Effect of wall location for guided wave attenuation
Since the tube now functions like a waveguide, I would argue that the shield need not be considered as all you need to be worried about is the radiation reflected and transmitted by the tube. So the tube in either configuration can be approximated by waveguide laws, and the effects from the position of the conductor would not affect the radiation passing through the waveguide. 


  $$ \lambda (meters) = \frac{300m/\mu s}{Frequency(MHz)}$$
  
  As the frequency increases, its wavelength gets smaller. Once the
  signal frequency increases to the point where its half-wavelength
  (l/2) is smaller than the spacing between the transmission-line
  conductors, a new propagation mode can exist, and the conductors can
  be configured as a hollow tube or waveguide. At these frequencies and
  higher, the RF energy can propagate through a hollow tube, essentially
  a single-conductor transmission line.
  
  Although there are several popular waveguide configurations, the
  rectangular cross-section is the easiest to visualize. In this case,
  the waveguide’s cutoff frequency ( Fco ) is determined by its width,
  which must be slightly wider than l/2.
  
  The height is independent of frequency, but a 2:1 width-to-height
  ratio is very popular. At this ratio, the RF energy entering the
  waveguide reflects at an angle from side to side and travels through
  the tube with very little loss.
  
  A waveguide is very efficient at passing all frequencies above its
  Fco, acting like a high-pass filter. For example, at 10-GHz,
  transmission-line losses in a 2-in. rectangular copper waveguide range
  from approximately 0.02 to 0.12 dB per meter length, depending on
  height.
  
  There are a number of possible RF transmission modes, but from a
  shielding perspective, it is the waveguide’s behavior below cutoff for
  its lowest frequency propagation mode, such as the TE10 mode, that is
  of interest. Below its Fco, a waveguide is not a good transmission
  line, and it has very high attenuation.
  
  For a circular waveguide:
  
  $$ F_{co}(MHz) = \frac{6920in/\mu s}{D}$$
  
  Below the Fco, the attenuation per unit length (inches) at any
  frequency (F) is given by:
  
  $$ A_{UL}(db/in) = 0.00463 F \sqrt{(F_{co}/F)^2-1}$$


Source: https://evaluationengineering.com/features/0101emc.htm