Answer:
[tex]T=92.16 \°C[/tex]
[tex]y_{et}=0.433\\\\y_w=0.567[/tex]
Explanation:
Hello,
In this case, the Raoult's law for this problem is:
[tex]y_{et}P=x_{et}P_{et}^{sat}\\\\y_{w}P=x_{w}P_{w}^{sat}[/tex]
Which can be written as:
[tex]P=x_{et}P_{et}^{sat}+x_{w}P_{w}^{sat}[/tex]
Thus, by using the Antoine equation, we can symbolically represent the the temperature at which such mixture boil:
[tex]1atm=0.2608*10^{(8.13484-\frac{ 1662.48}{238.131+T})}/760+0.7392*10^{(5.40221-\frac{1838.675}{-31.737+T-273.15})}/1.01325[/tex]
The solution, by numerical iteration process (there is not way to solve it analytically) is 92.16 °C considering the data extracted from NIST database. Next, vapor fractions are:
[tex]y_{et}=x_{et}*10^{(8.13484-\frac{ 1662.48}{238.131+T})}/760/P\\\\y_{et}=0.2608*10^{(8.13484-\frac{ 1662.48}{238.131+92.16})}/760/1atm\\\\y_{et}=0.433\\\\y_w=1-y_{et}=1-0.433\\\\y_w=0.567[/tex]
Regards.
Two technicians are discussing a starter that clicks once when turning the key to the crank position. Technician A says that it could be caused by worn solenoid contacts. Technician B says it could be caused by an open ignition switch. Who is correct?
A. Technician A
B. Technician B
C. Both Technicians A and B
D. Neither Technician A nor B
Answer:
Technician A is correct.
Two technicians are discussing a starter that clicks once when turning the key to the crank position. Technician A is correct. The correct option is A.
What is a starter?An electromagnet known as a starter solenoid is activated to turn on an internal combustion engine's starter motor. The starter solenoid then connects two contacts or metal points as the key is turned.
The starter motor receives electrical currents from the ignition through the solenoid in this way. This sets off a series of electrical reactions that ultimately start the combustion engine. The multifunctional crankshaft position sensor regulates ignition timing, measures RPM, and determines relative engine speed.
The solenoid is likely defective, and the starter may sustain serious damage as a result, if the starter engages but does not disengage when you release the key.
Therefore, the correct option is A. Technician A.
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Which of the following is NOT a part of an "I" statement?
a. Draw the logic gate implementation for the Boolean function F = W X(Y + Z).b. Write the DeMorgan equivalent Boolean statement in part (a) and draw its logic gate schematic.
Answer:
DeMorgan equivalent :
F = B + C
F' = ⁻B⁻+⁻C⁻ = ⁻BC⁻
Explanation:
Attached below is the logic gate implementation diagram and the DeMorgan equivalent Boolean statement as requested in part A and B
By balancing information security and access, a completely secure information system can be created.A. TrueB. False
Answer: true
Explanation:
Name four major categories of automotive parts and systems.
Answer:
The major systems of an automobile are the engine, fuel system, exhaust system, cooling system, lubrication system, electrical system, transmission, and the chassis. The chassis includes the wheels and tires, the brakes, the suspension system, and the body.
Explanation:
The Stokes-Oseen formula for drag force F on a sphere of diameter D in a fluid stream of low velocity V, density p and viscosity μ is
F=3πμDV+9π/16∗pV2d2
Is this formula dimensionally homogenous?
Answer:
[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]
Hence it is proved that Stokes-Oseen formula is dimensionally homogenous.
Explanation:
For equation to be dimensionally homogeneous both side of the equation must have same dimensions.
For given Equation:
F= Force, μ= viscosity, D = Diameter, V = velocity, ρ= Density
Dimensions:
[tex]F=\frac{ML}{T^2}[/tex]
[tex]\mu=\frac{M}{LT}[/tex]
[tex]D=L\\\\V=\frac{L}{T}\\ \\\rho=\frac{M}{L^3}[/tex]
Constants= 1
Now According to equation:
[tex]\frac{ML}{T^2}=[\frac{M}{LT}][L] [\frac{L}{T}] + [\frac{M}{L^3}][\frac{L^2}{T^2}][L^2][/tex]
Simplifying above equation, we will get:
[tex]\frac{ML}{T^2}=2*\frac{ML}{T^2}[/tex]
Ignore "2" as it is constant with no dimensions. Now:
[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]
Hence it is proved that Stokes-Oseen formula is dimensionally homogenous.
what are PAT&E tests on production sysems used for.
