A brass rod 12.0 cm long, a copper rod 18.0 cm long, and an aluminum rod 24.0 er

Question

A brass rod 12.0 cm long, a copper rod 18.0 cm long, and an aluminum rod 24.0 err long-each with cross-sectional area 2.30 cm^3-are welded together end to end to form a rod 54.0 cm long, with copper as the middle section. The free end of the brass section is maintained at 100.0degree C, and the free end of the aluminum section is maintained at 0.0degree C. Assume that there is no heat loss from the curved surfaces and that the steady-state heat current has been established. What is (a) the temperature T_1 at the junction of the brass and copper sections; (b) the temperature T_2 at the junction of the copper and aluminum sections; the heat current in the aluminum section?

Explanation / Answer

Solution: Length of brass rod, Lb = 12.0 cm = 12.0cm*(1m/100cm) = 0.12 m

Thermal conductivity of brass, kb = 109 W/m.oC

Length of copper rod, Lc = 18.0 cm = 0.18 m

Thermal conductivity of copper, kc = 401 W/m.oC

Length of aluminum rod, La = 24.0 cm = 0.24 m

Thermal conductivity of aluminum, ka = 235 W/m.oC

Cross sectional area of each of the rod is, A = 2.30 cm2 = 2.30 cm2 *(1m/100cm)2 = 2.3*10-4 m3

High temperature at the free end of brass section, TH = 100.0oC

Low temperature at the free end of aluminum section, TC = 0.0oC

There is no heat loss and the system is in thermal steady state. Thus rate of heat conduction Pcond is same everywhere in the composite rod

For the steady state, the rate of heat conduction through composite slabs (in our case the composite rods) is given by,

Pcond = A*(TH – TC) / (L/k)

For brass, copper and aluminum, we get,

Pcond = A*(TH – TC) / (Lb/kb + Lc/kc + La/ka)

Pcond = (2.3*10-4 m2)*( 100.0oC - 0.0oC) / [(0.12m/109W.moC) + (0.18m/401W.moC)+ (0.24m/235W.moC)]

Pcond = 8.9457 W

Part (a)

Now consider the brass-copper interface. Let T1 be the temperature at the brass-copper interface.

From the rate of conduction for these brass we have,

Pcond = kb*A(TH – T1)/Lb

8.9457 W = (109 W.m/oC)*( 2.3*10-6 m3)*(100oC – T1) / 0.12m

(100oC – T1) = (8.9457 W)*(0.12 m) / [(109 W.m/oC)*( 2.3*10-4 m2)]

(100oC – T1) = 42.82oC

T1 = 57.18 oC

Hence the answer to this part is 57.18 oC.

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Part (B) Now consider the copper-aluminum interface. Let T2 be the temperature at the copper-aluminum interface.

From the rate of conduction for the copper we have,

Pcond = kc*A(T1 – T2)/Lc

8.9457 W = (401 W.m/oC)*( 2.3*10-4 m2)*(57.18oC – T2) / 0.18m

(57.18oC – T2) = (8.9457 W)*(0.18 m) / [(401 W.m/oC)*( 2.3*10-4 m3)]

(57.18oC – T2) = 17.46oC

T2 = 39.72 oC

Hence the answer to this part is 39.72 oC.

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Part (c) Since the composite rod is in the steady state, the heat current is same everywhere in the rod. Thus, the heat current in aluminum section is, Pcond = 8.9457 W

Hence the answer is 8.95W

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