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[Fluid Mechanics for Chemical Engineers 유체역학] 05장 연습문제 풀이
| 자료번호 | e1334122 | ||
|---|---|---|---|
| 수정일 | 2015.04.22 | 등록일 | 2015.04.22 |
| 페이지수 | 15Page | 파일형식 |  워드(doc) |
| 판매자 | pa**** | 가격 | 1,200원 |
프리미엄자료
[유체역학] 나노유체 연구 조사 및 적용방안 연구 - 나노유체의 열전달- 유체 (Nano sized copper particle coated with Alkanethiol self assembled monolayer in nanofluids) 오일 기반의 고효율 나노유체 및 그 제조방법 (Oil-based Nanofluid with High Thermal Efficiency and Method for Preparation Thereof) (1) 나노유체를 이용한 냉각 (2) 원자력발전소 열교환
- [실험] 배관계의 유체 마찰손실 측정 실험
- (m) REFERENCE ★유체역학 - 박 이동 저서 ★Fudamentals of fluid mechanics - P 실험. 배관계의 유체 마찰손실 측정 실험 1 . 서론 ⑴목적 2 . 이론 (1) 유량의 개념 3.실험 (1)실험장치 (2)실험 방법 4. 결과 및 고찰 (1)결과 (2)고찰 5. 결론
소개글
Fluid Mechanics for Chemical Engineers, 5장 연습문제 풀이입니다.
목차
PROBLEM
5.2. Show that in the head form of B.E. each term has the dimension of a length.
5.5. Rework Example 5.2, calculating the density from the formula ρ=MP_avg/RT where P_avg is 0.5(P_1+P_atm). Compare the results with those shown in Table 5.1.
5.9. Repeat Prob. 5.7, except that now the horizontal cross-sectional area of the tank is 5 ft2.
5.14. The tank in Figure 5.24 is cylindrical with a diameter of 10m. The outlet is a cylindrical frictionless nozzle, with diameter 1m. The top of the tank is open to the atmosphere. When the level in the take is 10 m above the centerline of the outlet, how fast is the level in the tank falling?
5.18.* in the tank Fig 5.28, water is under a layer of compressed air that is at a pressure of 20 psig. The water is flowing out through a frictionless nozzle that 5 ft below the water surface. What is the velocity of the water?
5.20. The system in Fig 5.29 consists of a water reservoir with a layer of compressed air above the water and a large pipe and nozzle. The pressure of the air is 50 psig, and the effects of friction can be neglected. What is the velocity of the water flowing out through the nozzle?
5.21.*The tank in Fig. 5.30 has a later of mercury under a layer of water. The mercury is flowing out through a frictionless nozzle. What is the velocity of the fluid leaving the nozzle?
5.23.* An industrial centrifuge is sketched in Fig 5.32. The fluid in the basket is water. The radii are r_1=21 in and r_2=20 in. The basket is revolving at 2000 rpm. There is a small hole in the outer wall of the centrifuge through which the fluid is flowing in frictionless flow. What is the velocity of flow through this hole.
5.29. A pitot tube, connected to a bourdon-tube pressure gauge, is used to measure the speed of a boat. The tube is just below the waterline and faces directly forward. When the boat is going 60 km/h, what is the reading of the pressure gauge?
5.32. In the venture meter shown Fig 5.10, the flowing fluid is air, the manometer fluid is water, (D2/D1)=0.5, and the manometer reading is 1ft. Estimate the velocity at point 2.
5.33. Then ventrui meter in Example 5.8 is now set at 30o to the horizontal, as in Fig 5.12. The flowing fluid is gasoline. The gauge at point 1 read 7 psig, and the gauge at point 2 reads 5 psig. The difference in elevation between the gauges, (z_3-z_4), is 1 ft. What is the volumetric flow rate of gasoline?
5.41. Mercury is flowing at 1 ft/s in a 1-in diameter pipe. We want to select a drilled plate to insert in the pipe so that the pressure-drop signal across it will be 3 psig. What diameter should we select for the orifice hole?
