a metal cylinder containing water at 60ged C, has a thickness of 4mm.And the thermal conductivity 400W per m per K. it is lagged by felt of thickness 2mm and thermal conductivity 0.002W per m per K. the room temperature is 10deg C.Find the temperature gradient g for for the metal and the felt
q1 =- K1•A•(ΔT1/Δx1),
q2 =- K2•A•(ΔT2/Δx2),
q1=q2, =>
K1•A•(ΔT1/Δx1)= K2•A•(ΔT2/Δx2),
ΔT1= 60 –T, ΔT2 = T -10,
(60 –T)/(T-10) = (K2/K1) •(Δx1/Δx2) =
=(0.02/400) •(0.02/0.04) = 1•10^-4.
T =60.001/(1+1•10^-4) = 59.995 degr.
ΔT1/Δx1 = (60 –T)/Δx1 = 0.005/0.004 =1.25 degr/m,
ΔT2/Δx2 = (T-10)/Δx2 = 49.995/0.002 =24997.5 degr/m.
thanks :)
To find the temperature gradient for the metal and the felt, we need to calculate the rate of heat transfer through each material and use the formula for temperature gradient.
The rate of heat transfer through a material can be calculated using Fourier's law of heat conduction:
Q = (k * A * ΔT) / d
where:
Q is the rate of heat transfer (in watts),
k is the thermal conductivity of the material (in watts per meter per Kelvin),
A is the cross-sectional area (in square meters),
ΔT is the temperature difference (in Kelvin), and
d is the thickness of the material (in meters).
First, let's calculate the temperature difference for each material:
ΔT₁ = 60°C - 10°C = 50°C (temperature difference for metal)
ΔT₂ = 10°C - 10°C = 0°C (temperature difference for felt)
Next, let's calculate the rate of heat transfer for each material:
Q₁ = (400 W/m·K * A * 50 K) / 0.004 m (rate of heat transfer for metal)
Q₂ = (0.002 W/m·K * A * 0 K) / 0.002 m (rate of heat transfer for felt)
Since Q₂ is 0 (there is no temperature difference across the felt), we can ignore it for now.
Finally, let's calculate the temperature gradient:
g = ΔT₁ / d₁
where:
g is the temperature gradient (in Kelvin per meter),
ΔT₁ is the temperature difference for the metal (in Kelvin),
and d₁ is the thickness of the metal (in meters).
Given that the thickness of the metal is 4 mm (or 0.004 m), we can substitute the values into the formula:
g = 50 K / 0.004 m = 12,500 K/m
Therefore, the temperature gradient for the metal is 12,500 Kelvin per meter.
Please note that the temperature gradient for the felt is not needed in this case since there is no heat transfer across it (Q₂ = 0).