SERIE N°1: VISCOSIDAD Y PRESION (resolución)
PROBLEMA Nº1.- La viscosidad dinámica del agua a 20ºC es 0,01008 poises y la densidad
relativa es 0,998. Calcular la viscosidad cinemática (en m 2 / s).
𝑔
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑑𝑖𝑛𝑎𝑚𝑖𝑐𝑎 (𝜇) 0,01008 𝑐𝑚. 𝑠 𝑐𝑚2
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑐𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐𝑎 (𝜈) = = 𝑔 = 0,0101002004
𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑 𝑚𝑎𝑠𝑖𝑐𝑎 (𝜌) 0,998 𝑠
𝑐𝑚3
𝑐𝑚2 (0,01 𝑚)2 𝑚2
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑐𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐𝑎 (𝜈) = 0,0101002004 ∗ 2
= 1,01002004 ∗ 10−6
𝑠 𝑐𝑚 𝑠
𝑚2
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑐𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐𝑎 (𝜈) ≅ 1,01 ∗ 10−6
𝑠
PROBLEMA Nº2.- Calcular la viscosidad dinámica del aire a 20ºC y 1 at mediante la fórmula
de Wilke. DATOS:
Oxígeno: viscosidad dinámica = 2.031*10-7 poise Masa molecular = 32 g/mol
Nitrógeno: viscosidad dinámica = 1.754*10-7 poise Masa molecular = 28 g/mol
𝐿𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎 𝑝𝑢𝑟𝑎 𝑒𝑠 𝑎𝑖𝑟𝑒 (𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑑𝑜 𝑐𝑜𝑛 𝑚𝑢𝑦 𝑏𝑎𝑗𝑜 𝑐𝑜𝑛𝑡𝑒𝑛𝑖𝑑𝑜 𝑑𝑒 ℎ𝑢𝑚𝑒𝑑𝑎𝑑) 𝑎 20°𝐶 𝑦 1 𝑎𝑡𝑎
𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑑𝑎:
𝑂𝑋𝐼𝐺𝐸𝑁𝑂 (21%) 𝑦 𝑁𝐼𝑇𝑅𝑂𝐺𝐸𝑁𝑂 (79%), 𝑚𝑒𝑑𝑖𝑑𝑜𝑠 𝑒𝑛 𝑓𝑟𝑎𝑐𝑐𝑖𝑜𝑛𝑒𝑠 𝑚𝑜𝑙𝑎𝑟𝑒𝑠
𝑥𝑖 ∗ 𝜇𝑖
𝐹𝑂𝑅𝑀𝑈𝐿𝐴 𝐷𝐸 𝑊𝐼𝐿𝐾𝐸: 𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = ∑
∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗
𝑂𝑏𝑡𝑒𝑛𝑐𝑖𝑜𝑛 𝑑𝑒 𝑙𝑜𝑠 𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 (𝜙)
1 2
−
1 𝑀𝑖 2 𝜇𝑖 4 𝑀𝑗
𝜙𝑖𝑗 = ∗ [1 + ] ∗ [1 + √ ∗ √ ]
√8 𝑀𝑗 𝜇𝑗 𝑀𝑖
i Especie Fracción Molar (x) Masa Molecular (g/mol) Viscosidad (P)
1 Oxigeno (O2) 0,21 32 2031
2 Nitrógeno (N2) 0,79 28 1754
𝑀𝑖 𝜇𝑖
𝑖 𝑗 𝜙𝑖𝑗 ∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗
𝑀𝑗 𝜇𝑗
1 1 1 1
1 1,00462235
2 1,14285714 1,15792474 1,005851073
1 0,875 0,86361398 0,992762345
2 0,99848009
2 1 1 1
, 𝑖 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 1 𝑦 𝑗 = 1 ∶ =
𝑗 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 32 𝑚𝑜𝑙 𝜇𝑖 2.031 ∗ 10−7 𝑃
= =1 = =1
𝑀𝑗 32 𝑔 𝜇𝑗 2.031 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 32 −7 4 32
𝜙11 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √2.031 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1 ∗ 1 ∗ 4 = 1
𝑔 2.031 ∗ 10−7 𝑃 𝑔
√8 32 32 √8 √2
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
𝑖 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 1 𝑦 𝑗 = 2 ∶ =
𝑗 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 32 𝑚𝑜𝑙 𝜇𝑖 2.031 ∗ 10−7 𝑃
= = 1,142857143 = = 1,157924743
𝑀𝑗 28 𝑔 𝜇𝑗 1.754 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 32 −7 4 28
𝜙12 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √2.031 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1,005851073
