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A car air-conditioning unit has a 0.5-kg aluminum storage cylinder that is sealed with a valve. It contains 2 L of refrigerant R134a at 500 kPa, and both are at room temperature, 20C^{\circ} \mathrm{C}. It is now installed in a car sitting outside, where the whole system cools down to ambient temperature at −10C^{\circ} \mathrm{C}. What is the irreversibility of this process?


Answered 2 years ago
Answered 2 years ago
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We are given following data for air conditioner made of aluminum and contains R-134a:

T1=20 C=293 KT_{1}=20\text{ C}=293\text{ K}

T2=Tamb=10 C=263 KT_{2}=T_{amb}=-10\text{ C}=263\text{ K}

P=500 kPaP=500\text{ kPa}

mA=0.5 kgm_A=0.5\text{ kg}

VR=2 L=0.002 m3V_R=2\text{ L}=0.002\text{ m}^3

From Properties tables we can find Aluminum and R-134a specific heat:

CR=0.852 kJ kg KC_R=0.852\frac{\text{ kJ}}{\text{ kg K}}

CA=0.90 kJ kg KC_A=0.90\frac{\text{ kJ}}{\text{ kg K}}

ρR=4.20 kg m3\rho_R=4.20\frac{\text{ kg}}{\text{ m}^3}

Calculating R-134a mass:

mR=ρRVR=4.200.002=0.0084 kgm_R=\rho_R\cdot V_R=4.20\cdot 0.002=0.0084\text{ kg}

Heat transfer is equal to:

Q=mACA(T2T1)+mRCR(T2T1)=0.50.90(1020)+4.200.852(1020)=13.7 kJ\begin{align*} Q&=m_A\cdot C_A\cdot (T_2-T_1)+m_R\cdot C_R\cdot (T_2-T_1)\\\\ &=0.5\cdot 0.90\cdot (-10-20)+4.20\cdot 0.852\cdot (-10-20)\\\\ &=-13.7\text{ kJ} \end{align*}

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