ) 0.054015 0.2842 − 288 = − 0.0065 h h = 288 ( 1 − ( 99 101 ) 0.054015 0.2842 ) 0.0065 h ≈ 168.1 Thus, the height of the building is 168.1 m _ .

(a) To derive: The equation d p d z = − M g R t p and solve it for the “isothermal” case where T is constant to obtain the barometric pressure equation p ( z ) = p ( z 0 ) exp [ − M g ( z − z 0 ) R T ] . The drop in air pressure p between the top and bottom of a cylindrical volume element of height Δ z and cross section area A equals the weight of the air enclosed, per unit area: p ( z + Δ z ) − p ( z ) = − ρ ( z ) ( A Δ z ) g A = − ρ ( z ) g Δ z This can be written as, p ( z + Δ z ) − p ( z ) Δ z = − ρ ( z ) g . Let Δ z approaches 0, then lim Δ z → 0 p ( z + Δ z ) − p ( z ) Δ z = − lim Δ z → 0 ( ρ ( z ) g ) d p d z = − ρ g Given that, the density ρ = M p R T . Thus, the above differential equation becomes d p d z = − M p R T g . Here, R is the universal gas constant, M is the molar mass of the air, and g is gravitational constant. Suppose we assume that the absolute temperature T is the constant, then the term M g R T is a constant and take it as A . Thus, the differential equation becomes d p d z = − A p Separate the variables and rewrite the differential equation as follows. 1 p d p = − A d z Integrate both sides of the equation. ∫ 1 p d p = − ∫ A d z ln p = − A z + C p = e − A z + C p = k e − A z ( where k = e C ) For the initial height z 0 , the pressure is b . That is, p ( z 0 ) = b . Apply the initial condition p ( z 0 ) = b in the solution p = k e − A z to find the constant k . p ( z 0 ) = k e − A z 0 b = k e A z 0 k = b e A z 0 Substitute k = b e A z 0 in p = k e − A z . p ( z ) = b e A z 0 e − A z = p ( z 0 ) e − A ( z − z 0 ) ( ∵ p ( z 0 ) = b ) = p ( z 0 ) e − M g ( z − z 0 ) R T ( Substitute back, A = M g R T ) = p ( z 0 ) exp ( − M g ( z − z 0 ) R T ) Thus, the barometric pressure equation is p ( z ) = p ( z 0 ) exp [ − M g ( z − z 0 ) R T ] . (b) To derive: The solution p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } when the temperature also varies with altitude T = T ( z ) . From part (a), the differential equation is d p d z = − M p R T g . Here, R is the universal gas constant, M is the molar mass of the air, and g is gravitational constant. Suppose we assume that the temperature varies with altitude T = T ( z ) . Then, the differential equation becomes d p d z = − M g R p T ( z ) Separate the variables and rewrite the differential equation as follows. 1 p ( z ) d p = − M g R d z T ( z ) The value of z starts at z 0 because the initial condition is p ( z 0 ) = b . Also, the end value is unknown. Therefore, the limits of integration would be [ z 0 , z ] . Integrate both sides of the equation from ζ = z 0 to ζ = z where ζ be the variable of integration. ∫ z 0 z 1 p ( ζ ) d p = − ∫ z 0 z M g R d ζ T ( ζ ) [ ln p ( ζ ) ] z 0 z = − M g R ∫ z 0 z d ζ T ( ζ ) l n p ( z ) − l n p ( z 0 ) = − M g R ∫ z 0 z d ζ T ( ζ ) ln ( p ( z ) p ( z 0 ) ) = − M g R ∫ z 0 z d ζ T ( ζ ) Simplify further, p ( z ) p ( z 0 ) = exp { − M g R ∫ z 0 z d ζ T ( ζ ) } p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } Thus, the required solution p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } is derived. (c) The height of the building when the absolute temperature varies as T ( z ) = 288 − 0.0065 ( z − z 0 ) . The height of the building is 168.1 m _ . Given: The barometric pressure at the top of building is p ( z ) = 99 , 000 Pa . The barometric pressure at the base is p ( z 0 ) = 101 , 000 Pa . The absolute temperature varies as T ( z ) = 288 − 0.0065 ( z − z 0 ) . The universal gas constant R is, 8.31 N-m mol-K . The molar mass of the air M is 0.029 kg mol . The gravitational constant g is 9.8 m sec 2 Calculation: Given that, the absolute temperature varies as T ( z ) = 288 − 0.0065 ( z − z 0 ) . From part (b), if the temperature also varies with altitude T = T ( z ) , then the solution p ( z ) = p ( z 0 ) exp { − M g R ∫ z 0 z d ζ T ( ζ ) } . Substitute the known values in the solution as follows. 99 , 000 = 101 , 000 exp { − ( 0.029 ) ( 9.8 ) 8.31 ∫ z 0 z d ζ 288 − 0.0065 ( ζ − z 0 ) } 99 , 000 101 , 000 = exp { − 0.2842 8.31 ∫ z 0 z d u − 0.0065 u } ( Take u = 288 − 0.0065 ( ζ − z 0 ) , then d u = − 0.0065 d ζ ) ln 99 101 = 0.2842 0.054015 [ ln u ] z 0 z 0.054015 0.2842 ln 99 101 = [ ln ( 288 − 0.0065 ( ζ − z 0 ) ) ] z 0 z ( Substitute back, u = 288 − 0.0065 ( ζ − z 0 ) ) Simplify further, ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 ( z − z 0 ) ) − ln ( 288 − 0.0065 ( z 0 − z 0 ) ) ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 ( z − z 0 ) ) − ln ( 288 − 0.0065 ( 0 ) ) ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 h ) − ln ( 288 ) ( ∵ h = z − z 0 ) ln ( 99 101 ) 0.054015 0.2842 = ln ( 288 − 0.0065 h 288 ) Raise e to the power of both sides. ( 99 101 ) 0.054015 0.2842 = 288 − 0.0065 h 288 288 ( 99 101 ) 0.054015 0.2842 = 288 − 0.0065 h 288 ( 99 101 ) 0.054015 0.2842 − 288 = − 0.0065 h h = 288 ( 1 − ( 99 101 ) 0.054015 0.2842 ) 0.0065 h ≈ 168.1 Thus, the height of the building is 168.1 m _ .