Answer:
Portable appliance testing (PAT) is the term used to describe the examination of electrical appliances and equipment to ensure they are safe to use. Most electrical safety defects can be found by visual examination but some types of defect can only be found by testing.
please give brainlist
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( ꈍᴗꈍ)
some connecting rods have ____ to help lubricate the cylinder wall or piston pin.
Answer:
some connecting rods have spit holes
Some connecting rods have crankshaft to help lubricate the cylinder wall or piston pin.
What are crankshaft?A crankshaft is a mechanical component that converts reciprocating motion in a piston engine into rotational motion.
The crankshaft is a rotating shaft with one or more crankpins that drive the pistons via the connecting rods.
Reduced fuel economy is one issue you'll face if your car's crankshaft isn't working properly. The reason for this is that your vehicle's fuel injectors will not efficiently direct gas to the engine.
The connecting rods' big ends are lubricated with oil that travels up an oilway drilled through the crankshaft.
From the main bearings to the big end bearings, oil is pushed. It will lubricate the cylinder walls in some engines by passing through a passage drilled in the connecting rod.
Thus, the answer is crankshaft.
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MATLAB can solve a variety of engineering problems including those requiring simulating differential equations and iterative numerical calculations.A. TrueB. False
Answer:
A. True
Explanation:
MATLAB may be defined as a programming platform that is designed specifically for the engineers as well as the scientists to carry out different analysis and researches.
MATLAB makes use of a desktop environment which is tuned for certain iterative analysis and the design processes with a programming language which expresses matrix as well as array mathematics directly.
Thus the answer is true.
Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K. The constant B = 3. 56 times 1014 9cm -3 K-3/2) and the bandgap voltage Eg = 1. 42 eV.
This question is incomplete, the complete question is;
Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K.
The constant B = 3.56×10¹⁴ (cm⁻³ K^-3/2) and the bandgap voltage E = 1.42eV.
Answer: the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³
Explanation:
Given that;
T = 300k
B = 3.56×10¹⁴ (cm⁻³ K^-3/2)
Eg = 1.42 eV
we know that, the value of Boltzmann constant k = 8.617×10⁻⁵ eV/K
so to find the ni for gallium arsenide;
ni = B×T^(3/2) e^ ( -Eg/2kT)
we substitute
ni = (3.56×10¹⁴)(300^3/2) e^ ( -1.42 / (2× 8.617×10⁻⁵ 300))
ni = (3.56×10¹⁴)(5196.1524)e^-27.4651
ni = (3.56×10¹⁴)(5196.1524)(1.1805×10⁻¹²)
ni = 2.1837 × 10⁶ cm⁻³
Therefore the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³
A construction manager is looking for workers to build a series of walkways within a large garden apartment project. What type of
workmen should the manager employ?
Answer:
Casual workers
Explanation:
Casual workers are workers on a short-term employment contract that unlike non-permanent workers have restricted entitlement to benefits, little employment security, with the primary distinction of there being a lack of a stable connection or relationship between them and their employers for the to be referred to as employees.
In mechanics of materials, the bending stress of a beam in bending can be determined by the equation σ = MyIwhere expressed in terms of SI base units M is the bending moment in Newton-meters (N-m), y is the distance from the neutral axis in meters (m), and I is the moment of inertia in meters to the fourth power (m4). The bending stress σ has the same units as those of:_____.a) pressure.b) density.c) force.d) spring constant.
Answer:
Explanation:
In mechanics of materials, the bending stress of a beam in bending can be determined by the equation
When choosing a respirator for your job, you must conduct a _____ test.
A) Fit B) Practice C) Breathing D) Weight?
