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ad-work [718]
1 year ago
12

Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and

the work done, assuming the process takes place (a) in a piston-cylinder device and (b) in a steadyflow compressor. [15
Engineering
1 answer:
ra1l [238]1 year ago
4 0

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data :

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

Determine final temperature for an Isentropic  process

T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}  ----------- ( 1 )

T2 = 303 ( \frac{450}{90})^{(1.667- 1)/1.667} =  576.9K

Work done in a piston-cylinder device can be calculated using this formula

w_{in} = c_{v} ( T2 - T1 )    ------- ( 2 )

where : cv = 3.1156 kJ/kg.k  for helium gas

             T2 = 576.9K ,    T1 = 303 K

substitute given values Back to equation 2

w_{in}  = 853.4 kJ/kg

work done in a steady flow compressor can be calculated using this

w_{in} = c_{p} ( T2 - T1 )

where : cp ( constant pressure of helium gas )  = 5.1926 kJ/kg.K

             T2 = 576.9 k , T1 = 303 K

substitute values back to equation 3

w_{in} = 1422.3 kJ / kg

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Explanation:

When a steady potential is connected across the end points of the conductor, there will be a of current through the conductor and the internal resistance in the conductor will tend to resist the flow of the current

A good example is a simple circuit comprises of a battery, wires and a touch bulb.

When the battery is connected to the two ends of the torch bulb, current flow through and the bulb is lit up.

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See explaination

Explanation:

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2 years ago
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Problem 5) Water is pumped through a 60 m long, 0.3 m diameter pipe from a lower reservoir to a higher reservoir whose surface i
kap26 [50]

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2  =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV  =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f  = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

4 0
2 years ago
The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13
prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0
2 years ago
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