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.
3. 1 4 1 5 9
this is pi
Answer:
FOLLOWED BY... 2 6 5 3 5 8 9
UwU
Find R subscript C and R subscript B in the following circuit such that BJT would be in the active region with V subscript C E end subscript equals 5 V and I subscript C equals 25 m A V subscript C C end subscript equals 15 space V comma space V subscript D 0 end subscript equals 0.7 space V comma space beta equals 100 comma space V subscript A equals infinity.. Ignore the early effect in biasing calculations.
Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
The reversible and adiabatic process of a substance in a compressor begins with enthalpy equal to 1,350 kJ/kg, and ends with enthalpy equal to 3,412 kJ/kg. If the compressor efficiency is 0.85, find the actual specific work required by the compressor to operate, in kJ/kg.
Answer:
the actual specific work required by the compressor to operate is 2425.88 kJ/kg
Explanation:
Given that;
h₁ = 1350 kJ/kg
h₂₅ = 3412 kJ/kg
compressor efficiency П_ise = 0.85
we know that;
compressor efficiency П_ise = isentropic work / actual work
П_ise = (h₂₅ - h₁) / (h₂ - h₁ )
so
0.85 = (h₂₅ - h₁) / (actual work )
Actual work = (h₂₅ - h₁) / 0.85
Actual work = (3412 - 1350) / 0.85
Actual work = 2062 / 0.85
Actual work = 2425.88 kJ/kg
Therefore the actual specific work required by the compressor to operate is 2425.88 kJ/kg
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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According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.
Answer:
high school diploma
Explanation:
According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.
In other words, they do not seek to acquire a post-secondary school education.
Answer:
B
Explanation:
According to edge its answer B
associate's degree or on-the-job experience
got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.
Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
bachelor’s degree
associate's degree or on-the-job experience .......This is the correct answer.
some college, no degree
associate degree
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
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs. If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot fro the seesaw to be balanced.
Given :
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.
To Find :
If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.
Solution :
We know, for seesaw to balance :
[tex]m_1gd_1=m_2gd_2[/tex]
Here, [tex]d_1\ and \ d_2[/tex] is distance from origin from pivot point.
Putting all the values in the equation, we get :
[tex]50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet[/tex]
Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.
Hence, this is the required solution.
entor" by
What type of signal word is used in this sentence?
need and
en who was
generalization
description
thought
feeling
Answer:
generalization
Explanation:
Please mark me brainliest I need to level up
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 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
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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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
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
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 converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa.
Answer:
a) for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b) for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Explanation:
Given that;
A₂ = 0.001 m²
P₁ = 1 MPa
T₁ = 360 K
k = 1.4
P₂ = 500 Kpa
(1000/500)^(1.4-1 / 1.4) = 360 /T₂
2^(0.4/1.4) = 360/T₂
1.219 = 360 / T₂
T₂ = 360 / 1.219
T₂ = 295.32 K
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
we substitute
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
1.005 × 360 = 1.005 × 295.32 + v₂²/2000
v₂ = 360.56 m/s²
p₂v₂ = mRT₂
500 × (0.001 × 360.56) = m × 0.287 × 295.32
m = 2.127 kg/s
so Mach Number = V₂ / Vc
Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s
So Mach Number = V₂ / Vc = 360.56 / 344.47 = 1.046
Therefore for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b)
AT P₂ = 784 kPa
(1000/784)^(1.4-1 / 1.4) = 360/T₂
T₂ = 335.82 K
now
V₂²/2000 = 1.005( 360 - 335.82)
V₂ = 220.45 m/s
P₂V₂ = mRT₂
784 × (0.001 × 220.45) = m( 0.287) ( 335.82)
172.83 = 96.38 m
m = 172.83 / 96.38
m = 1.793 kg/s
just like in a)
Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s
Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6
Therefore for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Following are the
Given:
[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]
To find:
Flow rate of mass, and Mach number
Solution:
For point a)
Using formula:
[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]
[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]
Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]
[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]
[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]
For point b)
[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]
now
[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]
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Which of the following is a basic type of weld? O Groove O Lap O Edge O Corner
The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to carry a minimum load of 13 N applied evenly to the top of the span. The maximum load for each member is 4.5 Newtons in compression and tension. Each member is 4.2 cm long (ball center to ball center). All joints must be fully constrained. Determine the max value of P before the truss fails. The section that can hold the most weight will receive bonus points.
Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= [tex]\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12[/tex] = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
These are sites that allow you to upload and download media content such as images, audio, and video
using credit reduces future income
Answer:
lol
Explanation:r
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.
How many kg moles of Sodium Sulphate will contain 10 kg of
Sodium?
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
(as per my thinking)
Answer:
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
Technician A says that the B-pillars aid in resisting roof crush.
Technician B says that the IIHS rates vehicles to resist roof crush to five times the weight of the vehicle.
Who is right?
A only
O B only
Both A and B
Neither Anor B
Answer:
The answer is "Both A and B" are right
Explanation:
During the previous twenty years car producers have made significant advances in planning vehicle structures that give more noteworthy tenant insurance in planar accidents (Lund and Nolan 2003). Be that as it may, there has been little agreement with respect to the significance of rooftop strength in rollover crashes, just as the best strategy for surveying that strength. In 2006 one-fourth of lethally harmed traveler vehicle tenants were associated with crashes where vehicle rollover was considered the most hurtful occasion (Protection Establishment for Expressway Wellbeing, 2007). Numerous lethally harmed tenants in rollovers are unbelted, and some are totally or mostly launched out from the vehicle (Deutermann 2002).
There is difference concerning how underlying changes could influence launch hazard or the danger of injury for inhabitants who stay in the vehicle, paying little mind to belt use.
Answer:A Only
Explanation:
I Took the test
what is heat unit?in X ray machine
Answer:
joule
Explanation:
heat is express in joules in x Ray equipment
trevor moves a magnetic toy train away from a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement?
Answer:a
Ieieksdjd snsnsnsnsksks
2. What is the main job of a cylinder head?
OA. Contain the rapid increase in combustion chamber temperature
OB. Contain the rapid increase in combustion chamber pressure
OC. Prevent engine oil from getting past the pistons
OD. Hold the Head Gasket in place
Grade/Exit
Answer:
Explanation:
The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.
What is forging?Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.
Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.
The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.
Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.
Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.
Thus, the answer is cold forging.
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If a poems has a regular rhythm throughout the poem, it has: PLEASE HELP MEH I WILL GIVE YOU BRAINLEIST!!
A. tone
B. imagery
C. irony
D. meter
Answer:
D, meter.
Explanation:
Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.
If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.
What is meter in poem?Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.
The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.
Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.
Therefore, option D is correct.
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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].
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: