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Juli2301 [7.4K]
3 years ago
12

Consider a steady-state experiment in which the observed current due to reduction of Ox to R is 85 mA/cm2. What is the concentra

tion gradient at the electrode surface? Knowing that the concentration of the species being reduced is zero at the electrode surface, and 0.10 mM in the bulk, how thick is the diffusion layer in this experiment?
Engineering
1 answer:
Elanso [62]3 years ago
5 0

Answer: Yes

Explanation:

Surface Concentration: In electrochemistry, there is an important distinction between the concentration of a species at the electrode’s surface and its concentration at some distance from the electrode’s surface (in what we call the bulk solution). Suppose we place an electrode in a solution of Fe3+ and fix the potential at 1.00 V. At this potential Fe3+ is stable—the standard state reduction potential for Fe3+ to Fe2+ is +0.771 V, the concentration of Fe3+ remains the same at all distances from the electrode’s surface.

Bulk Concentraton: If we change the electrode’s potential to +0.500 V, the concentration of Fe3+ at the electrode’s surface decreases to approximately zero. The concentration of Fe3+ increases as we move away from the electrode’s surface until it equals the concentration of Fe3+ in bulk solution. The resulting concentration gradient causes additional Fe3+ from the bulk solution to diffuse to the electrode’s surface.

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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=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

7 0
3 years ago
Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an
fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

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