Homework #4 thorough answers, ungraded PDF

Preview

Summary

Homework #4 - 7.1, 7.2, 7.4, 7.5, 7.12, 7.14, 7.18, 7.20, 7.24, 7.26, 7.28, 7.31, ungraded. But I remember acing the assignment....

Description

EGR 320

Homework #4

7.1) The specific gravity of a liquid is 0.920. Find the density and specific weight of the liquid in SI and English units. DU = density of unknown liquid WU = specific weight of unknown liquid sg = specific gravity = 0.920 DW = density of water = 999.97 kg/m3 = 1.9403 slugs/ft3 WW = specific weight of water = 9.807 kN/m3 = 62.4 lb/ft3 SI units: DU = sg * DW = 0.920(999.97 kg/m3) = 919.9724 kg/m3 WU = sg * WW = 0.920(9.807 kN/m3) ≈ 9.0224 kN/m3 Imperial units: DU = sg * DW = 0.920(1.9403 slugs/ft3) ≈ 1.7851 slugs/ft3 WU = sg * WW = 0.920(62.4 lb/ft3) = 9.02 kN/m3 = 57.408 kN/m3

7.2) A 12-cm diameter cylindrical can is filled to a depth of 10 cm with motor oil (ρ = 878 kg/m3). Find the mass and weight of the motor oil. r = radius = 6 cm h = height = 10 cm V = volume = πr²h = π * 62 * 10 = π * 36 * 10 = π * 360 ≈ 1130.9734 ρ=m/V M = mass = density * volume = ρV = 0.878 g/cm3)(1130.9734 cm3) ≈ 992.9946 g ≈ 0.9930 kg W = weight = mass * gravity = mg ≈ (992.9946 g)(981 cm/s2) ≈ 974127.7026 dynes ≈ 974.1277 kilodynes ≈ 9.7413 N Answer: M ≈ 0.9930 kg W ≈ 9.7413 N

7.4) A 5-meter diameter spherical balloon contains hydrogen. If the density of the hydrogen is ρ = 0.0830 kg/m3, what is the mass and weight of the hydrogen in the balloon? volume of balloon = (4 / 3)πr3 = (4 / 3)π2.53 = (4 / 3)(15.625)π ≈ 65.4498 m3 M = D * V ≈ 0.0830 * 65.4498 ≈ 5.4323 kg W = M * G ≈ 5.4323 * 9.81 ≈ 53.2909 N Answer: M ≈ 5.4323 kg W ≈ 53.2909 N

7.5) Find the volume of mercury (sg = 13.55) that weighs the same as 0.04 m3 of ethyl alcohol (sg = 0.802). SGM = 13.55 SGA = 0.802 VA = 0.04 m3 M A = MM DW = density of water DA = density of ethyl alcohol DM = density of mercury D A = MA / VA M A = DA * V A D M = MM / VM M M = DM * VM Since MA = MM, you know that DA * VA = DM * VM VM = (DA * VA) / DM SGA = DA / DW SGM = DM / DW SGA / SGM = (DA / DW) / (DM / DW) = DA / DM VM = VA * (SGA / SGM) VM = 0.04 m3 * (0.802 / 13.55) VM = 0.00237 m3 = 2.37 liters

7.12)

Problem statement: The deepest known point in the oceans of the earth is the Mariana Trench, east of the Philippines, with a depth of approximately 10.9 km. Taking the specific gravity of seawater as sg = 1.030, what is the pressure at the bottom of the Mariana Trench? Express your answer in kPa, psi, and atmospheres. Diagram:

Assumptions: The depth is ≈10.9 km (the final answer will be an approximation). The specific gravity of seawater is exactly 1.030. Governing equations: P=D*G*H P = pressure D = density G = gravity H = height (depth) x kPa ≈ (9.8692 * 10-3 * x) atmospheres x atmospheres ≈ (x * 0.1450) psi 1 kPa = 1000 Pa 1 km = 1000 m Calculations:

