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Chapter 3
3.1 An examination of the standard temperature distribution through the atmosphere given in Figure 3.3 of the text shows that both 12 km and 18 km are in the same constant temperature region. Hence, the equations that apply are Eqs. (3.9) and (3.10) in the text. Since we are in the same isothermal region with therefore the same base values of p and ρ, these equations can be written as
ρρ21 = =pp21 e(g 0/RT h h)( 2 1)
where points 1 and 2 are any two arbitrary points in the region. Hence, with g_{0} = 9.8 m/sec^{2} and R = 287 joule/kgK, and letting points 1 and 2 correspond to 12 km and 18 km altitudes, respectively,
ρ2 = =p2 e (287)(216.66)9.8 (6000) = 0.3884
ρ1 p1
Hence:
p_{2 }= (0.3884)(1.9399 ´10 )^{4 }= 7.53´10^{3 }N/m^{2}
ρ_{2 }= (0.3884)(3.1194 ´10^{1}) = 0.121kg/m^{3}
and, of course,
T_{2 }= 216.66 K
These answers check the results listed in Appendix A of the text within roundoff error.
3.2 From Appendix A of the text, we see immediately that p = 2.65 × 10^{4} N/m^{2 }corresponds to 10,000 m, or 10 km, in the standard atmosphere. Hence,
pressure altitude = 10 km
The outside air density is
4
ρ= p = 2.65 ´10 = 0.419 kg/m3
RT (287)(220)
From Appendix A, this value of ρ corresponds to 9.88 km in the standard atmosphere.
Hence,
density altitude = 9.88 km
3.3 At 35,000 ft, from Appendix B, we find that p = 4.99 × 10^{2} = 499 lb/ft^{2}.
3.4 From Appendix B in the text,
33,500 ft corresponds to p = 535.89 lb/ft^{2}
32,000 ft corresponds to ρ = 8.2704 × 10^{−4} slug/ft^{3 }
Hence,
T = _{ρ}pR = (8.2704535.89´10^{4}_{)(1716) }= _{378 }R
h – h
3.5_{ }^{G }= 0.02 =1 – ^{hG} hh
From Eq. (3.6), the above equation becomes
1 çççèæ ^{r }^{+}_{r }^{hG }÷÷÷öø = 1 – 1 ^{h}_{r}^{G } = 0.02
h_{G }= 0.02 r = 0.02(6.357 ´10 )^{6}
h_{G }= 1.27 ´10^{5}m = 127 km
3.6 T = 15 − 0.0065h = 15 − 0.0065(5000) = −17.5°C = 255.5°K
a = ^{dT }= 0.0065 ^{dh }
From Eq. (3.12)
ö÷÷÷g aR/ æ pp1 ^{= }æèççççT^{T}1 ø÷ 0 = çèçç 255.5288 ÷÷÷øö(2.8)/( 0.0065)(287) = 0.533
p = 0.533 p_{1 }= 0.533 (1.01´10 )^{5 }= 5.38 ´10^{4}N/m^{2}
3.7 ln pp1 = – RTg (h – h1)
h – h_{1 }= – ^{1}g RT ln _{p}^{p}_{1 }= – _{24.9}^{1 }(4157)(150) ln0.5
Letting h_{1} = 0 (the surface)
h = 17,358 m = 17.358 km
3.8 A standard altitude of 25,000 ft falls within the first gradient region in the standard atmosphere. Hence, the variation of pressure and temperature are given by:
p = ççççèTT1 ÷÷÷÷öøaRg (1) æ
p1
and
T = T_{1} + a (h – h_{1}) (2)
Differentiating Eq. (1) with respect to time:
– g
p11 dpdt = æççççèT11 ÷ö÷ø aRèçç ARg ö÷÷÷øTæççèçaRg 1÷ö÷÷÷ø dTdt (3) ÷÷ æç
Differentiating Eq. (2) with respect to time:
dT dh
= a (4)
dt dt
Substitute Eq. (4) into (3)
g æ
dpdt = –p T_{1}( _{1})aRççæ öç g ÷÷TçççèaRg +1÷÷÷÷öø dhdt (5)
è øR ÷
In Eq. (5), dh/dt is the rateofclimb, given by dh/dt = 500 ft/sec. Also, in the first gradient region, the lapse rate can be calculated from the tabulations in Appendix B. For example, take 0 ft and 10,000 ft, we find
_{ a }= T2 – T1 = 483.04 – 518.69 = –_{0.00357 }°R
h_{2 }– h_{1 }10,000 – 0 ft
Also from Appendix B, p_{1}= 2116.2 lb/ft^{2} at sea level, and T = 429.64 °R at 25,000 ft.
