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Question: A rod CD of thermal resistance 5kW-1 is joined at the middle of an identical rod AB as shown in figu...

A rod CD of thermal resistance 5kW-1 is joined at the middle of an identical rod AB as shown in figure. The ends A,B and D are maintained at 100°C, 0°C, 25°C respectively. Find the heat current in CD.

Answer

4

Explanation

Solution

The problem involves heat conduction through a network of rods. We'll use the concept of thermal resistance and Kirchhoff's current law for heat flow.

1. Identify Thermal Resistances: Let R be the thermal resistance of a full rod (like CD or the entire AB). Given, thermal resistance of rod CD, RCD=5 kW−1R_{CD} = 5 \text{ kW}^{-1}. Since rod AB is identical to rod CD, the thermal resistance of the entire rod AB is also RAB=5 kW−1R_{AB} = 5 \text{ kW}^{-1}. Rod CD is joined at the middle of rod AB. This means rod AB is divided into two equal segments, AC and CB. The thermal resistance of a rod is directly proportional to its length. So, RAC=RAB2=52=2.5 kW−1R_{AC} = \frac{R_{AB}}{2} = \frac{5}{2} = 2.5 \text{ kW}^{-1} RCB=RAB2=52=2.5 kW−1R_{CB} = \frac{R_{AB}}{2} = \frac{5}{2} = 2.5 \text{ kW}^{-1}

2. Apply Kirchhoff's Current Law at Junction C: Let the temperature at junction C be TCT_C. The ends A, B, and D are maintained at TA=100∘CT_A = 100^\circ C, TB=0∘CT_B = 0^\circ C, and TD=25∘CT_D = 25^\circ C respectively. Heat current (HH) is given by the temperature difference divided by thermal resistance: H=ΔTRH = \frac{\Delta T}{R}. At junction C, the sum of heat currents entering the junction must equal the sum of heat currents leaving the junction. Assuming heat flows from higher to lower temperature: Heat current from A to C: HAC=TA−TCRAC=100−TC2.5H_{AC} = \frac{T_A - T_C}{R_{AC}} = \frac{100 - T_C}{2.5} Heat current from C to B: HCB=TC−TBRCB=TC−02.5H_{CB} = \frac{T_C - T_B}{R_{CB}} = \frac{T_C - 0}{2.5} Heat current from C to D: HCD=TC−TDRCD=TC−255H_{CD} = \frac{T_C - T_D}{R_{CD}} = \frac{T_C - 25}{5}

Applying the junction rule (HAC=HCB+HCDH_{AC} = H_{CB} + H_{CD}): 100−TC2.5=TC−02.5+TC−255\frac{100 - T_C}{2.5} = \frac{T_C - 0}{2.5} + \frac{T_C - 25}{5} To eliminate denominators, multiply the entire equation by 5: 2(100−TC)=2(TC−0)+(TC−25)2(100 - T_C) = 2(T_C - 0) + (T_C - 25) 200−2TC=2TC+TC−25200 - 2T_C = 2T_C + T_C - 25 200−2TC=3TC−25200 - 2T_C = 3T_C - 25 Rearrange the terms to solve for TCT_C: 200+25=3TC+2TC200 + 25 = 3T_C + 2T_C 225=5TC225 = 5T_C TC=2255=45∘CT_C = \frac{225}{5} = 45^\circ C

3. Calculate Heat Current in CD: Now that we have the temperature at junction C, we can find the heat current in CD: HCD=TC−TDRCDH_{CD} = \frac{T_C - T_D}{R_{CD}} HCD=45∘C−25∘C5 kW−1H_{CD} = \frac{45^\circ C - 25^\circ C}{5 \text{ kW}^{-1}} HCD=205H_{CD} = \frac{20}{5} HCD=4 WH_{CD} = 4 \text{ W} The positive value indicates that heat flows from C to D.

Solution:

  1. Determine thermal resistances: RCD=5 kW−1R_{CD} = 5 \text{ kW}^{-1}. Since AB is identical and CD is at its middle, RAC=RCB=RAB/2=RCD/2=2.5 kW−1R_{AC} = R_{CB} = R_{AB}/2 = R_{CD}/2 = 2.5 \text{ kW}^{-1}.
  2. Apply Kirchhoff's current law at junction C: TA−TCRAC=TC−TBRCB+TC−TDRCD\frac{T_A - T_C}{R_{AC}} = \frac{T_C - T_B}{R_{CB}} + \frac{T_C - T_D}{R_{CD}}.
  3. Substitute values: 100−TC2.5=TC−02.5+TC−255\frac{100 - T_C}{2.5} = \frac{T_C - 0}{2.5} + \frac{T_C - 25}{5}.
  4. Solve for TCT_C: 2(100−TC)=2TC+(TC−25)  ⟹  200−2TC=3TC−25  ⟹  5TC=225  ⟹  TC=45∘C2(100 - T_C) = 2T_C + (T_C - 25) \implies 200 - 2T_C = 3T_C - 25 \implies 5T_C = 225 \implies T_C = 45^\circ C.
  5. Calculate heat current in CD: HCD=TC−TDRCD=45−255=205=4 WH_{CD} = \frac{T_C - T_D}{R_{CD}} = \frac{45 - 25}{5} = \frac{20}{5} = 4 \text{ W}.