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1 month ago

5

Write a function named "total_population" that takes a string then a list as parameters where the string represents the name of

a CSV file containing city data in the format "CountryCode,CityName,Region,Population,Latitude,Longitude" and the second parameter is a list where each element is itself a list containing 3 strings as elements representing the CountryCode, CityName, and Region in this order. Return the total population of all cities in the list. Note that the city must match the country, name, and region to ensure that the correct city is being read.

1 answer:

mote1985 [299]1 month ago

5 0

Calculate the total population across all cities in the provided list. The function should process each item in cityinfo, accommodating multiple arrays.

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A liquid food with 12% total solids is being heated by steam injection using steam at a pressure of 232.1 kPa (Fig. E3.3). The p

iogann1982 [368]

Answer:

$m_{s}=20kg/min$

$H_{s}=1914kJ/kg$

Explanation:

A liquid food containing 12% total solids is heated via steam injection at a pressure of 232.1 kPa (see Fig. E3.3). The product starts at a temperature of 50°C and has a flow rate of 100 kg/min, being elevated to a temperature of 120°C. The specific heat of the product varies with its composition as follows:

$c_{p}=c_{pw}(mass fraction H_{2}0)+c_{ps}(mass fraction solid)$ and the

specific heat of the product at 12% total solids is 3.936 kJ/(kg°C). The goal is to calculate the quantity and minimum quality of steam required to ensure that the leaving product has 10% total solids.

Given

Product total solids in ( $X_{A}$ ) = 0.12

Product mass flow rate ( $m_{A}$ ) = 100 kg/min

Product total solids out ( $X_{B}$ ) = 0.1

Product temperature in ( $T_{A}$ ) = 50°C

Product temperature out ( $T_{B}$ ) = 120°C

Steam pressure = 232.1 kPa at ( $T_{S}$ ) = 125°C

Product specific heat in ( $C_{PA}$ ) = 3.936 kJ/(kg°C)

The mass equation is:

$m_{A}X_{A}=m_{B}X_{B}$

$100(0.12)=m_{B}(0.1)\\m_{B}=\frac{100(0.12)}{0.1} =120$

Also $m_{a}+m_{s}=m_{b}\\$

Therefore: $100}+m_{s}=120\\\\m_{s}=120-100=20$

The energy balance equation is:

$m_{A}C_{PA}(T_{A}-0)+m_{s}H_{s}=m_{B}C_{PB}(T_{B}-0)$

$3.936 = (4.178)(0.88) +C_{PS}(0.12)\\C_{PS}=\frac{3.936-3.677}{0.12} =2.161$

$C_{PB}= 4.232*0.9+0.1C_{PS}= 4.232*0.9+0.1*2.161=4.025$ kJ/(kg°C)

By substituting values into the energy equation:

$100(3.936)(50-0)+20H_{s}=120(4.025)}(120-0)$

$19680+20H_{s}=57960\\20H_{s}=57960-19680 \\20H_{s}=38280\\H_{s}=\frac{38280}{20} =1914$

$H_{s}=1914kJ/kg$

From the properties of saturated steam at 232.1 kPa,

$H_{c}$ = 524.99 kJ/kg

$H_{v}$ = 2713.5 kJ/kg

% quality = $\frac{1914-524.99}{2713.5-524.99} =63.5%$

Any steam quality above 63.5% will result in higher total solids in the heated product.

3 0

2 months ago

1. Mark ‘N’ if a wrong type of units is used. Mark ‘Y’ otherwise. (1 point each)

pantera1 [306]

Response:

bsbsdbsd

Clarification:

7 0

2 months ago

Water flows steadily through a horizontal nozzle, discharging to the atmosphere. At the nozzle inlet the diameter is D1; at the

mote1985 [299]

The inlet gauge pressure must be 61.627 Psi.

4 0

2 months ago

Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and

choli [298]

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data:

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

To determine final temperature in an Isentropic process

$T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}$ ----------- ( 1 )

T2 = 303 $( \frac{450}{90})^{(1.667- 1)/1.667}$ = 576.9K

The work performed in a piston-cylinder device is calculated using the subsequent formula

$w_{in} = c_{v} ( T2 - T1 )$ ------- ( 2 )

where: cv = 3.1156 kJ/kg.k for helium gas

T2 = 576.9K, T1 = 303 K

substituting values into equation 2

$w_{in}$ = 853.4 kJ/kg

the work done in a steady flow compressor is determined using this

$w_{in} = c_{p} ( T2 - T1 )$

where: cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K

T2 = 576.9 k, T1 = 303 K

plugging values back into equation 3

$w_{in}$ = 1422.3 kJ / kg

4 0

2 months ago