A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
transfer from well water at 9°C and discharges energy by heat transfer to the building at a rate of 120,000 kM/h. Over a period of 14 days, an electric meter records that 2000 kW · h of electricity is provided to the heat pump. Determine
a. the amount of energy that the heat pump recieves over the 14 dayperiod from the well water by ehat transer, in kJ.
b. the heat pump's coeffficient of performance
c. the coeffficient of performance of a reversible heat pump cycleoperating between hot and col reservoirs at 20°C and 10°C.
a. the amount of energy that the heat pump recieves over the 14 dayperiod from the well water by ehat transer, in kJ.
b. the heat pump's coeffficient of performance
c. the coeffficient of performance of a reversible heat pump cycleoperating between hot and col reservoirs at 20°C and 10°C.
1 answer:
Answer:
a) Ql=33120000 kJ
b) COP = 5.6
c) COPreversible= 29.3
Explanation:
a) of the attached figure we have:
HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:
W=Qh-Ql
Ql=Qh-W
where W=2000 kWh
Qh=120000 kJ/h
b) The coefficient of performance is:
c) The coefficient of performance of a reversible heat pump is:
Th=20+273=293 K
Tl=10+273=283K
Replacing:
You might be interested in
- Computer systems organization.
- Computing methodologies.
- General and reference.
- Networks.
- Software and its engineering.
- Theory of computation.
Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as
Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is
Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:
So we equate the COP of our heater with COP of Carnot heater
Rearrange the equation
Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C
Answer:
Hello some parts of your question is missing below is the missing part
Convection coefficient = 11 w/m^2. °c
answer : 44.83 watts
Explanation:
Given data :
surface emissivity ( ε )= 0.95
head ( sphere) diameter( D ) = 0.25 m
Temperature of sphere( T ) = 35° C
Temperature of surrounding ( T∞ ) = 25°C
Temperature of surrounding surface ( Ts ) = 15°C
б = ( 5.67 * 10^-8 )
Determine the total rate of heat loss
First we calculate the surface area of the sphere
As =
= = 0.2 m^2
next we calculate heat loss due to radiation
Qrad = ε * б * As( ) ---- ( 1 )
where ;
ε = 0.95
б = ( 5.67 * 10^-8 )
As = 0.2 m^2
T = 35 + 273 = 308 k
Ts = 15 + 273 = 288 k
input values into equation 1
Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )
= 22.83 watts
Qrad ( heat loss due to radiation ) = 22.83 watts
calculate the heat loss due to convection
Qconv = h* As ( ΔT )
= 11*0.2 ( 35 -25 ) = 22 watts
Hence total rate of heat loss
= 22 + 22.83
= 44.83 watts
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
Integrating both sides we get
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
Thus the remaining 125 meters will be covered with a constant speed of
in time equalling
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds