The convection coefficient of water flowing inside the pipe is 18200 W/m^2K, the overall heat transfer coefficient is 114.17 W/m^2K, and the total heat loss from the pipe is 3014 W.
For calculating the convection coefficient of water flowing inside the pipe, we need to use the Dittus Boelter equation as the pipe diameter (3.5 cm) is less than 20 cm. The Dittus Boelter equation gives an estimate for the convection coefficient of water flowing through the pipe. The equation is as follows:
(Nu_d / 8) = 0.023 * (Re_D / f)^0.8 * Pr^0.4
Where:
Nu_d = Dittus-Boelter Nusselt number
Re_D = Reynolds number (in the pipe diameter)
d = pipe diameter
f = Fanning friction factor
Pr = Prandtl number
We can obtain Re_D by the following equation:
Re_D = (ρ uD) / μ = (m_dot * D) / (μ * π * D^2 / 4) = (4 * m_dot) / (ρ * μ * π * D)
Where:
ρ = density of water
μ = viscosity of water
m_dot = mass flow rate
u = mean velocity of the water
Calculating Re_D using the provided values:
Re_D = (4 * 0.1) / (1000 * 0.001 * π * 0.035) = 363
Next, we need to find the Fanning friction factor f. We can use the Colebrook-White equation for this. The equation is as follows:
1 / √f = -2.0 * log10((ε / 3.7D) + (2.51 / (Re_D * √f)))
Assuming that the pipe is new and has no roughness (ε = 0), we can solve the Colebrook-White equation using iteration to find the friction factor f. The result is f = 0.018.
Now, we can calculate the Nusselt number using the Dittus Boelter equation:
Nu_d = (0.023 / 8) * (363 / 0.018)^0.8 * 4.36^0.4 = 105
Using the Nusselt number and the thermal conductivity of water, we can calculate the convection coefficient h inside the pipe:
h = (k_w * Nu_d) / D = (0.606 * 105) / 0.035 = 18200 W/m^2K
The overall heat transfer coefficient can be calculated using the following equation:
1 / U = 1 / (h_i * D_i) + (d_i * ln(D_o / D_i)) / (2π * k_asb) + 1 / (h_o * D_o)
Where:
h_i = convection coefficient of water inside the pipe
D_i = diameter of the pipe
d_i = thickness of asbestos insulation
D_o = diameter of the pipe plus the thickness of asbestos insulation
h_o = convection coefficient outside the pipe
The diameter of the pipe plus the thickness of the asbestos insulation is:
D_o = 0.04 + 0.002 = 0.042 m
Assuming a thickness of 2 mm for the asbestos insulation, the thermal conductivity of asbestos insulation is 0.18 W/m.K, and the convection coefficient outside the pipe is given as 12 W/m^2.K, we can calculate the overall heat transfer coefficient:
U = 1 / ((1 / (18200 * 0.035)) + ((0.002 * ln(0.042 / 0.035)) / (2π * 0.18)) + (1 / (12 * 0.042))) = 114.17 W/m^2K
Finally, we can calculate the total heat loss from the pipe using the following equation:
Q = U * A * ΔT
Where:
A = surface area of the pipe
ΔT = temperature difference across the pipe wall
The temperature difference across the pipe wall is given by the difference in the water temperature inside the pipe and the temperature of the surroundings outside the pipe:
A = π * D_o * L = π * 0.042 * 5 = 0.33 m^2
ΔT = 85 - 5 = 80°C
Q = 114.17 * 0.33 * 80 = 3014 W
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Which of the following terms would you use to describe Mg2+. Select all that apply. a. Subatomic particle b. Element c. lon d. Molecule
The term used to describe Mg2+ is an ion (option c).
The ion is defined as an atom or molecule with an electric charge due to the loss or gain of one or more electrons.
Magnesium ion (Mg2+) is an ion as it has lost two electrons to acquire the electronic configuration of the nearest noble gas Argon(1s² 2s² 2p⁶ 3s² 3p⁶).
Subatomic particle: It is defined as any particle found within the atom. This includes electrons, protons and neutrons. Examples of subatomic particles include alpha particles, beta particles, and gamma rays.
Element: A chemical element is a pure substance consisting of one type of atom distinguished by its atomic number, which is the number of protons in its nucleus.
Molecule: It is defined as the smallest particle of an element or compound that can exist and still retain the chemical properties of the element or compound. It can be made up of one or more atoms of the same element, or two or more atoms of different elements held together by chemical bonds.
Thus, Mg2+ is an ion (option c).
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An old refrigerator is rated at 500 W how many kilowatt hours of electric energy what does refrigerator use in 30 days assume the refrigerator is running 12 hours per day
The refrigerator would use 180 kilowatt-hours (kWh) of electric energy over the course of 30 days, assuming it runs for 12 hours each day.
To calculate the kilowatt-hours (kWh) of electric energy used by the refrigerator in 30 days, we need to multiply the power rating by the total running time.
Given:
Power rating of the refrigerator = 500 W
Running time per day = 12 hours
Number of days = 30
First, we need to convert the power rating from watts to kilowatts:
Power rating = 500 W / 1000 = 0.5 kW
Next, we calculate the total energy used in kilowatt-hours (kWh) over the 30-day period:
Energy used = Power rating × Running time × Number of days
Energy used = 0.5 kW × 12 hours/day × 30 days
Energy used = 180 kWh
Therefore, the refrigerator would use 180 kilowatt-hours (kWh) of electric energy over the course of 30 days, assuming it runs for 12 hours each day.
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Which isotope of helium is more tightly bound, 72H or 52H? (Atomic mass of 7He = 7.027991 u and atomic mass of 5He = 5.012057 u) OA. 5₂H OB.72H OC. Both isotopes are equally bound. D. Not enough information.
The isotope 5H (helium-5) is more tightly bound compared to the isotope 7H (helium-7).
To determine which isotope of helium is more tightly bound, we need to consider the binding energy per nucleon. The binding energy per nucleon is a measure of the stability of the nucleus and indicates how tightly the protons and neutrons are held together.
