Combination Of Metallic Rods

A combination of metallic rods plays a significant role in both engineering and everyday life, where they are used to create sturdy, functional structures. These rods, often made from materials like steel, aluminium, or copper, are essential in the construction of buildings, bridges, and vehicles due to their strength, durability, and ability to withstand various forces. In real life, the framework of a skyscraper or the chassis of a car relies on a carefully calculated combination of metallic rods to ensure safety and stability. Whether in massive infrastructures or simple household items, understanding the principles behind combining metallic rods is crucial for innovation and safety. In this article, we will cover the concept of Combination Of Metallic Rods. This concept falls under the broader category of Properties of Solids and Liquids.

Series Combination of Rod/Slabs in Heat Conduction

In heat conduction, the concept of series combination of rods or slabs is vital in understanding how heat transfers through materials layered in a sequence. When different metallic rods or slabs are connected end-to-end in a series, heat flows from the hotter end to the cooler end through each material. The rate at which heat is conducted depends on the thermal conductivity of each material and their individual lengths and cross-sectional areas. This principle is commonly seen in real-life applications such as the insulation of buildings, where multiple layers of different materials are used to control heat flow, ensuring energy efficiency.

Series Combination

Series combination refers to the arrangement of different materials (such as rods or slabs) in a sequential manner where heat must pass through one material before reaching the next. When materials are combined in series, the overall thermal resistance to heat flow is the sum of the individual resistances of each material.

Let n slabs each of cross-sectional area A , lengths l1,l2,l3……ln and conductivities K1,K2,K3……Kn respectively be connected in series

• Heat current: In the case of a series combination, the heat current is the same in all the conductors, So

Qt=H1=H2=H3………=Hn

So, by the law of thermal conductivity - K1A(θ1−θ2)l1=K2A(θ2−θ3)l2=KnA(θn−1−θn)ln

• Thermal resistance - Net thermal resistance is equal to the sum of thermal resistance of all the slabs/rods. So, -

Equivalent thermal resistance: R=R1+R2+……Rn

• Thermal conductivity - From the above equation of equivalent thermal resistance, equivalent thermal conductivity can be calculated as

From RS=R1+R2+R3+…l1+l2+…lnKeqAeq=l1K1A+l2K2A+…+lnKnA⇒Kequivalent =l1+l2+……lnl1K1+l2K2+……+lnKnAeqLeq=A1L1+A2L2+A3L3+…+AnLn

For series combination: Leq=L1+L2+L3+…+Ln (for each slab having a constant area of cross-section)
For parallel combination: Aeq=A1+A2+A3+…+An (for each slab having the same length of the slab)

• The temperature of the interface of the composite bar: For the calculation of this, let the two bars be arranged in series as shown in figure

i.., Qt=K1A(θ1−θ)l1=K2A(θ−θ2)l2

By solving, we get θ=K1l1θ1+K2l2θ2K1l1+K2l2

Parallel Combination of Rod/Slabs in Heat Conduction

A parallel combination refers to an arrangement where multiple rods or slabs are positioned side by side, allowing heat to flow simultaneously through each material. Unlike series combinations, where heat must pass sequentially through each material, in a parallel combination, the heat divides among the different paths based on the thermal conductivities and cross-sectional areas of the materials involved. This setup results in a lower overall thermal resistance, as heat can take multiple paths to transfer through the system.

Parallel Combination

A parallel combination refers to an arrangement where multiple rods or slabs are placed side by side, allowing heat to flow through each material simultaneously. Unlike in a series combination, where heat must pass through each material one after the other, in a parallel combination, the heat divides among the different paths, flowing more easily through materials with higher thermal conductivity.