Answer:
Fit test
Explanation:
it's on the automotive technology sight : )
what type of comptuer was name for the cabinet where the brains
Answer: Central Processing Unit
The steering column connects the steering wheel to the steering gear.
a. True
b. False
A reversible refrigerator operates between a low temperature reservoir at TL and a high temperature reservoir at TH . Its coefficient of performance is given by
Answer
TL/TH- TL
Because we know that power coefficient is. = QL/QH-QL
=so using this for performance we have
=>Perf= TL/(TH-TL)
2.24 Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state (a cycle). Process 1-2: Expansion from state 1 where p1 = 10 bar, V1 = 1 m3, to state 2 where V2 = 4 m3. During the process, pressure and pV1.5 volume are related by = constant. Process 2-3: Constant volume heating to state 3 where p3 = 10 bar. Process 3-1: Constant pressure compression to state 1. Sketch the processes on p –V coordinates and evaluate the work for each process, in kJ. What is the net work for the cycle, in kJ?
Answer:
Explanation:
Given that:
From process 1 → 2
[tex]P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3[/tex]
[tex]PV^{1.5} = \ constant[/tex]
[tex]\gamma = 1.5[/tex]
Process 2 → 3
The volume is constant i.e [tex]V_2 =V_3 = 4m^3[/tex]
[tex]P_3 = 10 \ bar[/tex]
Process 3 → 1
P = constant i.e the compression from state 1
Now, to start with 1 → 2
[tex]P_1V_1^{1.5} = P_2V_2^{1.5}[/tex]
[tex]P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}[/tex]
[tex]P_2 = 10 \times (\dfrac{1}{4})^{1.5}[/tex]
[tex]P_2 =1.25[/tex]
The work-done for the process 1 → 2 through adiabatic expansion is:
[tex]W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1][/tex]
We know that 1 bar = [tex]10^5 \ N/m^2[/tex]
∴
[tex]W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1][/tex]
[tex]W =1000000 \ J[/tex]
[tex]W_{1 \to 2} = 1000 kJ[/tex]
For process 2 → 3
Since V is constant
Thus:
W = PΔV = 0
[tex]W_{2 \to 3} = 0[/tex]
For process 3 → 1
W = PΔV
[tex]W _{3 \to 1} = P_3(V_1-V_3)[/tex]
[tex]W _{3 \to 1} = 10 \times 10^5 (1-4)[/tex]
[tex]W _{3 \to 1} = 10 \times 10^5 (-3)[/tex]
[tex]W _{3 \to 1} = -3 \times 10^6 \ J[/tex]
[tex]W _{3 \to 1} = -3000 \ kJ[/tex]
The net work-done now for the entire system is :
[tex]W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }[/tex]
[tex]W_{net} = (1000 + 0 + (-3000)) \ kJ[/tex]
[tex]W_{net} =-2000 \ kJ[/tex]
The sketch of the processes on p -V coordinates can be found in the image attached below.
A) The work done for each process are :
Process (1 - 2) = 1000 kJ Process (2 - 3) = 0 kJ process (3 - 1) = -3000 kJB) The net work for the cycle = -2000 kJ
Given Data :
For process (1 -2) For process ( 2 - 3 ) process ( 3 - 1 )
P₁ = 10 bar P₃ = 10 bar constant pressure compression
V₁ = 1 m³ constant volume heating
V₂ = 4 m³
PV[tex]^{1.5}[/tex] = constant
A) Determine work done for each process
Calculate work done for process (1 - 2)
W₁ ₋ ₂ = [tex]\frac{P_{1}V_{1} - P_{2}V_{2} }{n -1 }[/tex] * 100
= [ ( 10*1 ) - ( 1.25 * 4 ) ] / 1.5 - 1
= [ 10 - 5 ] / 0.5
= 10 * 100 = 1000 kJ
Calculate work done for process ( 2-3 )
given that there is constant volume heating
W₂₋₃ = 0 kJ
Calculate work done for process ( 3-1)
W₃₋₁ = P ( Δ V ) given that p = constant
= 10 * 100 ( -3 )
= - 3000 kJ
B) The net work for the cycle
W₁ ₋ ₂ + W₂₋₃ + W₃₋₁
= 1000 kJ + 0 kJ + - 3000 kJ
= - 2000 kJ
Hence we can conclude that the ) The work done for each process are :
Process (1 - 2) = 1000 kJ Process (2 - 3) = 0 kJ process (3 - 1) = -3000 kJand The net work for the cycle = -2000 kJ
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Attached below is the P-V sketch of the process
2.1 What is the minimum number of pins required for a so-called dual-op-amp IC package, one containing two op amps? What is the number of pins required for a so-called quad-op-amp package, one containing four op amps?