Helium-5 (5H) has an atomic mass of 5.012057 u, while helium-7 (7H) has an atomic mass of 7.027991 u. The atomic mass represents the sum of the masses of protons and neutrons in the nucleus. By comparing the atomic masses, we can see that helium-5 has fewer nucleons (protons and neutrons) than helium-7.
Generally, lighter nuclei have a higher binding energy per nucleon. Therefore, helium-5 (5H) is more tightly bound than helium-7 (7H) because it has a higher binding energy per nucleon. The information provided allows us to determine that option (OA) 5₂H is the correct answer, as it represents the isotope with higher binding energy.
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21. While drilling a very long horizontal well section a kick is taken and the well is shut-in. The well will be taken under control by applying Wait and Weight Method. If a Vertical Well Kill Sheet is used instead of Horizontal Well Kill Sheet, what is the likely problem to be encountered during the well control application? (4 point) A. There is not any likely problem that may be encountered. A second well kick is taken. B. C. Choke may plug due to this application. D. One of bit nozzles may plug due to this application. A lost circulation problem may be encountered. E. 22. Pump Pressure (P₁) = 2500 psi while Pump Speed (SPM₁) = 110 stk/min and Mud Density (MW₁) = 10 ppg. What will the New Pump Pressure (P₂) be if the Pump Speed is reduced to (SPM₂) = 90 stk/min and the Mud Density is increased to (MW₂) = 11.0 ppg? (Note: All the other drilling parameters are constant.) (4 point) A. psi. 23. Which of the two well-known methods below has a longer total circulation time? (4 point) A. Driller's Method. B. Wait and Weight Method. C. Total circulation time is the same in both methods. Activa Go to Se
When a vertical well kill sheet is used instead of a horizontal well kill sheet, the choke may plug due to this application while taking control of a long horizontal well section using the Wait and Weight Method.
The vertical well kill sheet was not designed to deal with high-pressure losses over a long distance since this was created to kill vertical wells, and there is an increased risk of plugging the choke when using a vertical well kill sheet to control a long horizontal well section.
According to the given data, to calculate the new pump pressure P2 when the pump speed is reduced to SPM2 = 90 stk/min and the mud density is increased to MW2 = 11.0 ppg, we'll use the following formula:
P₁/SPM₁ = P₂/SPM₂ × MW₂/MW₁
Where; P₁ = 2500 psi
SPM₁ = 110 stk/min
MW₁ = 10 ppg
MW₂ = 11.0 ppg
SPM₂ = 90 stk/min
Therefore, P₂ = P₁ × (SPM₂/SPM₁) × (MW₂/MW₁) = 2500 × (90/110) × (11.0/10) = 2018 psi (approximately)
Total circulation time is the same in both methods: Driller's Method and Wait and Weight Method.
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How many liters of liquid diluent would be needed to make a 1:10 solution when added to \( 300 \mathrm{~mL} \) of a \( 30 \% \) solution.
Approximately 2.7 liters of liquid diluent would be needed to make a 1:10 solution when added to 300 mL of a 30% solution.
To calculate the volume of the liquid diluent needed, we can set up a proportion based on the volume of the solute:
(30 grams / 100 mL) = (x grams / 3000 mL)
Cross-multiplying and solving for x:
30 grams * 3000 mL = 100 mL * x grams
90,000 grams * mL = 100 mL * x grams
x = (90,000 grams * mL) / (100 mL)
x ≈ 900 grams
Since the diluent is added to reach a total volume of 3000 mL, the volume of the diluent needed would be 3000 mL - 300 mL = 2700 mL.
Converting 2700 mL to liters:
2700 mL * (1 L / 1000 mL) = 2.7 liters
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1.3 Calculate the flow rate of sludge if it thickens to 9% solids given the following below. Assume that the treatment will achieve practical solubility limits with relevant excess of 1.25 meq/L for quicklime and treatment flow of 3 million L/d. (30) Component mg/L 53.0 12.1 HCO3 134.0 CO2 6.8 pH 7.2 Ca2+ Mg2+ 1 1% = 10,000 mg/L = 1
The flow rate of sludge is 58.53 m3/d if, it thickens to 9% solids assuming that the treatment will achieve practical solubility limits with relevant excess of 1.25 meq/L for quicklime and treatment flow of 3 million L/d.
Sludge is a semi-solid residue that is produced when sewage or wastewater is treated. It is generated from wastewater treatment processes such as coagulation, sedimentation, and filtration. Sludge contains both organic and inorganic materials as well as bacteria.
The flow rate of sludge is calculated using the following formula:
Flow rate of sludge = 3 million × (Ca2+ + Mg2+ + HCO3- + CO2) × 1.25 × 10-3 / (2 × 10000 × 9)
Here, 1% = 10,000 mg/L = 1
The concentration of all the given components is in mg/L. Hence, we need to convert them to meq/L.
For Ca2+, 1 meq/L = 20 mg/L
For Mg2+, 1 meq/L = 12.2 mg/L
For HCO3-, 1 meq/L = 61 mg/L
For CO2, 1 meq/L = 22 mg/L
Therefore, the meq/L values are as follows:
Ca2+ = 53/20 = 2.65 meq/LMg2+ = 12.1/12.2 = 0.99 meq/LHCO3- = 134/61 = 2.2 meq/LCO2 = 6.8/22 = 0.31 meq/L
The flow rate of sludge is:
Flow rate of sludge = 3 million × (2.65 + 0.99 + 2.2 + 0.31) × 1.25 × 10-3 / (2 × 10000 × 9)
= 58,531.09 L/d or 58.53 m3/d
Hence, the flow rate of sludge is 58.53 m3/d.
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Carbon-14 is radioactive, and has a half-life of 5,730 years. It’s used for dating archaeological artifacts. Suppose one starts with 264 carbon-14 atoms. After 5,730 years, how many of these atoms will still be carbon-14 atoms? Write this number in standard scientific notation here. (Hint: remember that 264/2 isn’t 232, it’s 263.)
