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Vitek1552 [10]
3 years ago
5

Exhaust gases entering a convergent nozzle have a total pressure (Pt) of 200 kPa and total temperature (Tt) of 800 K. The gases

exit the nozzle into ambient air at a static pressure (Po) of 101.3 kPa. Assuming that y 1.33 and R-287 J/(kg.K), determine the critical pressure ratio.
Engineering
2 answers:
madam [21]3 years ago
5 0

Answer:

critical pressure ratio = 108.07 k pa

Explanation:

Pt = 200 k Pa

Tt = 800 k

Po = 101.3 k Pa

y = 1.33

R = 287 J/(kg.k)

critical pressure ratio ( Pc )

\frac{Pc}{Pt} = (\frac{2}{y+1})^{\frac{y}{y-1} }  ------- equation 1

Pt given as 200 k Pa

Y = 1.33

back to equation 1

\frac{Pc}{Pt} = (\frac{2}{1.33 + 1})^{\frac{1.33}{1.33 - 1} }

Pc = 200 * 0.5404 = 108.07 k pa

Pavlova-9 [17]3 years ago
4 0

Answer:

0.351

Explanation:

n=1/(γ−1) = 1/ (1.33-1)= 3.03

critical pressure ratio

p_2/p_1 = \frac{2}{n+1} ^{(\frac{n}{n-1} )}\\\\=  \frac{2}{3.03+1} ^{(\frac{3.03}{3.03-1} )}\\\\= 0.496^{1.493}\\\\= 0.351

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Neporo4naja [7]

The examples of engineering controls is Biohazard waste containers and Spill clean up kits.

What is engineering controls?

An engineering controls is a workplace process that protect workers by removing hazardous conditions or by placing a barrier between the worker and the hazard.

An example of engineering controls is installation of exhaust ventilation to remove airborne emissions to shield the worker.

Hence, the examples of engineering controls is Biohazard waste containers and Spill clean up kits.

Therefore, the Option C and D is correct.

8 0
2 years ago
Given an integer k, a set C of n cities c1, . . . , cn, and the distances between these cities dij = d(ci , cj ), for 1 ⤠i &lt
Snezhnost [94]

Answer:

See explaination

Explanation:

2-Approximation Algorithm

Step 1: Choose any one city from the given set of cities C arbitrarily and put it in to a set H which is initially empty.

Step 2: For every city c in set C that is currently not present in set H compute min_distc = Minimum[ d(c, c1), d(c, c2), d(c, c3), ..... . . . . d(c, ci) ]

where c1, c2, ... ci are the cities in set H

and d(x, y) is the euclidean distance between city x and city y

Step 3: H = H ∪ {cx} where cx is the city have maximum value of min_dist over all possible cities c, computed in Step-2.

Step 4: Step-2 and Step-3 are iterated for k-1 times so that k cities are included int set H.

The set H is the required set of cities.

Example

Assume:-

C = {0, 1, 2, 3}

d(0,1) = 10, d(0,2) = 7, d(0,3) = 6, d(1,2) = 8, d(1,3) = 5, d(2,3) = 12

k = 3

Solution:-

Initially H = { }

Step-1: H = {0}

Step-2: Cities c \not\in H are {1, 2, 3}

min_dist1 = min{dist(0,1)} = min{10} = 10

min_dist2 = min{dist(0,2)} = min{7} = 7

min_dist3 = min{dist(0,3)} = min{6} = 6

Step-3: Max{10, 7, 6} = 10

Step-4: cx = 1

Step-5: H = H ∪ cx = {0} \cup {1} = {0, 1}

Step-6: Cities c \not\in H are {2, 3}

min_dist2 = min{dist(0,2), dist(1,2)} = min{7, 8} = 7

min_dist3 = min{dist(0,3), dist(1,3)} = min{6, 5} = 5

Step-7: Max{7, 5} = 7

Step-8: cx = 2

Step-9: H = H \cup cx = {0, 1} \cup {2} = {0, 1, 2}

Result: The set H is {0, 1, 2}.

6 0
3 years ago
An uncovered swimming pool loses 1.0 inch of water off its 1,000 ft^2 surface each week due to evaporation. The heat of vaporiza
soldi70 [24.7K]

Answer:

The affirmation is true, the cover will be worth buying

Explanation:

The equation necessary to use is

E = m*cv,

Where

cv: the heat of vaporization.  

Finding the rate at which the water evaporates (m^3/week).  

The swimming pool loses water at 1 inch/week off its 1,000 ft^2

Than,

1000 ft² * 1 in/wk * 1 ft/12 in = 83.33 ft³/week

To obtains the rate of mass loss it is necessary to multiply it for the density of water

83.33 ft³/week * 62.4 lb/ft³ = 5200 lb/week

Knowing the vaporization heat it is possible to find the rate of heat which is leaving the swimming pool  

5200 lb/week * 1050 BTU/lb = 5460000 btu/week

Over a 15-week period, the pool loses 81.9 million BTU.  

Knowing the cost of energy to heat the pool is $10.00 per million btu

The price = $819

This way, the affirmation is true, the cover will be worth buying

3 0
3 years ago
4. Use voltage divider concepts to find the voltages indicated in the following circuits. You may want to use some of your resul
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Hello, because there is not a circuit I'll explain the voltage divider and make an exercise, this way you can solve the problem using the method described here.

Answer with explanation:

A voltage divider uses the voltage distribution among components to find a voltage in a specific element of the circuit. If we have a source V1 connected to impedances Z1 and Z2 in series, we can use a voltage divider to find the voltage across Z1 or Z2 base on their value and the input voltage.

VZ1 = V1*Z1/(Z1+Z2)

VZ2 = V1*Z2/(Z1+Z2)

In the image, to find the voltage Vo across R2 we apply the following equation: Vo = (V1*R2)/(R1+R2).

To solve the exercise in the other image, we need to apply a voltage divider twice:

In-circuit 1 we are asked to find the voltage VAB that falls on R2 and R3 (the same voltage for both resistances because are in parallel), to do so we use a voltage divider using V1, R1 and RT where RT is the equivalent resistance RT = R2//R3 + R4, therefore, for circuit two VAC = (V1*R1)/(R1+RT). After finding VAC we apply voltage divider again to find VAB, see circuit 3, to do so we apply VAB = (VAC*R2//R3)/(R2//R3 + R4) = (VAC*R2//R3)/(RT)

4 0
3 years ago
What's the best way to find the load capacity of a crane? Select the best option. Call the manufacturer Ask co-workers Look at t
Orlov [11]

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