𝑔 1.754 ∗ 10−7 𝑃 𝑔
√8 28 32
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
2
∑ 𝑥𝑗 ∗ 𝜙1𝑗 = 𝑥1 ∗ 𝜙11 + 𝑥2 ∗ 𝜙12 = 0,21 ∗ 1 + 0,79 ∗ 1,005851073 = 1,004622348
𝑗=1
𝑖 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 2 𝑦 𝑗 = 1 ∶ =
𝑗 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 28 𝑚𝑜𝑙 𝜇𝑖 1.754 ∗ 10−7 𝑃
= =1 = =1
𝑀𝑗 32 𝑔 𝜇𝑗 2.031 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 28 −7 4 32
𝜙21 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √1.754 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 0,9927623449
𝑔 −7
2.031 ∗ 10 𝑃 𝑔
√8 32 28
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
𝑖 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 2 𝑦 𝑗 = 2 ∶ =
𝑗 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 28 𝑚𝑜𝑙 𝜇𝑖 1.754 ∗ 10−7 𝑃
= =1 = =1
𝑀𝑗 28 𝑔 𝜇𝑗 1.754 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 28 −7 4 28
𝜙22 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √1.754 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1 ∗ 1 ∗ 4 = 1
𝑔 1.754 ∗ 10−7 𝑃 𝑔
√8 28 28 √8 √2
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
2
∑ 𝑥𝑗 ∗ 𝜙2𝑗 = 𝑥1 ∗ 𝜙21 + 𝑥2 ∗ 𝜙22 = 0,21 ∗ 0,9927623449 + 0,79 ∗ 1 = 0,9984800924
𝑗=1
, 𝑂𝑏𝑡𝑒𝑛𝑐𝑖𝑜𝑛 𝑑𝑒 𝑙𝑜𝑠 𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 (𝜙)
1 2
−
1 𝑀𝑖 2 𝜇𝑖 4 𝑀𝑗 𝜇𝑗 𝑀𝑖
𝜙𝑖𝑗 = ∗ [1 + ] ∗ [1 + √ ∗ √ ] 𝜙𝑗𝑖 = 𝜙𝑖𝑗 ∗ ∗ (𝑟𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑜)
√8 𝑀𝑗 𝜇𝑗 𝑀𝑖 𝜇𝑖 𝑀𝑗
1 2
𝑔 −2 𝑔
1 32 −7 4 28
𝜙12 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √2.031 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1,005851073
𝑔 1.754 ∗ 10−7 𝑃 𝑔
√8 28 32
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
1 2
𝑔 −2 𝑔
1 28 −7 4 32
𝜙21 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √1.754 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 0,9927623449
𝑔 2.031 ∗ 10−7 𝑃 𝑔
√8 32 28
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
𝑔
𝜇2 𝑀1 1.754 ∗ 10−7 𝑃 32 𝑚𝑜𝑙
𝜙21 = 𝜙12 ∗ ∗ = 1,005851073 ∗ ∗ = 0,9927623449
𝜇1 𝑀2 2.031 ∗ 10−7 𝑃 28 𝑔
𝑚𝑜𝑙
𝑥𝑖 ∗ 𝜇𝑖 𝑥1 ∗ 𝜇1 𝑥2 ∗ 𝜇2
𝐹𝑂𝑅𝑀𝑈𝐿𝐴 𝐷𝐸 𝑊𝐼𝐿𝐾𝐸: 𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = ∑ = +
∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗 𝑥1 ∗ 𝜙11 + 𝑥2 ∗ 𝜙12 𝑥1 ∗ 𝜙21 + 𝑥2 ∗ 𝜙22
𝑥𝑖 ∗ 𝜇𝑖 0,21 ∗ 2.031 ∗ 10−7 𝑃 0,79 ∗ 1.754 ∗ 10−7 𝑃
𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = ∑ = +
∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗 ∑2𝑗=1 𝑥𝑗 ∗ 𝜙1𝑗 ∑2𝑗=1 𝑥𝑗 ∗ 𝜙2𝑗
0,21 ∗ 2.031 ∗ 10−7 𝑃 0,79 ∗ 1.754 ∗ 10−7 𝑃
𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = +
1,004622348 0,9984800924
𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = 4,245475933 ∗ 10−5 𝑃 + 1,387769281 ∗ 10−4 𝑃 = 1,812316874 ∗ 10−4 𝑃
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑑𝑒 𝑙𝑎 𝑚𝑒𝑧𝑐𝑙𝑎 ∶ 𝜇 (20°𝐶) ≅ 0,0181 𝑐𝑃
PROBLEMA Nº3.- Repetir el problema 2 usando el Nomograma. Comparar los resultados.