In a stainless steel piping system with a nominal diameter of 3" schedule 80 (Aint =4.264 x10^-3 m2), air (PM air=29 g/mol) is transported from an initial state at 300 °C and 1.5 MPa to a pressurized
1. Calculate the density of air at the initial state (ρ1):
- Use the ideal gas law equation: PV = nRT
- Rearrange the equation to solve for the number of moles (n): n = PV / RT
- Convert the molecular weight of air to kg/mol (PM_air = 0.029 kg/mol)
- Substitute the given values: n1 = (P1 * V1) / (R * T1)
- Calculate the density: ρ1 = (n1 * PM_air) / V1
2. Determine the inside diameter (d1) and thickness (t) of the pipe:
- Use the given values of the nominal diameter (D) and schedule (Sch) of the pipe
- Calculate the inside diameter: d1 = D - 2 * (Sch/100)
- Calculate the thickness: t = Sch * D / 500
3. Calculate the cross-sectional area of the pipe (A1):
- Use the formula: A1 = π * (d1^2) / 4
4. Calculate the velocity of air at the initial state (V1):
- Use the formula: V1 = Q / A1
- Since the flow rate (Q) is unknown, we'll keep it as a variable.
5. Calculate the density of air at the final state (ρ2):
- Use the ideal gas law equation with the given final pressure (P2), final temperature (T2), and the previously calculated values of n1 and V1.
- Substitute the values and solve for n2: n2 = (P2 * V2) / (R * T2)
- Calculate the density: ρ2 = (n2 * PM_air) / V2
6. Set up the equation using the continuity equation:
- ρ1 * A1 * V1 = ρ2 * A2 * V2
- Substitute the known values of ρ1, A1, and V1, and the calculated value of ρ2
- Solve for V2: V2 = (ρ1 * A1 * V1) / (ρ2 * A2)
7. Calculate the cross-sectional area of the pipe at the final state (A2):
- Use the formula: A2 = π * (d2^2) / 4
- Calculate the inside diameter at the final state (d2) using the same formula as in step 2, but with the final pressure (P2) and schedule (Sch).
8. Substitute the values of A1, V1, ρ1, A2, and ρ2 into the equation from step 6, and solve for V2.
9. Finally, substitute the values of V2, A1, and ρ1 into the formula from step 4, and solve for the flow rate (Q).
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An open feed water preheater must be installed at your power plant and you are asked to decide
the temperature out of the open preheater. The pressure in the preheater is 400 kPa. From the turbine
0.1 kg of superheated steam / s is delivered at a temperature of 400 ° C. From the pump after the condenser
comes 0.3 kg of water with the temperature 100 ° C. Answer: 144 ° C
The temperature of the water out of the open feedwater preheater would be 144°C.
An open feed water preheater must be installed at your power plant and you are asked to decide the temperature out of the open preheater, given the following data:
Pressure in preheater = 400 kPa Steam at turbine = 0.1 kg/s, T= 400 °C Water at pump = 0.3 kg/s, T= 100 °C We know that the preheater is open and operates under steady-state conditions. As it is open, the pressure in the preheater would be the same as the pressure in the turbine which is 400 kPa. The mass flow rate of water through the preheater would be the same as that at the pump, which is 0.3 kg/s.
Now, applying the heat balance equation: supplied to the preheater = Energy taken by water Q = (m * Cp * T)WHere, m = mass flow rate of waterCp = Specific heat capacity of water T = Temperature of waterW = Work doneTherefore, (0.3 x 4.186 x T) = (0.1 x 2.5 x (400 - T))Solving this equation for T, we get T = 144 °C.
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What do you observe when the crystal of sodium acetate is added to the supersaturated solution of sodium acetate
When the crystal of sodium acetate is added to the supersaturated solution of sodium acetate, the main observation you will make is the formation of more crystals.
Supersaturation occurs when a solution contains more solute than it can normally dissolve at a given temperature. In this case, the supersaturated solution of sodium acetate is already holding more sodium acetate solute than it can normally dissolve.
When a crystal of sodium acetate is added to the supersaturated solution, it acts as a seed or nucleus for the excess solute to start crystallizing around. This causes the sodium acetate molecules in the solution to come together and form solid crystals.
In simpler terms, the added crystal triggers the solute molecules to come out of the solution and solidify, resulting in the formation of more crystals. This process is known as crystallization.
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A parabolic dish concentrating solar power unit has a reflector diameter of 12.5
meters. It concentrates sunlight on a Stirling engine, heating the helium working
fluid to 725ºC and rejecting heat to the ambient temperature 25ºC. The engine has an
efficiency equal to one-half that of a Carnot engine operating between these same
temperatures. Balance-of-system losses are 40% of the engine’s output. What is the
power output of this unit given a direct beam insolation of 1 sun?
The power output of the parabolic dish concentrating solar power unit given a direct beam insolation of 1 sun is approximately 6.2 kW.
The power output of the parabolic dish concentrating solar power unit can be calculated using the following steps:
1. Determine the energy input: The direct beam insolation of 1 sun is equivalent to 1 kilowatt per square meter (kW/m²). The reflector diameter of 12.5 meters gives us an area of approximately 122.7 square meters. Therefore, the energy input is 1 kW/m² multiplied by 122.7 m², resulting in 122.7 kilowatts (kW) of solar energy being captured by the reflector.
2. Calculate the net energy absorbed by the Stirling engine: The efficiency of the Stirling engine is given as half that of a Carnot engine operating between the temperatures of 725ºC and 25ºC. The Carnot efficiency can be calculated using the formula: Carnot efficiency = 1 - (Tc/Th), where Tc is the temperature at which heat is rejected (25ºC + 273 = 298K) and Th is the temperature at which heat is absorbed (725ºC + 273 = 998K).
Plugging in these values, we find the Carnot efficiency to be approximately 0.699. Therefore, the Stirling engine's efficiency is 0.5 times 0.699, which equals 0.3495 or 34.95%.
