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swat32
3 years ago
8

Pumped-storage hydroelectricity is a type of hydroelectric energy storage used by electric power systems for load balancing. The

idea is to pump water from a lower elevation reservoir to a higher elevation reservoir when electricity demand is low and inexpensive, such as at night. During periods of high electrical demand, such as during the day, the stored water is released through turbines to produce power. (a) (5 points) If water of density ows by gravity from the higher elevation reservoir (a) to the lower one (b) at a steady ow rate Q over a height H, what is the loss in energy associated with this ow? (b) (5 points) If this same amount of loss is associated with pumping the water from the lower reservoir to the higher one at the same owrate, determine the power required to run the pump. 3

Engineering
1 answer:
NikAS [45]3 years ago
5 0

Answer:

A) energy loss E = pgQtH

Where p = density in kg/m3

g = gravity acceleration in m/s2

Q = flow rate in m3/s

t = time taken for flow in sec

H = height of flow in m

B) power required to run pump;

P = pgQH

Explanation:

Detailed explanation and calculation is shown in the image below

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A fluid has a dynamic viscosity of 0.048 Pa.s and a specific gravity of 0.913. For the flow of such a fluid over a flat solid su
sattari [20]

Answer:

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

\tau = \mu \cfrac{\partial v}{\partial y}

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at y=75 \, mm and the velocity at that point is 1.125 \, m/s

We can put that in mathematical terms as:

v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\

From the no-slip condition, we can deduce that A=0 and so we are left with just two terms:

v(y) = By + C y ^2 \\

We know that the vertex is at y= 75 \, mm and so we can rewrite the last equation as:

v(y) = k(y-75 \, mm) ^2+h

where k and h are constants to be determined. First we check that v( 75 \, mm) = 1.125 \,  m/s :

v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \,  m/s

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

v( 0) = k( -75 \, mm) ^2+  1.125 \,  m/s= 0 \quad (no \, \textendash slip)  \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= -  \cfrac{0.2}{mm \times s}

And thus we find that the final expression for the fluid's velocity is:

v( y) = 1125-  0.2 ( y -75 ) ^2

where v is in mm/s and y is in mm.

In SI units it would be:

v( y) = 1.125-  200 ( y -0.075 ) ^2

To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:

\tau = \mu \cfrac{\partial v}{\partial y}

\tau =0.048\,   \cdot  (-400) ( y-0.075   )

for y= 0.050 \, m we have:

\tau =0.048\,   \cdot  (-400) ( 0.050 -0.075   ) = 0.48\, Pa

Which is our final result

5 0
4 years ago
What is the lehr and what purpose does it serve?
levacccp [35]

Explanation:

Step1

Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.  

Step2

Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product.  So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.

5 0
3 years ago
Consider a system having p processes, where each process needs a maximum of m instances of resource type R1. Given that there ar
Genrish500 [490]

Answer:

Consider a system consisting of 4 resources of same type that are share by 3 processes each of which needs at most two resources.Now we will show that the system is deadlock free.

If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done.

Consider a system with m resources of same type being shared by n processes. Resources can be requested and released by processes only on at a time. The system is deadlock free if and only if The sum of all max needs is < m+n .

We can understand the notion of a deadlock from the following simple real-life example.To be able to write a letter one needs a letter pad and a pen. Suppose there in one letterpad and one pen on a table with two persons seated around the table. We shall identify these two persons as Mr. A and Ms. B. Both Mr. A and Ms. B are desirous of writing a letter. So both try to acquire the resources they need. Suppose Mr. A was able to get the letter pad. In the meantime, Ms. B was able to grab the pen. Note that each of them has one of the two resources they need to proceed to write a letter. If they hold on to the resource they possess and await the release of the resource by the other, then neither of them can proceed. They are deadlocked. We can transcribe this example for processes seeking resources to proceed with their execution. Consider an example in which process P1 needs three resources r1 ; r2, and r3 before it can make any further progress. Similarly, process P2 needs two resources r2 and r3 Also, let us assume that these resources are such that once granted, the permission to use is not withdrawn till the processes release these resources. The processes proceed to acquire these resources. Suppose process P1 gets resources r1 and r3 and process P2 is able to get resource r2 only. Now we have a situation in which process P1 is waiting for process P2 to release r2 before it can proceed. Similarly, process P2is waiting for process P1 to release resource r3 before it can proceed. Clearly, this situation can be recognized as a deadlock condition as neither process P1 nor process P2 can make progress. Formally, a deadlock is a condition that may involve two or more processes in a state such that each is waiting for release of a resource which is currently held by some other process.

5 0
3 years ago
A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e  = \frac{V_v}{V_s} =  \frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)

Specific gravity G_{s} = \frac{P_s}{P_w} =  \frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)

% Saturation S = \frac{V_w}{Vv} × \frac{100}{1} =  \frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right} × \frac{100}{1}--------(3)

water content w =  \frac{M_w}{M_s} = \frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

∴ w = \frac{Se}{G_s} -----(8)

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, G_{s} = 2.7, e= 0.9

∴w= \frac{0.9*1}{2.7} = 0.333

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

w= \frac{0.01*0.9 }{2.7} = 0.0033

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil (P_w=\frac{1_g}{cm^{3} } )

Given, V_{s} = 100 cm^{3 }, S = 50% = 0.5

%S = \frac{V_w}{V_v} × \frac{100}{1} = \frac{V_w}{eV_s} × \frac{100}{1}

0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

7 0
4 years ago
An inductor is energized as in the circuit of Fig. 2-4a. The circuit has L 10 mH and VCC 14 V. (a) Determine the required on tim
Savatey [412]

Answer:

A) 11.1 ms

B) 5.62 Ω

Explanation:

L ( inductance ) = 10 mH

Vcc = 14V

<u>A) determine the required on time of the switch such that the peak energy stored in the inductor is 1.2J </u>

first calculate for the current  ( i )  using the equation for energy stored in an inductor hence

i = \sqrt{\frac{2W}{L} }   ----- ( 1 )

where : W = 1.2j ,  L = 10 mH

Input values into equation 1  

i = 15.49 A

Now determine the time required  with expression below

i( t ) = 15.49 A

L = 10 mH, Vcc = 14

hence the time required ( T-on ) = 11.1 ms

attached below is detailed solution

B) <u>select the value of R such that switching cycle can be repeated every 20 ms </u>

using the expression below

τ = \frac{L}{R}  ---- ( 2 )

but first we will determine the value of τ

τ = t-off / 5 time constants

  = (20 - 11.1 ) / 5  = 1.78 ms

Back to equation 2

R = L / τ

  = (10 * 10^-3) / (1.78 * 10^-3)

  = 5.62 Ω

3 0
3 years ago
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