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The Mass Balance Along a Flowband of Width w I

The Mass Balance Along a Flowband of Width w I. Figure 1: Mass balance along an ice-sheet flowband. Mass balance Equation along x. (  h 1 /  t)x = ax - (u x h I - u 0 h 0 ) (1). Mass balance at x.  h I /  t = a -  (u x h I )/  x = a - u x dh I /dx - h I du x /dx (2).

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The Mass Balance Along a Flowband of Width w I

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  1. The Mass Balance Along a Flowband of Width wI

  2. Figure 1: Mass balance along an ice-sheet flowband

  3. Mass balance Equation along x • (h1/t)x = ax - (uxhI - u0h0) (1)

  4. Mass balance at x • hI/t = a - (uxhI)/x = a - uxdhI/dx - hIdux/dx (2)

  5. Solving (2) for dhI/dx • dhI/dx = (a - hI/t)/ux - hI(dux/dx)/ux • = (a - hI/t)/ux - hIxx/ux (3)

  6. Substituting for ux from (1) • dhI/dx = [hI(a -hI/t)]/[(a -hI/t)x + u0h0] -hI2xx/[(a - hI/t)x + u0h0] (4)

  7. Ignoring accumulation rate and thinning rate • dhI/dx = - hI2xx / u0h0 (5)

  8. Using the flow law • xx = R(/xx/A)n = (T/2A)n (6) where R = 1 and /xx = T/2 for tensile flow

  9. Figure 2: A flowband profile

  10. From Figure (2) • T = (1/2)ghI(1 - wPw/PI)2 = (1/2)IghI(1 - w)

  11. Substitute the flow law for xxand hI/xfor dhI/dx • dhI/dx = - hI2xx / u0h0 •  • hI/x = -[(g / 4A)(1-I/w)]3(hI5u0h0) (8) • where grounding line velocity u0 is negative • Equation (8) is solved for  for Byrd Glacier

  12. Figure 3: Byrd Glacier showingradio-echo flight line

  13. Figure 4: Results along flightlinefrom Reusch and Hughes (2003)

  14. Results from Equation (8)

  15. Results from Vance and Ashley

  16. Results for floating ice • The first two 5 km steps gave from Equation (8) because Byrd Glacier was afloat with a basal melting rate of 12 m/a and hI/x = 0. Then Equation (3) is: • dhI/dx = (a - hI/t)/ux - hI(dux/dx)/ux = (a - hI/t)/ux - hIxx/ux= 0 (9)

  17. Equation (9) for floating ice • Solving Equation (9) for hI/t gives: hI/t = a - hIxx= a + hIzz (10) a = -12 m/a (Kenneally and Hughes, 2004) hI = 725 m (Reusch and Hughes, 2003) xx = (-120 m/a)/(50 km) (Brecher, 1982) Equation (10) giveshI/t = -10 m/a

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