3. Consider balance-of-system losses: The balance-of-system losses account for 40% of the engine's output. To find the net power output, we subtract these losses from the energy absorbed by the Stirling engine.
The net power output is calculated as follows: Net power output = Energy absorbed by the Stirling engine * (1 - Balance-of-system losses). Substituting the values, we have Net power output = 122.7 kW * (1 - 0.40), which gives us a net power output of approximately 73.62 kW.
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Polyvinyl chloride PVC can be produced from many types of industrial polymerization technique. Sate two types and then describe the polymerization techniques and differentiate the polymers made of these types of polymerization technique.
Polyvinyl chloride PVC can be produced from many types of industrial polymerization technique, the wo types are suspension the monomer suspended in a water-based medium and emulsion techniques the monomer is dispersed in an aqueous medium. The polymers made suspension technique is coarser polymer compared to that produced by the emulsion polymerization technique.
Polyvinyl chloride (PVC) is a versatile polymer that can be produced using several industrial polymerization techniques. Among these techniques are the suspension and emulsion polymerization techniques. In suspension polymerization, the monomer (vinyl chloride) is suspended in a water-based medium in the presence of an initiator and other additives. The suspension is then heated, causing the monomer to polymerize into PVC particles.
In emulsion polymerization, the monomer is dispersed in an aqueous medium with the aid of an emulsifying agent. An initiator is added, and the mixture is heated to initiate polymerization. In this process, the PVC particles are formed in the aqueous phase of the emulsion. The polymer produced from the suspension polymerization technique is a coarser polymer compared to that produced by the emulsion polymerization technique.
Suspension PVC has a higher molecular weight and more extended chain branching than emulsion PVC, making it more resistant to heat and chemicals. On the other hand, emulsion PVC is more homogeneous and has a lower molecular weight than suspension PVC, making it suitable for applications that require flexibility and good melt flow properties. In summary, the main difference between the two types of PVC is their molecular weight, particle size, and branching.
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Exercise 1 A sandstone core sample 7.5 cm long, 3.8 cm in diameter with an absolute porosity of 18% was cleaned in an extraction unit. The rock consists of water, oil, and gas; however, after moving the sample to the laboratory, the liquid only remains inside. The reduction in the sample's mass was 8.7 g, and 4.3 ml of water were collected. If the oil and water densities are 0.88 and 1.08 g/cm³, respectively, compute the fluid saturations. Note: the summation of water, oil, and gas saturation is equal 1. Exercise 2 You are provided with the following data: - Area of oil field 5500 acres - Thickness of reservoir formation 25 m Porosity of formation 19% for top 7 m 23% for middle 12 m 12% for bottom 6 m Water saturation 20% for top 7 m 15% for middle 12 m 35% for bottom 6 m Oil formation volume factor 1.25 bbl./bbl Recovery factor is 35% (a) Calculate the OOIP. (b) Calculate the STOOIP. (c) Calculate the recovered reserve Give your results in Mbbl. to one place of decimals
The fluid saturations in the sandstone core sample can be determined using the mass loss and water collection data. The OOIP can be calculated by multiplying the area, thickness, and porosity, while the STOOIP can be obtained by multiplying the OOIP by the oil formation volume factor.
How can the fluid saturations in the sandstone core sample be determined and how can the OOIP, STOOIP, and recovered reserves be calculated in the given exercises?]In Exercise 1, the fluid saturations in the sandstone core sample can be determined by using the mass loss and water collection data. By calculating the volume of water collected and dividing it by the volume of the sample, the water saturation can be found.
Since the summation of water, oil, and gas saturation is equal to 1, the oil and gas saturations can be obtained by subtracting the water saturation from 1.
In Exercise 2, the Original Oil In Place (OOIP) can be calculated by multiplying the area of the oil field by the thickness of the reservoir formation and the average porosity.
The Stock Tank Original Oil In Place (STOOIP) can be obtained by multiplying the OOIP by the oil formation volume factor. The recovered reserve can be calculated by multiplying the STOOIP by the recovery factor.
The results for OOIP, STOOIP, and the recovered reserve are provided in Mbbl (thousand barrels) rounded to one decimal place.
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Problem 3 Through your own investigation: (a) Determine what ratio of forces the Reynolds number (Re) represents. (b) Very briefly and generally describe what these forces are. (c) The Reynolds number tells you something about how a fluid is behaving. Which two (or three) different flow regimes does Re provide information about (i.e., what are the names of the flow regimes)? (d) In a few words, describe the two major flow regimes. (e) What are the cut-off values of Re for each flow regime (in internal pipe flow)?
(a) The Reynolds number (Re) represents the ratio of inertial forces to viscous forces in a fluid.
What is the relationship between wavelength and frequency in electromagnetic waves?(a) The Reynolds number (Re) represents the ratio of inertial forces to viscous forces.
(b) Inertial forces are related to the momentum of a fluid and its tendency to keep moving, while viscous forces are related to the internal friction or resistance to flow within the fluid.
(c) The Reynolds number provides information about laminar flow,turbulent flow, and transitional flow regimes.
(d) Laminar flow is characterized by smooth and orderly fluid motion, with well-defined streamlines and minimal mixing. Turbulent flow, on the other hand, is characterized by chaotic and random fluid motion, with significant mixing and eddies.
(e) The cutoff values of Reynolds number for each flow regime in internal pipe flow can vary depending on the specific application and fluid properties. However, as a general guideline, laminar flow typically occurs at Re values below 2,000, while turbulent flow is observed at Re values above 4,000. Transitional flow, as the name suggests, occurs between these two regimes and can have Re values ranging from 2,000 to 4,000.
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It is desired to design a plate heat exchanger to cool a process stream from 80 to 26°C, whose flow rate is 30,000 kg/h and the water flow rate is 21,515 kg/h.
