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
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
The image as shown here can here can be used to describe charging by induction.
<h3>What is a charge?</h3>
A charge may be positive or negative. One of the methods of transferring a charge is by induction.
In this case, an objects induces an opposite charge on a material. The image as shown here can here can be used to describe charging by induction.
Learn more about charging by induction:brainly.com/question/10254645
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Answer:


Explanation:
Given Data:
Numbers of times Telescope cycled around the earth in 6 years=37,000 times
Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km
Find:
Kilometers in one Orbit=?
Solution:
Kilometers in 37,000 Orbits=1,280,000,000 Km
Kilometers in 1 Orbit=1,280,000,000/37,000
In Scientific Notation:


Kilometers in 1 Orbit=34594.594 Km
Kilometers in 1 Orbit in Scientific notation:

Time t = ?
<span>When wave is moving from
y = 0 to y =12 cm</span>
By using the formula,
y = 15cos [(π/12) t)] =
0,
cos [(π/12) t)] = 0 =
cos (π/2), so,
(π/12)t = π/2,
t = (π/2) (12/π)
t = 12/2
<span>t = 6 sec</span>
<span>so 6 sec is the least amount of time required</span>