P = pressure = density * gravity * depth ρ = density = 1.030 = 1030 kg/m3 g = gravity = 9.81 m/s2 d = depth = 10.9 km P = 1030 * 9.81 * 10900 = 110136870 Pa = 110136.87 kPa ≈ 110136.87 * 9.8692 * 10-3 ≈ 1086.9628 atmospheres ≈ 110136.87 * 0.1450 ≈ 15969.8462 psi Answer: P = 110136.87 kPa ≈ 1086.9628 atmospheres ≈ 15969.8462 psi Solution check: Doing my calculations backwards, I get: 1569.8462 / 0.1450 ≈ 110136.87 1086.9628 / (9.8692 * 10-3) ≈ 110136.87 110136870 / 10900 ≈ 1030 * 9.81 110136870 / 9.81 ≈ 1030 * 10900 110136870 / 1030 ≈ 9.81 * 10900 The answer checks out. Discussion: In the Mariana Trench, which has a depth of approximately 10.9 km, and the specific gravity of seawater being sg = 1.030, the pressure at the very bottom is approximately 110136.87 kPa, which is approximately 1086.9629 atmospheres, which is approximately 15969.8462 psi.

7.14) Problem statement: A storage tank containing heavy fuel oil (ρ = 906 kg/m3) is filled to a depth of 5 m. Find the gauge pressure at the bottom of the tank. Diagram:

Assumptions: The heavy fuel oil in the storage tank has a density of exactly 906 kg/m3. It is filled exactly 5 meters high. Governing equations: Pressure = ρ * G * H ρ = density G = gravity H = height 1 kPa = 1000 N/m2 Calculations: Pressure = ρ * g * h ≈ 906 * 9.81 * 5 ≈ 44439.3 N/m2 ≈ 44.4393 kPa Answer: ≈ 44.4393 kPa Solution check: Doing the calculations backwards, I get: 44.4393 * 1000 = 44439.3 44439.3 ≈ 906 * 9.81 * 5 The answer checks out.

Discussion: Given a storage tank filled to a depth of 5 meters of heavy fuel oil (ρ = 906 kg/m3), the gauge pressure at the bottom of the tank is approximately 44.4393 kPa.

7.18) Problem statement: Calculate the gauge pressure at the bottom of an open 2-liter container full of soft drink. Diagram:

Assumptions: The soft drink has the same density as water (999.97 kg/m3). The 2-liter container is 25 cm in height. Governing equations: P=D*G*H P = pressure D = density G = gravity H = height 1 kPa = 1000 Pa 1 m = 100 cm

Calculations: P=D*G*H P = 999.97 * 9.81 * 0.25 ≈ 2452.4264 Pa ≈ 2.4524 kPa Solution check: 2.4524 * 1000 = 2452.4 2452.4 ≈ 999.97 * 9.81 * 0.25 Answer checks out. Discussion: Assuming a soft drink has the same density as water (999.97 kg/m3), and it is in a 2-liter container that is 25 cm in height, then the gauge pressure at the bottom of the open 2-liter container, when full of the soft drink, is approximately 2.4524 kPa.

7.20) Problem statement: A glass tube carries mercury at an average velocity of 30 cm/s. If the inside diameter of the tube is 4.0 mm, find the volume flow rate and mass flow rate. Diagram:

Assumptions: The average velocity of mercury through the glass tube is 30 cm/s. The inside diameter of the tube is exactly 4.0 mm throughout the entire tube. The density of mercury is approximately 13534 (and so the mass flow rate, which the density is used to calculate, will be an approximation). Governing equations: Volume flow rate:

