Engineering, 10.12.2019 04:31 jazzzzhands21

Aprocess fluid having a specific heat of 3500 j/kg∙k and flowing at 2 kg/s is to be cooled from 80 °c to 50 °c with chilled water, which is supplied at a temperature of 15 °c and a flow rate of 2.5 kg/s.

assuming an overall heat transfer coefficient of 2000 w/m2∙k, calculate the required heat transfer areas for the following exchanger configurations:

(a) parallel flow;

(b) counter flow;

(c) a 1-2 shell and tube exchanger with the water on the shell side.

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Answer from: mir1425

a) $A_{s}=5.02m^{2}$

b) $A_{s}=4.75m^{2}$

c) $A_{s}=4.87m^{2}$

Explanation:

Water in: $m_{w}=2.5kg/s, C_{w}=4.2kJ/kg\°C, T_{w,in}=15\°C, T_{w,out}=?\°C$

Process fluid in: $m_{pf}=2kg/s, C_{pf}=3.5kJ/kg\°C, T_{pf,in}=80\°C, T_{pf,out}=50\°C$

$U=2000W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2}$

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

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

Q=2*3.5*(80-50)

Q=420kW

Step 2: Determine outlet temperature of water

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

$420=2.5*4.2*(15-T_{w,out})$

$T_{w,out}=25\°C$

Part a) Parallel Flow

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

$dT_{1}=T_{pf,in}-T_{w,in}$

$dT_{1}=80-15$

$dT_{1}=65\°C$

$dT_{2}=T_{pf,out}-T_{w,out}$

$dT_{2}=50-25$

$dT_{2}=25\°C$

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

$LMTD = \frac{65-25}{ln(\frac{65}{25})}$

$LMTD = \frac{40}{ln(2.6)}$

$LMTD = 41.86\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$420*10^{3}=2000*A_{s}*41.86$

$A_{s}=5.02m^{2}$

Part b) Counter Flow

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

$dT_{1}=T_{pf,in}-T_{w,out}$

$dT_{1}=80-25$

$dT_{1}=55\°C$

$dT_{2}=T_{pf,out}-T_{w,in}$

$dT_{2}=50-15$

$dT_{2}=35\°C$

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

$LMTD = \frac{55-35}{ln(\frac{55}{35})}$

$LMTD = \frac{20}{ln(1.57)}$

$LMTD = 44.25\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$420*10^{3}=2000*A_{s}*44.25$

$A_{s}=4.75m^{2}$

Part c) 1-2 shell and tube exchanger with the water on the shell side

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

$dT_{1}=T_{pf,in}-T_{w,out}$

$dT_{1}=80-25$

$dT_{1}=55\°C$

$dT_{2}=T_{pf,out}-T_{w,in}$

$dT_{2}=50-15$

$dT_{2}=35\°C$

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

$LMTD = \frac{55-35}{ln(\frac{55}{35})}$

$LMTD = \frac{20}{ln(1.57)}$

$LMTD = 44.25\°C$

Step 4: Determine LMTD Correction Factor and Corrected LMTD

$P=\frac{T_{pf,out}-T_{pf,in}}{T_{w,in}-T_{pf,in}}$

$P=\frac{50-80}{15-80}$

$P=\frac{-30}{-65}$

P=0.462

$R=\frac{T_{w,in}-T_{w,out}}{T_{pf,out}-T_{pf,in}}$

$R=\frac{15-25}{50-80}$

$R=\frac{-10}{-30}$

R=0.333

Correction factor F can be determined from attached graph using P and F values calculated above

F=0.975

$LMTD_{corrected}=F*LMTD$

$LMTD_{corrected}=0.975*44.25$

$LMTD_{corrected}=43.14\°C$

Step 5: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD_{corrected}$

$420*10^{3}=2000*A_{s}*43.14$

$A_{s}=4.87m^{2}$

Aprocess fluid having a specific heat of 3500 j/kg∙k and flowing at 2 kg/s is to be cooled from 80 °

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Answer from: Quest

Infrared sensor is most used sensor in wireless technology where remote controlling functions and detection of surrounding objects/ obstacles are involved. this post will discuss about what is infrared sensor, its working principle, how it works, types, applications, advantages and disadvantages. what is infrared sensor

ir sensor is a simple electronic device which emits and detects ir radiation in order to find out certain objects/obstacles in its range. some of its features are heat and motion sensing.

ir sensors use infrared radiation of wavelength between 0.75 to 1000µm which falls between visible and microwave regions of electromagnetic spectrum. ir region is not visible to human eyes. infrared spectrum is categorized into three regions based on its wavelength i.e. near infrared, mid infrared, far infrared.

wavelength regions of infrared spectrum

near ir – 0.75µm to 3 µm
mid ir – 3 µm to 6 µm
far ir – > 6 µm
1 introduction to infrared sensor

fig. 1 – introduction to infrared sensor

working principle of infrared sensor

infrared sensors works on three fundamental physics laws:

planck’s radiation law: any object whose temperature is not equal to absolute zero (0 kelvin) emits radiation.
stephan boltzmann law: the total energy emitted at all wavelengths by a black body is related to the absolute temperature.
wein’s displacement law: objects of different temperature emit spectra that peak at different wavelengths that is inversely proportional to temperature.
key elements of infrared detection system

the key elements of infrared detection system are:

ir transmitter

ir transmitter acts as source for ir radiation. according to plank’s radiation law, every object is a source of ir radiation at temp t above 0 kelvin. in most cases black body radiators, tungsten lamps, silicon carbide, infrared lasers, leds of infrared wavelength are used as sources.

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Aprocess fluid having a specific heat of 3500 j/kg∙k and flowing at 2 kg/s is to be cooled from 80 °...

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