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2 years ago

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A piston cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air at 300K. The air undergoes a const

ant pressure process to a final temp of 400K. During the process heat is transfered to the air by Q=12kJ. Assuming the ideal gas model with k=1.4 and negligible changes in kinetic and potential energy for the air, determine the work done by the paddle on the air and the work done by the air to displace the piston. Gas constant for air is 0.287 kJ/kg*K
note ideal gas model is pV=mRT, and with k=1.4 the total internal energy change of air canbe calculated by U2-U1=(mR/k-1)(delta T)

1 answer:

vova2212 [387]2 years ago

7 0

The answer & explanation for this question is given in the attachment below.

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A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.

AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass $m_{in}+m_{out}=\bigtriangleup m_{system}$

so, $m_e=m_1-m_2$

$Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1$

#Also, given the properties of water as;

$(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg$

#We assume constant properties for the steam at average temperatures: $h_e=\approx(h_1+h_2)/2$

#Replace known values in the equation above; $h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg$

#Using the mass and energy balance relations;

$m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg$

#We have $Q_i_n+m_eh_e=m_2u_2-m_1u_1$ : we replace the known values in the equation as;

$Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ$

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

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3 years ago

To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr

alukav5142 [94]

<u></u> $\ T_{c}$ has greater effect.

<u>Explanation</u>:

$\eta_{\max }=1-\frac{T_{c}}{T_{A}}$

$T_{c}\\$ = Temperature of cold reservoir

$T_{H}$ = Temperature of hot reservoir

when $T_{c}$ is decreased by 't',

$\eta_{\text {incre }}$ = $1-\frac{\left(\tau_{c}-t\right)}{T_{H}}$

$=n \ + \frac{t}{T_{n}}$ $-(i)$

when ${T_{H}}$ is increased by 'T'

$\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)$

$\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}$

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3 years ago

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vitfil [10]

Answer:

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Explanation:

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1) Overheating break systems

2) Use of wrong brake fluid

3) Broken or damaged drum brake backing plates, rotors, or calipers

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Elodia [21]

I think balance

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Marina CMI [18]

<u>Explanation:</u>

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2 years ago