Answer:
8 for dual-op-amp package, and 14 for quad-op-amp
Explanation;
This is because every op-amp has 2 input terminal 4 pns
So one output terminal that is 2 pins which are required for power
and the same for a minumum number of pins required by quad op amp which is 14
Two technicians are discussing the a series-parallel brake lamp circuit that does not illuminate. Technician A states that an open in one of the parallel branches could be the cause. Technician B states that a shorted (or stuck closed) brake lamp switch wired in series could be the cause. Which technician is correct? Group of answer choices
Answer:
Technician B is correct.
Explanation:
Two technicians are discussing the a series-parallel brake lamp circuit that does not illuminate.
The statement made by Technician A saying that an open in one of the parallel branches could be the cause has no ground because an opening in any branches has not to do with the issue since there is no disconnection The statement made by Technician B saying that a shorted (or stuck closed) brake lamp switch wired in series could be the cause is correct.
The correct area in sq. Inches and sq. Feet is: Select one: a. 966.76 sq. Inches and 8.056 sq. Feet b. 96.676 sq. Inches and 8.056 sq. Feet c. 96.676 sq. Inches and 0.671 sq. Feet
Answer:
c. 96.676 sq. Inches and 0.671 sq. Feet
Explanation:
From the list of the given option, we are told to chose the correct area in sq. inches that correspond to sq. Feet.
If we recall from the knowledge of our conversion table that,
1 sq feet = 144 sq inches
Then, let's confirm if the option were true.
a. 966.76 sq. Inches and 8.056 sq. Feet
Assuming
if 1 sq feet = 8.056
in sq inches, we have ( 8.056 × 144 ) sq inches
= 1160.064 sq. inches
So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 1 is wrong
b. 96.676 sq. Inches and 8.056 sq. Feet
if 1 sq feet = 8.056
in sq inches, we have ( 8.056 × 144 ) sq inches
= 1160.064 sq. inches
So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 2 is wrong/
c. 96.676 sq. Inches and 0.671 sq. Feet
if 1 sq feet = 0.671
in sq inches, we have ( 0.671 × 144 ) sq inches
= 96.624 sq. Inches which is closely equal to 96.676 sq. Inches
Therefore, this is the correct answer as it proves that 96.676 sq. Inches = 0.671 sq. Feet
1. Gas Metal Arc Welding is also known as
welding.
Answer: metal inert gas (MIG) welding
Explanation:
Gas metal arc welding (GMAW), also called metal inert gas (MIG) welding, is an arc welding process in which the heat for melting the metal is generated by an electric arc between a consumable electrode and the metal (Fig.
A rocket is shot straight up from the earth with a net acceleration (= acceleration by the rocket engine - gravitational pullback) of 7m/sec during the initial stage of flight until the engine cut out at t = 10 sec. How high will it go, air resistance neglected?
Answer:
599.7 m
approximately 600 m
Explanation:
initial speed of the rocket = 0
net acceleration upwards = 7 m/s^2
the engine cuts out 10 sec after take off
maximum height reached = ?
we neglect air resistance
To get the velocity of the rocket at the point where the engine cuts off, we use the equation
v = u + at
where
v is the velocity at this point where the engine stops = ?
u is the initial velocity of the rocket from rest = 0 m/s
a is the net acceleration upwards = 7 m/s^2
t is the time the engine runs = 10 s
substituting, we have
v = 0 + (7 x 10)
v = 70 m/s
to get the distance from the ground to this point, we use the equation
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
where
v is the final velocity at the the height where the engine is cut out = 70 m/s
u is the initial speed at the ground = 0 m/s
a is the net acceleration on the rocket = 7 m/s^2
s is the distance from the ground to this point
substituting, we have
[tex]70^{2}[/tex] = [tex]0^{2}[/tex] + 2(7 x s)
4900 = 14s
s = 4900/14 = 350 m
After this point when the engine cuts out, the rocket experiences an acceleration proportional to the acceleration due to gravity 9.81 m/s^2 downwards, and slows down gradually before coming to a stop at the maximum height.
