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aspifsGak5u

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2021-12-14

Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at $T_{\infty} = 1200^{\circ} C$ . and maintains a convection coefficient of $h = 250 \frac{W}{m^{2}} \cdot K$ over the blade.
The blades, which are fabricated from Inconel, $k \approx 20 \frac{W}{m} \cdot K$ , have a length of L=50 mm. The blade profile has a uniform cross-sectional area of $A_{c} = 6 \times 10^{- 4} m^{2}$ and a perimeter of P=110 mm.
A proposed blade-cooling scheme, which involves routing air through the supporting disc, is able to maintain the base of each blade at a temperature of $T_{b} = 300^{\circ} C$ .
a) If the maximum allowable blade temperature is $1050^{\circ} C$ and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory?
b) For the proposed cooling scheme, what is the rate at which heat is transferred from each blade to the coolant?

Answer & Explanation

psor32

Beginner2021-12-15Added 33 answers

Given: $L = 0.05 m, A_{c} = 6 x 10^{- 4}, P = 0.11 m, k = 20 \frac{W}{m} \cdot K, h = 250 \frac{W}{m^{2}} \cdot K, T_{b} = 573 K, T_{\infty} = 1473 K, T {(L)}_{m a z} = 1323 K$
Schematic of turbine blade mounted to a rorating disc as follows:
<img src="19610907981.jpgZSK
(a)
Temperature distribution of a fin with an adiabatic tip as follows
$\frac{θ}{θ_{b}} = \frac{\cos h (L - x)}{\cos h (m L)} h e r e, θ = T - T_{\infty}$
$A t, x = L, θ = T (L) - T_{\infty}$
$\frac{T (L) - T_{\infty}}{T_{b} - T_{\infty}} = \frac{\cos h (L - x)}{\cos h (m L)}$
$\frac{T (L) - T_{\infty}}{T_{b} - T_{\infty}} = \frac{1}{\cos h (m L)}$
Calculate the value of m as follows:
$m = \sqrt{\frac{h P}{k A_{c}}}$
$= \sqrt{\frac{250 \times 0.11}{20 \times 6 \times 10^{- 4}}}$
$m = 48.87 m^{- 1}$
Calculate the value of T(L) as folows
$\frac{T (L) - 1473}{573 - 1473} = \frac{1}{\cos h (47.87 \times 0.05)}$
$T (L) - 1473 = - 163$
$T (L) = 1310 K$
As $T (L) = 1310 K < T {(L)}_{m a z}$ . The cooling scheme is satisfactory
(b)
We have the fin heat transfer rate as
$q f = - q b$
$q f = \sqrt{h P k A_{c} θ_{b}} tanh (m L)$
$- q b = \sqrt{h P k A_{c} θ_{b}} (T_{b} - T_{\infty}) tanh (m L)$
$= \sqrt{250 \times 0.11 \times 20 \times 6 \times 106^{- 4}} \times (573 - 1473) \times tanh$

usaho4w

Beginner2021-12-16Added 39 answers

(a) With the maximum temperature existing at x = L, Eq. 3.75 yields

\frac{T (L) - T \infty}{T_{b} - T_{\infty}} = \frac{1}{\cos h m L}

m = {(h \frac{P}{k} A_{c})}^{\frac{1}{2}} = {(250 \frac{W}{m^{2}} \cdot K \times 0.11 \frac{m}{20} \frac{W}{m} \cdot K \times 6 \times 10^{- 4} m^{2})}^{\frac{1}{2}}

m = 47.87 m^{- 1} and m L = 47.87 m^{- 1} \times 0.05 m = 2.39

From Table

B .1, \cos h m L = 5.51

. Hence,

T (L) = 1200^{\circ} C + {(300 - 1200)}^{\circ} \frac{C}{5.51} = 1037^{\circ} C

And the operating conditions are acceptable. NS (b)With

M = {(h P k A_{c})}^{\frac{1}{2}} θ_{b} = {(250 W i m^{2} \cdot K \times 0.11 m \times 20 \frac{W}{m} \times K \times 6 \times 10^{- 4} m^{2})}^{\frac{1}{2}} (- 900 ° c) = - 517 W

Eq. 3.76 and Table B.1 yield

q f = M tanh m L = - 517 W (0.938) = - 508 W

Hence,

q b = - q f = 508 W

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