Ci 150ms Ki5774105 kwmk Vi8214105 m2s Pri69 Nui023Rei45 Pri1

C,i= 150m/s

K,i=5774*10^-5 kw/mk

V,i=8.214*10^-5 m^2/s

Pr,i=.69

Nu,i=.023(Re,i)^4/5 (Pr,i)^1/3

C,o=340 m/s

K,o=10.233*10^-5 kw/mk

V,o=32.34*10^-5 m^2/s

Pr,o=.73

Nu,o=.193(Re,o)^.618(Pr,o)^1/3

Baseline Case: We consider a hollow cylinder (to represent leading edge of a cooled vane) with length = 200mm (to represent span), OD = 25mm, and ID = 10mm (to represent cooling hole hydraulic diameter). Material = IN 718. The cylinder is coated with 1mm thick YSZ. External flow: T_o = 1900K, C = 340m/s, use average external heat transfer coefficient as given by: Nu-_od = 0.193 Re_od^0.618 Pr^1/3. For simplicity, assume recovery factor of 1.0. Internal flow: enters the cooling hole at 790K with a velocity of 150m/s. Assume fully developed flow. Calculate the maximum metal temperature for baseline case. Plot temperature variation through metal and TBC as a function of radius. Clearly indicate coolant and gas temperatures on the temperature axis. If internal heat transfer is enhanced over the baseline case by use of turbulators by a factor of (h/ho = 1.2), recalculate the maximum metal temperature If C is increased to 400 m/s as compared to the baseline case, recalculate the maximum metal temperature. If TBC spalls off as compared to the baseline case, recalculate the maximum metal temperature.

Solution

Given : d_o = 0.025 + 0.002 = 0.027 m, d_i = 0.01m, T_o = 1900K, T_i = 790K

a) Re_i = (150*0.01)/(8.214*10^-5) = 18261.5047

Nu_i = 52.1483

h_i = 301.104*10³ W/m²K

Re_o = (340*0.027)/(32.34*10^-5) = 28385.899

Nu_o = 98.178

h_o = 372.096 W/m²K

Now, h_i *A_i*(T_s-T_i) = h_i *A_i*(T_s-T_i)

where T_s = surface temperature

Substituting values & calculating, we get T_s = 793.684K

c) given: h/h_o = 1.2,

Therefore new h_i = 361.3248*10³ w/m²K

Now, substituting values in below equation and solving,

h_i *A_i*(T_s-T_i) = h_i *A_i*(T_s-T_i)

T_s = 792.949K

d) C= 400 m/s,

Re_o= 33395.17625

Nu_o = 108.5512

So h_o = 411.409 w/m²k

Now, substituting values in below equation and solving,

h_i *A_i*(T_s-T_i) = h_i *A_i*(T_s-T_i)

T_s = 794.0727 K

e) If TBC spalls off, then d_o = 0.025m

Re_o= 26283.24

Nu_o = 93.6181

So h_o = 383.197 w/m²k

Now, substituting values in below equation and solving,

h_i *A_i*(T_s-T_i) = h_i *A_i*(T_s-T_i)

T_s = 793.513 K