Answer:
See explanation
Explanation:
Notice that the condenser section includes both the hot water and space heater and station (3) is specified as being in the Quality region. Assume that 50°C is a reasonable maximum hot water temperature for home usage, thus at a high pressure of 1.6 MPa, the maximum power available for hot water heating will occur when the refrigerant at station (3) reaches the saturated liquid state. (Quick Quiz: justify this statement). Assume also that the refrigerant at station (4) reaches a subcooled liquid temperature of 20°C while heating the air.
Using the conditions shown on the diagram and assuming that station (3) is at the saturated liquid state
a) On the P-h diagram provided below carefully plot the five processes of the heat pump together with the following constant temperature lines: 50°C (hot water), 13°C (ground loop), and -10°C (outside air temperature)
b) Using the R134a property tables determine the enthalpies at all five stations and verify and indicate their values on the P-h diagram.
c) Determine the mass flow rate of the refrigerant R134a. [0.0127 kg/s]
d) Determine the power absorbed by the hot water heater [2.0 kW] and that absorbed by the space heater [0.72 kW].
e) Determine the time taken for 100 liters of water at an initial temperature of 20°C to reach the required hot water temperature of 50°C [105 minutes].
f) Determine the Coefficient of Performance of the hot water heater [COPHW = 4.0] (defined as the heat absorbed by the hot water divided by the work done on the compressor)
g) Determine the Coefficient of Performance of the heat pump [COPHP = 5.4] (defined as the total heat rejected by the refrigerant in the hot water and space heaters divided by the work done on the compressor)
h) What changes would be required of the system parameters if no geothermal water loop was used, and the evaporator was required to absorb its heat from the outside air at -10°C. Discuss the advantages of the geothermal heat pump system over other means of space and water heating
The answer is A.68 degrees
Answer:
14.2 m/s
Explanation:
Given data:
Speed of the stream, v₁ = 7.1 m/s
let the cross section area at initial point be A₁
now area at the second point, A₂ = (1/2)A₁ = 0.5A₁
now, from the continuity equation, we have
A₁v₁ = A₂v₂
where, v₂ is the velocity at the narrowed portion
thus, on substituting the values, we get
A₁ × 7.1 = 0.5A₁ × v₂
or
v₂ = 14.2 m/s
I would assume it is Oxidation since it would have an impact on metals t hat oxidizes in the weather
Answer:
Sound and light are similar in that both are forms of energy that travel in waves. They both have properties of wavelength, freqency and amplitude. Here are some differences: Sound can only travel through a medium (substance) while light can travel through empty space.