Answer:
acceleration = 24.23 ms⁻¹
Explanation:
Let's gather the data:
The acceleration of the car is given by a = 0.5 [tex]e^{t}[/tex]
The acceleration is the change in the speed in relation to time. In other words:
[tex]\frac{dv}{dt}[/tex] = a = a = 0.5 [tex]e^{t}[/tex] ...1
Solving the differential equation yields:
v = 0.5 [tex]e^{t}[/tex] + C₁
Initial conditions : 0 = 0.5 [tex]e^{0}[/tex] + C₁
C₁ = -5
at any time t, the velocity is: v= 0.5[tex]e^{t}[/tex]- 5
Solving for distance, s = 0. 5[tex]e^{t}[/tex] - 0.5 t - 0.5
18 = 0.5 [tex]e^{t}[/tex] - 0.5 t - 0.5
t = 3.71 s
Substitute t = 3.71 s
v= 0.5[tex]e^{t}[/tex]- 5
= 19.85 m/s
a = 0.5 [tex]e^{t}[/tex] ...1
= 20.3531
an = [tex]\frac{v^{2} }{p}[/tex]
= [tex]\frac{(19.853)^{2} }{30}[/tex]
= 13.1382
Magnitude of acceleration = [tex]\sqrt{ (a)^{2} + (an)^{2} }[/tex]
= [tex]\sqrt{(20.3531)^{2} +(13.1382)^{2} }[/tex]
= 24.23 ms⁻¹ Ans
Define a named tuple Player that describes an athlete on a sports team. Include the fields name, number, position, and team.
Answer:
Explanation:
we would be analyzing this question with the important code given
#code :
from collections import namedtuple
#creating a named tuple named 'Player' with field names name, number, position and team
Player = namedtuple('Player',['name','number','position', 'team'])
cam = Player('Cam Newton','1','Quarterback','Carolina Panthers')
lebron = Player('Lebron James','23','Small forward','Los Angeles Lakers')
print(cam.name+'(#'+cam.number+')'+' is a '+cam.position+' for the '+cam.team+'.')
print(lebron.name+'(#'+lebron.number+')'+' is a '+lebron.position+' for the '+lebron.team+'.')
NB:
Lebron James (#23) rep. Small forward for the LA lakers
Cam Newton(#1) rep. a Quaterback for the Carolina Panthers
cheers i hope this helps
One the eight safety systems that failed was deemed to be ____________.
Select one:
a. political
b. misinterpreted pressure tests to determine whether the well had been sealed
c. the water temperature
d. the weather temperature
Answer: B. misinterpreted pressure tests to determine whether the well has been sealed.
Explanation:
Pressure test is a non-destructive test carried out to ensure there is integrity of the pressure shell on an equipment that is new or pressure and piping equipment installed earlier that has undergone some repairs.
After a crew carried out a variety of pressure tests to determine whether the well was sealed or not. The results of these tests were misinterpreted, so they thought the well was under control. This caused oil to spill.
For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True
B. False
Answer:
B.
Explanation:
For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.
The truth table for a NAND gate with 2 inputs is as follows:
0 0 1
0 1 1
1 0 1
1 1 0
The truth table for an OR gate, is as follows:
0 0 0
0 1 1
1 0 1
1 1 1
If we add two extra columns for inverted inputs, the truth table will be this one:
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
which is the same as for the NAND gate, not the opposite, so the statement is false.
This means that the right choice is B.
At a point in the flow over a high-speed missile, the local velocity and temperature are 300 ft/s and 500oR, respectively. Calculate the Mach number M and the characteristic Mach number, M* at this point.
Answer:
0.267
Explanation:
The Mach number is given by the following formula:
[tex]Ma = \frac{v}{c}[/tex]
where v = speed of the object
c= speed of air
Ma = Mach number
From the data:
Velocity of the object = 91.44 m/s
Velocity of air = 343 m/s
Therefore, the Mach number is given by the following formula:
Ma = [tex]\frac{91.44}{343}[/tex]
= 0.267
In the kitchen of your house you would like to run a 360 W blender, a 60 W phone charger, a 840 W toaster oven, and four 120 W lights. Assuming your house is wired for 120 Volts, how much current will be pulled by these electrical devices
The kitchen appliances, including a blender, phone charger, toaster oven, and lights, will pull a total current of 15 Amperes when operating on a 120-Volt household electrical system.