𝑃𝑎𝑟𝑎 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑟 𝑒𝑙 𝑛𝑜𝑚𝑜𝑔𝑟𝑎𝑚𝑎 𝑑𝑒𝑏𝑒 𝑎𝑠𝑒𝑔𝑢𝑟𝑎𝑟𝑠𝑒 𝑒𝑙 𝑒𝑠𝑡𝑎𝑑𝑜 𝑑𝑒 𝑙𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎 (𝑔𝑎𝑠 𝑜 𝑙𝑖𝑞𝑢𝑖𝑑𝑜)
𝑈𝑛𝑎 𝑣𝑒𝑧 𝑠𝑒𝑙𝑒𝑐𝑐𝑖𝑜𝑛𝑎𝑑𝑜 𝑝𝑜𝑟 𝑒𝑙 𝑒𝑠𝑡𝑎𝑑𝑜 𝑓𝑖𝑠𝑖𝑐𝑜 𝑑𝑒 𝑙𝑎 𝑚𝑖𝑠𝑚𝑎, 𝑠𝑒 𝑏𝑢𝑠𝑐𝑎 𝑙𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎 𝑒𝑛 𝑙𝑎 𝑡𝑎𝑏𝑙𝑎
𝑎𝑑𝑗𝑢𝑛𝑡𝑎, 𝑒𝑛 𝑙𝑎 𝑞𝑢𝑒 𝑓𝑖𝑔𝑢𝑟𝑎𝑛 𝑢𝑛 𝑝𝑎𝑟 𝑑𝑒 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑎𝑠 (𝑥; 𝑦), 𝑎𝑙 𝑜𝑏𝑡𝑒𝑛𝑒𝑟 𝑙𝑎𝑠 𝑚𝑖𝑠𝑚𝑎𝑠, 𝑑𝑒𝑏𝑒
𝑑𝑖𝑟𝑖𝑔𝑖𝑟𝑠𝑒 𝑎𝑙 𝑔𝑟𝑎𝑓𝑖𝑐𝑜 𝑦 𝑢𝑏𝑖𝑐𝑎𝑟 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑒𝑛 𝑙𝑎 𝑐𝑢𝑎𝑑𝑟𝑖𝑐𝑢𝑙𝑎, 𝑙𝑜 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑒𝑠 𝑓𝑖𝑗𝑎𝑟 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒
𝑙𝑎 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑎 𝑙𝑎 𝑞𝑢𝑒 𝑠𝑒 𝑑𝑒𝑠𝑒𝑎 𝑐𝑜𝑛𝑜𝑐𝑒𝑟 𝑙𝑎 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑦 𝑠𝑒 𝑚𝑎𝑟𝑐𝑎 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑠𝑜𝑏𝑟𝑒 𝑒𝑙 𝑒𝑗𝑒
𝑖𝑧𝑞𝑢𝑖𝑒𝑟𝑑𝑜 𝑑𝑒 𝑙𝑎 𝑐𝑢𝑎𝑑𝑟𝑖𝑐𝑢𝑙𝑎. 𝐶𝑜𝑛 𝑒𝑠𝑡𝑜𝑠 2 𝑝𝑢𝑛𝑡𝑜𝑠 (𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑦 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎)𝑡𝑟𝑎𝑧𝑎𝑟 𝑢𝑛𝑎 𝑟𝑒𝑐𝑡𝑎
𝑞𝑢𝑒 𝑙𝑜𝑠 𝑢𝑛𝑎, 𝑦 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑟 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 𝑒𝑛 𝑒𝑙 𝑒𝑗𝑒 𝑑𝑒𝑟𝑒𝑐ℎ𝑜, 𝑒𝑛 𝑒𝑠𝑒 𝑠𝑒 𝑖𝑛𝑑𝑖𝑐𝑎𝑟𝑎 𝑒𝑙
𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑝𝑎𝑟𝑎 𝑒𝑠𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎, 𝑎 𝑒𝑠𝑎 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎
𝐴𝑖𝑟𝑒 (𝑔𝑎𝑠): (𝑥; 𝑦) = (11,0 ; 20,0) 𝑠𝑒 𝑚𝑎𝑟𝑐𝑎 𝑒𝑠𝑡𝑒 𝑝𝑢𝑛𝑡𝑜 𝑦 𝑠𝑒 𝑓𝑖𝑗𝑎𝑛 𝑙𝑜𝑠 20°𝐶
∆𝜇 0,0181 𝑐𝑃 − 0,0178 𝑐𝑃
𝜇 (20°𝐶) ≈ 0,0178 𝑐𝑃 ⇒ 𝐸𝑟𝑟𝑜𝑟 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑜 = = ≈ 0,01658
𝜇 0,0181 𝑐𝑃
PROBLEMA Nº1.- La viscosidad dinámica del agua a 20ºC es 0,01008 poises y la densidad
relativa es 0,998. Calcular la viscosidad cinemática (en m 2 / s).