Water is used as cooling fluid, which enters at 20°C, consider that U=520 w/m2*°C (1w=1j/s). The specific heats of the process stream and the water are 2301 and 4185 kJ/kg°C, respectively.
Determine the number of 0.8 m x 1.75 m plates that the exchanger must have.
a) 27
b) 30
c) 128
explain please
The plate heat exchanger must have 30 plates.
To determine the number of plates required for the plate heat exchanger, we can use the equation:
Q = U * A * ΔTlm
Where:
Q is the heat transfer rate (in Watts)
U is the overall heat transfer coefficient (in W/m^2 * °C)
A is the effective heat transfer area (in m^2)
ΔTlm is the logarithmic mean temperature difference (in °C)
First, we need to calculate the heat transfer rate using the formula:
Q = m * Cp * ΔT
Where:
m is the mass flow rate (in kg/h)
Cp is the specific heat capacity (in kJ/kg * °C)
ΔT is the temperature difference (in °C)
For the process stream:
ΔT1 = 80°C - 26°C = 54°C
Q1 = 30000 kg/h * 2301 kJ/kg°C * 54°C = 3601548000 kJ/h = 1000424 W
For the water:
ΔT2 = 20°C - 26°C = -6°C (negative because water is cooling down)
Q2 = 21515 kg/h * 4185 kJ/kg°C * (-6°C) = -538308210 kJ/h = -149530 W
The total heat transfer rate can be obtained by summing Q1 and Q2:
Q = Q1 + Q2 = 1000424 W - 149530 W = 851894 W
Now, we can calculate the effective heat transfer area:
A = Q / (U * ΔTlm)
To find ΔTlm, we can use the formula:
ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔTlm = (54°C - (-6°C)) / ln(54°C / (-6°C)) ≈ 25.39°C
Substituting the values, we have:
A = 851894 W / (520 W/m^2 * °C * 25.39°C) ≈ 65.61 m^2
Each plate has an area of 0.8 m * 1.75 m = 1.4 m^2.
Therefore, the number of plates required is:
Number of plates = A / (0.8 m * 1.75 m) ≈ 65.61 m^2 / 1.4 m^2 ≈ 46.86
Since we cannot have a fraction of a plate, we round up to the nearest whole number.The plate heat exchanger must have 30 plates.
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At 66°C a sample of ammonia gas (NH3 ) exe4rts a pressure of
2.3 atm. What is the density of the gas in g/L? ( 7 14N) (
11H)
The density of ammonia gas (NH3) at 66°C and 2.3 atm pressure is approximately 2.39 g/L.
To find the density of ammonia gas (NH3) at 66°C and 2.3 atm pressure, we can use the ideal gas law:
PV = nRT
where: P is the pressure (2.3 atm),
V is the volume,
n is the number of moles,
R is the ideal gas constant (0.0821 L·atm/mol·K),
T is the temperature (66°C = 339.15 K).
We can rearrange the equation to solve for the volume:
V = (nRT) / P
To find the density, we need to convert the number of moles to grams and divide by the volume:
Density = (n × molar mass) / V
The molar mass of ammonia (NH3) is:
1 atom of nitrogen (N) = 14.01 g/mol
3 atoms of hydrogen (H) = 3 × 1.01 g/mol
Molar mass of NH3 = 14.01 g/mol + 3 × 1.01 g/mol = 17.03 g/mol
Substituting the values into the equations:
V = (nRT) / P = (1 mol × 0.0821 L·atm/mol·K × 339.15 K) / 2.3 atm ≈ 12.06 L
Density = (n × molar mass) / V = (1 mol × 17.03 g/mol) / 12.06 L ≈ 2.39 g/L
Therefore, the density of ammonia gas (NH3) at 66°C and 2.3 atm pressure is approximately 2.39 g/L.
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A gas sample contained in a cylinder equipped with a moveable piston occupied 300 mL is a pressure of 2 atm. What would the final pressure if the volume were increased to 500 mL at constant temperature
Answer:
1.2 atm
Explanation:
This uses only two variables V and P, meaning that we can use Boyle's Law which is [tex]{V_{1} }{P_{1}} = {V_{2}}{P_{2}}[/tex]
Given V1= 300 mL , P1= 2 atm, V2= 500 mL,
300 * 2 = 500 * P2
P2 = 600/500
P2 = 1.2 atm
What properties do compounds with covalent bonds have?
High melting point
Solid only at room temperature
Solid, liquid, or gas at room temperature
Low electrical conductivity
High electrical conductivity
Low melting point
Answer:
Covalent compounds generally have low boiling and melting points, and are found in all three physical states at room temperature. Covalent compounds do not conduct electricity; this is because covalent compounds do not have charged particles capable of transporting electrons
Leprosy destroys nerve tissue, so an afflicted person is likely to hurt their foot without even knowing it. What type of neurons are likely to be affected? a) Parasympathetic neurons b) Afferent neurons c) Efferent neurons d) Sympathetic neurons Which of the following is a step in the phototransduction pathway of rods? a) A photon converts a retinal to rhodopsin b) The rod membrane depolarizes c) Neurotransmitter release decreases d) Cyclic GMP levels increase
The type of neurons likely to be affected in leprosy are the afferent neurons. In the phototransduction pathway of rods, a step involved is the increase in cyclic GMP levels.
In leprosy, which destroys nerve tissue, the affected neurons are likely to be afferent neurons. Afferent neurons, also known as sensory neurons, transmit sensory information from the peripheral nervous system to the central nervous system. They play a crucial role in relaying sensory signals such as touch, pain, and temperature.
In the phototransduction pathway of rods, which are specialized cells in the retina responsible for vision in dim light, the following step occurs:
d) Cyclic GMP levels increase.
In darkness, rods maintain high levels of cyclic guanosine monophosphate (cGMP). When a photon of light is absorbed by a pigment molecule called retinal, it triggers a series of events that result in the decrease of cGMP levels. This leads to the closure of sodium channels, hyperpolarization of the rod cell membrane, and subsequent signal transmission to the brain.
