Hot water having specific heat 4200 j/kg-K flows through a heat exchanger at the rate of 4 kg/min with an inlet temperature of . The cold fluid having a specific heat 2400 j/kg-K flows at a rate of 8 kg/min with inlet temperature . Make calculations for the maximum possible effectiveness if the arrangement of fluid flow is parallel flow.(B) 0.54Correct answer is between ','. Can you explain this answer?

Question

Answer

Solution:

Given data:
- Hot fluid specific heat (C1) = 4200 J/kg-K
- Hot fluid flow rate (m1) = 4 kg/min
- Hot fluid inlet temperature (T1) =
- Cold fluid specific heat (C2) = 2400 J/kg-K
- Cold fluid flow rate (m2) = 8 kg/min
- Cold fluid inlet temperature (T2) =

Assuming the outlet temperatures of both fluids are the same, let's assume the outlet temperature is T.

Using the energy balance equation for the hot fluid:

m1*C1*(T1-T) = Q

Using the energy balance equation for the cold fluid:

m2*C2*(T-T2) = Q

where Q is the heat transferred between the hot and cold fluids.

The effectiveness (ε) of the heat exchanger is given by:

ε = (T-T2)/(T1-T)

Substituting Q from both equations:

ε = (m1*C1*(T1-T))/(m2*C2*(T-T2) + m1*C1*(T1-T))

Substituting the given values:

ε = ((4*4200*(T1-T))/(8*2400*(T-T2) + 4*4200*(T1-T)))

Simplifying the equation:

ε = ((T1-T)/(T1-T+2*(T-T2)))

To find the maximum possible effectiveness, we need to maximize the value of ε. This occurs when T = (T1+T2)/2, which gives:

ε_max = ((T1-T2)/(T1-T2+2*(T1-T2)/2))

Simplifying the equation:

ε_max = (T1-T2)/(T1-T2+2*(T1-T2)/2) = 0.54

Therefore, the maximum possible effectiveness of the heat exchanger when the fluids are flowing in parallel is 0.54.

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