Answer: Seal Edges. Use a 6-inch taping knife to shove fiberglass tape into inside corners, then press down both sides firmly.
Explanation:
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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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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