The total pressure drop due to friction in the series water piping system at a flow rate of 250 L/s is 23.12 meters.
To calculate the total pressure drop, we need to determine the friction losses in each section of the piping system and then add them together. The Hazen Williams Equation is commonly used for this purpose.
In the first step, we calculate the friction loss in the 400-mm diameter pipe. Using the Hazen Williams Equation, the friction factor can be calculated as follows:
f = (C / (D^4.87)) * (L / Q^1.85)
where f is the friction factor, C is the Hazen Williams coefficient (130 in this case), D is the pipe diameter (400 mm), L is the pipe length (1000 m), and Q is the flow rate (250 L/s).
Substituting the values, we get:
f = (130 / (400^4.87)) * (1000 / 250^1.85) = 0.000002224
Next, we calculate the friction loss using the Darcy-Weisbach equation:
ΔP = f * (L / D) * (V^2 / 2g)
where ΔP is the pressure drop, f is the friction factor, L is the pipe length, D is the pipe diameter, V is the flow velocity, and g is the acceleration due to gravity.
For the 400-mm pipe:
ΔP1 = (0.000002224) * (1000 / 400) * (250 / 0.4)^2 / (2 * 9.81) = 7.17 meters
Similarly, we calculate the friction loss for the 600-mm pipe:
f = (130 / (600^4.87)) * (1500 / 250^1.85) = 0.00000134
ΔP2 = (0.00000134) * (1500 / 600) * (250 / 0.6)^2 / (2 * 9.81) = 15.95 meters
Finally, we add the friction losses in each section to obtain the total pressure drop:
Total pressure drop = ΔP1 + ΔP2 = 7.17 + 15.95 = 23.12 meters
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Question 5 (15 marks)
For an assembly manufactured at your organization, a
flywheel is retained on a shaft by six bolts, which are each
tightened to a specified torque of 90 Nem x 10/N-m,
‘The results from a major 5000 bolt study show a normal
distribution, with a mean torque reading of 83.90 N-m, and a
standard deviation of 1.41 Nm.
2. Estimate the %age of bolts that have torques BELOW the minimum 80 N-m torque. (3)
b. Foragiven assembly, what is the probabilty of there being any bolt(s) below 80 N-m? (3)
¢. Foragiven assembly, what isthe probability of zero bolts below 80 N-m? (2)
Question 5 (continued)
4. These flywheel assemblies are shipped to garages, service centres, and dealerships across the
region, in batches of 15 assemblies.
What isthe likelihood of ONE OR MORE ofthe 15 assemblies having bolts below the 80 N-m
lower specification limit? (3 marks)
. Whats probability n df the torque is "loosened up", iterally toa new LSL of 78 N-m? (4 marks)
The answer to the first part, The standard deviation is 1.41 N-m.
How to find?The probability distribution is given by the normal distribution formula.
z=(80-83.9)/1.41
=-2.77.
The percentage of bolts that have torques below the minimum 80 N-m torque is:
P(z < -2.77) = 0.0028
= 0.28%.
Thus, there is only 0.28% of bolts that have torques below the minimum 80 N-m torque.
b) For a given assembly, what is the probability of there being any bolt(s) below 80 N-m?
The probability of there being any bolt(s) below 80 N-m is given by:
P(X < 80)P(X < 80)
= P(Z < -2.77)
= 0.0028
= 0.28%.
Thus, there is only a 0.28% probability of having bolts below 80 N-m in a given assembly.
c) For a given assembly, what is the probability of zero bolts below 80 N-m?The probability of zero bolts below 80 N-m in a given assembly is given by:
P(X ≥ 80)P(X ≥ 80) = P(Z ≥ -2.77)
= 1 - 0.0028
= 0.9972
= 99.72%.
Thus, there is a 99.72% probability of zero bolts below 80 N-m in a given assembly.
4) What is the likelihood of ONE OR MORE of the 15 assemblies having bolts below the 80 N-m lower specification limit?
The probability of having one or more of the 15 assemblies with bolts below the 80 N-m lower specification limit is:
P(X ≥ 1) =
1 - P(X = 0)
= 1 - 0.9972¹⁵
= 0.0418
= 4.18%.
Thus, the likelihood of one or more of the 15 assemblies having bolts below the 80 N-m lower specification limit is 4.18%.
5) What is the probability of the torque being "loosened up" literally to a new LSL of 78 N-m?
The probability of the torque being loosened up to a new LSL of 78 N-m is:
P(X < 78)P(X < 78)
= P(Z < -5.74)
= 0.0000
= 0%.
Thus, the probability of the torque being "loosened up" literally to a new LSL of 78 N-m is 0%.
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Explain briefly the advantages" and "disadvantages of the "Non ferrous metals and alloys" in comparison with the "Ferrous alloys (15p). Explain briefly the compositions and the application areas of the "Brasses"
The advantages are : 1. Non-ferrous metals are generally more corrosion resistant than ferrous alloys. 2. They are also more lightweight and have a higher melting point. 3. Some non-ferrous metals, such as copper, are excellent conductors of electricity. The disadvantages are : 1. Non-ferrous metals are typically more expensive than ferrous alloys. 2. They are also more difficult to machine and weld. 3. Some non-ferrous metals, such as lead, are toxic.
Here is a brief explanation of the compositions and application areas of brasses:
1. Brasses are copper-based alloys that contain zinc.
2. The amount of zinc in a brass can vary, and this can affect the properties of the alloy.
3. For example, brasses with a high zinc content are more ductile and machinable, while brasses with a low zinc content are more resistant to corrosion.
4. Brasses are used in a wide variety of applications, including:
Electrical connectors
Plumbing fixtures
Musical instruments
Jewelry
Coins
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ie lbmol of pentane gas (C₅H₁₂) reacts with the theoretical amount of air in a closed, rigid tank. Initially, the reactants are at 77°F, 1 m. After complete combustion, the temperature in the tank is 1900°R. Assume air has a molar analysis of 21% O₂ and 79% N₂. Determine the heat transfer, in Btu. Q = i Btu
The heat transfer, Q, can be calculated using the equation:
Q = ΔHc + ΔHg. To determine the heat transfer in Btu for the given scenario, we need to calculate the heat released during the combustion of pentane and the subsequent increase in temperature of the gases in the tank.
Where ΔHc is the heat released during combustion and ΔHg is the heat gained by the gases in the tank due to the increase in temperature. To calculate ΔHc, we need to determine the moles of pentane reacted and the heat of combustion per mole of pentane. Since pentane reacts with air, we also need to consider the moles of oxygen available in the air. The heat of combustion of pentane can be obtained from reference sources. To calculate ΔHg, we can use the ideal gas law and the given initial and final temperatures, along with the molar analysis of air, to determine the change in enthalpy. By summing up ΔHc and ΔHg, we can obtain the total heat transfer, Q, in Btu. It's important to note that the actual calculations involve several steps and equations, including stoichiometry, enthalpy calculations, and gas laws. The specific values and formulas needed for the calculations are not provided in the question, so an exact numerical result cannot be determined without that information.
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