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Stella [2.4K]
3 years ago
10

Five kg of nitrogen gas (N2) in a rigid, insulated container fitted with a paddle wheel is initially at 300 K, 150 kPa. The N2 g

as receives work from the paddle wheel until the gas is at 500 K and 250 kPa. Assuming the ideal gas model with a constant specific heat (i.e., use a constant specific heat of N2 at 300K), neglect kinetic energy and potential energy effects
Required:
a. Draw a system schematic and set up states.
b. Determine the amount of work received from the paddle wheel (kJ).
c. Determine the amount of entropy generated (kJ/K).
Engineering
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

A) attached below

B) 743 KJ

C) 1.8983 KJ/K

Explanation:

A) Diagram of system schematic and set up states

attached below

<u>B) Calculate the amount of work received from the paddle wheel </u>

assuming ideal gas situation

v1 = v2 ( for a constant volume process )

work generated by paddle wheel = system internal energy

dw = mCv dT .     where ; Cv = 0.743 KJ/kgk

     = 5 * 0.743 * ( 500 - 300 )

     = 3.715 * 200 = 743 KJ

<u>C) calculate the amount of entropy generated  ( KJ/K )</u>

S2 - S1 = 1.8983 KJ/K

attached below is the detailed solution

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Explain why a hydraulic power system would be the best choice when building a device or vehicle that requires large amounts of p
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A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an
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Answer:

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Explanation:

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and,

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h_3=hg

    =2610.8 \ KJ/Kg

s_3=sg

    =5.3108 \ KJ/Kgh

At 45,

⇒ x_{45} = \frac{5.3108-1.0912}{6.5019}

          =0.66

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h_f=340.54

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V_f=0.001030 \ m^3/Kg

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⇒ h_2=340.54+0.001030(15\times 10^{3}-50)

        =355.94 \ kJ/kg

hence,

The isentropic efficiency of turbine will be:

⇒ n_T=\frac{h_3-h_4}{h_3-h_{45}}

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4 0
3 years ago
Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1
tia_tia [17]

Answer:

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Explanation:

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1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

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The formula for queueing delay of <em>n-th</em> packet is :  (n - 1) × L/R

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Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

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