A cylindrical rod of cross-sectional area 5.0mm2 is made by joining a 0.30m rod of silver to a 0.12m rod of nickel. The silver end is maintained at 290K and the nickel end at 440K. given the thermal conductivities of silver and nickel are 0.42kWm-1K-1 and 91Wm-1K-1 respectively, calculate:
the temperature of the join under steady conditions, and
b the rate of conduction of heat down the rod
State any assumptions you need to make.
A=5•10⁻⁶ mm²,
Silver: k₁= 420 W/m•K, L₁= 0.3 m,
Nickel: k₂=91 W/m•K, L₂=0.12 m,
T=? q=?
Q/t=q=k•A•ΔT/L
Resistance of the silver rod:
R₁=L₁/k₁A₁ =0.3/420•5•10⁻⁶=142.9 Ω.
Resistance of the nickel rod:
R₂=L₂/k₂A₂ = 0.12/91•5•10⁻⁶= 263.7 Ω.
Two rods are the conductotrs in series:
R=R1+R2=406.6 Ω.
q= ΔT/R= (440-290)/406.6=0.369 W.
ΔT₁=q•R₁=0.369•142.9=52.7 K
ΔT₂=q•R₂=0.369•263.7=97.3 K
T=ΔT₁+T₁=52.7+290=342.7 K
or
T= T₂-ΔT₂=440-97.3=342.7 K
= >temperature of the join is
T=342.7 K ~343 K,
the rate of conduction of heat is
q= 0.369 W ~ 0.37 W
To calculate the temperature at the join under steady conditions, we can use the equation for steady-state heat conduction:
Q = (kAΔT) / L
where:
Q is the rate of heat conduction
k is the thermal conductivity
A is the cross-sectional area
ΔT is the temperature difference
L is the length
First, let's calculate the temperature at the join:
The heat conduction rate in silver (Qs) is equal to the heat conduction rate in nickel (Qn) at the join:
Qs = Qn
The rate of heat conduction (Q) can be calculated as follows:
Qs = (k_silver * A * ΔT_silver) / L_silver
Qn = (k_nickel * A * ΔT_nickel) / L_nickel
Since Qs = Qn, we can equate the two equations:
(k_silver * A * ΔT_silver) / L_silver = (k_nickel * A * ΔT_nickel) / L_nickel
We are given the following values:
k_silver = 0.42 kW/m^2K
A = 5.0 mm^2 = 5.0 * 10^(-6) m^2
ΔT_silver = (290 - T_join)
L_silver = 0.30 m
k_nickel = 91 W/m^2K
ΔT_nickel = (T_join - 440)
L_nickel = 0.12 m
Substituting these values into the equation and solving for T_join:
(0.42 * 10^3 * 5.0 * 10^(-6) * (290 - T_join)) / 0.30 = (91 * 5.0 * 10^(-6) * (T_join - 440)) / 0.12
Simplifying, we have:
(0.14 * 10^(-3) * (290 - T_join)) = (0.38 * 10^(-3) * (T_join - 440))
0.14 * 10^(-3) * 290 - 0.14 * 10^(-3) * T_join = 0.38 * 10^(-3) * T_join - 0.38 * 10^(-3) * 440
0.14 * 10^(-3) * T_join + 0.38 * 10^(-3) * T_join = 0.14 * 10^(-3) * 290 + 0.38 * 10^(-3) * 440
0.52 * 10^(-3) * T_join = 67.6 * 10^(-3)
T_join = (67.6 * 10^(-3)) / (0.52 * 10^(-3))
T_join ≈ 130 K
Therefore, the temperature at the join under steady conditions is approximately 130 K.
Now, let's calculate the rate of conduction of heat down the rod:
We can use the same equation for Q:
Q = (kAΔT) / L
To find the rate of conduction down the rod, we need to calculate Q_silver and Q_nickel separately and then sum them up:
Q_silver = (k_silver * A * ΔT_silver) / L_silver
Q_nickel = (k_nickel * A * ΔT_nickel) / L_nickel
Substituting the given values:
Q_silver = (0.42 * 10^3 * 5.0 * 10^(-6) * (290 - T_join)) / 0.30
Q_nickel = (91 * 5.0 * 10^(-6) * (T_join - 440)) / 0.12
Calculating Q_silver and Q_nickel:
Q_silver = (0.42 * 10^3 * 5.0 * 10^(-6) * (290 - 130)) / 0.30 ≈ 2.1 W
Q_nickel = (91 * 5.0 * 10^(-6) * (130 - 440)) / 0.12 ≈ -1.7 W (negative due to the heat flowing in the opposite direction)
The rate of conduction of heat down the rod is the sum of Q_silver and Q_nickel:
Rate = Q_silver + Q_nickel = 2.1 W + (-1.7 W) ≈ 0.4 W
Therefore, the rate of conduction of heat down the rod is approximately 0.4 W.
Assumptions made:
1. The rod is in steady-state, meaning the temperature remains constant over time.
2. There is no heat loss to the surroundings.
3. The thermal conductivities of silver and nickel remain constant over the temperature range.
4. The cross-sectional area of the rod is constant along its length.
5. Heat conduction is the only mode of heat transfer occurring in the rod.