Explanation:To calculate the total current pulled by various electrical devices in the kitchen, you add up their power ratings and divide the sum by the voltage of the electrical system. These devices, which include a 360 W blender, a 60 W phone charger, an 840 W toaster oven, and four 120 W lights, all operate on a standard household voltage of 120 Volts. The formula for current (I) when power (P) and voltage (V) are known is I = P/V.
First, calculate the total power consumed by all devices:
360 W (blender) + 60 W (phone charger) + 840 W (toaster oven) + (4 × 120 W) (lights) = 1800 W.
Then, to find the current, we use the formula with the total power and the household voltage: I = 1800 W / 120 V = 15 A. Therefore, these appliances will pull a total current of 15 Amperes.
On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading be when you carry it up to the observation deck 500 ft above the base of the monument?
Answer:
The barometer reading will be 29.43 in
Explanation:
Using the formula of pressure variation
p2 - p1 = -yair * H
= 7.65 * [tex]10^{-2} \frac{lb}{ft^{3} } * 500 ft\\[/tex]
= 38.5 [tex]\frac{lb}{ft^{2} }[/tex]
According to the relationship between the pressure and the height of the mercury column
p = yHg * h --> where yHg and h is the barometer reading
yHg [tex](\frac{29.97}{12} ft)[/tex] - yHg * h1 = 38.5 [tex]\frac{lb}{ft^{2} }[/tex]
h1 = ([tex]\frac{29.97}{12} ft[/tex]) - [tex]\frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }[/tex]
[tex][(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in[/tex]
The barometer reading be "29.43 in".
According to the question,
Barometer reading,
[tex]H' = 29.97 \ in[/tex]Pressure variation for incompressible and static fluid will be:
→ [tex]P_2-P_1 = V_{air} H[/tex]
[tex]= 7.65\times 10^{-2}\times 500[/tex]
[tex]= 38.5 \ lb/ft^2[/tex]
Pressure and height relationship for mercury will be:
→ [tex]P = V_{Hg} H'[/tex]
→ [tex]V.Hg\times \frac{29.97}{12} - V_{Hg} h_1 = 38.5[/tex]
→ [tex]h_1 = \frac{29.97}{12} - \frac{38.5}{847}[/tex]
[tex]= [\frac{29.97}{12} - 0.0455][/tex]
[tex]= 29.42 \ in[/tex]
Thus the above answer is right.
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A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air enters the compressor at 520 R and the turbine at 2000 R. Accounting for the variation of specific heats with temperature, determine (a) the air temperature at the compressor exit, (b) the back work ratio, and (c) the thermal efficiency.
The question involves analyzing an ideal Brayton cycle to find the exit temperature from a compressor, the back work ratio, and the thermal efficiency of a gas turbine engine with air as the working fluid, accounting for the variation in specific heats with temperature.
Explanation:The student's question pertains to the ideal Brayton cycle, which is a thermodynamic cycle used to describe the workings of a gas turbine engine. Their question involves air with varying specific heats at different temperatures, typical in engineering thermodynamics analysis.
Although the appropriate formulas and calculations are not provided in the provided references, the typical approach would involve thermodynamic equations of state and specific heat relations to find the unknown temperatures, efficiency, and other parameters of interest. Analysis often includes using the Brayton cycle efficiency formula, temperature ratios, and the specific gas constant for air.
The back work ratio and thermal efficiency in particular can be calculated using the first and second laws of thermodynamics applied to each component of the cycle: the compressor and turbine. As specific heats vary with temperature, one typically uses mean values or functions that provide the specific heat values at the given temperatures.
Explanation of the Brayton cycle in gas turbines, calculation methods for compressor exit temperature, back work ratio, and thermal efficiency.
The Brayton cycle is a thermodynamic cycle that describes how gas turbines operate. It consists of four processes: two isentropic (reversible adiabatic) processes and two isobaric (constant-pressure) processes.
(a) The air temperature at the compressor exit can be calculated using the temperature ratios at the compressor and turbine, and the pressure ratio.
(b) The back work ratio is the ratio of the work required to drive the compressor to the work produced by the turbine.
(c) The thermal efficiency of the Brayton cycle can be determined by comparing the work output to the heat input.