To get this height, we use the equation
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] - 2gs (the negative sign is due to the downward direction of the acceleration g)
where
v is the final velocity at the maximum height = 0 m/s (it comes to a stop)
u is the speed at the instance that the engine is cut out = 70 m/s
g is the acceleration due to gravity = 9.81 m/s^2
s is the distance from this point to the maximum height
substituting values, we have
[tex]0^{2}[/tex] = [tex]70^{2}[/tex] - 2(9.81 x s)
0 = 4900 - 19.62s
4900 = 19.62s
s = 4900/19.62 = 249.7 m
The maximum height that will be reached = 350 m + 249.7 m = 599.7 m
approximately 600 m
How many cables enter the 2-gang wall switch box located next to the side garage door?
Answer:
According to the National Electric Code Paragraph 404 section 2, the cables entering into the 2-gang wall must be 4 in number and it must be fitted with a neutral conductor terminal as well.
Cheers
When storing used oil, it need to be kept in________ container?
Answer: In a clean plastic or metal container with a tightly sealed lid
The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is 225∘C and the change in plate temperature with time (dT/dt) is -0.022 K/s. The ambient air temperature is 25∘C and the plate measures 0.3×0.3with a mass of 3.75 kg and a specific heat of 2770J/kg⋅K.
Answer:
h = 6.35 W/m².k
Explanation:
In order to solve this problem, we will use energy balance, taking the thin hot plate as a system. According to energy balance, the rate of heat transfer to surrounding through convection must be equal to the energy stored in the plate:
Rate of Heat Transfer Through Convection = Energy Stored in Plate
- h A (Ts - T₀) = m C dT/dt
where,
h = convection heat transfer coefficient = ?
A = Surface area of plate through which heat transfer takes place = 2 x 0.3 m x 0.3 m (2 is multiplied for two sides of thin plate) = 0.18 m²
Ts = Surface Temperature of hot thin plate = 225⁰C
T₀ = Ambient Temperature = 25°C
m = mass of plate = 3.75 kg
C = Specific Heat = 2770 J/kg. k
dT/dt = rate of change in plate temperature = - 0.022 K/s
Therefore,
- h (0.18 m²)(225 - 25) k = (3.75 kg)(2770 J/kg.k)(- 0.022 k/s)
h = (- 228.525 W)/(- 36 m².k)
h = 6.35 W/m².k
A standard 20° pressure angle, 20 tooth pinion with a diametral pitch of 12 rotates at 1776 rpm driving a mating gear at 740 rpm.a) How many teeth are there on the gear?b) What are the pitch diameters of the pinion and gear?c) What are the standard addendum and dedendum diameters of the pinion and gear?d) What is the theoretically correct center distance?e) What are the radial, tangential and normal tooth loads when contact between the pinion and the gear occurs at the pitch point and the gearset is transmitting ½ hp?
Answer:
A) 48
B) Pitch diameters : pinion = 42.164 mm, Gear = 101.19 mm
C) standard addendum : pinion = 46.3804, Gear = 105.406
standard dedendum : pinion = 37.265 mm, Gear = 96.312 mm
D) 71.672 mm
E) 94.989 N , 101.0858 N, 34.573 N
Explanation:
Given Data :
∅ = 20⁰ , Tp = 20 ( tooth pinion ),
diameter pitch = 12, Np = 1776 rpm ,
Ng = 740 rpm,
attached below is the detailed solution of the given problems
Compute the average (root mean square) velocity (m/s) of Neon molecules at 356 Kelvins and 0.9 bars.
Answer:
V = 20.6 m/s
Explanation:
Given that the temperature of the neon molecules = 356 Kelvin
Pressure = 0.9 bar
The mass number of Neon = 21.