𝑔
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑑𝑖𝑛𝑎𝑚𝑖𝑐𝑎 (𝜇) 0,01008 𝑐𝑚. 𝑠 𝑐𝑚2
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑐𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐𝑎 (𝜈) = = 𝑔 = 0,0101002004
𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑 𝑚𝑎𝑠𝑖𝑐𝑎 (𝜌) 0,998 𝑠
𝑐𝑚3
𝑐𝑚2 (0,01 𝑚)2 𝑚2
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑐𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐𝑎 (𝜈) = 0,0101002004 ∗ 2
= 1,01002004 ∗ 10−6
𝑠 𝑐𝑚 𝑠
𝑚2
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑐𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐𝑎 (𝜈) ≅ 1,01 ∗ 10−6
𝑠
PROBLEMA Nº2.- Calcular la viscosidad dinámica del aire a 20ºC y 1 at mediante la fórmula
de Wilke. DATOS:
Oxígeno: viscosidad dinámica = 2.031*10-7 poise Masa molecular = 32 g/mol
Nitrógeno: viscosidad dinámica = 1.754*10-7 poise Masa molecular = 28 g/mol
𝐿𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎 𝑝𝑢𝑟𝑎 𝑒𝑠 𝑎𝑖𝑟𝑒 (𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑑𝑜 𝑐𝑜𝑛 𝑚𝑢𝑦 𝑏𝑎𝑗𝑜 𝑐𝑜𝑛𝑡𝑒𝑛𝑖𝑑𝑜 𝑑𝑒 ℎ𝑢𝑚𝑒𝑑𝑎𝑑) 𝑎 20°𝐶 𝑦 1 𝑎𝑡𝑎
𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑑𝑎:
𝑂𝑋𝐼𝐺𝐸𝑁𝑂 (21%) 𝑦 𝑁𝐼𝑇𝑅𝑂𝐺𝐸𝑁𝑂 (79%), 𝑚𝑒𝑑𝑖𝑑𝑜𝑠 𝑒𝑛 𝑓𝑟𝑎𝑐𝑐𝑖𝑜𝑛𝑒𝑠 𝑚𝑜𝑙𝑎𝑟𝑒𝑠
𝑥𝑖 ∗ 𝜇𝑖
𝐹𝑂𝑅𝑀𝑈𝐿𝐴 𝐷𝐸 𝑊𝐼𝐿𝐾𝐸: 𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = ∑
∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗
𝑂𝑏𝑡𝑒𝑛𝑐𝑖𝑜𝑛 𝑑𝑒 𝑙𝑜𝑠 𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 (𝜙)
1 2
−
1 𝑀𝑖 2 𝜇𝑖 4 𝑀𝑗
𝜙𝑖𝑗 = ∗ [1 + ] ∗ [1 + √ ∗ √ ]
√8 𝑀𝑗 𝜇𝑗 𝑀𝑖
i Especie Fracción Molar (x) Masa Molecular (g/mol) Viscosidad (P)
1 Oxigeno (O2) 0,21 32 2031
2 Nitrógeno (N2) 0,79 28 1754
𝑀𝑖 𝜇𝑖
𝑖 𝑗 𝜙𝑖𝑗 ∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗
𝑀𝑗 𝜇𝑗
1 1 1 1
1 1,00462235
2 1,14285714 1,15792474 1,005851073
