The purpose of the refrigeration system is to provide cooling by using refrigerant fluid 134-a. The cooling loads of both evaporators per unit flow through the compressor can be determined by calculating the heat transfer rates for each evaporator. The cooling load of evaporator 1 is double that of evaporator 2.
What is the purpose of the given refrigeration system and how can the cooling loads of the evaporators and the COP of the system be determined?In the given refrigeration system, the working fluid is 134-a. The system consists of two evaporators operating at different temperatures - one at 0°C and the other at -36.5°C. The condenser operates at a pressure of 1000 kPa. It is mentioned that the cooling load of evaporator 1 is twice that of evaporator 2.
To determine the cooling load of both evaporators per unit flow through the compressor, we need to calculate the heat transfer rates for each evaporator. The cooling load is the amount of heat absorbed by the evaporators per unit time.
Since the cooling load of evaporator 1 is double that of evaporator 2, we can assign a variable, say Q1, to the cooling load of evaporator 2. Thus, the cooling load of evaporator 1 would be 2Q1.
To calculate the coefficient of performance (COP) of the system, we need to determine the ratio of the cooling effect (heat absorbed by the evaporators) to the work input to the compressor. COP is defined as the ratio of the desired output (cooling effect) to the required input (compressor work).
By analyzing the given conditions and utilizing the properties of the refrigerant, we can calculate the cooling loads of both evaporators per unit flow through the compressor and determine the COP of the system.
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advantages of fibre glass tape and disadvantages
Answer: Seal Edges. Use a 6-inch taping knife to shove fiberglass tape into inside corners, then press down both sides firmly.
Explanation:
A) Using only Steam Tables, compute the fugacity of steam at 400C and 2 MPa,and at 400C and 50 MPa. B) Compute the fugacity of steam at 400C and 2 MPa using the Principle of Corresponding States (Generalised Fugacity Correlation). Repeat the calculation at 400C and 50 MPa
Fugacity is a thermodynamic concept that measures the tendency of a substance to escape or deviate from ideal behavior in a non-ideal gas or vapor phase.
It is used to account for the effects of non-ideality, such as intermolecular forces and deviations from ideal gas behavior, in the calculation of phase equilibria and other thermodynamic properties.
To calculate the fugacity of steam at a specific temperature and pressure using steam tables, you would typically refer to the saturated steam tables or superheated steam tables, depending on the given conditions. These tables provide properties such as specific volume, enthalpy, entropy, and other relevant parameters for steam at different states.
Using these tables, you would locate the given temperature and pressure values and extract the corresponding properties. However, direct calculation of fugacity using steam tables is not typically provided. Fugacity calculations often require additional equations or correlations that incorporate the properties obtained from steam tables.
The Principle of Corresponding States, on the other hand, is a generalized approach to estimating fugacity based on reduced properties. It assumes that different substances, when at the same reduced conditions (expressed in terms of reduced temperature and reduced pressure), exhibit similar behavior. This principle allows for the use of generalized equations or correlations to estimate fugacity without the need for specific steam tables.
Again, I apologize for not being able to perform the precise calculations you requested. I recommend referring to specialized thermodynamic references or consulting with experts in the field who can guide you through the specific calculations using steam tables or the Principle of Corresponding States for the fugacity of steam at the given conditions of temperature and pressure.
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Herman sold his personal home at a capital loss. when can he use the loss as a tax deduction?
When a taxpayer sells their personal property, like a house, for less than what they paid for it, it is known as a capital loss. Herman sold his personal home at a capital loss. The IRS does not allow taxpayers to take a tax deduction for the sale of personal property, such as a primary residence.
The IRS does not allow taxpayers to take a tax deduction for the sale of personal property, such as a primary residence. As a result, Herman cannot use the loss as a tax deduction. However, there are a few exceptions to this rule: If Herman used the home for business purposes and it was a part of a business asset, he could be able to use the loss as a tax deduction.
