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IgorC [24]
2 years ago
7

A compressor receives air at 290 K, 95 kPa and shaft work of 5.5 kW from a gasoline engine. It should deliver a mass flow rate o

f 0.01 kg/s air to a pipeline. Find the maximum possible exit pressure of the compressor.
Engineering
1 answer:
aev [14]2 years ago
3 0

Answer:

P2 = 3.9 MPa

Explanation:

Given that

T₁ = 290 K

P₁ = 95 KPa

Power P = 5.5 KW

mass flow rate  = 0.01 kg/s

solution

with the help of table A5

here air specific heat and adiabatic exponent is

Cp = 1.004 kJ/kg K

and k = 1.4

so

work rate will be

W = m × Cp × (T2 - T1)              ..........................1

here T2 = W ÷ ( m × Cp) + T1    

so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290

T2 = 838 k

so final pressure will be here

P2 = P1 × (\frac{T2}{T1})^\frac{k}{k-1}        ..............2

P2 = 95 × (\frac{838}{290})^\frac{1.4}{1.4-1}

P2 = 3.9 MPa

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A crankcase heater is often used to prevent refrigerant from mixing with compressor oil during periods of:
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Low ambient temperature

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What are Tresca and Von Mises yield criteria?
elena-s [515]

Answer

For isotropic material plastic yielding depends upon magnitude of the principle stress not on the direction.

Tresca and Von Mises yield criteria are the yield model which is widely used.

The Tresca yield criterion stated that yielding will occur in a material only when the greatest maximum shear stress reaches a critical value.

max{|σ₁ - σ₂|,|σ₂ - σ₃|,|σ₃ - σ₁|} = σ_f

under plane stress condition

  |σ₁ - σ₂| = σ_f

The Von mises yielding criteria stated that the yielding will occur when elastic energy of distortion reaches critical value.

σ₁² - σ₁ σ₂ + σ₂² =  σ²_f

5 0
3 years ago
The three resistors in the circuit shown have values of 4ohms 4ohms and 2ohms. The battery has a value of 12 volts. What is the
gtnhenbr [62]

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3 0
3 years ago
134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.
vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = - 10^{\circ}C = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = 60^{\circ}C = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T =60^{\circ}C:

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at  P = 140 kPa, T =- 10^{\circ}C:

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = s'_{s} and h'_{s} = 278.06 kJ/kg.K at 700kPa

(a) Isentropic efficiency of the compressor, \eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}

\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%

(b) The temperature of the environment, T_{e} = 27^{\circ}C = 273 + 27 = 300 K

Availability at state 1, \Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg

Similarly for state 2, \Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg

Now, the efficiency of the compressor as per the second law;

\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757 = 67.57%

4 0
3 years ago
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