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Elan Coil [88]
3 years ago
8

Voltage-regulated channels can be found a. at the motor end plate. b. on the surfaces of dendrites. c. in the membrane that cove

rs axons. d. on the soma of neurons. e. along the perikaryon of neurons.
Engineering
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

Option C

In the membrane that covers axons

Explanation:

Voltage-regulated channels allow for selective passage of different beneficial ions such as potassium and are  found on the surface of a wide variety of cells such as nerve, muscle, and secretory cells. They mainly regulate cell membrane excitability, repetitive low frequency firing in some neurons, and recover the nerve fiber membrane.

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The flow rate in the pipe system below is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. Point 1 is 0.60 m higher
DedPeter [7]

Answer:

Explanation:

The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.

The answer to the above question is

The pressure at point 2 = 75.959 kPa

Explanation:

Bernoulli's equation with losses gives

hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)

Between points 1 and 2, z₁ = z₃ + 0.6 m therefore

hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

But v = Q/A

or  since A = π×D²/4 we have

A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²  

Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃  = 1.315 m/s  from there we find the friction coefficient from Moody Diagram as follows

ε = \frac{Roughness _. value}{ Diameter} Which gives

the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175

Substituting he above values into the h_{l} equation we get h_{l} = 19.761 m

Combined head loss = 19.761 m

Hence 19.743 m  = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)

or 260 kPa-18.82 m × 9.81 m/s²×ρ=  P₃

Where ρ = density of water, we have

260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa

6 0
3 years ago
A stream of liquid n-pentane flows at a rate of 50.4 L/min into a heating chamber, where it evaporates into a stream of air 15%
allochka39001 [22]

Answer:

(a) the fractional conversion of pentane achieved in the furnace is  90% conversion

(b) the volumetric flow rates (Umin) of the feed air  is 256 x 10³ 1/m

(c) the volumetric flow rates (Lmin) of the gas leaving the condenser is 404.9  x 103 l/min

Explanation:

a)   Molecular weight of pentane = 72.15 g/mol

density of liquid pentane = 626 kg/m3

Flow rate of feed liquid nitrogen = 50.4 l/min

                                                  = 626*50.4*10-3

                                                   = 31.55 kg/min

                                                   = 31.55/72.15 kmol/min

                                                    = 0.4372 kmol/min

Pentane existng the burner = 3.175 kg/min

Fractional conversion = (31.55 - 3.175)/ 31.55

= 0.9 = 90% conversion

b)

C₂H₅ + 8O₂ ----------->   5CO₂ + 6H₂O

From the Stoichiometric reaction,

8 mol of O2 are used for combustion of 1 mol of pentane

for 0.4372 kmol/min of pentane = 8 * 0.4372 kmol/min of Oxygen will be required

                                                  = 3.49 kmol/min of O2

amount of air will be = 3.49/0.21 = 16.62 kmol/min

15% excess air = 16.62*1.15 = 19.12 kmol/min

assuming air to be ideal gas

V = nRT / P.........(1)

V = 19.12 X 8.314 X 336 / 208.6

  = 256 m³ = 256 x 10³ 1/m

c)  Oxygen:

Amount of pentane consumed = (31.55 - 3.175) = 28.375 kg/min = 28.375/72.15 = 0.3932 kmol/min......(2)

Amount of O2 consumed = 8*0.3932 = 3.146 kmol/min

Amount of O2 fed by air = 0.21*19.12 = 4.0215 kmol/mim

unused O2 left = 4.0215 - 3.146

= 0.8755 kmol/min = 19.75*103 l/min............ (using (1))

Carbon Dioxide:

1 mol of pentane = 5 mol of CO2

0.3932 kmol/min of pentane = 5*0.3932 kmol/min of CO2..................(from (2)

                                                    = 1.966 kmol/min

= 44.362*103 l/min..........................(using (1))  

Nitrogen:

v = 0.79 x 19.12 x 8.314 x 275 / 100.325

   = 340.8 m³ / min = 340.8 x 10³ 1/min

Total volumetric flow rate of gases leaving the condenser = (340.8 + 44.36 + 19.75) x 103 l/min

= 404.9  x 103 l/min

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