+1 862 207 3288 Learning Outcome 3.2
3.2: Estimate heat transfer performance
An exhaust pipe is 85 mm diameter and it is cooled by surrounding it with a water jacket. The exhaust gas enters at 450oC and the
water enters at 10oC. The surface heat transfer coefficients for the gas and water are 300 and 1500 W/m2 K respectively. The wall is thin
so the temperature drop due to conduction is negligible. The gasses have a mean specific heat capacity Cp of 1130 J/kg K and they must be
cooled to 100oC. The specific heat capacity of the water is 4190 J/kg K. The flow rate of the gas and water is 250 and 1450 kg/h
respectively.
Calculate:
the required length of pipe for Parallel flow
the required length of pipe for Contra flow
estimate the performance of the heat transfer in both cases
Calculate and explain the effective difference if a cross flow heat exchanger was used, in not more than 250 words.
Answer:
Overall heat transfer coefficient is U:
Where 1/U=1/h_g +1/h_w +X/K
If the wall is too thin, so x ≈0
1/U=1/h_g +1/h_w =1/0.3+1/1.5=4→U=0.25
Parallel flow:
Heat lost by the gas is:
φ=mc_p ∆T=250×1.13(450-100)=98875kJ/h
Heat gained by the water is:
φ=98875=mc_p ∆T
98875=1450×4.19x(θ-10)
98875=6075.5(θ-10)
98875=6075.5θ-60755
159630=6075.5θ
Therefore, θ=159630/6075.50=〖26.27〗^0 C
∆T_i=440K ,∆T_o=73.73K
φ=UA ((∆T_o-∆T_i ))/(ln (∆T_o)/(∆T_i ))=98875/3600=27.47kW
27.47=0.25A (73.73-440)/ln〖73.73/440〗 =91.57A/1.79=51.2A→A=27.47/51.2=0.537m^2
If A= πDL therefore,L=A/πD=0.537/(π(0.085))=2.01m
Contra flow:
∆T_i=423.73K , ∆T_o=90K
φ=27.47=UA (∆T_o-∆T_i)/(ln (∆T_o)/(∆T_i ))
27.47= 0.25A (90-423.73)/ln〖(90/423.73)〗 =83.43A/1.55=53.83A
Therefore,A=27.47/53.83=0.51m^2
If,A= πDL→ Therefore,L=A/πD=0.51/(π(0.085))=1.91m
