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frutty [35]
2 years ago
5

A 2 stage piston compressor has a(n) ____ where compressed air is cooled while passing to the second stage.

Engineering
1 answer:
Nikitich [7]2 years ago
8 0

A 2 stage piston compressor has an intercooler where compressed air is cooled while passing to the second stage.

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P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame
eimsori [14]

Answer:

a)  m=0.17kg/s

b)  Ma=0.89

Explanation:

From the question we are told that:

Pressure P=60kPa

Diameter d=3cm

Generally at sea level

T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4

Generally the Power series equation for Mach number is mathematically given by

\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}

\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}

Ma=0.89

Therefore

Mass flow rate

\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\rho=0.848kg/m^3

Generally the equation for Velocity at throat is mathematically given by

V=Ma(r*T_0\sqrt{T_e})

Where

T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}

T_e=248

Therefore

V=0.89(1.4*288\sqrt{248})\\\\V=284

Generally the equation for Mass flow rate is mathematically given by

m=\rho*A*V

m=0.84*\frac{\pi}{4}*3*10^{-2}*284

m=0.17kg/s

6 0
3 years ago
Which phrase best describes a safety-critical system? A. a system that faces a very high risk of failure B. a system isolated fr
KIM [24]

Answer:

B.

Explanation:

A safety-critical system (SCS) or life-critical system is a system whose failure or malfunction may result in one (or more) of the following outcomes: death or serious injury to people. loss or severe damage to equipment/property.

7 0
3 years ago
A large building will need several different types of workmen to install and repair pipes for water, heating,
siniylev [52]

Answer:

Plumber and pipefitters

Explanation:

3 0
3 years ago
Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun
Alex17521 [72]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19,  24,  13 , 29,  19,  21,  14

It is given that,

∑x = 299

∑y = 167

∑x^{2} = 11887

∑y^{2} = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 +  24 + 13 + 29 + 19 +  21 +  14

∑y = 169

No, it is not equal to the given value.

3. ∑x^{2} = 11887

Let's verify it:

For this to find,  first we need to square all the value of x individually and then add them together to verify.

∑x^{2} = 27^{2} + 44^{2} + 32^{2} + 47^{2} + 23^{2} + 40^{2} + 34^{2} + 52^{2}

∑x^{2} = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑y^{2} = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑y^{2} = 30^{2} + 19^{2} +  24^{2} + 13^{2} + 29^{2} + 19^{2} +  21^{2} +  14^{2}

∑y^{2}  = 3,845

No, it is not equal to the given value.

4 0
3 years ago
PythonA group of statisticians at a local college has asked you to create a set of functionsthat compute the median and mode of
skelet666 [1.2K]

Answer:

  1. def median(l):
  2.    if(len(l) == 0):
  3.       return 0
  4.    else:
  5.        l.sort()
  6.        if(len(l)%2 == 0):
  7.            index = int(len(l)/2)
  8.            mid = (l[index-1] + l[index]) / 2
  9.        else:
  10.            mid = l[len(l)//2]  
  11.        return mid  
  12. def mode(l):
  13.    if(len(l)==0):
  14.        return 0
  15.    mode = max(set(l), key=l.count)
  16.    return mode  
  17. def mean(l):
  18.    if(len(l)==0):
  19.        return 0
  20.    sum = 0
  21.    for x in l:
  22.        sum += x
  23.    mean = sum / len(l)
  24.    return mean
  25. lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
  26. print(mean(lst))
  27. print(median(lst))
  28. print(mode(lst))

Explanation:

Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).

In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.

In mean function,  after checking the length of list,  we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).

In the main program, we test the three functions using a sample list and we shall get

20.5

12.5

12

3 0
3 years ago
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