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MaRussiya [10]
3 years ago
14

What is made in heaven?​

Engineering
1 answer:
kramer3 years ago
3 0

Answer:

Babies come from heaven didn't you know?

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Your company is planning to build a pipeline to transport gasoline from the refinery to a field of storage tanks. The parameters
AleksandrR [38]

Answer:

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.

Explanation:

The Reynolds number (Re_{D}) is a dimensionless criterion use for flow regime of fluids, which is defined as:

Re_{D} = \frac{\rho \cdot v\cdot D}{\mu} (Eq. 1)

Where:

\rho - Density, measured in kilograms per cubic meter.

\mu - Dynamic viscosity, measured in kilograms per meter-second.

v - Average flow velocity, measured in meters per second.

D - Pipe diameter, measured in meters.

We need to find the equivalent velocity of water used in the prototype system. In this case, we assume that Re_{D,gas} = Re_{D,w}. That is:

\frac{\rho_{w}\cdot v_{w}\cdot D_{w}}{\mu_{w}} = \frac{\rho_{gas}\cdot v_{gas}\cdot D_{gas}}{\mu_{gas}} (Eq. 2)

Where subindex w is used for water and gas for gasoline.

If we know that \rho_{gas} = 690\,\frac{kg}{m^{2}}, \mu_{gas} = 0.006\,\frac{kg}{m\cdot s}, v_{gas} = 0.5\,\frac{m}{s}, D_{gas} = 1\,m, \rho_{w} = 1000\,\frac{kg}{m^{3}}, \mu_{w} = 0.0018\,\frac{kg}{m\cdot s} and D_{w} = 0.05\,m, then we get the following formula:

57500 = 27777.778\cdot v_{w}

The fluid velocity for the prototype system is:

v_{w} = 2.07\,\frac{m}{y}

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.

3 0
3 years ago
The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24C, and a wet-bulb temperature of 17C. Using the psychrome
TEA [102]

Answer:

(a) Relative Humidity = 48%,

Specific humidity = 0.0095

(b) Enthalpy = 65 KJ/Kg of dry sir

Specific volume = 0.86 m^3/Kg of dry air

(c/d) 12.78 degree C

(e) Specific volume = 0.86 m^3/Kg of dry air

8 0
3 years ago
pump delivers 11000 lb/hr of water from an elevation 30 ft below the pump to an elevation 50 ft above the pump through various d
mojhsa [17]

Answer:

good luck

Explanation:

4 0
4 years ago
Design a lead compensator for the closed-loop (CL) system whose open loop transfer function is given below. Design objectives: r
Montano1993 [528]

Answer:

see attached bellow

Explanation:

5 0
3 years ago
On a nonprecision approach, what is the maximum acceptable descent rate during the final stages of the approach (below 1,000 ft.
Dovator [93]

Answer: For non-precision approaches, the maximum acceptable descent rate acceptable should be one that ensures the aircraft reaches the minimum descent altitude at a distance from the threshold that allows landing in the touch down zone. Otherwise, a decent rate greater than 1000fpm is unacceptable.

Explanation: For non-precision approaches, a descent rate should be used that ensures the aircraft reaches the minimum decent altitude at a distance from the threshold that allows landing in the touchdown zone (TDZ) . On many instrument approach procedures, this distance is annotated by a visual descent point (VDP) If no VDP is annotated, calculate a normal descent point to the TDZ. To determine the required rate of descent, subtract the TDZ elevation (TDZE) from the final approach fix (FAF) altitude and divide this by the time inbound. For illustration, if the FAF altitude is 1,000 feet mean sea level (MSL), the TDZE is 200 feet MSL and the time inbound is two minutes, an 400 fpm rate of descent should be applied.

A descent rate greater than approximately 1,000 fpm is unacceptable during the final stages of an approach (below 1,000 feet AGL). Operational experience and research shows that this is largely due to a human perceptual limitation that is independent of the airplane or helicopter type. As a result, operational practices and techniques must ensure that descent rates greater than 1,000 fpm are not permitted in either the instrument or visual portions of an approach and landing operation.

3 0
4 years ago
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