Answer to Question #241273 in Mechanical Engineering for Sweeja

Question

1: 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

Answer 1

(part-a)

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)

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

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

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

$1000 - 25Tw = 1500 - 30Tw$

$5Tw = 500$

$Tw =100$

(part-b)

$V = 20m/s$

from --(1)

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

from 2 & 4

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

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

$4500 = 135T_{th}$

$T_{th} = 33.3\degree$