A 10-V-emf battery is connected in series with the following: a 2-µF capacitor, a 2.0-Ω resistor, an ammeter, and a switch, initially open; a voltmeter is connected in parallel across the capacitor. At the instant the switch is closed, what are the current and capacitor voltage readings, respectively?
Answer:
I = 5.0 A V=0
Explanation:
Assuming that the ammeter is an ideal one (which means that the internal resistance is negligible compared with other resistors in the circuit), as the voltage through the capacitor can't change instantaneously, just after the switch is closed, will behave like a short, so applying Ohm's law, the current in the circuit will be as follows:
[tex]I =\frac{V}{R} = \frac{10 V}{2.0 \Omega} = 5.0 A[/tex]
As the voltage in the capacitor, can't change instantaneously, assuming an ideal voltmeter (infinite resistance) , the lecture on the voltmeter across the capacitor will be just 0.
As time goes by, the voltage measured will follow the following equation:
[tex]V = V0*( 1-e^{-\frac{t}{R*C}} )[/tex]
We see that when t=0, V=0.
Assume that the conversion of energy into mechanical work (at the wheel) in an internal combustion engine is 20%. Calculate gallons of gasoline required to deliver 30 horsepower at the wheel, for one hour. 1 HP = 746 Watts
1 HP for 1 hour is 0.746 kWh
1 Kwh = 3 412 BTU
To deliver 30 horsepower at the wheel for one hour with a 20% efficient internal combustion engine, approximately 2.88 gallons of gasoline are required.
Explanation:The question at hand involves calculating the quantity of gasoline required to deliver 30 horsepower (HP) at the wheel for one hour, given that the energy conversion efficiency of an internal combustion engine is 20%. First, understand that 1 HP is equivalent to 746 Watts, and therefore, 30 HP equals 22,380 Watts or 22.38 Kilowatts. Since 1 HP for 1 hour is 0.746 kWh, 30 HP for 1 hour is 22.38 kWh. Knowing the energy content of gasoline and the engine's efficiency will allow us to calculate the required gallons of gasoline.
Given that a gallon of gasoline releases about 140 MJ (or 38.89 kWh) of energy when burned and considering the 20% efficiency, the useful energy per gallon of gasoline is approximately 7.778 kWh. To find out how many gallons of gasoline are needed to produce 22.38 kWh at 20% efficiency, we divide the total required energy output (22.38 kWh) by the useful energy per gallon (7.778 kWh), resulting in approximately 2.88 gallons of gasoline needed.
In the frequency domain, you view a signal whose highest frequency is fH=4.3kHz, and lowest frequency is fL=300Hz. What is the signal’s frequency bandwidth?
Answer:
[tex]BW=\Delta f= f_H-f_L=4300Hz-300Hz=4000Hz=4kHz[/tex]
Explanation:
Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. The bandwidth may be determined by use of the following formula:
[tex]BW=f_H-f_L\\\\Where:\\\\f_H=Upper\hspace{3}cutoff\hspace{3}frequency=4.3kHz=4300Hz\\f_L=Lower\hspace{3}cutoff\hspace{3}frequency=300Hz[/tex]
Hence the signal’s frequency bandwidth is:
[tex]BW=4300-300=4000Hz=4kHz[/tex]
Select the correct verb form to complete each of the following sentences.Members of the committee ____ issued a mobile phone and one roll of stamps. Neither the CEO nor the CFO _____any recollection of the illegal funds transfer. Mr. Adams and his assistant______ planning to attend the annual technology showcase. The finance committee _______planning their vacations around the annual June meeting.Everything in my home office _____ to stop functioning when I need it the most.
The correct verb forms to complete the given sentences are:
Members of the committee issued a mobile phone and one roll of stamps.
Neither the CEO nor the CFO has any recollection of the illegal funds transfer.
Mr. Adams and his assistant are planning to attend the annual technology showcase.
The finance committee is planning their vacations around the annual June meeting.
Everything in my home office seems to stop functioning when I need it the most.
In the first sentence, the verb "issued" is correct because it is in the past tense and agrees with the subject of the sentence, "Members of the committee."
In the second sentence, the verb "has" is correct because it is in the present tense and agrees with the subject of the sentence, "Neither the CEO nor the CFO." The verb "has" is also used when the subject is a singular noun that refers to two or more people or things joined by "neither" or "nor."
