Respuesta :
Answer:
[tex]L=107.6m[/tex]
Explanation:
Cold water in: [tex]m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C[/tex]
Hot water in: [tex]m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C[/tex]
[tex]D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m[/tex]
Step 1: Determine the rate of heat transfer in the heat exchanger
[tex]Q=m_{c}C_{c}(T_{c,out}-T_{c,in})[/tex]
[tex]Q=1.2*4.18*(80-20)[/tex]
[tex]Q=1.2*4.18*(80-20)[/tex]
[tex]Q=300.96kW[/tex]
Step 2: Determine outlet temperature of hot water
[tex]Q=m_{h}C_{h}(T_{h,in}-T_{h,out})[/tex]
[tex]300.96=2*4.18*(160-T_{h,out})[/tex]
[tex]T_{h,out}=124\°C[/tex]
Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)
[tex]dT_{1}=T_{h,in}-T_{c,out}[/tex]
[tex]dT_{1}=160-80[/tex]
[tex]dT_{1}=80\°C[/tex]
[tex]dT_{2}=T_{h,out}-T_{c,in}[/tex]
[tex]dT_{2}=124-20[/tex]
[tex]dT_{2}=104\°C[/tex]
[tex]LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}[/tex]
[tex]LMTD = \frac{104-80}{ln(\frac{104}{80})}[/tex]
[tex]LMTD = \frac{24}{ln(1.3)}[/tex]
[tex]LMTD = 91.48\°C[/tex]
Step 4: Determine required surface area of heat exchanger
[tex]Q=UA_{s}LMTD[/tex]
[tex]300.96*10^{3}=649*A_{s}*91.48[/tex]
[tex]A_{s}=5.07m^{2}[/tex]
Step 5: Determine length of heat exchanger
[tex]A_{s}=piDL[/tex]
[tex]5.07=pi*0.015*L[/tex]
[tex]L=107.57m[/tex]