5.43. For a siphon similar to that sketched in Fig 5.16, we want the fluid to have a velocity of 10ft/s in siphon pipe. If we assume that flow is frictionless and that the minimum pressure allowable is 1 psia, what is the maximum height that the top of the siphon may have above the liquid surface level?
5.58. For frictionless pumps and compressors pumping constant-density fluids, the required work is given by Eq. 5.11. If the fluid is an ideal gas, then that equation becomes
5.66. A meteorologist, discussing a record-breaking hurricane said, “It had a pressure of 850 millibars in the center, so it had winds of 250 miles an hour!” Explain this statement in terms of B.E.
5.2. Show that in the head form of B.E. each term has the dimension of a length.
5.5. Rework Example 5.2, calculating the density from the formula ρ=MP_avg/RT where P_avg is 0.5(P_1+P_atm). Compare the results with those shown in Table 5.1.
5.9. Repeat Prob. 5.7, except that now the horizontal cross-sectional area of the tank is 5 ft2.
5.14. The tank in Figure 5.24 is cylindrical with a diameter of 10m. The outlet is a cylindrical frictionless nozzle, with diameter 1m. The top of the tank is open to the atmosphere. When the level in the take is 10 m above the centerline of the outlet, how fast is the level in the tank falling?
5.18.* in the tank Fig 5.28, water is under a layer of compressed air that is at a pressure of 20 psig. The water is flowing out through a frictionless nozzle that 5 ft below the water surface. What is the velocity of the water?
5.20. The system in Fig 5.29 consists of a water reservoir with a layer of compressed air above the water and a large pipe and nozzle. The pressure of the air is 50 psig, and the effects of friction can be neglected. What is the velocity of the water flowing out through the nozzle?
5.21.*The tank in Fig. 5.30 has a later of mercury under a layer of water. The mercury is flowing out through a frictionless nozzle. What is the velocity of the fluid leaving the nozzle?
5.23.* An industrial centrifuge is sketched in Fig 5.32. The fluid in the basket is water. The radii are r_1=21 in and r_2=20 in. The basket is revolving at 2000 rpm. There is a small hole in the outer wall of the centrifuge through which the fluid is flowing in frictionless flow. What is the velocity of flow through this hole.
5.29. A pitot tube, connected to a bourdon-tube pressure gauge, is used to measure the speed of a boat. The tube is just below the waterline and faces directly forward. When the boat is going 60 km/h, what is the reading of the pressure gauge?
5.32. In the venture meter shown Fig 5.10, the flowing fluid is air, the manometer fluid is water, (D2/D1)=0.5, and the manometer reading is 1ft. Estimate the velocity at point 2.
5.33. Then ventrui meter in Example 5.8 is now set at 30o to the horizontal, as in Fig 5.12. The flowing fluid is gasoline. The gauge at point 1 read 7 psig, and the gauge at point 2 reads 5 psig. The difference in elevation between the gauges, (z_3-z_4), is 1 ft. What is the volumetric flow rate of gasoline?
5.41. Mercury is flowing at 1 ft/s in a 1-in diameter pipe. We want to select a drilled plate to insert in the pipe so that the pressure-drop signal across it will be 3 psig. What diameter should we select for the orifice hole?
5.43. For a siphon similar to that sketched in Fig 5.16, we want the fluid to have a velocity of 10ft/s in siphon pipe. If we assume that flow is frictionless and that the minimum pressure allowable is 1 psia, what is the maximum height that the top of the siphon may have above the liquid surface level?
5.58. For frictionless pumps and compressors pumping constant-density fluids, the required work is given by Eq. 5.11. If the fluid is an ideal gas, then that equation becomes
5.66. A meteorologist, discussing a record-breaking hurricane said, “It had a pressure of 850 millibars in the center, so it had winds of 250 miles an hour!” Explain this statement in terms of B.E.
