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3 years ago

Explain the effect of heat transfer, reversibilities and irreversibilities on changes in entropy

Engineering

1 answer:

Natali5045456 [20]3 years ago

5 0

Answer is given below

Explanation:

The second law of thermodynamics states that in the reversible process, the entropy of the universe is constant, while in the unchanging process, such as the transfer of heat from a hot object to a cold object, the reflection of the universe increases.

ΔS = ΔQ ÷ T

The irreversible process increases the entropy of the universe. Since entropy is a state function, the change in entropy of the system is the same, the process is reversible or irreversible. The second law of thermodynamics can be used to determine whether a process is reversible.

for an irreversible process, ΔS is not same as reversible process, because more than 1 reversible process is required for 2 ΔS to be equal.

if the initial and final states are the same, ΔS for an irreversible process can be calculated as if it were a reversible process.

If the initial and final condition is same then ΔS can find out for irreversible process even if it were reversible process

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A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

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Answer:

$L=107.6m$

Explanation:

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Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

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