Answer to Question #144953 in Molecular Physics | Thermodynamics for SUSHIL SHENDE

Question

The compressor of a heat pump that is used to maintain the temperature of a space at 298 K consumes 1600 J/s power. The heat pump absorbs heat from hot water flowing into the evaporator at the rate of 3.9 kg/min. The temperature of water at the evaporator inlet and outlet are 333 K and 40°C, respectively. The refrigerant used in the heat pump is R134a, it enters the evaporate with a temperature of 285 K with 15% quality and exits as a saturated vapour. Find (a) refrigerant mass flow rate (b) rate of heat supplied to the space (c) the COP (d) the the minimum power input to the compressor for the same rate of heat supply.

Answer 1

As per the given question,

Temperature of the space $(T)=298K$

Power of heat engine $(P)=1600 J/s$

Flow of vapor=3.9kg/min

Inlet temperature $(T_i)=333K$

Outlet temperature $(T_o)=40^\circ C=(273+40)=313K$

Entering temperature of evaporate = 285K

Quality =15%

Specific heat of saturated vapor is represented by $h_1$ and specific heat of compressed vapor is represented by $h_2$

a)$\dfrac{dW}{dt}=\frac{dm}{dt}(h_2-h_1)\times 0.15$

Now, substituting the values in the above equation,

$\dfrac{dm}{dt}=\frac{1600}{0.15(423.9-392.7)}=341.88 kg/s$

b)$\frac{dQ}{dt}= \frac{3.9kg}{60s}(392.7-241.7)=9.81J/s$

c)$E_f=1-\frac{T_c}{T_h}$

$E_f=1-\frac{313}{333}=\frac{20}{333}$

$\Rightarrow COP= \frac{1}{E_f}$

$\Rightarrow COP = \frac{333}{20}=16.65$