Let n slabs each of lengths L , cross-sectional area A1,A2,A3…….An and conductivities K1,K2,K3……Kn respectively be connected in parallel

Heat current: If each slab has different thermal conductivity, then the Net heat current will be the sum of heat currents through individual slabs. i.e.,
H=H1+H2+H3+…Hn

So, by the law of thermal conductivity -
K(A1+A2+…,+An)(θ1−θ2)l=K1A1(θ1−θ2)l+K2A2(θ1−θ2)l+…+K1An(θ1−θ2)l
Equivalent Thermal resistance - Net thermal resistance in parallel combination -
1Rs=1R1+1R2+1R3+……1Rn
Thermal conductivity - From the above equation of equivalent thermal resistance, equivalent thermal conductivity can be calculated as
K(A1+A2+…,+An)(θ1−θ2)l=K1A1(θ1−θ2)l+K2A2(θ1−θ2)l+…+K1An(θ1−θ2)l⇒Kequivalent =K1A1+K2A2+K3A3+….KnAnA1+A2+A3+….An

• The temperature of the interface of the composite bar: The temperature gradient Same across each slab.

Solved Examples Based on Combination of Metallic Rods

Example 1: In a composite rod, when two rods of different lengths and of the same cross-section area are joined end to end, and K is the effective coefficient of thermal conductivity, then l1+l2K is equal to

l1K1−l2K2 2) l1K2−l2K1 3) l1K1+l2K2 4) l1K1+l2K2

Solution

For a series of combinations R=R1+R2
⇒l1+l2KA=l1K1A+l2K2A⇒l1+l2K=l1K1+l2K2

Hence, the answer is the option (4).

Example 2: Three identical rods A, B and C of equal length and equal diameter are joined in series. Their thermal conductivities are 2K, K and K2 respectively. The temperature of the two junction points are

1)85.7, 57.1^oC

2)80.85, 50.3^oC

3)77.3, 48.3^oC

4)75.8, 49.3^oC

Solution

Equivalent Thermal Conductivity
Ks=2K1K2K1+K2
wherein
For two slabs of equal length.
Heat current is the same in A,B and C
⇒100−T1l2KA=T1−T2lKA=T2−0l12A
⇒T1−T2=T22 or T1=3T222
2(100−T1)=T22
or 100−T1=4T2 or 100−3T22=T24
⇒100=(32+14)T2=74T2
or T2=4007=57.1∘C and T1=85.7∘C

Hence, the answer is the option (1).

Example 3: Two walls of thickness d1 and d2 and thermal conductivities K1 and K2 are in contact. In the steady state, if the temperature at the inner surface is T₁ and the temperature at the outer surface is T2, the temperature at the common wall is

1) K1T1d2+K2T2d1K1d2+K2d1
2) K1T1+K2T2d2+d1
3) K1d2+K2d1T1+T2
4) K1T1d1+K2T2d2K1d1+K2d2

Solution

Temperature of Interface (Junction Temperature)

θ=K1l1θ1+K2l2θ2K1l1+K2l2

Let the temperature at the junction be T
⇒K1(T1−T)Ad1=K2(T−T2)Ad2⇒K1T1d2−K1Td2=K2d1T−K2d1T2⇒T=K1T1d2+K2d1T2K1d2+d1K2

Hence, the answer is the option (1).

Example 4: Three rods of identical cross-sections and lengths are made of three different materials of thermal conductivity K1,K2 and K3

respectively. They are joined together at their ends to make a long rod (See figure). One end of the long rod is maintained at 100∘C and the other at 0∘C (see figure). If the joints of the rod are at 70∘C and 20∘C in a steady state and there is no loss of energy from the surface of the rod, the correct relationship between K1,K2 and K3 is :

1) K1:K3=2:3K2:K3=2:5

2) K1<K3<K3

3) K1:K2=5:2,K1:K3=3:5

4) K1<K2<K3

Solution

Rods are identical and have the same length (l) and area of cross-section (A)

Combinations are in series, so the heat current is the same for all Rods

(ΔQΔt)AB=(ΔQΔt)BC=(ΔQΔt)C= Heat current (100−70)K1 Aℓ=(70−20)K2 Aℓ=(20−0)K3 Aℓ30 K1=50 K2=20 K33 K1=2 K3K1 K3=23=2:35 K2=2 K3K2 K3=25=2:5

Hence, the answer is the option (1).

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Summary

This article explores the principles of heat conduction in metallic rods arranged in series and parallel combinations. It explains how thermal conductivity, resistance, and heat current behave in different setups, using real-life examples like building insulation and electronic cooling systems. Through solved examples, the article demonstrates the calculation of equivalent thermal conductivity and temperature distribution in composite rods, providing a practical understanding of thermal management in engineering applications.