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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

6 0

3 years ago

Obtain a relation for the logarithmic mean temperature difference for use in the LMTD method?

kolezko [41]

Answer:

The log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

Explanation:

To evaluate the equivalent average temperature difference between two fluids we consider a parallel-flow double-pipe heat exchanger (see attached diagram). The temperature of the hot and cold fluids is large at the inlet of the heat exchanger and decreases exponentially toward the outlet.

We can assume that the outer surface of the heat exchanger is well insulated and that heat transfer only occurs between the two fluids. We can also assume negligible kinetic and potential. The energy balance on each fluid can be written as the rate of heat loss from the hot fluid is equal to the rate of heat gained by the cold fluid in any section of the heat exchanger:

Q = -m,h×c,ph×dT,h (1)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the hot fluid, dT,h= differential temperature of the hot fluid

Q = m,c×c,pc×T.c (2)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the cold fluid, dT,h= differential temperature of the cold fluid

The temperature of the hot fluid change is negative and is added to make Q positive. Solving equations 1 and 2 in terms of dT:

dT.h = - Q/(m,h×c,ph)

dT.c = Q/(m,c×c,pc)

and taking the difference:

dT,h-dT,c= d(T,h - T,c) = -Q(1/(m,h×c,ph) + 1/(m,c×c,pc)) (3)

The heat transfer rate in the differential section of the heat exchanger can be expressed as:

Q = U(T,h-T,c)×dA,s (4)

where U=overall heat transfer coefficients, dA,s = differential sectional area. Substitute equation 4 into 3:

d(T,h - T,c)/(T,h - T,c) = -U×dA,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (5)

Integrating equation 5:

㏑((T,h out - T,c out)/(T,h in - T,c in)) = -U×A,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (6)

The first law of thermodynamics requires the rate of heat transfer from hot and cold fluid to be equal.

Q= m×c, pc×(T, c out-T, c in) (7)

Q= m×c, ph×(T,h out-T, h in) (8)

Solve equations 7 and 8 for m,c×c, pc and m,h×c, ph and substituting into equation 6:

Q = U×A,s×ΔT,lm

Where the log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

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8 0

3 years ago

A series resistive circuit has two resistors. R1 is 570 ohms and R2 is 560 ohms.

ZanzabumX [31]

Answer:

10.203 Volts

Explanation:

For this problem, we need to understand that a series resistive circuit is simply a circuit with some type of voltage source and some resistors, in this case, R1 and R2.

First, we need to find the voltage in the circuit. To do this, we need to find the total resistance of the circuit. When two resistors are in series, you sum the resistance. So we can say the following:

R_Total = R1 + R2

R_Total = 570 Ω + 560 Ω

R_Total = 1130 Ω

Now that we have R_Total for the circuit, we can find the voltage of the circuit by using Ohm's law, V = IR.

V_Total = I_Total * R_Total

V_Total = 17.9 mA * 1130 Ω

V_Total = 20.227 V

Now that we have V_Total, we can find the voltage drop across each resistor by using Ohm's law once more. Note, that since our circuit is series, both resistors will have the same current (I.e., I_Total = I_1 = I_2).

V_Total = V_1 + V_2

V_Total = V_1 + I_2*R2

V_Total - I_2*R2 = V_1

20.227 V - (17.9 mA * 560 Ω) = V_1

20.227 V - (10.024 V) = V_1

10.203 V = V_1

Hence, the voltage drop across R1 is 10.203 Volts.

Cheers.

3 0

3 years ago

Determine the convention heat transfer coefficient inside the for the flow of (a) air and (b) water at a velocity of 2 m/s in an

Veronika [31]

Answer:

Find the attachments sequence wise for complete solution.

7 0

3 years ago

When is the output of an XOR gate HIGH? explain

erma4kov [3.2K]

Answer:

The output of a NOR gate is LOW whenever one or more inputs are HIGH. The output of an XOR gate is HIGH whenever the two inputs are different. The output of an XNOR gate is HIGH whenever the two inputs are identical

5 0

3 years ago