1 0,875 0,86361398 0,992762345
2 0,99848009
2 1 1 1
, 𝑖 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 1 𝑦 𝑗 = 1 ∶ =
𝑗 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 32 𝑚𝑜𝑙 𝜇𝑖 2.031 ∗ 10−7 𝑃
= =1 = =1
𝑀𝑗 32 𝑔 𝜇𝑗 2.031 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 32 −7 4 32
𝜙11 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √2.031 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1 ∗ 1 ∗ 4 = 1
𝑔 2.031 ∗ 10−7 𝑃 𝑔
√8 32 32 √8 √2
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
𝑖 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 1 𝑦 𝑗 = 2 ∶ =
𝑗 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 32 𝑚𝑜𝑙 𝜇𝑖 2.031 ∗ 10−7 𝑃
= = 1,142857143 = = 1,157924743
𝑀𝑗 28 𝑔 𝜇𝑗 1.754 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 32 −7 4 28
𝜙12 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √2.031 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1,005851073
𝑔 1.754 ∗ 10−7 𝑃 𝑔
√8 28 32
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
2
∑ 𝑥𝑗 ∗ 𝜙1𝑗 = 𝑥1 ∗ 𝜙11 + 𝑥2 ∗ 𝜙12 = 0,21 ∗ 1 + 0,79 ∗ 1,005851073 = 1,004622348
𝑗=1
𝑖 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 2 𝑦 𝑗 = 1 ∶ =
𝑗 𝑜𝑥𝑖𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 28 𝑚𝑜𝑙 𝜇𝑖 1.754 ∗ 10−7 𝑃
= =1 = =1
𝑀𝑗 32 𝑔 𝜇𝑗 2.031 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 28 −7 4 32
𝜙21 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √1.754 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 0,9927623449
𝑔 −7
2.031 ∗ 10 𝑃 𝑔
√8 32 28
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
𝑖 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑃𝑎𝑟𝑎 𝑖 = 2 𝑦 𝑗 = 2 ∶ =
𝑗 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑜
𝑔
𝑀𝑖 28 𝑚𝑜𝑙 𝜇𝑖 1.754 ∗ 10−7 𝑃
= =1 = =1
𝑀𝑗 28 𝑔 𝜇𝑗 1.754 ∗ 10−7 𝑃
𝑚𝑜𝑙
1 2
𝑔 −2 𝑔
