Hi,

I cannot contribute much, but it appear that if the variable name, in your example "acc" or "volume", is in lexicographical ordering small, the result is 2.38994, if it is later in the order it is 1.35899. So I played a while and noted that the transition seems to be somewhere close to "h". If your variable is anything up to, say gzzzzzzzz the result will be 2.38994, for haaaaaa and later will give 1.35899.

So

a = 80; le = 62; g = 9.8; tf = 50; b = 3.4; r = 5.86/le; ppa = 32*22;

ans = Flatten@
   NDSolve[{b*u[t, x]*Derivative[0, 1][h][t, x] + 
       b*h[t, x]*Derivative[0, 1][u][t, x] + 
       b*Derivative[1, 0][h][t, x] == 
      r, (r*u[t, x])/(b*g*
          h[t, x]) + (Abs[u[t, x]]*(b + 2*h[t, x])^(4/3)*
          u[t, x])/(a^2*(b*h[t, x])^(4/3)) + 
       Derivative[0, 1][h][t, 
        x] + (u[t, x]*Derivative[0, 1][u][t, x])/g + 
       Derivative[1, 0][u][t, x]/g == 0, 
     Derivative[1, 0][gzzzzz][t, x] == b*h[t, x]*u[t, x], 
     h[0, x] == 1*ppa/ppa, u[0, x] == 0, 
     h[t, le] == gzzzzz[t, le]/ppa, u[t, 0] == 0, 
     gzzzzz[0, x] == 1*ppa}, {u, h, gzzzzz}, {t, 0, tf}, {x, 0, le}];

(h /. ans)[tf, 0]

gives 2.38994 and

a = 80; le = 62; g = 9.8; tf = 50; b = 3.4; r = 5.86/le; ppa = 32*22;

ans = Flatten@
   NDSolve[{b*u[t, x]*Derivative[0, 1][h][t, x] + 
       b*h[t, x]*Derivative[0, 1][u][t, x] + 
       b*Derivative[1, 0][h][t, x] == 
      r, (r*u[t, x])/(b*g*
          h[t, x]) + (Abs[u[t, x]]*(b + 2*h[t, x])^(4/3)*
          u[t, x])/(a^2*(b*h[t, x])^(4/3)) + 
       Derivative[0, 1][h][t, 
        x] + (u[t, x]*Derivative[0, 1][u][t, x])/g + 
       Derivative[1, 0][u][t, x]/g == 0, 
     Derivative[1, 0][haaa][t, x] == b*h[t, x]*u[t, x], 
     h[0, x] == 1*ppa/ppa, u[0, x] == 0, h[t, le] == haaa[t, le]/ppa, 
     u[t, 0] == 0, haaa[0, x] == 1*ppa}, {u, h, haaa}, {t, 0, tf}, {x,
      0, le}];

(h /. ans)[tf, 0]

gives 1.35899.

Note that one of the integration variables is called h.

Cheers, Marco