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

Baffles are used to deflect the flow.

Air leaked into the condensor can be easily extracted.

As given value is between these two, so cross flow is the best option.

Here, C_h = C_c.

Heat lost by fluid = Heat gained by cold fluid

Here,

Log mean temperature difference,


Q = mc_pwΔT

= 87.22 × 10³ Joule/s
Again
Q = UAθ_m
or 87.22 × 10³ = 2000 × A × 61.65
∴ A = 0.707 m²

Th₂ = 100°C

C_h = C_c = 0.1 × 4200 = 420 J/s
C_c (T_c1 – T_c2) = C_h (Th₁ – Th₂)
(70 – 20) = 90° T_h2

Counter flow heat exchanged Surface Area A = 20 m², mass flow rate = 0.4 k g/s
 Temperature T_ci = 280K

Since m is same for both flow = 0.4 kg/s
Assume C_p is same = 100 J/kg.K
Hence

Heat exchanger is a counterflow type.




= 21.83 m².

∴ ΔT₁ = 70 - 50 = 20°C,   ΔT₂ = 40 - 30 = 10°C

Heat capacity of hot fluid
= 1 × 10 = 10 kJ / k – s
Heat capacity of cold fluid
= 2 × 5 = 10 kJ / k – s
Since heat capacity is same, so LMTD is difference of temperature at either end
i.e. LMTD = 60º – 30º = 30º C

θ₁ = θ₂
∴ θ_m = 20°C
T_h1 – T_c2 = 20°C
100 – T_c2 = 20°C
T_c2 = 80°C

Flow configuration in condenser as shown below.


= 21.6° C

∴ 
⇒ t_h2 = 40°C

⇒ 

∴ LMTD = 10°C

The temperature difference is not for a given fluid but across the fluids and maximum heat transfer occurs for C_min and the temperature difference is equal to .

∴ 
or 
Alternately

From the relation

Now, 

ε of condenser is given by 1-exp(-NTU)
Because