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Ratling [72]
3 years ago
14

The gage pressure measured as 2.2 atm, the absolute pressure of gas is 3.2 bar. Please determine the local atmospheric pressure

in kPa.
Engineering
1 answer:
LiRa [457]3 years ago
8 0

Answer:

97.085\ \text{kPa}

Explanation:

P_{g} = Gauge pressure = 2.2 atm = 2.2\times 101325=222915\ \text{Pa}

P_{abs} = Absolute pressure = 3.2\ \text{bar}=3.2\times 10^5\ \text{Pa}

P_{atm} = Local atmospheric pressure

Absolute pressure is given by

P_{abs}=P_{atm}+P_g\\\Rightarrow P_{atm}=P_{abs}-P_g\\\Rightarrow P_{atm}=3.2\times 10^5-222915\\\Rightarrow P_{atm}=97085\ \text{Pa}=97.085\ \text{kPa}

The local atmospheric pressure is 97.085\ \text{kPa}.

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For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho
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Answer:d

Explanation:

Given

Temperature=200^{\circ}\approc 473 K

Also \gamma for air=1.4

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.=\frac{V}{\sqrt{\gamma RT}}

(a)1000 km/h\approx 277.78 m/s

Mach no.=\frac{277.78}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.63

(b)500 km/h\approx 138.89 m/s

Mach no.=\frac{138.89}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.31

(c)2000 km/h\approx 555.55 m/s

Mach no.=\frac{555.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=1.27

(d)200 km/h\approx 55.55 m/s

Mach no.=\frac{55.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0
3 years ago
Define factors that can change the performance of a polymer, such are additives
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Answer:

 The performance of the polymer is basically change by the various type of factors like shape, tensile strength and color.

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A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg
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Answer:

\dot{w}= -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}

Given that tank is insulated so\dot{Q}=0 and no mass is leaving so

\dot{m_e}=0

\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt

m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t

Mass conservation m_2-m_1=m_e-m_i

m_1,m_2 is the initial and final mass in the system respectively.

Initially tank is evacuated so m_1=0

We know that for air u=C_vT ,h=C_p T,P_2v_2=m_2RT_2

m_2=0.42 kg

So now putting values

0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300

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