Using the formula below
1/2 m v^2 = (3kT)/2
Where
T = temperature = 356 k
K = Bolzmann constant
= 1.38 × 10^-23jk^-1
NA = 6.02214076×10²³ mol⁻¹
Where NA = Avogadro number
Substitute all the parameters Into the formula
1/2 × 21/NA × v^2 = (3 × k × 356)/2
1.744×10^-23V^2 = 7.3692 × 10^-21
Make V^2 the subject of formula
V^2 = (7.37×10^-21)/1.744×10^-23
V^2 = 422.55
V = sqrt( 422.55)
V = 20.55
Therefore, the average (root mean square) velocity (m/s) of Neon molecules at 356 Kelvins and 0.9 bars is 20.55 m/s
The wall is 15 ft high and consists of 2 x 4 in. studs, plastered on one side. On the other side there is 4-in. clay brick. Determine the average load in lb>ft of length of wall that the wall exerts on the floor.
Answer:
645 lb/ft
Explanation:
The height of the wall = 15 ft
For the 2 x 4 in. studs:
The minimum dead load for a 2 x 4 in. studs (from table 1-3) = 4 psf = 4 lb/ft²
Therefore for a 15 ft tall wall, the dead load for a 2 x 4 in. studs = 15 ft × 4 lb/ft² = 60 lb/ft
For the 4 in. clay brick:
The minimum dead load for a 4 in. clay brick (from table 1-3) = 39 psf = 39 lb/ft²
Therefore for a 15 ft tall wall, the dead load for a 4 in. clay brick = 15 ft × 39 lb/ft² = 585 lb/ft
Therefore the average load = load for the 4 in. clay brick + load for the 2 x 4 in. studs = 585 lb/ft + 60 lb/ft = 645 lb/ft
A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejecting its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.
Answer:
hello your question is incomplete attached below is the missing part and also attached is the solution
answer: a) 0.4801
b) 5.398 kw
c) 2.14
d) 12.72
Explanation:
The quality of the refrigerant at the evaporator inlet
h4 = hf4 + x4(hfx4)
Refrigeration load
Ql = m(h1-h4)
COP of the refrigerator
Ql / m(h2-h1) - Qm
Theoretical maximum refrigeration load
( Ql )max = COPr.rev * [m(h2-h1) - Qin]
The quality that will exist at the inlet of the refrigerant's evaporator would be:
a). [tex]0.4801[/tex]
The load of the refrigeration would be
b). [tex]5.398 kW[/tex]
The refrigerator's COP would be:
c). [tex]2.14[/tex]
The maximum refrigeration load would be as follows:
d) [tex]12.72[/tex]
a). The determination of the quality of the refrigerant at the inlet of the evaporator would be:
[tex]h4 = hf_{4} + x_{4}(hfx_{4})[/tex]
As given,
[tex]hf_{4}[/tex] [tex]= 3.841[/tex]
[tex]hf_{2,4}[/tex][tex]= 223.95[/tex]
[tex]h_{4} = 111.37[/tex]
Now,
solving for [tex]x_{4}[/tex] [tex]= (111.37 - 3.841)/223.95[/tex]
[tex]= 0.4801[/tex]
b). Refrigeration load
[tex]Q_{l}[/tex] [tex]= m(h_{1} - h_{4})[/tex]
[tex]= 0.0455(230.01-111.37)[/tex]
[tex]= 5.398 kW[/tex]
c). COP of the refrigerator
[tex]Q_{l}[/tex]/[tex]m(h_{2} - h_{1}) - Q_{m}[/tex]
by putting the values, we get
∵ COP [tex]= 2.14[/tex]
d). Theoretical maximum refrigeration load
[tex](Q_{l})[/tex]max [tex]= COPr.rev[/tex] × [tex]{m(h_{2} - h_{1}) - Q_{in}[/tex]
by putting the values, we get
∵ [tex]Q_{L}max = 12.72[/tex]
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A proposed piping and pumping system has 20-psig static pressure, and the piping discharges to atmosphere 160 ft above the pump. If the piping friction loss is 20 ft head, the minimum pressure rating (psi) of the piping system is most nearly:
(A) 50
(B) 100
(C) 150
(D) 250
Answer: (B) 100
Explanation:
Given that;
Pstatic = 20 psig , hz = 160ft, hf = 20ft
Now total head will be;
T.h = hz + hf
T.h= 160 + 20
T.h = 180ft
Minimum pressure = Psatic + egh
we know that specific weight of water is 62.4 (lb/ft3)
so
P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr
P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)
P.min = 20 + 78
P.min = 98 lbf/in²
Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100