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
Answer:
0.002699 m or 2.699 mm
Explanation:
y = Fringe distance
d= Distance between slits = 0.310mm
L = Screen distance = 4.40m
λ= Wavelength
Given from question
λ₁= 660 nm = 6.6 x 10^-9 m
λ₂= 470 nm = 4.7 x 10^-9 m
d = 0.340 mm = 3.4 x 10^-3 m
L = 4.40 m
In the case of constructive interference, we use below formula
y/L = mλ/d
For first order wavelength
(y₁/4.40) =(1×660x10⁻⁹)/(0.310*10⁻³)
y₁= (0.310*10⁻³)×(4.40)/(0.310*10⁻³)
y₁=0.00937m
(y2/4.40) =(1×470x10⁻⁹)/(0.310*10⁻³)
y2= =(1×470x10⁻⁹)×(4.40)/(0.310*10⁻³)
y2=0.00667m
distance between the fringes is given by (y₁ -y2)
=0.00937-0.00667=0.002699m
Therefore, distance on the screen between the first-order bright fringes for the two wavelengths is 0.002699 m or 2.699 mm
Answer:
No, the magnitude of the magnetic field won't change.
Explanation:
The magnetic field produced by a wire with a constant current is circular and its flow is given by the right-hand rule. Since this field is circular with center on the wire the magnitude of the magnetic field around the wire will be given by B = [(\mi_0)*I]/(2\pi*r) where (\mi_0) is a constant, I is the current that goes through the conductor and r is the distance from the wire. If the field sensor will move around the wire with a fixed radius the distance from the wire won't change so the magnitude of the field won't change.
The relationship between electricity and magnetism is called<span> electromagnetism.</span>
Answer:
The probability of an incorrect report is found to be 0.03 or 3%.
Explanation:
We will get an incorrect report in both the cases of false alarm or missing excessive radiation. Since, both are mutually exclusive events. Therefore, the probability of both events to occur simultaneously will be 0. Thus, the probability of an incorrect report will be the sum of the probability of false alarm and the probability of a missing radiation.
P (False Alarm) = 0.02
P (Missing Radiation) = 0.01
P(Incorrect Report) = P (False Alarm) + P(Missing Radiation)
P (Incorrect Report) = 0.02 +0.01
P(Incorrect Report) = 0.03 = 3%