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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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Carbon dioxide (CO2) expands isothermally at steady state with no irreversibilities through a turbine from 10 bar, 500 K to 2 ba
xenn [34]

Answer:

The answer to the question above is = 152.02 KJ/Kg

Explanation:

Given:

Temperature at first state, (T1)= 500k

Temperature at second state, (T2)= 500k

The above explains an isothermal process as a thermodynamic process,in which the temperature of the system remains constant

Pressure at first state, (p1) = 10 bar

Pressure at second state, (p2) = 2 bar

The heat transfer=

Qrev/m= T x [s(T2) - s(T1) - R ㏑ (p2 ÷ p1)]

Isothermal means the temperature does not change, while Expansion means the volume has increased.

For the internal isothermal process:

Qrev/m=  T x [- R ㏑ (p2 ÷ p1)]

= 500 x  - (8.314 ÷ 44.01) x In (2 ÷ 10) = 152.02 KJ/Kg

Energy equation at turbine is for the internally reversible isothermal process is:

Q-w = m [( (V2²- V1²) ÷ 2) + g ( Z2 - Z1)]

where w= the most efficient work possible in  J

Neglecting the effect of both potential and kinetic energy

(w/ m) = Qrev/ m

= 152.02 KJ/Kg

4 0
3 years ago
The two types of outlets that are found in an electrical system are:______
Dmitry [639]

Answer:

a. lighting and receptacle outlets

Explanation:

The two types of outlets that are found in an electrical system are

a. lighting and receptacle outlets

Outlets allow electrical equipment to connect to the electrical grid. The electrical grid provides alternating current to the outlet.

3 0
3 years ago
A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac
Alex Ar [27]

Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

 ⇒d_2=28.60 mm

So throat diameter d_2=28.60 mm

     

4 0
3 years ago
Going back the b beginning of the process is common in engineering true or false?
Yuliya22 [10]

Answer:

true !!

Explanation:

hope i helped :)! brainliest ?

6 0
3 years ago
A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s
grandymaker [24]

Answer:

the endurance strength  S_e = 421.24  MPa

Explanation:

From the given information; The objective is to estimate the endurance strength, Se, in MPa .

To do that; let's for see the expression that shows the relationship between the ultimate tensile strength and Brinell hardness number .

It is expressed as:

200 \leq H_B \leq 450

S_{ut} = 3.41 H_B

where;

H_B = Brinell hardness number

S_{ut} =  Ultimate tensile strength

From ;

S_{ut} = 3.41 H_B; replace 290 for H_B ; we have

S_{ut} = 3.41 (290)

S_{ut} = 988.9 MPa

We can see that the derived value for the ultimate tensile strength when the Brinell harness number = 290 is less than 1400 MPa ( i.e it is 988.9 MPa)

So; we can say

S_{ut} < 1400

The Endurance limit can be represented by the formula:

S_e ' = 0.5 S_{ut}

S_e ' = 0.5 (988.9)

S_e ' = 494.45 MPa

Using Table 6.2 for parameter for Marin Surface modification factor. The value for a and b are derived; which are :

a = 1.58

b =  -0.085

The value of the surface factor can be calculate by using the equation

k_a = aS^b_{ut}

K_a = 1.58 (988.9)^{-0.085

K_a = 0.8792

The formula that is used to determine the value of  k_b for the rotating shaft of size factor d = 10 mm is as follows:

k_b = 1.24d^{-0.107}

k_b = 1.24(10)^{-0.107}

k_b = 0.969

Finally; the the endurance strength, Se, in MPa if the rod is used in rotating bending is determined by using the expression;

S_e =k_ak_b S' _e

S_e= 0.8792×0.969×494.45

S_e = 421.24  MPa

Thus; the endurance strength  S_e = 421.24  MPa

8 0
3 years ago
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