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Ber [7]
3 years ago
11

Air enters a control volume operating at steady state at 1.2 bar, 300K, and leaves at 12 bar, 440K, witha volumetric flow rate o

f 1.3 m3/min. The work input to the control volume is 240 kJ per kg of air flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in kW.
Engineering
1 answer:
topjm [15]3 years ago
4 0

Answer:

Heat transfer = 2.617 Kw

Explanation:

Given:

T1 = 300 k

T2 = 440 k

h1 = 300.19 KJ/kg

h2 = 441.61 KJ/kg

Density = 1.225 kg/m²

Find:

Mass flow rate = 1.225 x [1.3/60]

Mass flow rate = 0.02654 kg/s

mh1 + mw = mh2 + Q

0.02654(300.19 + 240) = 0.02654(441.61) + Q

Q = 2.617 Kw

Heat transfer = 2.617 Kw

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Explanation:

<u>Filler:</u>

  Filler is the material rod is used when we are joining two material by using welding process.If thickness of work piece is more so it will become compulsory to provide some filler material for making the welding join to withstand high stresses.

<u>Electrode:</u>

  Electrode is the element which is used to complete the electric circuit in welding .Some time electrode is connected with positive terminal and some time with negative terminal ,it depends on the requirement of welding process.In Tungsten inert gas welding electrode is connected negative terminal but on the other hand Metal inert gas welding electrode is connected with positive terminal.Electrode can be consumable non-consumable depends on the condition.

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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

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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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