Solution Summary: The author explains how to solve the barometric pressure equation p(z) and the drop in air pressure between the top and bottom of a cylindrical volume element.

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Fundamentals of Differential Equations (9th Edition)

Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 716, solve the equation. 7. xdydx=1y3 Ch. 2.2 - In Problems 716, solve the equation. 8. dxdt=3xt2 Ch. 2.2 - In Problems 716, solve the equation. 9....Ch. 2.2 - In Problems 716, solve the equation. 10....

Ch. 2.2 - In Problems 716, solve the equation. 11....Ch. 2.2 - In Problems 716, solve the equation. 12....Ch. 2.2 - In Problems 716, solve the equation. 13....Ch. 2.2 - In Problems 716, solve the equation. 14. dxdtx3=x Ch. 2.2 - In Problems 716, solve the equation. 15....Ch. 2.2 - In Problems 716, solve the equation. 16. y1 dy +...Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 23E Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 27E Ch. 2.2 - Sketch the solution to the initial value problem...Ch. 2.2 - Uniqueness Questions. In Chapter 1 we indicated...Ch. 2.2 - As stated in this section, the separation of...Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Mixing. Suppose a brine containing 0.3 kilogram...Ch. 2.2 - Newtons Law of Cooling. According to Newtons law...Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Compound Interest. If P(t) is the amount of...Ch. 2.2 - Free Fall. In Section 2.1, we discussed a model...Ch. 2.2 - Grand Prix Race. Driver A had been leading...Ch. 2.2 - Prob. 40E Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - Radioactive Decay. In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. The equation (18) dydx+2y=xy2...Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 9E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 15E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 17E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 25E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 27E Ch. 2.4 - For each of the following equations, find the most...Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Orthogonal Trajectories. A geometric problem...Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.5 - Prob. 1E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Verify that when the linear differential equation...Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 8E Ch. 2.6 - Use the method discussed under Homogeneous...Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Use the method discussed under Bernoulli Equations...Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Use the method discussed under Equations with...Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - In Problems 3340, solve the equation given in: 36....Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Show that equation (13) reduces to an equation of...Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2 - In Problems 130, solve the equation. 1....Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - In Problems 130, solve the equation. 6. 2xy3 dx ...Ch. 2 - In Problems 130, solve the equation. 7. t3y2 dt +...Ch. 2 - Prob. 8RP Ch. 2 - In Problems 130, solve the equation. 9. (x2 + y2)...Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP Ch. 2 - Prob. 23RP Ch. 2 - In Problems 130, solve the equation. 24. (y/x +...Ch. 2 - Prob. 25RP Ch. 2 - Prob. 26RP Ch. 2 - Prob. 27RP Ch. 2 - Prob. 28RP Ch. 2 - Prob. 29RP Ch. 2 - Prob. 30RP Ch. 2 - Prob. 31RP Ch. 2 - Prob. 32RP Ch. 2 - Prob. 33RP Ch. 2 - Prob. 34RP Ch. 2 - Prob. 35RP Ch. 2 - Prob. 36RP Ch. 2 - Prob. 37RP Ch. 2 - Prob. 38RP Ch. 2 - Prob. 39RP Ch. 2 - Prob. 40RP Ch. 2 - Prob. 41RP

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