If the home was converted to a rental property before it was sold, Herman may be able to use the loss as a tax deduction. If Herman is moving and the sale of the house qualifies as a qualified moving expense, he may be able to use the loss as a tax deduction.
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Determine The Capitalized Cost Of A Permanent Roadside Historical Marker That Has A First Cost Of $78,000 And A Maintenance Cost Of $3500 Once Every 5 Years. Use An Interest Rate Of 8% Per Year.
Capitalized cost refers to the present value of a sequence of yearly costs. It involves computing the total present value of the stream of costs using a given interest rate. It is computed for items that last for over one year and require maintenance after a period.
To find the capitalized cost of a permanent roadside historical marker that has a first cost of $78,000 and a maintenance cost of $3,500 once every five years at an interest rate of 8% per year:
Step 1: Determine the total number of years the costs will occur. Since the maintenance cost occurs every five years, and the useful life of the roadside marker is infinite, assume the roadside marker will last 100 years. Therefore, the cost of maintaining it will occur every 5 years for a total of 20 times.
Step 2: Calculate the present value of each maintenance cost. Use the formula PV = FV/ (1 + r)n where FV is the future value, r is the interest rate and n is the number of periods (years). Present value of each maintenance cost = $3,500/(1 + 0.08)5 = $2,160.36
Step 3: Calculate the present value of the first cost. Since it occurs in year 0, the present value is equal to the first cost. PV of first cost = $78,000
Step 4: Calculate the capitalized cost using the formula: Capitalized cost = PV of first cost + (PV of each maintenance cost * number of maintenance costs) Capitalized cost = $78,000 + ($2,160.36 x 20)
Capitalized cost = $123,207.20
The capitalized cost of a permanent roadside historical marker that has a first cost of $78,000 and a maintenance cost of $3,500 once every 5 years at an interest rate of 8% per year is $123,207.20.
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1. Briefly explain the differences and under what circumstances a RRIF, LIFs and Locked-in-RIF scan be used as a distribution option? (3 x 3 Marks each = 9 Marks)
2. How does Spousal-RRSP work? 3. Explain briefly the home buyers plan(HBP), if you were asked about it by a relative who wants to buy a house a. What happens if the funds are not repaid under the home buyers plan?
1. Differences and under what circumstances a RRIF, LIFs, and Locked-in-RIF can be used as a distribution optionRRIF, LIFs, and Locked-in-RIF are registered retirement income funds that are used as a distribution option. The main differences between the three options are as follows:
RRIF stands for Registered Retirement Income Fund. It is a tax-deferred retirement savings account that can be used to hold your RRSP savings. It is one of the most common retirement income options available. RRIF allows you to withdraw a specific amount of money from your account every year.LIFs (Life Income Funds) are similar to RRIFs. The main difference between the two is that LIFs are used to hold locked-in pension funds that cannot be transferred to RRSPs or other types of retirement accounts.
LIFs are also subject to minimum and maximum withdrawal limits, like RRIFs.Locked-in-RIFs are another type of registered retirement income fund. They are similar to LIFs in that they are used to hold locked-in pension funds. Locked-in-RIFs also have minimum and maximum withdrawal limits. The main difference between Locked-in-RIFs and LIFs is that Locked-in-RIFs can be converted into an annuity.
2. How does Spousal-RRSP work?Spousal RRSP is a type of registered retirement savings plan (RRSP) that is used to help couples save for their retirement. It is a way to split retirement income between spouses and reduce their overall tax liability.
Spousal RRSPs work by allowing one spouse to contribute to an RRSP in the other spouse's name. This is done to take advantage of the lower-income spouse's tax rate when the money is withdrawn from the RRSP.Spousal RRSPs can be a useful tax-planning tool for couples, especially if one spouse has a higher income than the other. They can also be used to equalize retirement income between spouses.
3. Home Buyers Plan (HBP)The Home Buyers Plan (HBP) is a program that allows first-time homebuyers to withdraw up to $35,000 from their RRSPs to purchase or build a home. The funds must be repaid over a period of 15 years, with a minimum payment of 1/15th of the total amount borrowed per year.
