山东科技大学
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- Crude oil (s=0.855) at 50℃ flows at 300L/s through a 450-mm-diameter pipe (e=0.054mm) 1500m long. Find the kilowatt loss.
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- Air at 150℃ and a pressure of 240 kPa abs flows at a velocity of 16 m/s through an 180-mm-diameter pipe. What is the Reynolds number?
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- Oil of viscosity of 0.000 38 m2/s flows in a 100 mm diameter pipe at a rate of 0.64 L/s. Find the head loss per unit length.
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- Gas (=5.25kg/m3,v=2.0×10-5m2/s)is flowing in a 20-mm-diameter pipe. When a gas flowmeter measures the flow as being 0.064 kg/s, it registers a pressure drop of 8.5 kpa. Investigators plan to test an enlarged model that is geometrically similar in a 180-mm-diameter pipe. What would the pressure drop across the water meter be?
- A 40 mm-diameter jet has a velocity of 25 m/s. If this jet were to strike a large flat plate normally, what would be the resultant force on the plate?
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- If the atmospheric pressure is 780 mb abs and a gage attached to a tank reads 330 mmHg vacuum, what is the page pressure?
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- A certain missile travels at 2500 km/h through the atmosphere at an elevation of5 km. Air at 20C and standard atmospheric conditions will flow around a model of the missile in a wind tunnel. What air speed in the wind tunnel will achieve dynamic similarity?
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- The relationship among absolute pressure and relative pressure, vacuum pressure, and the local atmospheric pressure is （ ）
- The baseline for relative pressure is : （ ）
- In horizontal divergent tube, ignoring the head loss, the core pressure in the section, has the following relationship: （ ）
- The Bernoulli equation
express （ ） - Uniform flow is: （ ）
答案:
答案:
答案:The kilowatt loss is approximately 267 kW.
答案:
答案:The Reynolds number (Re) for the given fluid flow conditions can be calculated using the formula: \[ Re = \frac{\rho \cdot v \cdot D}{\mu} \] where: - \(\rho\) is the density of the air, - \(v\) is the velocity of the air, - \(D\) is the diameter of the pipe, and - \(\mu\) is the dynamic viscosity of the air. However, to directly provide the Reynolds number without analysis, we need to use properties of air at the given temperature and an estimated pressure since the density and viscosity depend on these conditions. Typically, for simplification in such calculations and lacking specific gas properties, we might approximate air as an ideal gas or use standard values for room temperature. But since we're at 150°C, we should adjust our approach. Given the complexity and the need for property lookups or calculations which aren't directly executable here, a typical approach would involve these steps: 1. **Density (\(\rho\))**: Use the ideal gas law \( PV = mRT \) or a steam table to find the density of air at 150°C and 240 kPa abs, knowing that \( \rho = \frac{P}{RT} \), where \(R\) is the specific gas constant for dry air (approximately 287 J/kg*K). 2. **Viscosity (\(\mu\))**: Air viscosity varies with temperature; it can be interpolated or calculated from charts or equations provided for the Sutherland's formula, but this requires specific knowledge of how viscosity changes with temperature. For a quick estimation without actual calculation: - Assume standard air properties if the exact calculation isn't crucial (not advisable for precise work). - Recognize that at 150°C, the density will be significantly lower and viscosity somewhat decreased compared to room temperature, affecting the Reynolds number. Since I'm unable to perform the necessary lookups or complex calculations here, providing a direct Reynolds number value without analysis or access to specific data isn't feasible. For an accurate answer, you'd typically use engineering software, calculators, or refer to thermodynamic tables to find the exact properties of air at 150°C and 240 kPa abs, then apply them in the Reynolds number formula.
答案:
答案:The head loss per unit length for the given fluid flow conditions can be calculated using the Darcy-Weisbach equation: \[ h_f = \frac{f L (V^2/2g)}{D} \] Where: - \( h_f \) is the head loss per unit length, - \( f \) is the Darcy friction factor, - \( L \) is the length of the pipe (since we're looking for head loss per unit length, we consider \( L = 1 \) unit), - \( V \) is the flow velocity, - \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), - \( D \) is the diameter of the pipe. First, we need to find the flow velocity \( V \). Given the flow rate \( Q = 0.64 \, \text{L/s} = 0.64 \times 10^{-3} \, \text{m}^3/\text{s} \) and pipe diameter \( D = 100 \, \text{mm} = 0.1 \, \text{m} \), we can calculate the cross-sectional area \( A \) of the pipe and then find \( V \): \[ A = \pi(D/2)^2 = \pi(0.1/2)^2 = \pi \times 0.0025 \, \text{m}^2 \] \[ V = \frac{Q}{A} = \frac{0.64 \times 10^{-3}}{\pi \times 0.0025} = \frac{0.64}{\pi \times 25} \, \text{m/s} \] \[ V \approx \frac{0.64}{7.854} \, \text{m/s} \approx 0.0815 \, \text{m/s} \] For turbulent flow in smooth pipes, the Darcy friction factor \( f \) can be estimated using the Blasius equation: \[ f = \frac{0.0791}{Re^{0.25}} \] Where \( Re \) is the Reynolds number: \[ Re = \frac{VD}{\nu} \] Given the viscosity \( \nu = 0.00038 \, \text{m}^2/\text{s} \), we calculate \( Re \): \[ Re = \frac{0.0815 \times 0.1}{0.00038} \approx 2170 \] Now, calculate \( f \): \[ f = \frac{0.0791}{2170^{0.25}} \approx \frac{0.0791}{6.87} \approx 0.0115 \] Finally, compute the head loss per unit length \( h_f \): \[ h_f = \frac{f (V^2/2g)}{D} = \frac{0.0115 \times (0.0815^2/2 \times 9.81)}{0.1} \] \[ h_f \approx \frac{0.0115 \times (0.00664225/19.62)}{0.1} \] \[ h_f \approx \frac{0.0115 \times 0.003386}{0.1} \approx 0.00040 \, \text{m}^{-1} \] So, the head loss per unit length is approximately \( 0.00040 \, \text{m}^{-1} \).
答案:
答案:The pressure drop across the enlarged model in the 180-mm-diameter pipe would be 120 Pa.
A:
B:
C:
D:
A:liquid surface pressure. B:the local atmospheric pressure; C:one standard atmospheric pressure; D:absolute vacuum;
A:
B:
C:Indetermination
D:
A:total mechanical energy of the fluid through flow section. B:mechanical energy of the fluid in unit volume C:mechanical energy of the fluid in unit mass; D:mechanical energy of the fluid in unit weight;
A:Total acceleration is zero. B:mitigation acceleration is zero; C:the local acceleration is zero; D:The centripetal acceleration is zero;
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