In the third sentence, the verb "are" is correct because it is in the present tense and agrees with the subject of the sentence, "Mr. Adams and his assistant." The verb "are" is also used when the subject is a plural noun.
In the fourth sentence, the verb "is" is correct because it is in the present tense and agrees with the subject of the sentence, "The finance committee." The verb "is" is also used when the subject is a singular noun that refers to a group of people or things.
In the fifth sentence, the verb "seems" is correct because it is in the present tense and agrees with the subject of the sentence, "Everything in my home office." The verb "seems" is also used to express an opinion or belief.
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Consider a composite wall that includes an 8-mm-thick hardboard siding, 40-mm by 170-mm hardwood studs on 0.65-m centers with glass fiber insulation (paper faced, 28 kg/m3 ), and a 12-mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high)?
Answer:
dimensions and materials information that are provided about composite wall (2.5mx6.5m 10 studs each 2.5m high)
to find:wall thermal resistance=??
properties:
Ka=0.094W/m.K
Kb=0.16W/m.K
Kc=0.17W/m.K
Kd=0.038W/m.K
using isothermal surface assumption:
(La/KaAa)=0.008/0.094(0.65x2.5)
=0.0524K/W
(LB/KbAb)=0.13/0.16(0.04x2.5)
=8.125K/W
(Ld/KdAd)=0.13/0.38(0.6x2.5)
=2.243K/W
core's equivalent resistance is:
Req=(1/Rb+1/Rd)⁻¹
=(1/8.125+1/2.243)⁻¹
=1.758K/W
For total resistance we will find sum of
Rtot=Ra+Req+Rc
=1.854K/W
Whereas 10 studs are used so:
Rtotal=(10x1/Rtot)⁻¹
=0.1854K/W
D *4.80 It is required to use a peak rectifier to design a dc power supply that provides an average dc output voltage of 12 V on which a maximum of ±1-V ripple is allowed. The rectifier feeds a load of 200. The rectifier is fed from the line voltage (120 V rms, 60 Hz) through a transformer. The diodes available have 0.7-V drop when conducting. If the designer opts for the half-wave circuit: (a) Specify the rms voltage that must appear across the transformer secondary. (b) Find the required value of the filter capacitor. (c) Find the maximum reverse voltage that will appear across the diode, and specify the PIV rating of the diode. (d) Calculate the average current through the diode during conduction. (e) Calculate the peak diode current.
Designing a DC power supply with a half-wave rectifier involves calculating the necessary transformer secondary RMS voltage, filter capacitor value, diode's peak inverse voltage, and average and peak diode currents considering the specific design requirements like output voltage and ripple.
Explanation:The question relates to designing a DC power supply using a half-wave rectifier, including calculations for transformer secondary voltage, filter capacitor value, peak inverse voltage (PIV), and diode current parameters. Solving such a design problem involves understanding and applying principles of electrical engineering, particularly those regarding rectification, filtering, and electrical power supply designs.
To calculate the RMS voltage required across the transformer secondary, one must consider the average output voltage required, the diode forward voltage drop, and the peak-to-peak voltage required to maintain the ripple within the specified limits.The filter capacitor value is crucial to minimize ripple and can be determined based on the allowable ripple voltage, load resistance, and the frequency of the rectified output.The PIV rating of the diode is determined by considering the maximum reverse voltage that the diode can withstand during operation, which is affected by the transformer's secondary RMS voltage and the rectifier configuration.Calculating the average and peak diode current involves understanding the load requirements and the electrical characteristics of the diode and the circuit.A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 10 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.28 m/s2, determine the coefficient of static friction between the trunk and the turntable.
Answer:
The coefficient of static friction, μₛ, between the trunk and turntable = 0.32
Explanation:
For this motion of the trunk B,
Initial velocity, v₀ = 0
Tangential Acceleration, a = 0.28 m/s²
Time taken, t = 10s
Using equations of motion,
v = v₀ + at
v = 0 + 0.28 × 10 = 2.8 m/s
Frictional force, Fᵣ = μₛN
μₛ = coefficient of static friction,
N = Normal reaction exerted on the trunk B as a result of its weight = mg
Doing a force balance on the trunk B,
Force keeping the trunk B in circular motion must balance the frictional forces.
