## Related questions with answers

A two-stage compression refrigeration system operates with refrigerant-134a between the pressure limits of 1.4 and 0.10 MPa. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.4 MPa. The refrigerant leaving the low-pressure compressor at 0.4 MPa is also routed to the flash chamber. The vapor in the flash chamber is then compressed to the condenser pressure by the high-pressure compressor, and the liquid is throttled to the evaporator pressure. Assuming the refrigerant leaves the evaporator as saturated vapor and both compressors are isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the rate of heat removed from the refrigerated space for a mass flow rate of 0.25 kg/s through the condenser, and (c) the coefficient of performance.

Solution

VerifiedThe enthalpy and entropy at state 1 are determined from the given pressure $P_{1}=100\:\text{kPa}$ and the values for the saturated vapor state from A-12:

$\begin{align*} &h_{1}=234.46\:\dfrac{\text{kJ}}{\text{kg}}\\ &s_{1}=0.95191\:\dfrac{\text{kJ}}{\text{kg}\text{K}} \end{align*}$

The enthalpy at state 2 is determined from the pressure $P_{2}=400\:\text{kPa}$ and the condition $s_{2}=s_{1}$ with data from A-13 using interpolation:

$\begin{align*} h_{2}=262.74\:\dfrac{\text{kJ}}{\text{kg}} \end{align*}$

The enthalpy at state 3 is determined from the pressure $P_{3}=400\:\text{kPa}$ and the saturated vapor values from A-12:

$\begin{align*} h_{3}=255.61\:\dfrac{\text{kJ}}{\text{kg}} \end{align*}$

The enthalpy at 5 and 6 is determined from the pressure $P_{5}=1400\:\text{kPa}$ and the saturated liquid values from A-12:

$\begin{align*} h_{5}=h_{6}=127.25\:\dfrac{\text{kJ}}{\text{kg}} \end{align*}$

The enthalpy at 7 and 8 is determined from the pressure $P_{7}=400\:\text{kPa}$ and the saturated liquid values from A-12:

$\begin{align*} h_{7}=h_{8}=63.92\:\dfrac{\text{kJ}}{\text{kg}} \end{align*}$

$\textbf{\large Part A}$

$\text{The amount of evaporated refrigerant is simply the quality at state 6:}$

$\begin{align*} q&=\dfrac{h_{6}-h_{\text{liq, 400}}}{h_{\text{evap, 400}}}\\ &=\dfrac{127.25-63.92}{191.68}\\ &=\boxed{0.33} \end{align*}$

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