LM340T15 current limiter
Note that maximum Vout is about
\$V_{out_{max}} = V_{in} - V_{reg} - 3V\$ (due to ~~~= 3V across regulator.)
The Higher the valuie of \$V_{reg}\$ the lower the available \$V_{out}\$ range.
With \$V_{in} = 30V\$ and \$V_{reg} = 15V \$ max \$V_{out}\$ is about 12V ( = 30-15-3)
\$V_{out} = I\times R = I_{put} \times R_{load}\$.
\$V_{reg_{drop}} \sim Vin - V_{reg_{out}} - Vout\$,(across R9) so
\$V_{reg_{drop}} = V_{in} - V_{reg_{out}} - I_{reg} \times R_{load}\$.
As \$P_{reg} = V_{reg_{drop}} \times I_{reg}\$ and as \$I_{reg}\$ is constant,
then regulator dissipation is maximum when \$V_{out}\$ is 0 (as there is no dissipation in R10.
So - Pregulator  is max at Rload =0 and decreases as Vout (= R10 x Ireg) increases to the maximum allowed.
\$P_{reg_{max}} =   (V_{in} - V_{reg}) \times I_{reg} = (V_{in} - V_{R10}) \times I_{reg} = (30-15) \times 0.15 = 2.25 W\$
Lower voltage regulators will allow a higher Vout range and incur higher worst case regulator dissipations when the output is shorted, and lower (constant) R9 dissipations. .
Power in R9 is a constant \$P_{R9} = V \times I = V_{reg} \times I_{reg}\$.
Here \$P_{R9} = 15V \times 150 mA = 2.25 Watt\$