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zysi [14]
3 years ago
14

A student lives in an apartment with a floor area of 60 m2 and ceiling height of 1.8 m. The apartment has a fresh (outdoor) air

exchange rate of 0.5/hr. The stove in the apartment heats by natural gas. The student cooks a meal using two gas burners that each emit carbon monoxide (CO) at a rate of 100 mg/hr. The outdoor CO concentration can be assumed to be negligible (0 ppm). The initial (time = 0) indoor CO concentration can be assumed to be 0 ppm (except for problem 4). Carbon monoxide can be considered as an inert gas, i.e., it does not stick to or react with any surfaces or other gases in air.
1. Assume that the student cooks for a long enough period of time to achieve a steady-state CO concentration in the apartment. What is that concentration in ppb?
2. Assume that the student cooks for only 45 minutes and turns off both burners at that time. What is the CO concentration in ppb at the end of 45 minutes?
3. Repeat problem 2 for air exchange rates that vary from 0.1 to 1/hr and plot the concentration at 45 minutes (in ppb) versus air exchange rate.
4. Assume that for the conditions of problem 2, the student waits 25 minutes after turning the burners off and then starts cooking again with two burners on. How long will it take to reach a concentration that is 95% of steady-state under this condition?
Note that you can actually address this question with an eloquent mathematical derivation (preferred) or simply by crunching the concentration profile in a spreadsheet.
What is the concentration at 95% of steady-state?
Compare your result with the time that would be required to reach 95% of steady-state had the initial indoor CO concentration been 0 ppm.
Engineering
1 answer:
USPshnik [31]3 years ago
5 0

Answer:

4

Explanation:

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Andrei [34K]

Answer:

I am in 6th grade why am i in high school things.

Explanation:

8 0
3 years ago
Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the
Nata [24]

<u>Explanation</u>:

For series

\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)

\(I=I_{1}=I_{2}=I_{n}\) (current is the same)

V=I R(\text {voltage is directly proportional to } R)

R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }

For parallel

\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })

I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})

\(I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)\)

\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}

3 0
3 years ago
Based on the graphs of stress-strain from the V-MSE site, how would you characterize the general differences between polymers an
Pepsi [2]

Answer:

Option A

Explanation:

Alloys are metal compounds with two or more metals or non metals to create new compounds that exhibit superior structural properties. Alloys have high level of hardness that resists deformation thereby making it less ductile compared to polymers. This is due to the varying difference in the chemical and physical characteristics of the constituent metals in the alloy.

6 0
3 years ago
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
Mekhanik [1.2K]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( \frac{M1}{100} )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

5 0
3 years ago
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