1 28 −7 4 28
𝜙22 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √1.754 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1 ∗ 1 ∗ 4 = 1
𝑔 1.754 ∗ 10−7 𝑃 𝑔
√8 28 28 √8 √2
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
2
∑ 𝑥𝑗 ∗ 𝜙2𝑗 = 𝑥1 ∗ 𝜙21 + 𝑥2 ∗ 𝜙22 = 0,21 ∗ 0,9927623449 + 0,79 ∗ 1 = 0,9984800924
𝑗=1
, 𝑂𝑏𝑡𝑒𝑛𝑐𝑖𝑜𝑛 𝑑𝑒 𝑙𝑜𝑠 𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 (𝜙)
1 2
−
1 𝑀𝑖 2 𝜇𝑖 4 𝑀𝑗 𝜇𝑗 𝑀𝑖
𝜙𝑖𝑗 = ∗ [1 + ] ∗ [1 + √ ∗ √ ] 𝜙𝑗𝑖 = 𝜙𝑖𝑗 ∗ ∗ (𝑟𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑜)
√8 𝑀𝑗 𝜇𝑗 𝑀𝑖 𝜇𝑖 𝑀𝑗
1 2
𝑔 −2 𝑔
1 32 −7 4 28
𝜙12 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √2.031 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 1,005851073
𝑔 1.754 ∗ 10−7 𝑃 𝑔
√8 28 32
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
1 2
𝑔 −2 𝑔
1 28 −7 4 32
𝜙21 = ∗ [1 + 𝑚𝑜𝑙 ] ∗ 1 + √1.754 ∗ 10 𝑃 ∗ √ 𝑚𝑜𝑙 = 0,9927623449
𝑔 2.031 ∗ 10−7 𝑃 𝑔
√8 32 28
𝑚𝑜𝑙 [ 𝑚𝑜𝑙 ]
𝑔
𝜇2 𝑀1 1.754 ∗ 10−7 𝑃 32 𝑚𝑜𝑙
𝜙21 = 𝜙12 ∗ ∗ = 1,005851073 ∗ ∗ = 0,9927623449
𝜇1 𝑀2 2.031 ∗ 10−7 𝑃 28 𝑔
𝑚𝑜𝑙
𝑥𝑖 ∗ 𝜇𝑖 𝑥1 ∗ 𝜇1 𝑥2 ∗ 𝜇2
𝐹𝑂𝑅𝑀𝑈𝐿𝐴 𝐷𝐸 𝑊𝐼𝐿𝐾𝐸: 𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = ∑ = +
∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗 𝑥1 ∗ 𝜙11 + 𝑥2 ∗ 𝜙12 𝑥1 ∗ 𝜙21 + 𝑥2 ∗ 𝜙22
𝑥𝑖 ∗ 𝜇𝑖 0,21 ∗ 2.031 ∗ 10−7 𝑃 0,79 ∗ 1.754 ∗ 10−7 𝑃
𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = ∑ = +
∑ 𝑥𝑗 ∗ 𝜙𝑖𝑗 ∑2𝑗=1 𝑥𝑗 ∗ 𝜙1𝑗 ∑2𝑗=1 𝑥𝑗 ∗ 𝜙2𝑗
0,21 ∗ 2.031 ∗ 10−7 𝑃 0,79 ∗ 1.754 ∗ 10−7 𝑃
𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = +
1,004622348 0,9984800924
𝜇 (𝑚𝑒𝑧𝑐𝑙𝑎) = 4,245475933 ∗ 10−5 𝑃 + 1,387769281 ∗ 10−4 𝑃 = 1,812316874 ∗ 10−4 𝑃
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑑𝑒 𝑙𝑎 𝑚𝑒𝑧𝑐𝑙𝑎 ∶ 𝜇 (20°𝐶) ≅ 0,0181 𝑐𝑃
PROBLEMA Nº3.- Repetir el problema 2 usando el Nomograma. Comparar los resultados.