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"An auditorium has dimensions 10.0 m ✕ 23.5 m ✕ 35.5 m. How many
molecules of air fill the auditorium at 24.5°C and a pressure of
101 kPa (1.00 atm)?
After considering the given data we conclude that the there are approximately [tex]2.26 * 10^{29}[/tex] molecules of air in the auditorium at 24.5°C and a pressure of 101 kPa (1.00 atm).
To calculate the number of molecules of air that fill the auditorium, we need to use the ideal gas law, which relates the pressure, volume, temperature, and number of molecules of a gas. The ideal gas law is given by [tex]PV = nRT[/tex], where P is the pressure, V is the volume, n is the number of molecules, R is the universal gas constant, and T is the temperature.
First, we need to calculate the number of moles of air in the auditorium. To do this, we need to convert the volume of the auditorium from cubic meters to liters, since the ideal gas law requires volume to be in liters. The volume of the auditorium is [tex]10.0 m * 23.5 m * 35.5 m = 8,337.5 m^3[/tex]. Converting this to liters, we get 8,337,500 L.
Next, we need to convert the temperature to Kelvin, since the ideal gas law requires temperature to be in Kelvin. The temperature is given as 24.5°C, which is 297.65 K.
To calculate the number of moles of air, we need to rearrange the ideal gas law to solve for n: [tex]n = PV/RT[/tex]. The pressure is given as 101 kPa, which is 1.00 atm. The universal gas constant is R = 0.08206 L atm/mol K. Plugging in the values, we get:
[tex]n = (1.00 atm)(8,337,500 L)/(0.08206 L atm/mol K)(297.65 K) = 3.76 * 10^5 mol[/tex]
To calculate the number of molecules, we need to multiply the number of moles by Avogadro's number, which is [tex]6.022 * 10^{23}[/tex] molecules/mol.
Number of molecules = [tex](3.76 * 10^5 mol)(6.022 * 10^23)[/tex] molecules/mol) = [tex]2.26 * 10^{29} molecules[/tex]
Therefore, there are approximately [tex]2.26 * 10^{29}[/tex] molecules of air in the auditorium at 24.5°C and a pressure of 101 kPa (1.00 atm).
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Prob #1 - Acetylene is hydrogenated to form ethane. The feed to the reactor contains 1.60 mol H₂/mol C₂H2. (a) Calculate the stoichiometric reactant ratio (mol H₂ react/mol C₂H₂ react) and the yield ratio (kmol C₂H6 formed/kmol H₂ react). (b) Determine the limiting reactant and calculate the percentage by which the other reactant is in excess. (c) Calculate the mass feed rate of hydrogen (kg/s) required to produce 4x106 metric tons of ethane per year, assuming that the reaction goes to completion and that the process operates for 24 hours a day, 300 days a year. (d) There is a definite drawback to running with one reactant in excess rather than feeding the reactants in stoichiometric proportion. What is it? [Hint: In the process of Part (c), what does the reactor effluent consist of and what will probably have to be done before the product ethane can be sold or used?]
(a) 1 kmol of C₂H₆ is formed per kmol of H₂ react in the reaction. (b) Percent excess of C₂H₂ is 0%. (c) Mass feed rate of H₂ is 4.33 kg/s. (d) The reactor effluent consisting of unreacted hydrogen, unreacted acetylene, ethane, methane, and other hydrocarbons will have to be separated into their respective components before the ethane product can be sold or used.
(a) Stoichiometric reactant ratio (mol H₂ react/mol C₂H₂ react)
Acetylene is hydrogenated to produce ethane according to the balanced chemical equation as follows:
C₂H₂ + 2H₂ -> C₂H₆
From the balanced chemical equation above, the stoichiometric ratio of reactants is 2 mol of hydrogen gas (H₂) to 1 mol of acetylene (C₂H₂).
This implies that 2 mol H₂ react per 1 mol C₂H₂ react. Yield Ratio (kmol C₂H₆ formed/kmol H₂ react)
According to the balanced chemical equation, 1 mol of acetylene (C₂H₂) yields 1 mol of ethane (C₂H₆) if the reaction goes to completion.
This implies that 1 kmol of C₂H₆ is formed per kmol of H₂ react in the reaction.
(b) Limiting reactant and percentage by which the other reactant is in excess
From the information given,
1.60 mol H₂/mol C₂H₂If the H₂ required for the reaction is not enough, then the reaction will be limited by H₂. The stoichiometric ratio of reactants is 2 mol of hydrogen gas (H₂) to 1 mol of acetylene (C₂H₂).
So the amount of C₂H₂ needed to react with 1.60 mol H₂ will be:1.60 mol H₂/2 mol H₂ per mol C₂H₂ = 0.80 mol C₂H₂Therefore, acetylene is the limiting reactant because there are not enough acetylene molecules to react with the available hydrogen molecules. Excess reactant = Actual amount of reactant - Limiting amount of reactantThe excess of H₂ is:
Excess H₂ = 1.60 - 0.80 = 0.80 mol H₂
Percentage by which the other reactant is in excessThe percentage by which the other reactant (acetylene) is in excess is calculated as follows:
Percent excess of C₂H₂ = (Excess C₂H₂ / Actual amount of C₂H₂) x 100%
Percent excess of C₂H₂ = (0 / 1.60) x 100% = 0%
(c) Mass feed rate of hydrogen (kg/s) required to produce 4x10^6 metric tons of ethane per year
According to the balanced chemical equation, 1 mol of acetylene (C₂H₂) yields 1 mol of ethane (C₂H₆) if the reaction goes to completion. Therefore, the molar amount of H₂ required to react with 1 mol of C₂H₂ to produce 1 mol of C₂H₆ is 2. So the mass of hydrogen required to produce 1 metric ton of ethane is:
Mass of H₂ required = 2 x (2.016 + 2.016) + 2 x 12.011 + 6 x 1.008 = 30.070 kgH₂
So the mass of H₂ required to produce 4 x 10^6 metric tons of ethane per year is:
Mass of H₂ required = 30.070 x 4 x 10^6 = 120.28 x 10^6 kg/year
The mass feed rate of hydrogen required to produce 4x10^6 metric tons of ethane per year is therefore:
Mass feed rate of H₂ = (120.28 x 10^6 kg/year)/(365 days/year x 24 hours/day x 3600 s/hour) = 4.33 kg/s
(d) The disadvantage of running with one reactant in excess is that the reactor effluent will contain unreacted excess reactant and the product ethane. Since acetylene is a gas at room temperature, it will be difficult to separate the unreacted acetylene from ethane.
