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3 years ago

How does the clearance volume affect the efficiency of the Otto cycle?

Engineering

1 answer:

eduard3 years ago

7 0

Answer:

Explanation:

A smaller clearance volume means a higher compression. A higher compression means better thermal efficiency. However a compression ratio too high might be troublesome, as it can cause accidental ignition of the fuel-air mix. This is the reason why Otto cycle engines have lower compressions that Diesel engines. In a Diesel engine the mix ignites by compression instead of a spark.

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A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

Dafna1 [17]

Answer:

a) $\dot W = 1.062\,kW$

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}$

$COP_{HP} = 15.692$

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

$\dot Q_{H} = 60000\,\frac{kJ}{h}$

The minimum power supplied to the heat pump is:

$\dot W = \frac{\dot Q_{H}}{COP}$

$\dot W = \frac{\left(60000\,\frac{kJ}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}{15.692}$

$\dot W = 1.062\,kW$

5 0

3 years ago

Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured

aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

$CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]$

If $\bar{x}$ is 190, we can find the value of $z\sigma_{\bar{x}}$ :

$\bar{x} - z\sigma_{\bar{x}} = 188.29$

$190 - z\sigma_{\bar{x}} = 188.29$

$z\sigma_{\bar{x}} = 1.71$

Now we need to find the value of $\sigma_{\bar{x}}$ :

$\sigma_{\bar{x}} = s / \sqrt{n}$

$\sigma_{\bar{x}} = 5/ \sqrt{25}$

$\sigma_{\bar{x}} = 1$

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

$CI = 1 - 2*0.0436 = 0.9128 = 91.28\%$

The nearest CI in the options is 90%, so the correct option is B.

4 0

4 years ago

Write multiple if statements:

lora16 [44]

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3 0

3 years ago

Unit for trigonometric functions is always "radian". 1. 10 points: Do NOT submit your MATLAB code for this problem (a) Given f(x

RoseWind [281]

Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

clear all;

clc;

x0=input('Initial guess:\n');

x=x0;

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

i=i+1;

temp=x;

x=x-(f/g);

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=abs(x-temp);

fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

end

fprintf('The solution x = %6f \n',x);

%% End of MATLAB Program

Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

Guess x=3.0:

Initial guess:

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

Guess x=4.7:

Initial guess:

4.7

Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

7 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:

def median(l):
if(len(l) == 0):
return 0
else:
l.sort()
if(len(l)%2 == 0):
index = int(len(l)/2)
mid = (l[index-1] + l[index]) / 2
else:
mid = l[len(l)//2]
return mid
def mode(l):
if(len(l)==0):
return 0
mode = max(set(l), key=l.count)
return mode
def mean(l):
if(len(l)==0):
return 0
sum = 0
for x in l:
sum += x
mean = sum / len(l)
return mean
lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
print(mean(lst))
print(median(lst))
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

3 0

3 years ago