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 velocity of tennis racket after collision is 14.96m/s
<u>Explanation:</u>
Given-
Mass, m = 0.311kg
u1 = 30.3m/s
m2 = 0.057kg
u2 = 19.2m/s
Since m2 is moving in opposite direction, u2 = -19.2m/s
Velocity of m1 after collision = ?
Let the velocity of m1 after collision be v
After collision the momentum is conserved.
Therefore,
m1u1 - m2u2 = m1v1 + m2v2


Therefore, the velocity of tennis racket after collision is 14.96m/s
Answer:
fb = 240.35 Hz
Explanation:
In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.
Open tube:
(1)
vs: speed of sound = 343m/s
L: length of the open tube = 0.47328m
You replace in the equation (1):
Closed tube:

L': length of the closed tube = 0.702821m

Next, you use the following formula for the beat frequency:

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz
Answer:

Explanation:
As we know that the electrostatic force is a based upon inverse square law
so we have

now since it depends inverse on the square of the distance so we can say

now we know that


also we know that

now from above equation we have


