Answer:
generalization
Explanation:
Please mark me brainliest I need to level up
A heat pump is to be used to heat a house during the winter, as shown in Fig. 6–52. The house is to be maintained at 21 ℃ at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5 ℃. Determine the minimum power required to drive this heat pump.
Answer:
[tex]3.32\ \text{kW}[/tex]
Explanation:
[tex]T_c[/tex] = Outside temperature = [tex]-5^{\circ}\text{C}[/tex]
[tex]T_h[/tex] = Temperature of room = [tex]21^{\circ}\text{C}[/tex]
[tex]Q_h[/tex] = Heat loss = 135000 kJ/h = [tex]\dfrac{135000}{3600}=37.5\ \text{kW}[/tex]
Coefficient of performance of heat pump
[tex]\text{COP}=\dfrac{1}{1-\dfrac{T_c}{T_h}}\\\Rightarrow \text{COP}=\dfrac{1}{1-\dfrac{273.15-5}{273.15+21}}\\\Rightarrow \text{COP}=11.3[/tex]
Input power
[tex]W_i=\dfrac{Q_h}{\text{COP}}\\\Rightarrow W_i=\dfrac{37.5}{11.3}\\\Rightarrow W_i=3.32\ \text{kW}[/tex]
The minimum power required to drive this heat pump is [tex]3.32\ \text{kW}[/tex].
These are sites that allow you to upload and download media content such as images, audio, and video
The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect the total quantity of bricks used in the wall
True or False
Answer:
false
Explanation:
3. 1 4 1 5 9
this is pi
Answer:
FOLLOWED BY... 2 6 5 3 5 8 9
UwU
What is the name given to the vehicles that warn motorists about oversized loads/vehicles?
a) Pilot Car
b) Advanced Car
c) Trail Car
d) Leader Car
The car that is used to warn drivers about oversized loads is the Pilot Car.
What is a Pilot Car?The Pilot Car is also called Escort Car. It is a vehicle used to warn other vehicles of the presence of an over-sized vehicle.
The role of Pilot vehicle operators is to warn road users (motorists) to be cautious of over-sized loads or vehicles.
The cars are used to guide motorists that are making use of roads in construction sites.
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Which tool is used for cutting bricks and other masonry materials with precision?
Answer:
A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.
Explanation:
Answer:
Masonry Saw
Explanation:
Masonry Saws can be used to cut brick, ceramic, tile and or stone.
A structural component is fabricated from an alloy that has a plane strain fracture toughness of 45 MPa.m1/2. It has been determined that this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane-strain fracture toughness of 57.5 MPa.m1/2
Answer:
1.5510 mm
Explanation:
Plane strain fracture toughness = 45 MPa√m
failing stress ( б ) = 300 MPa
maximum length of surface crack ( a )= 0.95 mm
Determine maximum allowable surface crack length ( in mm )
we will make use of this relationship for Design stress equation to determine the value of Y
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 1 )
k = 45 MPa√m
б = 300 MPa
a = 0.95 mm
y = ?
From equation 1 make Y subject of the equation ( also substitute values into equation 1 above )
hence ; y = 2.7457
Now determine maximum allowable surface crack when component is exposed to a stress of 300 MPa and made from another alloy with plane-strain fracture toughness of 57.5 MPa√m
we will apply the equation
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 2 )
K = 57.5 MPa√m
б = 300 MPa
y = 2.7457
a ( maximum allowable surface crack ) = ?
from equation make a subject of the equation
a = [tex]\frac{1}{\pi } (\frac{k}{\alpha y} )^{2}[/tex]
a = [tex]\frac{1}{\pi } (\frac{57.5}{300*2.7457} )^2[/tex] = 1.5510 mm
Which of the following is a disadvantage of using a resistor in place of an inductor in a power-supply filter.
A The output DC voltage will be lower
B. The life of the capacitors will be shorter
C. A resistor will cost much more than an inductor
D. A resistor weighs much more than an inductor
Answer: A. The output DC voltage will be lower
Explanation:
Using a resistor in a power-supply filter instead of an inductor will lead to a lower DC voltage output as resistors reduce voltage.
It would therefore be ideal to use an inductor as it does not lower DC voltage but inductors are expensive and can be quite large which is why it is more common to see resistors used in power-supply filter circuits.
During her soccer game, Brittany hears her coach tell her team to look for better shot selection. The offensive strategy her coach is attempting to get Brittany and her teammates to use more is shooting when the:
Answer:
B: Team has the ball near center line
Explanation:
i guessed so...
A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a velocity of 0.2 m/s .A jet of water exits at 20 C, 1 atm, with a velocity of 20 m/s an elevation of 5 m. At steady state, the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input. Determine the power input to the motor, kW.
A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.
Answer:
Net power of 1.2 KW is being extracted
Explanation:
We are given;
Mass flow rate; m' = 0.1kg/s
Inlet temperature; T1 = 20°C = 293K
Inlet pressure; P1 = 1 atm = 10^(5) pa
Inlet velocity; v1 = 0.2 m/s
Exit Pressure; P2 = 1 atm = 10^(5) pa
Exit Temperature; T2 = 1 atm = 296K
Exit velocity; V2 = 20m/s
Change in elevation; h = Z2 - Z1 = 5m
We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.
