On a cold day a heat pump has a measured work input of 198 kW and receives heat

Question

On a cold day a heat pump has a measured work input of 198 kW and receives heat at a rate of 1184 kW from the surroundings in order to maintain a dwelling at 72 degree C. The coefficient of performance of the cycle is 85% of that of an equivalent reversible heat pump operating between the same two thermal reservoirs. Determine the temperature, in degree C, of the surroundings. 29.1 degree C A device called a desuperheater has 15 kg/s of steam entering at 20 bar, 320 degree C where it is mixed with liquid water at 20 bar, 210 degree C, 0.13 m3/min to produce saturated vapor at 20 bar. Determine The volumetric flow rate of the steam inlet stream, m3/min The mass flow rate of the liquid water inlet stream, kg/s The minimum diameter, in cm, of the exit pipe so that the exit velocity does not exceed 150 m/s. 117.72 m3/min, 1.85 kg/s, 11.9 cm Refrigerant-134a enters a diffuser operating at steady state as saturated vapor at 800 kPa with a velocity of 120 m/s, and it leaves at 900 kPa and 40 °degree C with a velocity of 60 m/s. The refrigerant is gaining heat at a rate of 2.5 kW as it passes through the diffuser. Determine the mass flow rate, in kg/s. How much larger than the inlet is the exit? 1.47 kg/s, 82.4% An insulated, vertical piston-cylinder device initially contains 10 kg of water, 6 kg of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of 200 kPa inside the cylinder. Now steam at 0.5 MPa and 360 degree C is allowed to enter the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine the mass, in kg, of the steam that has entered. 19.45 kg

Explanation / Answer

1) w in = 198 Kw

Q in = 1184 KW

so heat rejected = Q2 = Q in + w in = 1184 + 198 = 1382 KW

so, COP = Q2/Win = 1382/198 = 6.97

T1 = 72 C = (72 + 273) = 345 K

so, as per the question :

0.85 x T1/(T1 - T2) = 6.97

so, 0.85 x 345/( 345 - T2) = 6.97

so, T2 = 302.926 K = 29.2 C = temp of suurounding

2)

at 20 bar and 320 C, from stean tabel :

enthalpy = 3070.15 KJ/Kg

and specific volume = 0.131 m^3/Kg

a) voumetric flow rate = mass flow rate x specific volume

= 15 x 0.131 = 1.965 m^3/s = 1.965 x 60

= 117.72 m^3/min

b) at 20 bar and 210 C, enthalpy = 897.76 KJ/Kg

at 20 bar : enthalpy = hg at 20 bar = 2797 KJ/Kg

so, 15 x 3070.15 + m x897.76 = (15 + m2) x 2797

so, m2 = mass flow of water = 1.85 kg/s

c) at exit ,

mass = area x velocity /specific volume

so, 1.85= pi.4 x d^2 x 150 x (60/0.13)

so, diameter = d = 11.9 cm

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.