Total exposed surface 2760m2, of which say 70% may be used as effective surface for incident solar radiation. Total Solar radiation S is 900w/m2 max in the infra red. Assume pipe surface absorptivity = 0.8, and water velocity = v.
Over an incremental length dL of pipe traversed, heat absorbed per sec dQ = S x 0.8 x .7 x pi x D x dL
Heat absorbed by Mass of water traversing through pipe per sec, dQ = dL/dt . pi. (D2/4). rho. Cp. dT
Equating the 2 expressions and simplifying, D/4 . rho . Cp. dT/dt = S . 0.56, which can be integrated easily for the total traversal time. So for 4000m at 1.5m/sec, t = 4000/1.5 = 2670secs, and T-30 = 5.8degC, for Cp = 4180J/kg/degC. So T = 36degC. You can adjust the traversal time t if your water velocity is not 1.5m/sec.
Natural or forced convection from hot air to pipe is small at low wind velocity in comparison and neglected.