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skelet666 [1.2K]
3 years ago
9

A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with

no internal irreversibilities, the specific entropy at the exit is __________.
a. greater than the specific entropy of the gas at the inlet.
b. equal to the specific entropy of the gas at the inlet.
c. less than the specific entropy of the gas at the inlet.
Engineering
1 answer:
Hunter-Best [27]3 years ago
3 0

Answer:

b. equal to the specific entropy of the gas at the inlet.

Explanation:

Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.

If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

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The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the n
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Explanation:

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What is the real name of the town that Bud & Bugs stayed in while they were waiting for the train?
Leona [35]

Answer: Hooverville

Explanation:

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At a ski resort, water at 40 °F is pumped through a 3-in.-diameter, 2001-ft-long steel pipe from a pond at an elevation of 4286
deff fn [24]

Answer:

P = 24.38 hp

Explanation:

Given data:

diameter of steel pipe = 3 inc

length of pipe = 2001 ft

elevation of pond = 4286 ft

elevation of snow making machine = 4620 ft

Rate of water = 0.26 ft^3/s

Applying bernouli equation on bioth side

\frac{P}{\rho} + \frac{v_1^2}{2g} + z_1 + h_p =\frac{P_2}{\rho} + \frac{v_2^2}{2g} + z_2 + \frac{flv^2}{2gD}.....1

where P_2 = 182 psi

P_1 = 0, v_1 = 0

V =V_2 = \frac{Q}{\frac{\pi}{4} D^2}= \frac{0.26}{\frac{\pi}{4} \times (3/12)^2} = 5.30 ft/s

calculation fro friction factor

from standard table we have

\frac{\epsilon}{D} = \frac{0.00015}{(3/12)} = 6\times 10^{-4}

Re =\frac{VD}[\nu} = \frac{5.30 \times (3/12)}{1.66 \times 10^{-5}} = 7.96 \times 10^4

so for calculated R and\frac{\epsilon}{D}, F value = 0.0212

from equation 1

h_p = \frac{P_2}{\rho} + Z_2 -Z_1 +(1 + f\frac{l}{D}) \frac{V^2}{2g}

h_p = \frac{182 \times 144/ lb /ft^2}{62.4} + 4620 - 4286 + (1 + 0.0212 \frac{2001}{(3/12)}) \frac{5.30}{2\times 32.2}

h_p = 826.76ft so that

P =  \rho Q h_p = 62.4 \times 0.26 \times 826.76 = 13413.38 ft lb/s = 24.38 hp

3 0
3 years ago
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