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3 years ago

Match each titration term with its definition.

Engineering

1 answer:

Illusion [34]3 years ago

3 0

Answer:

1) titration

2) titrand

3) equivalence point

4) titrant

5) Burette

6) Indicator

Explanation:

The process in which a known volume of a standard solution is added to another solution so that the standard solution can react with the solution of unknown concentration such that its concentration is determined can be referred to as titration.

The solution which is added to another solution is called the titrant. The titrand is the solution of unknown concentration

A burette is a glassware used to slowly add a known volume of the titrant to the titrand.

The indicator used signals the point when the reaction is complete by a color change. At this point, a stoichiometric amount of titrant has been added to the titrand. This is also referred to as the equivalence point.

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Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

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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

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

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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