Thus,
g ^{32.2}
= = 5.256
aR ( 0.00357)(1716)
Hence, from Eq. (5) dp = (2116.2)(518.69)^{5.256}æçççè_{1716}^{32.2 }öø÷÷÷ (429.64)^{4.256}(500)
_{dt}
dp lb
dt = 17.17 ft sec_{2 }
3.9 From the hydrostatic equation, Eq. (3.2) or (3.3),
dp = ρg dh_{0 }
or dp = ρg_{0 }^{dh} dt dt
The upward speed of the elevator is dh/dt, which is
dp dp dt/
dt ^{= }ρg0
At sea level, ρ = 1.225 kg/m^{3}. Also, a onepercent change in pressure per minute starting from sea level is
dp ^{5 3 2}
= (1.01´10 )(0.01) = 1.01´10 N/m perminute
dt
Hence,
^{dh }= –^{1.01}´^{103 }= 84.1meterperminute
dt (1.225)(9.8)
3.10 From Appendix B:
At 35,500 ft: p = 535.89 lb/ft^{2}
At 34,000 ft: p = 523.47 lb/ft^{2}
For a pressure of 530 lb/ft^{2}, the pressure altitude is 33,500 + 500ççèæç 535.89535.89–523.47530 ø÷÷ö÷ = 33737ft
The density at the altitude at which the airplane is flying is
^{ρ}= ^{ρ }= ^{530 }= 7.919 ´10^{4 }slug/ft^{3}
RT (1716)(390)
From Appendix B:
At 33,000 ft: ^{ρ}= 7.9656 ´10^{4 }slug/ft^{3}
_{ }
At 33,500 ft: ^{ρ}= 7.8165 ´10^{4 }slug/ft^{3}
Hence, the density altitude is
33,000 + 500æçèçç 7.96567.9656–7.81657.919 öø÷÷÷ = 33,156 ft
3.11 Let ℓ be the length of one wall of the tank, ℓ = 30 ft. Let d be the depth of the pool, d = 10 ft. At the water surface, the pressure is atmospheric pressure, p_{a}. The water pressure increases with increasing depth; the pressure as a function of distance below the surface, h, is given by the hydrostatic equation
dp = ρ g dh (1)
Note: The hydrostatic equation given by Eq. (3.2) in the text has a minus sign because h_{G} is measured positive in the upward direction. In Eq. (1), h is measured positive in the downward direction, with h = 0 at the surface of the water. Hence, no minus sign appears in Eq. (1); as h increases (as we go deeper into the water), p increases. Eq. (1) is consistent with this fact. Integrating Eq. (1) from h = 0 where p = p_{a} to some local depth h where the pressure is p, and noting that ρ is constant for water, we have ρ h
dp = ρgò dh
pa 0
or,  p − p_{a} = ρ g h  
or,  p = ρ g h + p_{a}  (2) 
Eq. (2) gives the water pressure exerted on the wall at an arbitrary depth h.
