Coefficient Of Thermal Conductivity
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Card # 38 Coefficient of Thermal Conductivity The amount of heat Q flowing from one face of rod to the other face depends on the following factors.
Q
α
A
A= Area of cross-section
Q α (θ
1
− θ
2
)
( θ 1 − θ 2 ) = temperature difference of faces of the rod. Q α t Where t = time for heating 1 Where d = length of the rod Q α d On combining
Α (θ
Q α
1
− θ d
2
)t
Or Q
=
Κ Α (θ
1
− θ
2
)t
d
K = coefficient of thermal conductivity depends on the nature of rod.
Q t
=
K
=
Κ Α (θ
t (θ
1
1
d Q d − θ
− θ
2
2
)
= Rate of flow of heat
)Α
Temperature gradient The fall of temperature with distance between the ends of the rod in the direction of flow of heat is called temperature gradient θ1 = temperature of hot end
θ 2 = temperature of cold end θ1 − θ 2
Temperature gradient =
d
where d = length of the rod
Temperature at a Junction Temperature at the junction of the two bars at different temperatures can be found under steady state of conditions. Let us have two bars of Iron and Brass as shown in the figure. Under the steady state of condition the heat current Q1 through iron is equal to the heat current Q2 through brass bar, Q = Q1 = Q2
K1 A1 (T1 − T0 ) K 2 A2 (T1 − T0 ) = L1 L2 In case A1 = A2 and L1 = L2
K1 A1 (T1 − T0 ) = K 2 A2 (T1 − T0 ) K T + K 2T2 T0 = 1 1 K1 + K 2
Q
α
A
A= Area of cross-section
Q α (θ
1
− θ
2
)
( θ 1 − θ 2 ) = temperature difference of faces of the rod. Q α t Where t = time for heating 1 Where d = length of the rod Q α d On combining
Α (θ
Q α
1
− θ d
2
)t
Or Q
=
Κ Α (θ
1
− θ
2
)t
d
K = coefficient of thermal conductivity depends on the nature of rod.
Q t
=
K
=
Κ Α (θ
t (θ
1
1
d Q d − θ
− θ
2
2
)
= Rate of flow of heat
)Α
Temperature gradient The fall of temperature with distance between the ends of the rod in the direction of flow of heat is called temperature gradient θ1 = temperature of hot end
θ 2 = temperature of cold end θ1 − θ 2
Temperature gradient =
d
where d = length of the rod
Temperature at a Junction Temperature at the junction of the two bars at different temperatures can be found under steady state of conditions. Let us have two bars of Iron and Brass as shown in the figure. Under the steady state of condition the heat current Q1 through iron is equal to the heat current Q2 through brass bar, Q = Q1 = Q2
K1 A1 (T1 − T0 ) K 2 A2 (T1 − T0 ) = L1 L2 In case A1 = A2 and L1 = L2
K1 A1 (T1 − T0 ) = K 2 A2 (T1 − T0 ) K T + K 2T2 T0 = 1 1 K1 + K 2
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