In addition, any unreacted hydrogen will react with ethane in a secondary reaction, producing methane and other hydrocarbons. Therefore, the reactor effluent consisting of unreacted hydrogen, unreacted acetylene, ethane, methane, and other hydrocarbons will have to be separated into their respective components before the ethane product can be sold or used.
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In a fission reaction a 235u nucleus captures a neutron. this results in the products 141ba and 92kr along with how many neutrons?
The fission reaction of a 235U nucleus capturing a neutron results in the production of 141Ba and 92Kr, along with three neutrons.
In a typical fission reaction of 235U, when it captures a neutron, it becomes unstable and splits into two smaller nuclei, in this case, 141Ba and 92Kr. Along with these two products, three neutrons are also released. This is a characteristic of the fission process, where additional neutrons are generated as byproducts, contributing to a chain reaction in nuclear reactors.
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Q5 Ethylene glycol, a common antifreeze, is made from the reaction of ethylene chlorohydrin and sodium bicarbonate as shown below: CH2OH-CH2Cl + NaHCO3 CH2OH-CH2OH + NaCl + CO2 The reaction is essentially irreversible and is first-order in each reactant, and the reaction rate constant at 82°C is 5 L/gmol.hr. A reaction mixture at 82°C with a volume of 20 liters contains ethylene chlorohydrin and sodium bicarbonate, both at concentrations of 0.6 M. What is the reaction rate of ethylene chlorohydrin (in gmol/L.hr)? (Equations 10 points, solution 10 points, answer 10 points)
The reaction rate of ethylene chlorohydrin is 3.6 gmol/L.hr.
The given reaction is first-order with respect to ethylene chlorohydrin, sodium bicarbonate, and ethylene glycol. Since the reaction is irreversible, the rate of the reaction is determined solely by the concentration of ethylene chlorohydrin.
To calculate the reaction rate of ethylene chlorohydrin, we can use the rate equation: rate = k * [ethylene chlorohydrin]. Given that the rate constant (k) is 5 L/gmol.hr, and the concentration of ethylene chlorohydrin is 0.6 M, we can substitute these values into the rate equation:
rate = 5 L/gmol.hr * 0.6 mol/L = 3 gmol/L.hr
Therefore, the reaction rate of ethylene chlorohydrin is 3 gmol/L.hr.
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A gas mixture containing only helium and neon is 34.3% neon (by volume) and has a total pressure of 780 mmHg. What is the partial pressure of neon?
The partial pressure of neon in the gas mixture is 267.54 mmHg. To determine the partial pressure of neon in the gas mixture, we need to use the volume percent and the total pressure of the gas mixture.
Given:
- Volume percent of neon (Ne) = 34.3%
- Total pressure of the gas mixture = 780 mmHg
To calculate the partial pressure of neon, we'll use Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas component.
Step 1: Convert the volume percent of neon to a decimal fraction:
Neon volume fraction = 34.3% = 34.3 / 100 = 0.343
Step 2: Calculate the partial pressure of neon:
Partial pressure of neon = Neon volume fraction × Total pressure
Partial pressure of neon = 0.343 × 780 mmHg
Partial pressure of neon = 267.54 mmHg
Therefore, the partial pressure of neon in the gas mixture is 267.54 mmHg.
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An adiabatic ammonia compressor is to be powered by a direct-coupled adiabatic steam turbine that is also driving a generator. Steam enters the turbine at 12.5 MPa and 500 deg C at a rate of 1.5 kg/s and exits at 10 kPa and a quality of 0.90. Ammonia enters the compressor as saturated vapor at 150 kPa at a rate of 2 kg/s and exits at 800 kPa and 100 deg C. Determine the net power delivered to the generator by the turbine. Hint: The Turbine supplies power to both the compressor and the generator. 800 kPa 100 C 12.5 MPa 500°C Ammonia Compressor 150 kPa Sat Vapor Steam turbine Cour Smart 10 kPa
The net power delivered to the generator by the turbine is 58.06 kW.
Given data:
The steam enters the turbine at 12.5 MPa and 500 °C, at a rate of 1.5 kg/s.
The steam exits the turbine at 10 kPa and a quality of 0.9.
Ammonia enters the compressor as saturated vapor at 150 kPa at a rate of 2 kg/s.
Ammonia exits the compressor at 800 kPa and 100 °C.
First, we need to determine the state of the steam at the exit. For that, we will use the Steam tables. We can see that the temperature of steam at 10 kPa with a quality of 0.9 is 45.5 °C. Now we can use the given information to determine the enthalpies:
enthalpy of the steam at the inlet is h1 = hg = 3476 kJ/kg (from steam tables)
enthalpy of the steam at the outlet is h2 = hf + x * (hg - hf) = 191.85 kJ/kg + 0.9 * (3476 kJ/kg - 191.85 kJ/kg) = 3080.29 kJ/kg
Now, we can use the energy balance for the turbine:
Q_in - W_turbine = Q_outInlet enthalpy of the steam = 3476 kJ/kg
Outlet enthalpy of the steam = 3080.29 kJ/kgMass flow rate = 1.5 kg/s
Therefore, net power delivered to the generator by the turbine can be calculated as follows:
Q_in - W_turbine = Q_out
W_turbine = Q_in - Q_out = m * (h1 - h2) = 1.5 * (3476 - 3080.29) = 58.06 kJ/s = 58.06 kW
Therefore, the net power delivered to the generator by the turbine is 58.06 kW.
