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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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Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

\frac{P_{1} }{\gamma } +\frac{V_{1}^{2}  }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2}  }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}

For the pipe 1, the flow velocity is:

V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4

\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

h_{1} =\frac{V_{1}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m

For the pipe 2, the flow velocity is:

V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4

\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087

The head of pipe 1 is:

h_{2} =\frac{V_{2}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m

The required pumping power is:

P=Q\rho *g*h_{pump}  =0.018*999.1*9.8*1344.55=236965.16W=236.96kW

8 0
3 years ago
One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter
lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150×10^{-3} m^{3}

m = 1 Kg

P_{1} = 2 MPa

T_{2}  = 40°C

The waters specific volume is calculated:

v_{1} = V/m

Here, the waters specific volume at initial condition is v_{1}, the containers volume is V, waters mass is m.

v_{1} = 150×10^{-3} m^{3}/1

v_{1} = 0.15 m^{3}/ Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 m^{3}/ Kg and 0.13 m^{3}/ Kg.

T_{1}= 350+(400-350) \frac{0.15-0.13}{0.1522-0.1386}

T_{1} = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

v_{2} = v_{1}= 0.15 m^{3}/ Kg

In saturated water labels for T_{2}  = 40°C.

v_{f} = 0.001008 m^{3}/ Kg

v_{g} = 19.515 m^{3}/ Kg

The final state is two phase region v_{f} < v_{2} < v_{g}.

In saturated water labels for T_{2}  = 40°C.

P_{2} = P_{Sat} = 7.3851 kPa

P_{2} = 7.3851 kPa

7 0
3 years ago
Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity
Digiron [165]

Answer: D) All of the above

Explanation:

Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.

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3 years ago
Which explanation best summarizes what went wrong during Paul’s cost analysis?
Valentin [98]

Answer:

wut is it

Explanation:

4 0
2 years ago
How to design a solar panel<br>​
artcher [175]

Answer:

#1) Find out how much power you need

#2 Calculate the amount of batteries you need.

#3 Calculate the number of solar panels needed for your location and time of year.

#4 Select a solar charge controller.

#5 Select an inverter.

#6 Balance of system

Explanation: To design solar panel, consider the following steps

1.) Find the power consumption demands

The first step in designing a solar PV system is to find out the total power and energy consumption of all loads that need to be supplied by the solar PV system as follows:

Calculate total Watt-hours per day for each appliance used.

 Add the Watt-hours needed for all appliances together to get the total Watt-hours per day which must be delivered to the appliances.

Calculate total Watt-hours per day needed from the PV modules.

Multiply the total appliances Watt-hours per day times 1.3 (the energy lost in the system) to get the total Watt-hours per day which must be provided by the panels.

2. Size the PV modules

Different size of PV modules will produce different amount of power. To find out the sizing of PV module, the total peak watt produced needs. The peak watt (Wp) produced depends on size of the PV module and climate of site location. We have to consider panel generation factor which is different in each site location. For Thailand, the panel generation factor is 3.43. To determine the sizing of PV modules, calculate as follows:

2.1 Calculate the total Watt-peak rating needed for PV modules

Divide the total Watt-hours per day needed from the PV modules (from item 1.2) by 3.43 to get the total Watt-peak rating needed for the PV panels needed to operate the appliances.

Calculate the number of PV panels for the system

Divide the answer obtained in item 2.1 by the rated output Watt-peak of the PV modules available to you. Increase any fractional part of result to the next highest full number and that will be the 

number of PV modules required.

Result of the calculation is the minimum number of PV panels. If more PV modules are installed, the system will perform better and battery life will be improved. If fewer PV modules are used, the system may not work at all during cloudy periods and battery life will be shortened.

3. Inverter sizing

An inverter is used in the system where AC power output is needed. The input rating of the inverter should never be lower than the total watt of appliances. The inverter must have the same nominal voltage as your battery.

For stand-alone systems, the inverter must be large enough to handle the total amount of Watts you will be using at one time. The inverter size should be 25-30% bigger than total Watts of appliances. In case of appliance type is motor or compressor then inverter size should be minimum 3 times the capacity of those appliances and must be added to the inverter capacity to handle surge current during starting.

For grid tie systems or grid connected systems, the input rating of the inverter should be same as PV array rating to allow for safe and efficient operation.

4. Battery sizing

The battery type recommended for using in solar PV system is deep cycle battery. Deep cycle battery is specifically designed for to be discharged to low energy level and rapid recharged or cycle charged and discharged day after day for years. The battery should be large enough to store sufficient energy to operate the appliances at night and cloudy days. To find out the size of battery, calculate as follows:

     4.1 Calculate total Watt-hours per day used by appliances.

     4.2 Divide the total Watt-hours per day used by 0.85 for battery loss.

     4.3 Divide the answer obtained in item 4.2 by 0.6 for depth of discharge.

     4.4 Divide the answer obtained in item 4.3 by the nominal battery voltage.

     4.5 Multiply the answer obtained in item 4.4 with days of autonomy (the number of days that you need the system to operate when there is no power produced by PV panels) to get the required Ampere-hour capacity of deep-cycle battery.

Battery Capacity (Ah) = Total Watt-hours per day used by appliancesx Days of autonomy

(0.85 x 0.6 x nominal battery voltage)

5. Solar charge controller sizing

The solar charge controller is typically rated against Amperage and Voltage capacities. Select the solar charge controller to match the voltage of PV array and batteries and then identify which type of solar charge controller is right for your application. Make sure that solar charge controller has enough capacity to handle the current from PV array.

For the series charge controller type, the sizing of controller depends on the total PV input current which is delivered to the controller and also depends on PV panel configuration (series or parallel configuration).

According to standard practice, the sizing of solar charge controller is to take the short circuit current (Isc) of the PV array, and multiply it by 1.3

Solar charge controller rating = Total short circuit current of PV array x 1.3

5 0
3 years ago
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