Question

materials A, B, C have R-values of 1, 4, and 8ft^2-°F-hr/Btu, respectively. If the rate of heat transfer through material A alone is 20 Btu/hr, then the rate of heat transfer through the combination of A, B, and C is

Answer 1

Solution:

Given that materials A,B & C have R-values is given as 1,4 & 8.

If the rate of heat transfer through material A alone is 20Btu/hr., So Q1=20Btu/hr.

A+B+C=?

In the above sum, as materials are connecting in'series' materials have the same rate of heat transfer then answer is '20Btu/hr'.

If they are connected in parallel,then

Q=Q1+Q2+Q3

Then R values are

1/R=1/R1+1/R2+1/R3

= 1/1+1/4+1/8=11/8

Therefore,

1/R=1.375

So, R=0.7272

Again R1Q1=R2Q2=R3Q3=RQ

Now, We need to equate to each other

R1Q1=R2Q2

1*20=4*Q2 (Since Q1=20)

So, from above

Q2=5 Btu/hr

In same way Q3 Will be calculated

R1Q1=R3Q3

1*20=8*Q3

So, from above

Q3=2.5Btu/hr

Therefore, the total Q =Q1+Q2+Q3

So,Q= 20+5+2.5

Finally, Total Q=27.5Btu/hr.

Therefore the rate of heat transfer through the combination of A,B &C is Q=27.5Btu/hr.