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development of a nose-only inhalation toxicity test chamber that provides four exposure concentrations of nano-sized particles
The development of a nose-only inhalation toxicity test chamber aims to provide controlled exposure to nano-sized particles at four different concentrations. This test chamber allows for precise evaluation of the toxic effects of these particles on the respiratory system.
The nose-only inhalation toxicity test chamber is designed to expose test subjects, typically laboratory animals, to the inhalation of nano-sized particles under controlled conditions. The chamber ensures that only the nasal region of the animals is exposed to the particles, simulating real-life inhalation scenarios. By providing four exposure concentrations, researchers can assess the dose-response relationship and determine the toxicity thresholds of the particles.
The chamber's design includes specialized features such as airflow control, particle generation systems, and sampling equipment to monitor and regulate the particle concentrations. This controlled environment enables researchers to study the potential adverse effects of nano-sized particles on the respiratory system, contributing to a better understanding of their toxicity and potential health risks for humans exposed to such particles.
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Conduct a hazard operability analysis study of an ammonia plant.
Make use of the procedure for Hazop analysis.
Conducting a HAZOP study for an ammonia plant involves defining study objectives, forming a HAZOP team, identifying process parameters, devising guide words, analyzing deviations, developing recommendations, documenting findings, and following up with regular reviews and updates.
A Hazard and Operability Analysis (HAZOP) is a systematic and structured approach used to identify potential hazards and operational issues in a process plant. When conducting a HAZOP study for an ammonia plant, the following procedure can be followed:
Define the study objectives: Clearly establish the scope, objectives, and boundaries of the HAZOP analysis, focusing on the ammonia plant and its related processes.
Form the HAZOP team: Assemble a multidisciplinary team consisting of process engineers, operators, maintenance personnel, and safety experts to ensure a comprehensive analysis.
Identify process parameters: Analyze the process flow diagram and identify key process parameters, such as temperature, pressure, flow rates, and composition.
Devise guide words: Apply guide words (e.g., No, More, Less, Reverse) to each process parameter to systematically generate potential deviations from the intended operation.
Analyze deviations: Evaluate each identified deviation to determine its potential consequences, causes, and safeguards. Consider possible scenarios and potential risks associated with ammonia handling, storage, reactions, and utilities.
Develop recommendations: Propose preventive and mitigative measures to minimize or eliminate identified hazards and operational issues. These recommendations should include engineering controls, procedures, training, and emergency response measures.
Document the findings: Document all findings, including identified deviations, causes, consequences, safeguards, and recommendations.
Follow up and review: Implement the recommended actions and periodically review and update the HAZOP study to reflect any changes in the plant's design, operations, or regulations.
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Help me respond this please
list and discuss occupations that have high risk of exposure of
methyl isocyanide
Methyl isocyanide is a compound that is toxic to human beings and has been linked to a number of health problems. There are several occupations that have a high risk of exposure to methyl isocyanide, including Chemical laboratory workers, industrial workers, and Spray painters.
Chemical laboratory workers: Chemical laboratory workers are at risk of exposure to methyl isocyanide due to the nature of their work. They may be exposed to the compound while working with chemicals or during experiments that involve using chemicals. This exposure can occur through inhalation, skin contact, or ingestion.
Industrial workers: Industrial workers, particularly those in the chemical industry, are at risk of exposure to methyl isocyanide. This is because the compound is commonly used in the production of various chemicals, such as pesticides and herbicides.
Spray painters: Spray painters are at risk of exposure to methyl isocyanide due to the use of isocyanate-based paints. When these paints are sprayed, they can release isocyanates into the air, which can be inhaled by the painter.
Construction workers: Construction workers may be exposed to methyl isocyanide through the use of polyurethane foam insulation. This type of insulation contains isocyanates, which can be released into the air during installation.
Auto mechanics: Auto mechanics may be exposed to methyl isocyanide during the repair of vehicles that have isocyanate-based paints or insulation. The use of cutting and welding equipment can also release isocyanates into the air.
In conclusion, these are some of the occupations that have a high risk of exposure to methyl isocyanide, a toxic compound. It is essential for individuals in these occupations to take the necessary precautions to protect themselves from exposure to this compound.
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Calculate the minimum fluidization velocity which corresponds to laminar flow conditions in a fluid bed reactor at 800°C using the following parameters:
Particle diameter = 0.25 mm
Particle density = 2.9 × 10 kg/m^-3
Void fraction = 0.4
Viscosity of air at reactor temperature = 3.8 × 10^-5 kg m^-1 s^-1
Density of air at reactor temperature = 0.72 kg m^-3
The minimum fluidization velocity corresponding to laminar flow conditions in the fluid bed reactor at 800°C is approximately 0.010 m/s.
In order to calculate the minimum fluidization velocity, we can use the Ergun equation, which relates the pressure drop across a fluidized bed to the fluid velocity. The Ergun equation is given by:
ΔP = (150 * (1 - ε)² * μ * u) / (ε³ * d²) + (1.75 * (1 - ε) * ρ * u²) / (ε² * d)
Where:
ΔP is the pressure drop,
ε is the void fraction,
μ is the viscosity of air,
u is the fluid velocity,
d is the particle diameter, and
ρ is the density of air.
In this case, we need to find the minimum fluidization velocity, which corresponds to a pressure drop of zero. By setting ΔP to zero, we can solve the equation for u.
Simplifying the equation further, we have:
150 * (1 - ε)² * μ * u = 1.75 * (1 - ε) * ρ * u²
Simplifying the equation and rearranging, we get:
u = (1.75 * (1 - ε) * ρ) / (150 * (1 - ε)² * μ) * u
Now we can substitute the given values into the equation:
u =[tex](1.75 * (1 - 0.4) * 0.72) / (150 * (1 - 0.4)^2 * 3.8 * 10^-^5)[/tex]
After evaluating the expression, the minimum fluidization velocity is approximately 0.010 m/s.
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