Aside from the values is there anything wrong with the physical design of the psu?
No, it is fine.
(1) Off-load the output voltage from a bridge rectifier as you have drawn is.
(1.1) Vout DC= V secondary RMS * 1.414 (root 2) - 1.2V (two off-load diode drops)
(2) The peak to peak ripple voltage across a reservoir capacitor = (I * T)/C
Where
I = current drain in Amps
T = period of mains voltage/2 in seconds
C = the reservoir capacitor value in Farads
The big unknown, in any power supply like this, is the characteristics of the transformer which controls what the output voltage will be under load.
As a very rough rule of thumb, with 50Hz/60H mains transformers, is that the peak voltage of the secondary under the load current specified for the transformer will be equal to the specified off load RMS Voltage of the secondary.
For example, if you had a transformer specified with a secondary of 12V RMS at a current of 1 Amp (12Watt transformer) and you used it in the circuit you posted with a current drain of 1A, the out put voltage would be 12V- 2V (two diode drop) = 10V. The reason why the diodes now drop 2V instead of 1.2V as when off-load is because the diodes are now conducting a heavy current. In fact the diodes charge up the reservoir capacitor with huge bursts of current at the negative and positive peaks of the secondary voltage sine wave.
If the reservoir capacitor was 5,000uF and the mains frequency was 50Hz, then the peak to peak ripple Voltage across the reservoir capacitor would be:
(1A * 10 mili seconds)/ 5,000 micro Farads = 2V peak to peak.
spec
PS: the above is very simplified but it will get you by for the vast majority of power supply designs.