Question 417: Properties of Matter

Question

A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then, in steady-state

Accepted Answer

Let T
1
, T
2
, T
3
and T
4
be the temperatures at slab interfaces as shown in the figure.
The rate of heat flow across a slab, with thermal conductivity $$\kappa $$, width w, thickness l, length x, and temperature difference $$\Delta$$T, is given by
$$dQ/dt = \kappa lw(\Delta T/x)$$.
Thus, the heat flow rates through the given slabs are
$${{d{Q_A}} \over {dt}} = {{(2K)(4Lw)({T_1} - {T_2})} \over L} = 8Kw({T_1} - {T_2})$$ ...... (1)
$${{d{Q_B}} \over {dt}} = {{(3K)(Lw)({T_2} - {T_3})} \over {4L}} = {3 \over 4}Kw({T_2} - {T_3})$$ ...... (2)
$${{d{Q_C}} \over {dt}} = {{(4K)(2Lw)({T_2} - {T_3})} \over {4L}} = 2Kw({T_2} - {T_3})$$ ...... (3)
$${{d{Q_D}} \over {dt}} = {{(5K)(Lw)({T_2} - {T_3})} \over {4L}} = {5 \over 4}Kw({T_2} - {T_3})$$ ..... (4)
$${{d{Q_E}} \over {dt}} = {{(6K)(4Lw)({T_3} - {T_4})} \over L} = 24Kw({T_3} - {T_4})$$ ..... (5)
In the steady state, heat flow into the system is equal to the heat flow out of the system i.e., $$d{Q_A}/dt = d{Q_E}/dt$$.
Use equations (1) and (5) to get
$${T_1} - {T_2} = 3({T_3} - {T_4})$$ ..... (6)
Similarly, the steady state condition for the interface at temperature T
2
is
$$d{Q_A}/dt = d{Q_B}/dt + d{Q_C}/dt + d{Q_D}/dt$$,
which gives (by using equations (1)-(4))
$$2({T_1} - {T_2}) = {T_2} - {T_3}$$ ..... (7)
Use equations (6) and (7) to get
$${T_3} - {T_4} = (1/3)({T_1} - {T_2}) = (1/6)({T_2}/{T_3})$$.
Also, equations (2)-(4) give $$d{Q_C}/dt = d{Q_B}/dt + d{Q_D}/dt$$.

Answer

heat flow through A and E slabs are same.

Answer

heat flow through slab E is maximum.

Answer

temperature difference across slab E is smallest.

Answer

heat flow through C = heat flow through B + heat flow through D.

Properties of Matter

A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then, in steady-state

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