Consider an elementary small sliver of wall surface of length ℓ and height dh. The water force on this sliver of area is dF = p ℓ dh
Total force, F, on the wall is
F d
F =dF = ò p dhl (3)
0 0
where p is given by Eq. (2). Inserting Eq. (2) into (3),
d
F =^{(}ρg h + p_{a}) ldh
0
æ
or, F = ρg lççççç ^{d}_{2}^{2 }öø÷÷÷÷÷+ p_{a}ld (4)
è
In Eq. (4), the product ρ g is the specific weight (weight per unit volume) of water; ρ g = 62.4 lb/ft^{3}. From Eq. (4),
(10)^{2}
F = (62.4)(30) + (2116)(30)(10)
2
F = 93,600 + 634,800 = 728,400lb
Note: This force is the combined effect of the force due to the weight of the water, 93,600 lb, and the force due to atmospheric pressure transmitted through the water, 634,800 Ib. In this example, the latter is the larger contribution to the force on the wall. If the wall were freestanding with atmospheric pressure exerted on the opposite side, then the net force exerted on the wall would be that due to the weight of the water only, i.e., 93,600 Ib. In tons, the force on the side of the wall in contact with the water is
F = = 364.2tons
In the case of a freestanding wall, the net force, that due only to the weight of the water, is
F = = 46.8tons
3.12 For the exponential atmosphere model,
^{ρ }–g h RT_{0 }/( )
ρ_{0 }= _{e}
ρρ0 = (9.8)(45,000)/(287)(240) = e6.402 e
Hence,
2.03 ´10^{3 }kg/m^{3} 
_{ }^{ρ ρ}= _{0}e–^{6.402 }= (1225)(1.6575 ´10^{3}) =
From the standard atmosphere, at 45 km, p = 2.02 × 10^{−3} kg/m^{3}. The exponential atmosphere model gives a remarkably accurate value for the density at 45 km when a value of 240 K is used for the temperature.
3.13 At 3 km, from App. A, T = 268.67 K, p = 7.0121 × 10^{4} N/m^{2}, and = 0.90926 kg/m^{3}. At 3.1 km, from App. A, T = 268.02 K, p = 6.9235 × 10^{4} N/m^{2}, and = 0.89994 kg/m^{3}. At h = 3.035 km, using linear interpolation:
T = 268.67 −(268.67 − 268.02)^{}_{} ^{0.035}0.1 ^{}_{} = 268.44 K
p= 7.0121 10× ^{4 }−(7.0121 10× ^{4 }−6.9235×10 )^{4 }^{}_{} ^{0.035}0.1 ^{}_{}
= 6.98099×10^{4 }^{N }m2 
kg
0.906 m3 
ρ= 0.09026 −(0.90926 −0.89994)^{}_{ }^{0.035}_{0.1 }^{}_{ }=
3.14 The altitude of 3.035 km is in the gradient region. From Example 3.1, a = –0.0065 K/m. From
Eq. (3.14),
From Eq. (3.12), 
T = +T_{1 }a h( − = +h_{1}) T_{s }a h( − =0) 288.16 −(0.0065)(3035) = 268.43 K  
So 
−g aR/ p T ^{0}
= T1 p_{1} −g0 = −9.8 = 5.25328 aR ( 0.0065)(28.7)− 

5.25328 pps =TTs = 288.16268.435.25328 = 0.688946 
p= 0.688946(1.01325×10 )^{5 }= 6.9807×10^{4 }^{N}
m2
By comparison the approximate result from the solution of Prob. 3.13 differs from the exact result above by
6.980990 −06.9807 ^{−}^{5 }= 0.00415%
= 4.15×10
6.9807
A very small amount. From Eq. (3.13),
ρ _{ }_{T }_{}−[( / ) 1]g aR_{0 }+
=T_{1}
ρ_{1}
^{g}^{0 }+ = −1 5.25328+ = −1 4.25328
aR ρ T 4.2538 268.434.24328 ρ_{s }=_{Ts} = 288.16 = 0.73958
kg
0.90599 m3 
ρ= 0.73598 ρ_{s }= 0.73958 (1.2250) =
3.15 We want to calculate h_{G} when
hG − h = 0.01= −1 h hG ^{h}G
From Eq. (3.6)
h ^{= }^{rh}^{G }, where r = 6.356766×10^{6} m
r + h_{G}
Thus,
0.01 = −1 ^{r }r h+ G
r
= −1 0.01= 0.99 r h+ G
0.99(^{r h}+ _{G}^{) }=r
hG =1−0.990.99 r = 0.0101 (6.356766×10 )5
h_{G }= 64.21 10× ^{3} m = 64.21 km
3.16 From Eq. (3.1), 2
r
g g= 0r h+ G
where r = 6.356766 × 10^{6} m, g_{0} = 9.8 m/sec^{2}, and
0.3048 m
h_{G }= (70,000 ft)^{}_{ }_{1 ft }^{}_{ }= 21,336 m Thus,
g= 9.8 6.356766×106×+1021,3366 = 9.767 secm2
6.356766
9.8 −9.767 = 0.34%
98 100
Thus, at 70,000 ft, the acceleration of gravity has decreased only by 0.34% compared to its sea level value.