Question

A heat pump with refrigerant-134a as the working fluid is used to keep a space at 22 C by absorbing hear from a geothermal water that enters the evaporator at 60 C at a rate of 0.065 kg/s and leaves at 40 C. Refrigerant enters the evaporator at 60 C at a rate of 0.065 kg/s and leaves at 40 C. Refrigerant enters the evaporator at 12 C with a quality of 15 percent and leaves at the same pressure as a saturated vapor. If the compressor consumes 1.6 kW of power, Determine (a) the mass flow rate of the heat supply, (b) the rate of heat supply, (c) the actual and ideal COP of the system, and (d) the minimum power input to the compressor for the same rate of heat supply.

Answer 1

A) m'r = 0.0338 kg/s

B) Q'_H = 7.033 KW

C) COP = 4.4

D) W'_min = 0.8026 KW

Step-by-step explanation:

A) from energy balance, we have;

Q'_w = Q'r

This, when expanded gives;

m'w•c_p•ΔT_w = m'r(h2 - h1)

Where;

m'w is mass flow rate of water = 0.065 kg/s

c_p is specific heat of water = 4.18 KJ/Kg.k

ΔT_w = change in temperature = (60 + 273)K - (40 + 273)K = 20 K

m'r is mass flow rate of refrigerant

h2 is gotten from the table attached at 12°C, hg = h2 = 257.27 KJ/Kg

h1 = h_liq + q•h_evap. So, from the table attached at 12°C, h_f = h_liq = 68.18 KJ/Kg and h_fg = h_evap = 189.09 KJ/Kg. We are given quality(q) = 15% = 0.15

Thus, h1 = 68.18 + 0.15(189.09) = 96.54 KJ/Kg

So, we recall that, m'w•c_p•ΔT_w = m'r(h2 - h1)

Let's make m'r the subject;

m'r = m'w•c_p•ΔT_w/(h2 - h1)

Plugging in the relevant values to obtain;

m'r = 0.065 * 4.18 * 20/(257.27 - 96.54)

m'r = 0.0338 kg/s

B) The rate of heat supply is gotten from the formula;

Q'_H = Q'_L + W

Where;

Q'_L = m'r(h2 - h1)

W is power consumed = 1.6KW

Thus;

Q'_H = m'r(h2 - h1) + W

Plugging in the relevant values to obtain ;

Q'_H = 0.0338(257.27 - 96.54) + 1.6

Q'_H = 7.033 KW

C) formula for COP is;

COP = Q'_H/W

Thus, COP = 7.033/1.6

COP = 4.4

D) The minimum power is gotten from;

W'_min = Q'_H/COP_max

Now, COP_max is given by the expression;

COP_max = 1/[1 - (T_L/T_H)]

T_L = 22°C = 22 + 273K = 295K

T_H = 60°C = 60 + 273K = 333K

Thus;

COP_max = 1/[1 - (295/333)]

COP_max = 8.763

Thus,W'_min = 7.033/8.763

W'_min = 0.8026 KW