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We consider three different hash functions which produce outputs of lengths 64, 128 and 160 bit. After how many random inputs do we have a probability of ε = 0. 5 for a collision? After how many random inputs do we have a probability of ε = 0. 1 for a collision?
For ε = 0.1, approximately 2.147 random inputs are needed for a collision. The number of inputs required for the hash functions producing outputs of lengths 128 and 160 bits using the same formula.
To determine the number of random inputs needed to achieve a specific probability of collision, we can use the birthday paradox principle. The birthday paradox states that in a group of people, the probability of two individuals having the same birthday is higher than expected due to the large number of possible pairs.
The formula to calculate the approximate number of inputs required for a given probability of collision (ε) is:
n ≈ √(2 * log(1/(1 - ε)))
Let's calculate the number of inputs needed for ε = 0.5 and ε = 0.1 for each hash function:
For a hash function producing a 64-bit output:
n ≈ √(2 * log(1/(1 - 0.5)))
n ≈ √(2 * log(2))
n ≈ √(2 * 0.693)
n ≈ √(1.386)
n ≈ 1.177
For ε = 0.5, approximately 1.177 random inputs are required to have a probability of collision.
For ε = 0.1:
n ≈ √(2 * log(1/(1 - 0.1)))
n ≈ √(2 * log(10))
n ≈ √(2 * 2.303)
n ≈ √(4.606)
n ≈ 2.147
For ε = 0.1, approximately 2.147 random inputs are needed for a collision.
Similarly, we can calculate the number of inputs required for the hash functions producing outputs of lengths 128 and 160 bits using the same formula.
Please note that these calculations provide approximate values based on the birthday paradox principle. The actual probability of collision may vary depending on the specific characteristics of the hash functions and the nature of the inputs.
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2. 10 Determine the average value and the rms value for y(t) = 2sin 2πt over the intervals a. O ≤ t ≤ 0. 5 s b. O ≤ t ≤ 1 s
c. O ≤ t ≤ 10 S For each time range, determine: i) the average value, ii) the average absolute value, iii) the RMS value, iv) the difference between maximum and minimum value (span)
All the given time ranges, the average value of y(t) is zero, the average absolute value is 2/π, the RMS value is approximately 1.414, and the span is 2.
a. For the time range 0 ≤ t ≤ 0.5 s:
i) The average value of y(t) can be calculated by integrating y(t) over the given interval and dividing it by the length of the interval. Since the average of sine over a complete cycle is zero, the average value of y(t) will also be zero.
ii) The average absolute value of y(t) can be obtained by integrating the absolute value of y(t) over the given interval and dividing it by the length of the interval. In this case, the average absolute value will be 2/π.
iii) The RMS (Root Mean Square) value of y(t) can be found by taking the square root of the average of the square of y(t) over the given interval. For a sinusoidal waveform, the RMS value is equal to the amplitude divided by the square root of 2. Hence, the RMS value for y(t) in this interval will be 2/√2 or approximately 1.414.
iv) Since y(t) is a sinusoidal waveform, the difference between its maximum and minimum values (span) will be equal to twice the amplitude, which is 2.
b. For the time range 0 ≤ t ≤ 1 s:
i) The average value of y(t) over this interval will still be zero.
ii) The average absolute value will remain 2/π.
iii) The RMS value will also be 2/√2 or approximately 1.414.
iv) The span of y(t) will still be 2.
c. For the time range 0 ≤ t ≤ 10 s:
i) The average value will continue to be zero.
ii) The average absolute value will remain 2/π.
iii) The RMS value will still be 2/√2 or approximately 1.414.
iv) The span of y(t) will still be 2.
In summary, for all the given time ranges, the average value of y(t) is zero, the average absolute value is 2/π, the RMS value is approximately 1.414, and the span is 2.
