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Semmy [17]
3 years ago
7

The inside temperature of a wall in a dwelling is 15°C. If the air in the room is at 21°C, what is the maximum relative humidity

, in percent, the air can have before condensation occurs on the wall?
Engineering
1 answer:
Sveta_85 [38]3 years ago
8 0

Given:

Wall's inside temperature, T_{wa} = 15^{\circ}C

Room air temperature,  T_{air} = 21^{\circ}C

Solution:

To calculate percentage max relative humidity, we make use of steam table for saturated pressure of wall and air:

From steam table:

At T_{wa} = 15^{\circ}C:

P_{wa(sat)} = 1.7057 kPa

At  T_{air} = 21^{\circ}C:

P_{air(sat)} = 2.487 kPa

Now,

% Relative Humidity (RH)_{max} = \frac{P_{wa(sat)}}{P_{air(sat)}} \times 100

(RH)_{max} = \frac{1.7057}{2.487}\times 100

(RH)_{max} = 68.58%

Therefore, max Relative Humidity of the air before the occurrence of condensation in wall is 68.85%

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vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

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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:

A.

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.

B.

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 .

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3 years ago
Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could
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Answer:

4265.04\ \text{m}

2.38\times 10^{10}\ \text{W}

Explanation:

PE = Energy of food = 500 cal = 500\times4184=2.092\times10^6\ \text{J}

m = Mass of object = 50 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Potential energy of food is given by

PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}

Nancy could raise the weight to a maximum height of 4265.04\ \text{m}.

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Energy of H_2 = \dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}

Power

P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}

The power requirement is 2.38\times 10^{10}\ \text{W}.

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