Thus;
Q = -0.1W
From Bernoulli equation;
Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy
Where;
∆Potential energy = mg(z2 - z1)
∆Kinetic energy = ½m(v2² - v1²)
∆Pressure energy = mc_p(T2 - T1)
Thus;
-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]
Where C_p is specific heat capacity of water = 4200 J/Kg.k
Plugging in the relevant values, we have;
-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))
-1.1W = 4.905 + 19.998 + 1260
-1.1W = 1284.903
W = -1284.903/1.1
W ≈ -1168 J/s ≈ -1.2 KW
The negative sign means that work is extracted from the system.
Waste that is generated by a business is called a _____________.
A) Waste stream
B) Surplus
C) Hazard assessment
D) Trash stream
(This is for my Automotive class)
Answer:
waste stream
Explanation:
i got it right on sp2
Problem 10.012 SI A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at 2.6 bar, and saturated liquid exits the condenser at 12 bar. The isentropic compressor efficiency is 80%. The mass flow rate of refrigerant is 7 kg/min. Determine: (a) the compressor power, in kW. (b) the refrigeration capacity, in tons. (c) the coefficient of performance.
Answer:
a) 4.1 kw
b) 4.68 tons
c) 4.02
Explanation:
Saturated vapor enters compressor at ( p1 ) = 2.6 bar
Saturated liquid exits the condenser at ( p2 ) = 12 bar
Isentropic compressor efficiency = 80%
Mass flow rate = 7 kg/min
A) Determine compressor power in KW
compressor power = m ( h2 - h1 )
= 7 / 60 ( 283.71 - 248.545 )
= 4.1 kw
B) Determine refrigeration capacity in tons = m ( h1 - h4 )
= 7/60 ( 248.545 - 107.34 )
= 16.47 kw = 4.68 tons
C) coefficient of performance ( COP )
= Refrigeration capacity / compressor power
= 16.47 / 4.1 = 4.02
Attached below is the beginning part of the solution
determine the values of the viscous damping coefficient c for which the system has a damping ratio of 0.5 and 1.5
Answer:
7.47 lb. s/ft
22.42 lb. s/ft
Explanation:
Without mincing words let us dive straight into the solution to the above problem.
STEP ONE: calculate or determine the frequency.
The frequency can be calculated by making use of the formula given below;
frequency, w = √k/m. Thus, frequency = √ 12 × 15/ [40/32.2] = 12 rad/s.
STEP TWO: Determine or calculate for the viscous damping coefficient, c for damping ratio of 0.5 and 1.5 respectively.
The viscous damping coefficient, c for 0.5 = [ 12 × 0.5 × 2 ×[ 40/32.2] / 2= 7.47 lb. s/ft.
The viscous damping coefficient, c for 1.5= [ 12 × 1.5 × 2 ×[ 40/32.2] / 2 = 22.42 lb. s/ft.
An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into the tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45 mm of water below atmospheric. Assume that the air is incompressible, and at 25 C, 100 kPa absolute. Calculate the air speed in the wind-tunnel test section
Answer:
Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s
Explanation:
Given data :
Manometer reading ; p1 - p2 = 45 mm of water
Pressure at section ( I ) p1 = 100 kPa ( abs )
temperature ( T1 ) = 25°C
Pw ( density of water ) = 999 kg/m3
g = 9.81 m/s^2
next we apply Bernoulli equation at section 1 and section 2
p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex] ---------- ( 1 )
considering ideal gas equation
Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )
R ( constant ) = 287 NM/kg.k
T = 25 + 273.15 = 298.15 k
P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2
substitute values into equation ( 2 )
= 100 * 10^3 / (287 * 298.15)
= 1.17 kg/m^3
Also note ; p1 - p2 = PwgΔh ------- ( 3 )
finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )
[tex]\frac{PairV^{2} _{2} }{2}[/tex] = PwgΔh
[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex] = 753.86
[tex]v_{2}[/tex] = 27.5 m/s
Thermoplastic parts are
A.commonly used for outer mirror housings.
B. commonly used for grilles
C.formed by stamping a plastic sheet in a mold
D.formed by forcing a molten solution into a mold.
Answer:
c
Explanation:
Thermoplastic parts are formed by forcing a molten solution into a mold.
Thus option D is correct.
Here,
Thermoplastic are plastic parts made from thermoplastic materials. These materials have the ability to be melted and remolded several times without undergoing any chemical change.
The thermoplastic parts are commonly used for different purposes such as in automotive industries, construction, medical, consumer goods, and much more. These parts are easily moldable and can be made into different shapes and sizes. They are also lightweight, strong, and durable, making them ideal for a wide range of applications.
They are commonly used for applications that require high strength and durability, such as in the automotive and aerospace industries.
Therefore option D is correct.
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