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
Given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.
<h3>
What is Speed?</h3>
Speed is simply referred to as distance traveled per unit time.
Mathematically, Speed = Distance ÷ time.
Given the data in the question;
- Speed of the Ultraviolet light c = 3.0 × 10⁸m/s = 1.08 × 10⁹km/h
- Distance it must cover d = 4.0 × 10¹³km
We substitute our given values into the expression above.
Speed = Distance ÷ time
1.08 × 10⁹km/h = 4.0 × 10¹³km ÷ t
t = 4.0 × 10¹³km ÷ 1.08 × 10⁹km/h
t = 3.7 × 10⁴ hrs
Therefore, given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.
Learn more about speed here: brainly.com/question/7359669
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Answer:
D.-4.798m/s
Explanation:
Greetings !
Given values
Solve for V of the given expression
Firstly, recall the velocity-time equation
plug in known values to the equation
solve for final velocity
Hope it helps!
Answer:
Explanation:
which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator. For us:
so
a = .92 m/s/s (NOT negative because you're speeding up)