101地特三等流力第二題 - 高考
By Suhail Hany
at 2012-12-18T21:15
at 2012-12-18T21:15
Table of Contents
這類需要迭代的題目，近年來開始出現
從100年高考、今年高考，至今年地特又出現了
小弟在此提供一下個人解法
1.先取 Bernoulli equ
P1/r + z1 + v1^2/2g = P2/r + z2 + v2^2/2g + (0.3+0.2)*v2^2/2g + f*L/D*v2^2/2g
0 + 314 + 0 = 0 + 290 + [1.5+f*(100/0.05)]*v2^2/2g
470.4 = (1.5+2000f)*v2^2/2g
v2=[470.4/(1.5+2000f)]^0.5-------(*)
2.進行迭代
1/f^0.5 = 1.93 log [(Re*f^0.5)]-0.537
= 1.93 log{[470.4/(1.5+2000f)]^0.5*0.05/10^-6*f^0.5}-0.537
f 左邊 右邊
10^-3 31.6 7.69
10^-2 10 7.89
2*10^-2 7.07 7.91
0.015 8.16 7.90
0.016 7.9 7.90
故可知f約為0.016
代入(*)式
得v2=3.74 m/sec
Q=pi/4*D^2*v2
=pi/4*(0.05)^2*3.74
=7.34*10^-3 m^3/sec
=440.39 L/min
以上個人答案若有錯誤，還望各位大大鞭小力點XDDD
--
從100年高考、今年高考，至今年地特又出現了
小弟在此提供一下個人解法
1.先取 Bernoulli equ
P1/r + z1 + v1^2/2g = P2/r + z2 + v2^2/2g + (0.3+0.2)*v2^2/2g + f*L/D*v2^2/2g
0 + 314 + 0 = 0 + 290 + [1.5+f*(100/0.05)]*v2^2/2g
470.4 = (1.5+2000f)*v2^2/2g
v2=[470.4/(1.5+2000f)]^0.5-------(*)
2.進行迭代
1/f^0.5 = 1.93 log [(Re*f^0.5)]-0.537
= 1.93 log{[470.4/(1.5+2000f)]^0.5*0.05/10^-6*f^0.5}-0.537
f 左邊 右邊
10^-3 31.6 7.69
10^-2 10 7.89
2*10^-2 7.07 7.91
0.015 8.16 7.90
0.016 7.9 7.90
故可知f約為0.016
代入(*)式
得v2=3.74 m/sec
Q=pi/4*D^2*v2
=pi/4*(0.05)^2*3.74
=7.34*10^-3 m^3/sec
=440.39 L/min
以上個人答案若有錯誤，還望各位大大鞭小力點XDDD
--
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