VFR = V * A V = velocity A = cross-sectional area MFR = ṁ = mass flow rate = D * A * V D = density A = area V = velocity A = πr2 A = cross-sectional area r = radius 1 m = 100 cm = 1000 mm Calculations: velocity = 30 cm/s = 0.3 m/s diameter = 4 mm radius = diameter / 2 mm = 4 mm / 2 mm = 2 mm = 0.002 m area (cross-sectional) = πr2 = π * 0.0022 ≈ 1.2566 * 10-5 m2 volume flow rate = 0.3 * 1.2566 * 10-5 ≈ 3.7699 * 10-6 m3/s mass flow rate = ṁ = density * area * velocity ≈ 13534 * 1.2566 * 10-5 * 0.3 ≈ 0.0510 kg/s (density of mercury is approximately 13534 kg/m3) Answer: volume flow rate ≈ 3.7699 * 10-6 m3/s mass flow rate = ṁ ≈ 0.0510 kg/s Solution check: The final answers seem reasonable for this scenario. Doing the calculations backwards, this is what I get: 0.0510 ≈ 13534 * 1.2566 * 10-5 * 0.3 3.7699 * 10- 6 ≈ 0.3 * 1.2566 * 10-5 1.2566 * 10-5 ≈ π * 0.0022 0.002 m = 2 mm 2 mm = 4mm / 2mm 30 cm = 0.3 m Answer checks out. Discussion: If a glass tube carries mercury at an average velocity of 30 cm/s and the inside diameter of the tube is 4.0 mm consistently throughout, then the volume flow rate is approximately 3.7699 * 10-6 m3/s and the mass flow rate is approximately 0.0510 kg/s. This is assuming the density of mercury is approximately

13534.

7.24) Problem statement: A ventilation duct provides fresh filtered air to a clean room where semiconductor devices are manufactured. The cross section of the filter medium is 1.30 m x 1.1 m. If the volume flow rate of air to the clean room is 3.75 m3/s, find the average velocity of the air as it passes through the filter. If ρ = 1.194 kg/m3, for the air, find the mass flow rate. Diagram:

Assumptions: The cross-section of the filter medium is exactly 1.30 m x 1.1 m. The volume flow rate of air to the clean room is exactly 3.75 m3/s. The density of the air is exactly 1.194 kg/m3. Governing equations: volume flow rate = VFR = velocity * area velocity = VFR / area MFR = ṁ = D * A * V D = density A = area V = velocity Calculations:

V = velocity cross section area = 1.30 m * 1.1 m = 1.43 m2 volume flow rate = velocity * area 3.75 = V * 1.43 V ≈ 2.6224 m/s density of air = 1 kg/m3 If density = 1.194 kg/m3: mass flow rate = density * area * velocity ≈ 1.194 * 1.43 * 2.6224 ≈ 4.4775 kg/s Answer: velocity ≈ 2.6224 m/s ṁ ≈ 4.4775 kg/s Solution check: The answers seem reasonable given the scenario. Doing the calculations backwards, this is what I get: 4.4775 ≈ 1.194 * 1.43 * 2.6224 3.75 ≈ 2.6224 * 1.43 1.43 = 1.30 * 1.1 Answer checks out. Discussion: Given a ventilation duct whose filter medium has across section of 1.30 m x 1.1 m, and whose volume flow rate of air to the room is 3.75 m3/s, the average velocity of air through the room is approximately 2.6224 m/s. The mass flow rate is approximately 4.4775 kg/s when the density of the air is 1.194 kg/m3.

7.26) Problem statement: A nozzle is a device that accelerates the flow of a fluid. A circular nozzle that converges from an inside diameter of 5 cm to 3.5 cm carries a gas at a volume flow rate of 0.10 m3/s. Find the change in average velocity of the gas. Diagram:

Assumptions: The circular nozzle converges from an inside diameter of exactly 5 cm to exactly 3.5 cm. The volume flow rate of the gas is exactly 0.10 m3/s. Governing equations: R=D/2 R = radius D = diameter A = πr2 A = cross-sectional area r = radius volume flow rate = VFR = velocity * area velocity = VFR / area Calculations: 0.035 / 2 = 0.0175 0.05 / 2 = 0.025 π * 0.01752 ≈ 9.6211 * 10-4 π * 0.0252 ≈ 1.9635 * 10-3 0.1 / (9.6211 * 10-4) – 0.1 / (1.9635 * 10-3) ≈ 53.0088 m/s

average velocity ≈ 53.0088 m/s Solution check: The answer seems reasonable given the scenario. Doing my calculations backwards, I get: 53.0088 ≈ 0.1 / (9.6211 * 10-4) – 0.1 / (1.9635 * 10-3) 1.9635 * 10-3 ≈ π * 0.01752 9.6211 * 10-4 ≈ π * 0.0252 0.025 * 2 = 0.05 0.0175 * 2 = 0.035 Answer checks out. Discussion: Given a circular nozzle that converges from an inside diameter of 5 cm to 3.5 cm that carries a gas at a volume flow rate of exactly 0.10 m3/s, the change in average velocity of the gas is approximately 53.0088 m/s.