Force keeping the trunk B in circular motion, F = mv²/r
Fᵣ = F
μₛN = mv²/r but N = mg
μₛmg = mv²/r
μₛg = v²/r
μₛ = v²/gr
μₛ = 2.8²/(9.8 × 2.5) = 0.32
Hope this helps!!!
The coefficient of static friction between the trunk and the turntable is 1.
Explanation:To determine the coefficient of static friction between the trunk and the turntable, we can use the equation:
fs = m * at
where fs is the static friction, m is the mass of the trunk, and at is the tangential acceleration of the trunk.
Using the given values, we have:
0.28 = m * 0.28
Solving for m, we find:
m = 1 kg
Therefore, the coefficient of static friction between the trunk and the turntable is 1.
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A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is then heated until its temperature is 180°C. The final state is superheated water and the internal energy at this state should be obtained by interpolation. Calculate the heat transfer required for this process. Use data from the steam tables.
Answer:
98.13kJ
Explanation:
Given that;
The rigid 10-L vessel initially contains a mixture of liquid water & vapor at
[tex]T_1 =100^0C\\T_2 = 180^0C[/tex]
We are to calculate the heat transfer required during the process by obtaining our data from the steam tables.
In order to do that, let start with our Energy Balance
So, Energy Balance for closed rigid tank system is given as:
[tex]\delta E_{system} = E_{in} - E_{out}[/tex]
Since the K.E and P.E are insignificant;
∴ K.E = P.E = 0
[tex]Q_{in}= \delta U + W\\Q_{in} = m(u_2-u_1)+ W[/tex]
Where;
m = mass flow rate of the mixture
[tex](u_2-u_1)[/tex] = corresponding change in the internal energy at state point 2 and 1
However, since we are informed that the vessel is rigid, then there is no work done in the system, then W turn out to be equal to zero .i,e
W = 0
we have our above equation re-written as:
[tex]Q_{in}= m (u_2-u_1)+0\\[/tex]
[tex]Q_{in}= m(u_2-u_1)[/tex]
We were told to obtain our data from the steam table, so were going to do just that
∴ At inlet temperature [tex]T_1 = 100^0C[/tex], the given quality of mixture of liquid water and vapor [tex](x_1)[/tex] = 123% = 0.123
Using the equation:
[tex]v_1 = v_f + x_1v_{fg}\\v_1 = v_f + x_1(v_g-v_f)[/tex]
where;
[tex]v_1[/tex] = specific volume at state 1
[tex]v_f[/tex] = specific volume of the liquid
[tex]v_g[/tex] = specific volume of the liquid vapor mixture
The above data from the steam table is given as;
[tex]v_f[/tex] = 0.001043 m³/kg
[tex]v_g[/tex] = 1.6720 m³/kg
so; we have
[tex]v_1 = 0.001043 + 0.123(1.6720-0.001043)[/tex]
[tex]v_1 = 0.001043+0.123(1.670957)[/tex]
[tex]v_1 =0.001043+0.205527711[/tex]
[tex]v_1= 0.206570711m^3/kg[/tex]
[tex]v_1=0.2065m^3/kg[/tex]
At [tex]T_1[/tex] = 100°C and [tex]x_1=0.123[/tex];
the following steam data from the tables were still obtained for the internal energy; which is given as:
Internal Energy [tex](u_1)[/tex] at the state 1
[tex]u_1= u_f + xu_{fg}[/tex]
where;
specific internal energy of the liquid [tex](u_f)[/tex] = 419.06 kJ/kg
The specific internal energy of the liquid vapor mixture [tex](u_{fg})[/tex] = 2087.0 kJ/kg
∴ since ; [tex]u_1= u_f + xu_{fg}[/tex]
[tex](u_1)[/tex] = 419.06 + (0.123 × 2087.0)
[tex](u_1)[/tex] = 675.761 kJ/kg
As the tank is rigid, so as the volume which is kept constant:
[tex]v_2=v_1\\=0.2065 m^3/kg[/tex]
Now, let take a look at when [tex]T_2 = 180^0C[/tex] from the data in the steam tables
Specific volume of the liquid [tex](v_f)[/tex] = 0.00113 m³/kg
specific volume of the liquid vapor mixture [tex](v_g)[/tex] = 0.19384 m³/kg
