Unless otherwise posted, the maximum speed limit is ______mph on a two-lane undivided highways and for vehicles towing trailers.

Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr

maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

$P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}$

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

$\eta_t$ = The efficiency of the pump = 78%

$\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}$

$Q_{max}$ = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump $Q_{max}$ ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0

2 years ago

assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

4 0

3 years ago

What’s the difference between engineering stress and strain and true stress and strain

Nana76 [90]

True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimenAnswer:

Explanation:

4 0

3 years ago

Read 2 more answers

Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient press

lutik1710 [3]

Solution :

Methods for selling pressure of a distillation column :

a). Set, $\text{based on the pressure required to condensed}$ the overhead stream using cooling water.

(minimum of approximate 45°C condenser temperature)

b). Set, $\text{based on highest temperature}$ of bottom product that avoids decomposition or reaction.

c). Set, $\text{based on available highest }$ not utility for reboiler.

Running the distillation column above the ambient pressure because :

The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.

Run the reactor at an evaluated temperature because :

a). The rate of reaction is taster. This results in a small reactor or high phase conversion.

b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.

Run the reaction at an evaluated pressure because :

The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.

The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.

7 0

3 years ago

The present worth of income from an investment that follows an arithmetic gradient is projected to be $475,000. The income in ye

Nikitich [7]

Answer:

G = $37,805.65

Explanation:

I found this on another site:

475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)

475,000 = 25,000(4.3553) + G(9.6842)

9.6842G = 366,117.50

G = $37,805.65

4 0

3 years ago