7.28) Problem statement: A pipe branch is illustrated in schematic form in Figure P7.28. Find the mass flow rate ṁ4. Does the fluid in branch 4 enter or leave the junction? Diagram:

Assumptions: The mass flow rates shown in the diagram are exact, including the 10 kg/s, 6 kg/s and 18 kg/s. Governing equations: Simple addition/subtraction algebra required to find flow rate. That and the only other concept needed to solve this problem is knowing the quantity of mass flow rates entering a system is equal to that exiting the system. Calculations: (10 + 6) = 18 + ṁ4 16 = 18 + ṁ4 -2 = ṁ4 ṁ4 = 2 kg/s inward Solution check: The result seems reasonable given the system. After doing the calculations backwards, the answer is consistent: 10 + 6 = 18 + (-2) 16 = 16 Checks out. Discussion: Given a pipe branch with two mass flow rates inward of 10 kg/s and 6 kg/s, and one outward of 18 kg/s, the last pipe must have an inward flow of 2 kg/s. Therefore, the fluid enters the junction at 2 kg/s.

7.31) Problem statement: The mixing chamber shown in P7.31 facilitates the blending of three liquids. Find the volume flow rate and mass flow rate of the mixture at the exit of the chamber. Diagram:

Assumptions: The mass flow rates entering the mixing chamber are exactly 40 kg/s, 25 kg/s, and one initially unknown value. The fluid with an unknown mass flow rate has a velocity of exactly 7 m/s and specific gravity of 0.85. The cylinder the fluid enters the chamber in has a diameter of exactly 6 cm. The density of the fluid going through the exit chamber is exactly 925 kg/m3. There are no other entrances or exits of fluid for this chamber other than the four just mentioned. Governing equations: R=D/2 R = radius D = diameter A = πr2

A = area r = radius volume flow rate = VFR = A * V A = area V = velocity mass flow rate = VFR * sg VFR = volume flow rate sg = specific gravity 1 m = 100 cm Calculations: ṁ1 = 40 kg/s ṁ3 = 25 kg/s R2 = D2 / 2 = 0.06 / 2 = 0.03 A2 = area of flow = π * (R2)2 = 0.032π = 0.0009π ≈ 2.8274 * 10-3 m2 Volume flow rate = A2 * velocity ≈ 2.8274 * 10-3 * 7 ≈ 1.9792 * 10-2 m3/s Mass flow rate = ṁ2 = VFR2 * sg2 ≈ 1.9792 * 10-2 * 0.85 ≈ 1.6823 * 10-2 ≈ 16.823 kg/s Exit mass flow rate ≈ 40 + 16.823 + 25 ≈ 81.823 kg/s Exit volume flow rate = exit mass flow rate / density ≈ 81.823 / 925 ≈ 8.8457 * 10-2 m3/s Answer: Exit mass flow rate ≈ 81.823 kg/s Exit volume flow rate ≈ 8.8457 * 10-2 m3/s Solution check: The answers seem reasonable given the scenario. When doing the calculations backwards, this is what I get: 8.8457 * 10-2 ≈ 81.823 / 925 81.823 = 40 + 16.823 + 25 1.6823 * 10-2 ≈ 1.9792 * 10-2 * 0.85 1.9792 * 10-2 = 2.8274 * 10-3 * 7 2.8274 * 10-3 = 0.032π 0.03 * 2 = 0.06 The answer checks out. Discussion: Given a mixing chamber where three liquids are mixed and sent through one exit, the first entrance of which has a mass flow rate of 40 kg/s and the third of which has a mass flow rate of 25 kg/s, and the

second of which has an initially unknown mass flow rate, but has a velocity of 7 m/s, a diameter (of the cylinder the fluid passes through) of 6 cm, and a specific gravity of 0.85, and the fluid exits the chamber through the only exit at a density of 925 kg/m3, the unknown mass flow rate (for the second entrance) is approximately 16.823 kg/s and the exiting volume flow rate is approximately 8.8457 * 10-2 m3/s. This is all under the assumption there are only these three entrances and one exit to the mixing chamber....