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lorasvet [3.4K]
3 years ago
11

Many households in developing countries prepare food over indoor cook stoves with no ducting system to exhaust the combustion pr

oducts outside. A woman in a household lights a fire in her cookstove at 4 pm. The fire emits benzo(a)pyrene, a carcinogenic compound, at a rate of 0.01 ng/min. The house has a size of 40 m3 and a ventilation rate of 1 m3/h. The ambient indoor air concentration of benzo(a)pyrene was 0.2 ng/m3 when she started the fire, but no significant concentration of benzo(a)pyrene is found in the outdoor air. What is the indoor benzo(a)pyrene concentration in the household at 6 pm?
Engineering
1 answer:
lorasvet [3.4K]3 years ago
7 0

Answer:

   C = 0.22857 ng / m³

Explanation:

Let's solve this problem for part the total time in the kitchen is

          t = 2h (60 min / 1h) = 120 min

The concentration (C) quantity of benzol pyrene is the initial quantity plus the quantity generated per area minus the quantity eliminated by the air flow. The amount removed can be calculated assuming that an amount of extra air that must be filled with the pollutant

amount generated

         C = co + time_generation rate / (area_house + area_flow)

         C = 0.2 + 0.01 120 / (40+ 2)

         C = 0.22857 ng / m³

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For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta
Crank

Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

c=\sqrt(\gamma*R*T)

Where R is the gas's particular constant defined in terms of Cp and Cv:

R=Cp-Cv

For particular values given:

R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}

R=188.9 \frac{J}{Kg-K}

The gamma undimensional constant is also expressed as a function of Cv and Cp:

\gamma=Cp/Cv

\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}

\gamma=1.29

And the variable T is the temperature in Kelvin. Thus for the known temperature:

c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)

c=\sqrt(91867.73 \frac{J}{Kg})

The Jules unit can expressing by:

J=N.m=\frac{Kg.m}{s^2}* m

J=\frac{Kg.m^2}{s^2}

Replacing the new units for the speed of the sound:

c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})

c=\sqrt(91867.73 \frac{m^2}{s^2})

c=313 m/s

3 0
2 years ago
Read 2 more answers
(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\dot Q), measured in BTU per hour, is represented by this formula:

\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

Where:

\epsilon - Emissivity, dimensionless.

A - Surface area of the student, measured in square feet.

\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

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2 years ago
A segment of four-lane freeway (two lanes in each direction) has a 3% upgrade that is 1500 ft long followed by a 1000-ft 4% upgr
Dennis_Churaev [7]

Answer:

The level of service of of compound grade freeway is LOSB.

Explanation:

Find the provided attachments for explanation

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3 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

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What is the name of the first federation vessel in star trek hint 17?
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Answer:

The Fesarius

Explanation:

It's the first vessel

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