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Sergio039 [100]
3 years ago
9

What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?

Engineering
1 answer:
irga5000 [103]3 years ago
7 0

Answer:

\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency  increase.

 Now from T-S diagram

    W_{net}=W_{out}-W_{in}

   W_{net}=(h_3-h_4)-(h_2-h_1)

  Q_{in}=h_3-h_5

  Due to generation (h_5-h_2) amount of energy has been saved.

  Q_{generation}=Q_{saved}

So efficiency of cycle \eta =\frac{W_{net}}{Q_{in}}

  \eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

Effectiveness of re-generator

  \varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}

So the efficiency of regenerative cycle

\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

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a sprue is 12 in long and has a diameter of 5 in at the top. The molten metal level in the pouring basing is taken to be 3 in fr
vampirchik [111]

Answer:

See explaination

Explanation:

We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.

See the attached file for detailed solution of the given problem.

8 0
3 years ago
Read 2 more answers
Examples of reciprocating motion in daily life
bonufazy [111]

Answer:

Examples of reciprocating motion in daily life are;

1) The needles of a sewing machine

2) Electric powered reciprocating saw blade

3) The motion of a manual tire pump

Explanation:

A reciprocating motion is a motion that consists of motion of a part in an upward and downwards (\updownarrow) or in a backward and forward (↔) direction repetitively

Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump

In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.

8 0
2 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
. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C
MrRissso [65]

Answer:

The inside temperature, T_{in} is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×(T_s - T_{\infty) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

P = \dfrac{K \times A \times \Delta T}{L}

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

2456.25  = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}

T_{in} = 250 - \dfrac{2456.25  \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C

The inside temperature, T_{in} = 247.99 °C  ≈ 248 °C.

3 0
3 years ago
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
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