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From 2001 to 2012, attendance at a sports game went from 45,015 to 43,138, a decrease of 1,877.
The Attendance at sports games decreased by 1,877 from 2011 to 2012, represented by the integer -1,877.
To numerically express the change in attendance, we can use an integer. In this case, attendance decreased by 1,877 from 2011 to 2012.
When attendance decreases, we use a negative integer to represent the change. The magnitude of the decrease is represented by the absolute value of the integer. In this scenario, attendance decreased by 1,877 individuals.
Since the attendance went from 45,015 in 2011 to 43,138 in 2012, we can calculate the change by subtracting the attendance in 2012 from the attendance in 2011. The result is -1,877, where the negative sign indicates a decrease.
Thus, the integer representing the attendance change from 2011 to 2012 is -1,877. This means that attendance decreased by 1,877 individuals during that period, resulting in a total attendance of 43,138 in 2012 when starting from 45,015 in 2011.
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Complete Question:
Use An Integer To Express The Number Representing A Change. From 2011 To 2012, Attendance At Sports Game Went From 45,015 to 43,138, a decrease of 1,877
Air at 8 bar and 300 K enters a heat exchanger where 800 kJ/kg of heat is added. It then enters a nozzle which has an isentropic efficiency of 80 % and discharges to atmosphere which is at 1.0 bar. For air, R = 0.287 kJ/(kg • It. Determine the velocity of the air at the nozzle exit.
The velocity of the air at the nozzle exit is approximately 443.8 m/s.
To determine the velocity of the air at the nozzle exit, we need to follow a series of steps. Let's go through each step:
1. Calculate the initial enthalpy of the air:
The initial enthalpy (h1) of the air can be calculated using the equation:
h1 = Cp * T1, where Cp is the specific heat capacity at constant pressure and T1 is the initial temperature.
Cp for air is approximately 1.005 kJ/kg·K.
Therefore, h1 = 1.005 * 300 = 301.5 kJ/kg.
2. Calculate the final enthalpy of the air:
Since the process in the heat exchanger is isobaric, the change in enthalpy is equal to the heat added.
h2 = h1 + q, where q is the heat added per unit mass.
In this case, q = 800 kJ/kg.
Therefore, h2 = 301.5 + 800 = 1101.5 kJ/kg.
3. Calculate the exit enthalpy of the air:
The exit enthalpy (h3) can be determined by assuming an isentropic process, using the isentropic efficiency (η) of the nozzle.
h3s = h2 + (h2s - h2) / η, where h2s is the theoretical exit enthalpy for an isentropic process.
h2s can be calculated using the equation for isentropic expansion:
h2s = h1 + (v2^2 - v1^2) / 2, where v1 and v2 are the specific volumes at the inlet and exit, respectively.
Since the process is adiabatic, the specific volumes can be related using the ideal gas equation:
v1 = R * T1 / P1, and v2 = R * T3 / P3, where T3 is the final temperature and P3 is the final pressure (1.0 bar).
Rearranging and substituting the values, we have:
h2s = h1 + (R * T3 / P3 - R * T1 / P1)^2 / 2.
Substituting the values, h2s = 301.5 + (0.287 * 300 / 1.0 - 0.287 * 300 / 8.0)^2 / 2.
Solving the equation gives h2s = 489.8 kJ/kg.
Now, substituting the values in the equation for h3s:
h3s = 1101.5 + (489.8 - 1101.5) / 0.8.
Solving the equation gives h3s = 1208.4 kJ/kg.
4. Calculate the exit velocity of the air:
The exit velocity (V3) can be calculated using the specific enthalpies:
h3 = Cp * T3 + V3^2 / 2.
Rearranging the equation, we have:
V3^2 = 2 * (h3 - Cp * T3).
Substituting the values, V3^2 = 2 * (1208.4 - 1.005 * 300).
Solving the equation gives V3 = 443.8 m/s.
Therefore, the velocity of the air at the nozzle exit is approximately 443.8 m/s.
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