The quality of the mixture at the final state 2 can be determined by using the equation shown below:
[tex]v_2=v_f+x_2v_{fg}[/tex]
[tex]x_2=\frac{v_2-v_f}{v_{fg}}[/tex]
[tex]x_2=\frac{v_2-v_f}{v_g-v_f}[/tex]
[tex]x_2=\frac{0.2065-0.00113}{0.19384-0.00113}[/tex]
[tex]x_2=\frac{0.20537}{0.19271}[/tex]
= 1.0657
From our usual steam table; we still obtained data for the Internal Energy when [tex]T_2=180^0C[/tex]
Specific internal energy of the liquid [tex](u_f)[/tex] = 761.92 kJ/kg
Specific internal energy of the liquid vapor mixture [tex]u_{fg}[/tex] = 1820.88 kJ/kg
Calculating the internal energy at finsl state point 2 ; we have:
[tex]u_2=u_f+u_{fg}[/tex]
= 761.92 + (1.0657 × 1820.88)
= 761.92 + 1940.511816
= 2702.431816
[tex]u_2[/tex] ≅ 2702.43 kJ/kg
Furthermore, let us calculate the mass in the system; we have:
[tex]m= \frac{V_1}{v_1}[/tex]
where V₁ = the volume 10 - L given by the system and v₁ = specific volume at state 1 as 0.2065
V₁ = the volume 10 - L = 10 × ( 0.001 m³/L)
v₁ = 0.2065
∴
mass (m) = [tex]\frac{10(0.001m^3/L)}{0.2065}[/tex]
= 0.04842 kg
Now, we gotten all we nee do calculate for the heat transfer that is required during the process:
[tex]Q_{in}= m(u_2-u_1)[/tex]
[tex]Q_{in}= 0.04842(2702.43-675.761)[/tex]
[tex]Q_{in}= 0.04842(2026.669)[/tex]
[tex]Q_{in}= 98.13 kJ[/tex]
Therefore, the heat transfer that is required during the process = 98.13 kJ
There you have it!, I hope this really helps alot!
The heat transfer required for this process is 129.168 kilojoules.
According to steam tables, the initial state of the water inside the vessel is:
[tex]P = 101.42\,kPa[/tex], [tex]T = 100\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u = 675.761\,\frac{kJ}{kg}[/tex], [tex]x = 0.123[/tex] (Liquid-Vapor Mixture)
Given that process is isochoric, that is, at constant volume, we know the following information:
[tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], Superheated vapor
We need to know the pressure and the internal energy must be determined by using linear interpolation from steam tables twice. First, we use [tex]T = 180\,^{\circ}C[/tex] as our pivot, then we construct the following information by linear interpolation:
Lower bound
[tex]P = 800\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.24700\,\frac{m^{3}}{kg}[/tex], [tex]u = 2593.9\,\frac{kJ}{kg}[/tex]
Upper bound
[tex]P = 1000\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.19444\,\frac{m^{3}}{kg}[/tex], [tex]u = 2583.0\,\frac{kJ}{kg}[/tex]
Lastly, we find the expected pressure and internal energy by linear interpolation where specific volume is our pivot:
[tex]P = 952.207\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u = 2585.605\,\frac{kJ}{kg}[/tex]
Once found all the required information, we can find the required heat transfer ([tex]Q[/tex]), in kilojoules, by means of this formula:
[tex]Q = \frac{V}{\nu}\cdot (u_{2}-u_{1})[/tex] (1)
Where:
[tex]V[/tex] - Volume of the vessel, in cubic meters,[tex]\nu[/tex] - Specific volume of water, in cubic meters per kilogram.[tex]u_{1}[/tex], [tex]u_{2}[/tex] - Initial and final internal energy, in kilojoules per kilogram.If we know that [tex]V = 0.014\,m^{3}[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u_{1} = 675.761\,\frac{kJ}{kg}[/tex] and [tex]u_{2} = 2585.605\,\frac{kJ}{kg}[/tex], then the heat transfer required for this process is:
[tex]Q = \left(\frac{0.014\,m^{3}}{0.207\,\frac{m^{3}}{kg} } \right)\cdot \left(2585.605\,\frac{kJ}{kg}-675.761\,\frac{kJ}{kg} \right)[/tex]
[tex]Q = 129.168\,kJ[/tex]
The heat transfer required for this process is 129.168 kilojoules.
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