Help with the equation
显示 更早的评论
Hi, can anyone help me solve this one?
contents:
ρ=1.025
L=130.2608788
B=20.48
δ= 0.695904983
D BAL=9220.372163
T=8.348645215
solving for T BAL.
it should equal something around 4.5 from what my proffesor told me :)
采纳的回答
John D'Errico
2023-12-20
So why not try it? First, use valid variable names, not greek letters for the variables.
rho = 1.025;
L = 130.2608788;
B = 20.48;
delta = 0.695904983;
D_BAL = 9220.372163;
T = 8.348645215;
You can use symbolic tools. Or you can use function handles and fzero.
syms T_BAL
Remember that MATLAB uses log to represent the latural log. NOT ln.
Tbalsol = solve(D_BAL == rho*L*B*T_BAL*(delta + 0.1*log(T_BAL/T)))
The w in there will be the Wright-Omega function, which often arises in problems like this. You can resolve the call to the omega function by use of double, or vpa.
The result will be not quite 4.5, but close. You can convince yourself this is a solution by plotting it.
fplot(-D_BAL + rho*L*B*T_BAL*(delta + 0.1*log(T_BAL/T)),[0,10])
hold on
grid on
plot(Tbalsol,0,'rx')
Or you could have used fzero.
更多回答（1 个）
Check whether the density input is correct or not to get a value of 4.5 forTBAL
syms T_BAL
Rho =1.225; % density ?? % standard air density
L=130.2608788;
B=20.48;
Delta = 0.695904983;
D_BAL=9220.372163;
T=8.348645215;
eqn = D_BAL == Rho*L*B*T_BAL*(Delta+0.1*log(T_BAL/T));
sol = solve(eqn,T_BAL)
vpa(sol)
1 个评论
See my corrected answer above, it is around 4.45 which is ~ 4.5 as what your professor told it. Another thing is if all the parameters have follow same unit system, the density of water is 1000 kg/m^3 and if i use your value to solve the equation it shows TBAL is around 0.027.
syms T_BAL
Rho =1025; % density ?? % standard air density
L=130.2608788;
B=20.48;
Delta = 0.695904983;
D_BAL=9220.372163;
T=8.348645215;
eqn = D_BAL == Rho*L*B*T_BAL*(Delta+0.1*log(T_BAL/T));
sol = solve(eqn,T_BAL)
vpa(sol)
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