𝑃𝑎𝑟𝑎 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑟 𝑒𝑙 𝑛𝑜𝑚𝑜𝑔𝑟𝑎𝑚𝑎 𝑑𝑒𝑏𝑒 𝑎𝑠𝑒𝑔𝑢𝑟𝑎𝑟𝑠𝑒 𝑒𝑙 𝑒𝑠𝑡𝑎𝑑𝑜 𝑑𝑒 𝑙𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎 (𝑔𝑎𝑠 𝑜 𝑙𝑖𝑞𝑢𝑖𝑑𝑜)
𝑈𝑛𝑎 𝑣𝑒𝑧 𝑠𝑒𝑙𝑒𝑐𝑐𝑖𝑜𝑛𝑎𝑑𝑜 𝑝𝑜𝑟 𝑒𝑙 𝑒𝑠𝑡𝑎𝑑𝑜 𝑓𝑖𝑠𝑖𝑐𝑜 𝑑𝑒 𝑙𝑎 𝑚𝑖𝑠𝑚𝑎, 𝑠𝑒 𝑏𝑢𝑠𝑐𝑎 𝑙𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎 𝑒𝑛 𝑙𝑎 𝑡𝑎𝑏𝑙𝑎
𝑎𝑑𝑗𝑢𝑛𝑡𝑎, 𝑒𝑛 𝑙𝑎 𝑞𝑢𝑒 𝑓𝑖𝑔𝑢𝑟𝑎𝑛 𝑢𝑛 𝑝𝑎𝑟 𝑑𝑒 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑎𝑠 (𝑥; 𝑦), 𝑎𝑙 𝑜𝑏𝑡𝑒𝑛𝑒𝑟 𝑙𝑎𝑠 𝑚𝑖𝑠𝑚𝑎𝑠, 𝑑𝑒𝑏𝑒
𝑑𝑖𝑟𝑖𝑔𝑖𝑟𝑠𝑒 𝑎𝑙 𝑔𝑟𝑎𝑓𝑖𝑐𝑜 𝑦 𝑢𝑏𝑖𝑐𝑎𝑟 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑒𝑛 𝑙𝑎 𝑐𝑢𝑎𝑑𝑟𝑖𝑐𝑢𝑙𝑎, 𝑙𝑜 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑒𝑠 𝑓𝑖𝑗𝑎𝑟 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒
𝑙𝑎 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑎 𝑙𝑎 𝑞𝑢𝑒 𝑠𝑒 𝑑𝑒𝑠𝑒𝑎 𝑐𝑜𝑛𝑜𝑐𝑒𝑟 𝑙𝑎 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑦 𝑠𝑒 𝑚𝑎𝑟𝑐𝑎 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑠𝑜𝑏𝑟𝑒 𝑒𝑙 𝑒𝑗𝑒
𝑖𝑧𝑞𝑢𝑖𝑒𝑟𝑑𝑜 𝑑𝑒 𝑙𝑎 𝑐𝑢𝑎𝑑𝑟𝑖𝑐𝑢𝑙𝑎. 𝐶𝑜𝑛 𝑒𝑠𝑡𝑜𝑠 2 𝑝𝑢𝑛𝑡𝑜𝑠 (𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑦 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎)𝑡𝑟𝑎𝑧𝑎𝑟 𝑢𝑛𝑎 𝑟𝑒𝑐𝑡𝑎
𝑞𝑢𝑒 𝑙𝑜𝑠 𝑢𝑛𝑎, 𝑦 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑟 𝑒𝑙 𝑝𝑢𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖𝑜𝑛 𝑒𝑛 𝑒𝑙 𝑒𝑗𝑒 𝑑𝑒𝑟𝑒𝑐ℎ𝑜, 𝑒𝑛 𝑒𝑠𝑒 𝑠𝑒 𝑖𝑛𝑑𝑖𝑐𝑎𝑟𝑎 𝑒𝑙
𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑑𝑎𝑑 𝑝𝑎𝑟𝑎 𝑒𝑠𝑎 𝑠𝑢𝑠𝑡𝑎𝑛𝑐𝑖𝑎, 𝑎 𝑒𝑠𝑎 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎
𝐴𝑖𝑟𝑒 (𝑔𝑎𝑠): (𝑥; 𝑦) = (11,0 ; 20,0) 𝑠𝑒 𝑚𝑎𝑟𝑐𝑎 𝑒𝑠𝑡𝑒 𝑝𝑢𝑛𝑡𝑜 𝑦 𝑠𝑒 𝑓𝑖𝑗𝑎𝑛 𝑙𝑜𝑠 20°𝐶
∆𝜇 0,0181 𝑐𝑃 − 0,0178 𝑐𝑃
𝜇 (20°𝐶) ≈ 0,0178 𝑐𝑃 ⇒ 𝐸𝑟𝑟𝑜𝑟 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑜 = = ≈ 0,01658
𝜇 0,0181 𝑐𝑃
