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Maslowich
3 years ago
14

A mixture of air and methane is formed in the inlet manifold of a natural gas-fueled internal combustion engine. The mole fracti

on of the methane is 15 percent. This engine is operated at 3000 rpm and has a 5-L displacement. Determine the mass flow rate of this mixture in the manifold where the pressure and temperature are 80 kPa and 208C.

Engineering
1 answer:
german3 years ago
4 0

Answer:

The mass flow rate of the mixture in the manifold is 6.654 kg/min

Explanation;

In this question, we are asked to calculate mass flow rate of the mixture in the manifold

Please check attachment for complete solution and step by step explanation.

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A brake caliper is considered a suspension item.<br> True<br> False
user100 [1]
True


Suspension is the system of tires, tire air, springs, shock absorbers and linkages that connects a vehicle to its wheels and allows relative motion between the two.[1] Suspension systems must support both road holding/handling and ride quality
5 0
3 years ago
Write a SELECT statement that returns these column names and data from the Products table: product_name The product_name column
I am Lyosha [343]

Answer:

 SELECT  

   product_name, list_price, discount_percent,    

   ROUND(list_price * (discount_percent / 100), 2) AS discount_amount,

   ROUND(list_price - (discount_percent / 100 * list_price), 2)  AS discount_price  

FROM

Products

ORDER BY (list_price - (discount_percent / 100 * list_price)) DESC

LIMIT 5;

Explanation:

In the above SELECT statement is used to select columns from Products table.

The columns are  product_name,list_price and discount_percent

Next a column discount_amount is selected which is the calculated from previous two columns that are list_price and discount_percent. The discount amount is basically calculated by multiplying list_price with discount_percent/100 and the resultant column is named as discount_amount using Alias which is used to give a temporary name to set of columns or a table.

Next another column discount_price is obtained from previous three columns that are  list_price , discount_percent and discount_amount as: list_price - (discount_percent / 100 * list_price) This as a whole is given a column name discount_price.

FROM is used to refer to a table from which these columns are to be selected. The table name is Product.

The result set is sorted in descending order by discount_price so ORDER BY is used to order the resultant records and DESC is used to sort these records according to the discount_price in descending order.

LIMIT statement is used to extract the records from Product and limit the number of rows returned as a result based on a limit value which is 5 here.

5 0
3 years ago
A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0
3 years ago
How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0
3 years ago
Anyone help me please ?
Degger [83]

Answer:

I can help but I need to know what it looking for

5 0
3 years ago
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