Consider the situation shown in the figure . The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.
Solution
`R_{BC} = 1/{KA }= 1/{KA}`
`R_{CD} = 1/{KA }= 60/{KA}`
`R_{CD} = 5/{KA} , R_{AB} = 20/M , R_{EF} = 20/{KA}`
Let ;
`R_1 = R_{BC} + R_{CD} + R_{DE} = 70/{KA}`
Let :
`R_{BE} = 60/{KA} = R_2
q = q1 + q2 ...............(1)
R1 and R2 are in parallel, so total heat across R1 and R2 will be same.
⇒ `q_1R_2 = q_2R_2`
`q_1xx 70/{KA} = q_2xx 60/{KA}`
`7q_1 = 6q_2`
`{7q_1}/{6}= q_2`
From equation (1) and (2),
q = q_1 + {7q_1}/6`
`q = {13q_1}/6`
`q=130 J`
`130 = 13Qq_1`
q1 = 60 J/sec