본문내용
[Fluid Mechanics for Chemical Engineers 유체역학] 05장 연습문제 풀이
PROBLEM
5.2. Show that in the head form of B.E. each term has the dimension of a length.
The head form of Bernoulli’s equation
∆(P/ρg+z+V^2/2g)=dW_(n.f.)/gdm-F/g (5.6)
P/ρg 의 dimension=[N/m^2 ]/([kg/m^3 ]∙[m/s^2 ] )=[(kg∙m)/s^2 ∙1/m^2 ]/([kg/m^3 ]∙[m/s^2 ] )=[m]
z의 dimension=[m]
V^2/2g 의 dimension=([m/s^2 ])^2/[m/s^2 ] =[m]
dW_(n.f.)/gdm 의 dimension=([J])/([m/s^2 ]∙[kg] )=[N∙m]/([m/s^2 ]∙[kg] )=[(kg∙m)/s^2 ∙m]/([m/s^2 ]∙[kg] )=[m]
≪ … 중 략 … ≫
5.9. Repeat Prob. 5.7, except that now the horizontal cross-sectional area of the tank is 5 ft2.
문제의 조건
The area of the outlet nozzle=2[ft^2]
The cross-sectional area of the tank = 5[ft^2]
∆(P/ρ+zg+V^2/2)=dW_(n.f.)/dm-F (5.5)
non-flow work는 이 system에 작용하지 않는다, dW_(n.f.)=0
The frictionless nozzle, F=0
The pressure at location 1 and 2 are the local atmospheric pressure. So it is practically the same, ∆P=0
∆(gz+V^2/2)=0
g(z_2-z_1 )+(V_2^2-V_1^2)/2=0
≪ … 중 략 … ≫
5.18.* in the tank Fig 5.28, water is under a layer of compressed air that is at a pressure of 20 psig. The water is flowing out through a frictionless nozzle that 5 ft below the water surface. What is the velocity of the water?
≪ 그 림 ≫
∆(P/ρ+zg+V^2/2)=dW_(n.f.)/dm-F (5.5)
non-flow work는 이 system에 작용하지 않는다, dW_(n.f.)=0
The frictionless nozzle, F=0
At point 1 the velocity is negligible, V_1=0
The pressure at location 2 is the local atmospheric pressure, P_2=P_atm
The pressure at location 1 is the air pressure, P_1=P_air
PROBLEM
5.2. Show that in the head form of B.E. each term has the dimension of a length.
The head form of Bernoulli’s equation
∆(P/ρg+z+V^2/2g)=dW_(n.f.)/gdm-F/g (5.6)
P/ρg 의 dimension=[N/m^2 ]/([kg/m^3 ]∙[m/s^2 ] )=[(kg∙m)/s^2 ∙1/m^2 ]/([kg/m^3 ]∙[m/s^2 ] )=[m]
z의 dimension=[m]
V^2/2g 의 dimension=([m/s^2 ])^2/[m/s^2 ] =[m]
dW_(n.f.)/gdm 의 dimension=([J])/([m/s^2 ]∙[kg] )=[N∙m]/([m/s^2 ]∙[kg] )=[(kg∙m)/s^2 ∙m]/([m/s^2 ]∙[kg] )=[m]
≪ … 중 략 … ≫
5.9. Repeat Prob. 5.7, except that now the horizontal cross-sectional area of the tank is 5 ft2.
문제의 조건
The area of the outlet nozzle=2[ft^2]
The cross-sectional area of the tank = 5[ft^2]
∆(P/ρ+zg+V^2/2)=dW_(n.f.)/dm-F (5.5)
non-flow work는 이 system에 작용하지 않는다, dW_(n.f.)=0
The frictionless nozzle, F=0
The pressure at location 1 and 2 are the local atmospheric pressure. So it is practically the same, ∆P=0
∆(gz+V^2/2)=0
g(z_2-z_1 )+(V_2^2-V_1^2)/2=0
≪ … 중 략 … ≫
5.18.* in the tank Fig 5.28, water is under a layer of compressed air that is at a pressure of 20 psig. The water is flowing out through a frictionless nozzle that 5 ft below the water surface. What is the velocity of the water?
≪ 그 림 ≫
∆(P/ρ+zg+V^2/2)=dW_(n.f.)/dm-F (5.5)
non-flow work는 이 system에 작용하지 않는다, dW_(n.f.)=0
The frictionless nozzle, F=0
At point 1 the velocity is negligible, V_1=0
The pressure at location 2 is the local atmospheric pressure, P_2=P_atm
The pressure at location 1 is the air pressure, P_1=P_air
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