All categories

3 years ago

A single-cylinder pump feeds a boiler through a delivery

Engineering

1 answer:

Studentka2010 [4]3 years ago

8 0

Answer:

Net discharge per hour will be 3.5325 $m^3/hr$

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius $r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m$

We know that area is given by

$A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2$

We know that discharge is given by $Q=AV$ , here A is area and V is velocity

So $Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec$

So net discharge in 1 hour = $981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour$

You might be interested in

A student wants to restate some ideas she found in a journal article by a prominent expert in economics. She combines her own wo

kipiarov [429]

Explanation:

Althought she referenced the article at the end, is imposible to know which part of the article is hers and which part is the expert's so that would be plagiarism.

If she used quotation marks in the words of the expert it would be clear and no plagiarism could be accused.

4 0

3 years ago

Sin x +√3 coax= √2<br>

Ber [7]

Answer:

The two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12

Explanation:

The given equation is

Sin x +√3 Cosx= √2

Upon dividing the equation by 2 we get

$\frac{1}{2}Sinx + \frac{\sqrt{3} }{2}Cosx = \frac{\sqrt{2} }{2}$

Sin( $\frac{pi}{6}$ )*Sinx + Cos( $\frac{pi}{6}$ )*Cosx = $\frac{1}{\sqrt{2} }$

This makes the formula of

CosACosB + SinASinB = Cos(A-B)

Cos(x- $\frac{pi}{6}$ ) = $\frac{1}{\sqrt{2} }$

cos(x- pi/6) = cos(pi/4)

upon writing the general equation we get

x-pi/6 = 2n*pi ± pi/4

x = 2n*pi ± pi/4 -pi/6

so we will have two solutions

x = 2n*pi + pi/4 -pi/6

= 2n*pi + pi/12

and

x = 2n*pi - pi/4 -pi/6

= 2n*pi -5pi/12

Therefore the two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12.

4 0

3 years ago

Determine the speed of sound in air at 400 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3

Reika [66]

Answer:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

Explanation:

For this case we have given the following data:

$T= 400 K$ represent the temperature for the air

$v = 310 m/s$ represent the velocity of the air

$k = 1.4$ represent the specific heat ratio at the room

$R = 0.287 KJ/ Kg K$ represent the gas constant for the air

And we want to find the velocity of the air under these conditions.

We can calculate the spped of the sound with the Newton-Laplace Equation given by this equation:

$\alpha = \sqrt{\frac{K}{\rho}}=\sqrt{k RT}$

Where K = is the Bulk Modulus of air, k is the adiabatic index of air= 1.4, R = the gas constant for the air, $\rho$ the density of the air and T the temperature in K

So on this case we can replace and we got:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

The Mach number by definition is "a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound" and is defined as:

$Ma=\frac{v}{\alpha}$

Where v is the flow velocity and $\alpha$ the volocity of the sound in the medium and if we replace we got:

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

And since the Ma<0.8 we can classify the regime as subsonic.

7 0

3 years ago

Water is being added to a storage tank at the rate of 500 gal/min. Water also flows out of the bottom through a 2.0-in-inside di

melomori [17]

Answer:

From the answer, the water level is falling (since rate of outflow is more than that of inflow), and the rate at which the water level in the storage tank is falling is

(dh/dt) = - 0.000753

Units of m/s

Explanation:

Let the volume of the system at any time be V.

V = Ah

where A = Cross sectional Area of the storage tank, h = height of water level in the tank

Let the rate of flow of water into the tank be Fᵢ.

Take note that Fᵢ is given in the question as 500 gal/min = 0.0315 m³/s

Let the rate of flow of water out of the storage tank be simply F.

F is given in the form of (cross sectional area of outflow × velocity)

Cross sectional Area of outflow = πr²

r = 2 inches/2 = 1 inch = 0.0254 m

Cross sectional Area of outflow = πr² = π(0.0254)² = 0.00203 m²

velocity of outflow = 60 ft/s = 18.288 m/s

Rate of flow of water from the storage tank = 0.0203 × 18.288 = 0.0371 m³/s

We take an overall volumetric balance for the system

The rate of change of the system's volume = (Rate of flow of water into the storage tank) - (Rate of flow of water out of the storage tank)

(dV/dt) = Fᵢ - F

V = Ah (since A is constant)

dV/dt = (d/dt) (Ah) = A (dh/dt)

dV/dt = A (dh/dt) = Fᵢ - F

Divide through by A

dh/dt = (Fᵢ - F)/A

Fi = 0.0315 m³/s

F = 0.0371 m³/s

A = Cross sectional Area of the storage tank = πD²/4

D = 10 ft = 3.048 m

A = π(3.048)²/4 = 7.30 m²

(dh/dt) = (0.0315 - 0.0370)/7.3 = - 0.000753

(dh/dt) = - 0.000753

4 0

3 years ago

which systems engineering support discipline has the goal to ensure that support considerations are an integral part of the syst

astraxan [27]

Answer:what's the question

Explanation:

8 0

3 years ago