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3 years ago

Hello Averyone Subs Chanel Me please "RezaDarmawangsa "Thanks !!

Engineering

1 answer:

Yanka [14]3 years ago

4 0

Answer:

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The enthalpy of the water entering an actual pump is 500 kJ/kg and the enthalpy of the water leaving it is 550 kJ/kg. The pump h

n200080 [17]

Answer:500,551.02

Explanation:

Given

Initial enthaly of pump \left ( h_1\right )=500KJ/kg

Final enthaly of pump \left ( h_2\right )=550KJ/kg

Final enthaly of pump when efficiency is 100%= $h_2^{'}$

Now pump efficiency is 98%

$\eta$ = $\frac{h_2-h_1}{h_2^{'}-h_1}$

0.98= $\frac{550-500}{h_2-500}$

$h_2=551.02KJ/kg$

therefore initial and final enthalpy of pump for 100 % efficiency

initial=500KJ/kg

Final=551.02KJ/kg

3 0

3 years ago

g Calculate a better value for the convection coefficient using resources from heat transfer. Assume forced convection of the ai

ollegr [7]

Answer:results for h should on the same order of magnitude of the value I provided. If it isn't, check your units. Update to you calculated values for h and resolve your model.

Explanation:

4 0

3 years ago

A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re

Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

4 0

3 years ago

Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the

ra1l [238]

The evaporation rate of the n-Hexane is $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

$N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}$

$\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}$

$P_{t}=1 a t m=101325 P a\\$

$\text { so, } P_{A 1}=$ the vapor pressure at hexane $25 \mathrm{C}$ $=20158.2 \mathrm{Pa}$

For wind, assume negligible hexane is present, hence $P_{A 2}=0$

Now,

$\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}$

$\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}$

$P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\$

$=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}$

$=90873.57 \mathrm{Pa}$

$R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

$z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\$

where T = 298 K

substituting all in the equation, we get

$\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}$

$=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\$

Now,Flux $\times$ area = Molar rate of evaporation

Evaporation rate = $0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)$