Defination of rolling

A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a

almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water ( $\dot m$ ), in kilograms per second, is determined by following formula:

$\dot m = \frac{\eta \cdot \dot W}{g\cdot H}$ (1)

Where:

$\dot W$ - Pump power, in watts.

$\eta$ - Efficiency, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$H$ - Hydrostatic column, in meters.

If we know that $\eta = 0.72$ , $\dot W = 5000\,W$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the mass flow rate of water is:

$\dot m = 14.683\,\frac{kg}{s}$

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump ( $\Delta P$ ), in pascals, is determined by this equation:

$\Delta P = \rho\cdot g\cdot H$ (2)

Where $\rho$ is the density of water, in kilograms per cubic meter.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the pressure difference is:

$\Delta P = 245175\,Pa$

The pressure difference across the pump is 245.175 kilopascals.

4 0

3 years ago

A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni

Dvinal [7]

Answer:

Tc = = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

$\rho = 8000 kg/m^3$

k = 50 w/m . k

c = 500 j/kg.k

$h_{\infty} = 1000 w/m^2$

$t_{\infity} = 750 k$

$t_w = 500 K$

$surface area = As = \pi dL$

$\frac{As}{L} = \pi D = \pi \timeS 0.06$

HEAT FLOW Q is

$Q = h_{\infty} As (T_[\infty} - Tw)$

$= 1000 \pi\times 0.06 (750-500)$

= 47123.88 w per unit length of rod

volumetric heat rate

$q = \frac{Q}{LAs}$

$= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}$

$q = 1.66\times 10^{7} w/m^3$

$Tc = \frac{- qR^2}{4K} + Tw$

$= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} + 500$

= 424.85 K

7 0

3 years ago

Which statement below can be used to read data from a file one character at a time?

Paraphin [41]

Answer:

b

Explanation: because it says input file.read

8 0

2 years ago

From your cooling load (8890.007 Btu/hr = 2.605kW, determine mass flow rate of refrigerants. Use the following "rule of thumb" e

NeX [460]

Answer:

0.740833917 ton/hr

Explanation:

Given:

Cooling load, 8890.007 Btu/hr = 2.605 kW

Room size = 180 $ft^{2}$

According to the thumb rule

1 ton of refrigerant = 12000Btu

Hence for 8890.007 Btu/hr,

the mass flow rate of the refrigerant is =8890.007 / 12000

= 0.740833917 ton per hr

Hence, mass flow rate is 0.740833917 ton/hr

7 0

3 years ago

Ethan is an engineer who is trying to create a totally quiet fluid power system. Which part of the fluid power system will he ne

creativ13 [48]

Answer:

C: compressor

Explanation:

As it states in the text, Unfortunately, the pump or compressor in a fluid power system is often noisy and heavy. This aspect of the fluid power system is a critical area of interest for engineers and scientists who seek to improve fluid power.

3 0

3 years ago

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