If sea-level pressure is 101,350 Pa, compute the standard pressure at an altitude of 5000 m, using (a) the exact formula and (b) an isothermal assumption at a standard sea-level temperature of 15°C. Is the isothermal approximation adequate?

Question

Accepted Answer

Part (a)

Use absolute temperature in the exact formula, Eq. (2.20):

[latex]p=p_a\left(1-\frac{Bz}{T_0}\right)^{g/(RB)} where \frac{g}{RB}=5.26(air)[/latex]

[latex]\rho=\rho_0\left(1-\frac{Bz}{T_0}\right)^{\frac{g}{RB}-1} where \rho_0=1.2255\frac{kg}{m^3}, p_0=101,350 Pa[/latex] (2.20)

[latex]p=p_a\left[1-\frac{(0.00650 K/m)(5000 m)}{288.16 K}\right]^{5.26}=(101,350 Pa)(0.8872)^{5.26}=101,350(0.5328)=54,000 Pa[/latex]

This is the standard-pressure result given at z = 5000 m in Table A.6.

Table A.6 Properties of the Standard Atmosphere
z, m	T, K	p, Pa	ρ, kg/[latex]m^3[/latex]	a, m/s
-500	291.41	107,508	1.2854	342.2
0	288.16	101,350	1.2255	340.3
500	284.91	95,480	1.1677	338.4
1000	281.66	89,889	1.1120	336.5
1500	278.41	84,565	1.0583	334.5
2000	275.16	79,500	1.0067	332.6
2500	271.91	74,684	0.9570	330.6
3000	268.66	70,107	0.9092	328.6
3500	265.41	65,759	0.8633	326.6
4000	262.16	61,633	0.8191	324.6
4500	258.91	57,718	0.7768	322.6
5000	255.66	54,008	0.7361	320.6
5500	252.41	50,493	0.6970	318.5
6000	249.16	47,166	0.6596	316.5
6500	245.91	44,018	0.6237	314.4
7000	242.66	41,043	0.5893	312.3
7500	239.41	38,233	0.5564	310.2
8000	236.16	35,581	0.5250	308.1
8500	232.91	33,080	0.4949	306.0
9000	229.66	30,723	0.4661	303.8
9500	226.41	28,504	0.4387	301.7
10,000	223.16	26,416	0.4125	299.5
10,500	219.91	24,455	0.3875	297.3
11,000	216.66	22,612	0.3637	295.1
11,500	216.66	20,897	0.3361	295.1
12,000	216.66	19,312	0.3106	295.1
12,500	216.66	17,847	0.2870	295.1
13,000	216.66	16,494	0.2652	295.1
13,500	216.66	15,243	0.2451	295.1
14,000	216.66	14,087	0.2265	295.1
14,500	216.66	13,018	0.2094	295.1
15,000	216.66	12,031	0.1935	295.1
15,500	216.66	11,118	0.1788	295.1
16,000	216.66	10,275	0.1652	295.1
16,500	216.66	9496	0.1527	295.1
17,000	216.66	8775	0.1411	295.1
17,500	216.66	8110	0.1304	295.1
18,000	216.66	7495	0.1205	295.1
18,500	216.66	6926	0.1114	295.1
19,000	216.66	6401	0.1029	295.1
19,500	216.66	5915	0.0951	295.1
20,000	216.66	5467	0.0879	295.1
22,000	218.6	4048	0.0645	296.4
24,000	220.6	2972	0.0469	297.8
26,000	222.5	2189	0.0343	299.1
28,000	224.5	1616	0.0251	300.4
30,000	226.5	1197	0.0184	301.7
40,000	250.4	287	0.0040	317.2
50,000	270.7	80	0.0010	329.9
60,000	255.7	22	0.0003	320.6
70,000	219.7	6	0.0001	297.2

Part (b)

If the atmosphere were isothermal at 288.16 K, Eq. (2.18) would apply:

[latex]p_2=p_1 \exp \left[-\frac{g(z_2-z_1)}{RT_0}\right][/latex] (2.18)

[latex]p \approx p_a \exp \left(-\frac{gz}{RT}\right) = (101,350 Pa) \exp \left\{-\frac{(9.807 m/s^2)(5000 m)}{[287 m^2/(s^2 \cdot K)](288.16 K)}\right\} = (101,350 Pa) \exp (-0.5929) \approx 56,000 Pa[/latex]

This is 4 percent higher than the exact result. The isothermal formula is inaccurate in the troposphere.

Question 2.2: If sea-level pressure is 101,350 Pa, compute the standard pr...

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