GATE Past Year Questions: Conduction | Heat Transfer - Mechanical Engineering PDF Download

So, glass wool has minimum thermal conductivity and therefore maximum temperature difference.

Where K is thermal conductivity
V is mean particle speed
λ is mean free path
C_V is molar head capacity
N_A is Avogadro's number
n is particles per unit volume
Gases transfer heat by direct collisions between molecules. As the temperature increases, the thermal conductivity increases due to increase in speed, movement and collisions in the molecules.
From the above expression, by increasing mean particle speed, the thermal conductivity increases.

where Slope = 
i.e.,  slope is more for less value of K, therefore

∴ K₁ < K₂

Temperature of water = 48ºC Temperature of glass plate = 40ºC Thermal conductivity of water = k_w = 0.6 W/mK Thermal conductivity of glass = k_G = 1.2 W/mK
The spatial gradient of temperature in water at the water-glass interface


At water glass interface

For convection
Heat flux, q = h(T_f – T_i)

∴ 
= 750 W/m²K

Given:
d₁ = 2 cm
r₁ = 1 cm
d₂ = 5 cm
r₂ = 2.5 cm
d₃ = 5 + 2 × 3 = 11
r₃ = 5.5 cm


∴ Heat transfer rate per unit length

Case -I:

Case-II:

For caseI-:

where 
For Case-II: 

where

As R₂ + R₃ > R₁ at constant temperature difference in both cases.
Then 
For Case-II:
R₂ > R₃ ∵ k_cu > k_S

One dimensional heat conduction equation is,

∴ 

By applying energy balance, we get Rate of heat generation = Rate of convective heat transfer
q_G × Volume of fuel rod = h₀ A (T_s – T₀)

T_s – 75 = 12.5
T_s = 87.5°C or 360.5K

Direction of heat flow is always normal to surface of constant temperature
So for surface P₁, 
From energy conservation,
Heat rate at P = Heat rate at Q

⇒

Total thermal resistance 

∴  r_c > r₀
⇒ The heat transfer from the wire to the ambient will increase

We know that one-dimensional steady state heat conduction equation with no heat generation is given by,

∵ A = Cx + B (linear variation)

Logarithmic temperature distribution.

Given data:
d₁ = 0.03 m


Thermal resistance

A_r = 2πr L
From Fourier ’s Law

Boundary conditions:

As per question, heat transfer in radial direction and contact resistance are negligible.
In series connection,

⇒ 

For maximum heat dissipation, Critical radius,


= 25 mm
Insulation thickness = r_c – r = 25 – 10
= 15 mm

Critical radius of insulation,

Radius of min = 0.5 mm
∴ Thickness of coating = 1 – 0.5 = 0.5 mm.

With an increases in the thickness of insulation around a circular pipe, heat loss to surroundings due to convection increases and due to conduction decreases.

r_c = 8mm
∴ the heat lost increases to maximum and then decreases.

At x = 0, T = 433 K
∴ C₂ = 433
At x = 20 × 10^–3 m, T = 393 K.
∴ C1 = 200
For maximum temperature,


x = 5 × 10^–3 = 5 mm

T_max = 165º [putting x = 5 × 10^–3 is equation (1)]

Under steady state conditionsHeat transfer by conduction = Heat transfer by convection
∴ 
or 

But y = 0

or h(100–30) = k x (100 - 30)
or h = k = 1 W/m²K

k = k₀ + b_T, k = f (T)
From energy equation:

⇒  ...(1)

Using B.C’s
x = 0, T = 0
x = 1, T = 100
A = 0

from equation (1)

at x = 0, T₁ = 0
x = l , T₂ = 100

⇒ 
∴ 
For maximum temp:

Given data: r = 0.005 m
d = 2r = 2 × 0.005 = 0.010 m

For steady state,
Rate of heat generation in the rod} = Rate of convection H.T. from rod surface to fluid

T_s – 25 = 100
T_s = 125° c or 398 K