Cecil Moore wrote:
If you change the source to a Thevenin equivalent 141.42V
and 50 ohm source impedance and specify Z=50 ohms feedline
impedance then the problem becomes solvable.

Going with this configuration which is similar to the 1/2
WL stub suggested by W7EL in his Food For Thought #1:

SW
Source nc c Pfor=100W-- --Pref=100W
141.4V---o---o---one second long lossless 50 ohm stub--open
50 ohm
o---/\/\/\/\/\--Gnd
no 50 ohms

For the first two seconds after power up, the Thevenin
source delivers 200 joules into the stub while dissipating
200 joules as heat in the source resistance. Those 200
joules of energy in the stub must be conserved.

During steady-state, assuming the stub is an exact integral
number of wavelengths, the source will see an open circuit
and be delivering zero power. A wattmeter calibrated for
50 ohms will read 100 watts forward power and 100 watts
reflected power during steady-state.

After steady-state is reached, we throw the switch and connect
the stub to a 50 ohm dummy load. 100 watts will be supplied
to the dummy load for two seconds for a total of 200 joules.

Who was it who said that reflected power cannot be
recovered? (Food For Thought #1 on www.eznec.com).
--
73, Cecil http://www.qsl.net/w5dxp