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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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Answer:

it is reducely very iloretable chance for a software engineer to give an end to this question

7 0
3 years ago
For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration
Jobisdone [24]

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

4 0
3 years ago
Air at atmospheric pressure and at 300K flows with a velocity of 1.5m/s over a flat plate. The transition from laminar to turbul
Savatey [412]

Answer:3.47 m

Explanation:

Given

Temperature(T)=300 K

velocity(v)=1.5 m/s

At 300 K

\mu =1.846 \times 10^{-5} Pa-s

\rho =1.77 kg/m^3

And reynold's number is given by

Re.=\frac{\rho v\time x}{\mu }

5\times 10^5=\frac{1.77\times 1.5\times x}{1.846\times 10^{-5}}

x=\frac{5\times 10^5\times 1.846\times 10^{-5}}{1.77\times 1.5}

x=3.47 m

5 0
3 years ago
Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area
dybincka [34]

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

8 0
2 years ago
A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

COP=\dfrac{Qr}{W}

2.5=\dfrac{5}{W}

2.5=\dfrac{5}{W}

W=2 MW

For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0
3 years ago
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