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prohojiy [21]
4 years ago
8

4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and

expands reversibly and adiabatically to 2 lbf/in2. Ignore kinetic and potential energy effects. Determine the mass flow rate, in lb/s, and the power developed by the turbine, in hp. Determine the mass flow rate, in lb/s. Determine the power developed by the turbine, in hp.
Engineering
1 answer:
hodyreva [135]4 years ago
6 0
Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
You might be interested in
Which of the following is a safety device meant to protect an operator from the point of operation?
Art [367]

Answer:

machine guarding

Explanation:

Machine guarding protects an operator from the point of operation.

6 0
4 years ago
slader A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy
nikklg [1K]

Answer:

a. 33120MJ

b. 5.6

c. 48

Explanation:

∆U= 120,000 KJ/h

Since

1 day =  24 hrs

14 days =24 x  14 hrs

14 days = 336 hrs

∆U = 120000 x 336 KJ

=40320000KJ

K = 1000

M = 1000000

∆U = 403200 MJ

Work done

W= 2000 KW.h    ( 1 h = 3600 s)

W= 7200 MJ

According to the first law of thermodynamics

∆U = Q+W

Q= 40320 - 7200 MJ

1)Qa=33120 MJ

Coefficient of performance

COP = ∆U/W

COP= 40320 / 7200

2)COP = 5.6

3)COP of ideal heat pump

Th/(Th - Tl)

Th = 15°C

Tl = 9°C

Convert Celsius to Kelvin

273 + 15 = 288

288/(15-9)

288/6

48

COP= 48

7 0
3 years ago
Read 2 more answers
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator
MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\

Carrier frequency 100kHz

Upper side band

100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

100k, 99.9k, 99.8k\ \ and \ \99.6kHz

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz

3 0
3 years ago
MODIFIED-BOTTOM-UP-CUT-ROD(p, n, c) to return not only the value but the actual solution, too. Hint: It is similar to how array
Vaselesa [24]

Answer:

b.

Matrix chain multiplication

M[i,j] = M[i,k] + M[(k+1),j] + p[i-1]*p[k]*p[j] i<=k<j

p[] = {5,10,3,12,5,50}

M[0][0] = 0,M[1][1] = 0,M[2][2] = 0,M[3][3] = 0,M[4][4] = 0,M[5][5] = 0,

M[1][2] = M[1][1]+M[2][2]+p[0]*p[1]*p[2] = 0+0+5*10*3 = 150

M[2][3] = M[3][3]+M[2][2]+p[1]*p[2]*p[3] = 0+0+10*3*12 = 360

M[3][4] = M[3][3]+M[4][4]+p[2]*p[3]*p[4] = 0+0+3*12*5 = 180

M[4][5] = M[4][4]+M[5][5]+p[3]*p[4]*p[5] = 0+0+12*5*50 = 3000

M[1][3] = min{M[1][1]+M[2][3]+p[0]*p[1]*p[3] , M[1][2]+M[3][3]+p[0]*p[2]*p[3]}

= {0 + 360 + 600 , 150+0+180} = {960,330} = 330

M[2][4] = min{M[2][2]+M[3][4]+p[1]*p[2]*p[4] , M[2][3]+M[4][4]+p[1]*p[3]*p[4]}

= {0 + 180 + 150 , 360+0+600} = {960,330} = 330

M[3][5] = min{M[3][3]+M[4][5]+p[2]*p[3]*p[5] , M[3][4]+M[5][5]+p[2]*p[4]*p[5]}

= {0 + 3000 + 1800 , 180+0+750} = {4800,930} = 930

M[1][4] = min{M[1][1] + M[2][4] +p[0]*p[1]*p[4] ,M[1][2] + M[3][4] +p[0]*p[2]*p[4] ,

M[1][3] + M[4][4] +p[0]*p[3]*p[4]}

{0+330+250 , 150+180+75 , 330+0+300} = 405

M[2][5] = min{M[2][2] + M[3][5] +p[1]*p[2]*p[5] ,M[2][3] + M[4][5] +p[1]*p[3]*p[5] ,

M[2][4] + M[5][5] +p[1]*p[4]*p[5]}

{0+930+1500 , 360+3000+6000,330+0+2500} = 2430

M[1][5] = min{M[1][1] +M[2][5]+p[0]*p[1]*p[5] , M[1][2] +M[3][5]+p[0]*p[2]*p[5],

M[1][3] +M[4][5]+p[0]*p[3]*p[5] , M[1][4] +M[5][5]+p[0]*p[4]*p[5]}

{0+2430+2500 , 150+930+750 , 330+3000+3000 , 405+0+1250} = 1655

(a)

MemoizedCutRod(p, n)

r: array(0..n) := (0 => 0, others =>MinInt)

return MemoizedCutRodAux(p, n, r)

MemoizedCutRodAux(p, n, r)

if r(n) = 0 and then n /= 0 then -- check if need to calculate a new solution

q: int := MinInt

for i in 1 .. n loop

q := max(q, p(i) + MemoizedCutRodAux(p, n-i, r))

end loop

end if

r(n) := q

end if

return r(n)

8 0
3 years ago
4. When the ESC system senses oversteer,
bagirrra123 [75]
I’m pretty sure the answer is b I’m not 100% but I think
5 0
2 years ago
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