Answer to Question #241216 in Mechanical Engineering for shyamnanda

Question

A thermocouple is used to measure the temperature of air flowing through a duct. The heat transfer coefficient between the air and the thermocouple is proportional to v0.8, where v is the average velocity of the air within the duct. Assume that the heat transfer by radiation from the duct walls to the thermocouple bead is linearly proportional to the temperature difference between them. For v = 5 m/s, the thermocouple reads 50 °C while when the air flows at 10 m/s, it reads 40 "C. (a) If the actual temperature of the air is 25 °C, estimate the wall

temperature. (b) What temperature will the thermocouple indicate when the gas velocity is 20 m/s.

Answer 1

given data

$v=5m/s$

$T=50\degree C$

air flow=$10 m/s$

T=$40\degree$

actual air temprature Ta =$25\degree$

from the data

$h\alpha 0.8 V$

$h=k(0.8V)$

in the thermocouple read the radiation from the wall is proportional to $\Delta T$

$q_{radiation}$ $\alpha$ $\Delta T$

$q_{radiation} = C(\Delta T)$

=C(Tb-Tw)

consider energy balance

$q_{conduction}$ = $q_{radiation}$

$h\Delta T=C \Delta (Tb -Tw)$

$k(0.8V)\Delta T = C (Tb - Tw)$ -------(1)

from equation 1

$0.8 (5) k (25 - 50) = C (50 - Tw)$ ------(2)

$0.8 (10) k (25 - 40) = C (40 - Tw)$-------(3)

$2/3 =(0.8) (5) k (-25) /(0.8)(10) k(15) = C (50 - Tw) / C ( 40 - Tw )$

$25 / 30 = 50 - Tw / 40 - Tw$

$25 (40 - Tw ) = 30 (50 - Tw)$

$1000 - 25Tw = 1500 - 30Tw$

$5Tw = 500$

$Tw =100$

(b)

$V = 20m/s$

$T_{th}$

from 1

$(0.8)(20) k (25 - T_{th} ) = C (T_{th} - Tw)$ --------(4)

from 2 & 4

$(0.8) 10k (-15) / (0.8) 20k (25 - T_{th}) = 40-100/T_{th} - 100$

$-15T_{th} + 1500 = -3000 + 120T_{th}$

$4500 = 135T_{th}$

$T_{th} = 33.3\degree$