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Inessa [10]
3 years ago
9

A water pump delivers 6 hp of shaft power when operating. The pressure differential between the outlet and the inlet of the pump

is measured to be 1.2 psi when the flow rate is 10 ft3/s, and the velocity changes to 5 ft/s from 2 ft/s as the water passes through the pump. Determine the mechanical efficiency of this pump assuming the water density to be 62.4 lbm/ft3 (written in decimal form with 3 significant figures).
Engineering
1 answer:
frez [133]3 years ago
5 0

Answer: Pump efficiency = 0.585 = 58.5%

Explanation: Pump efficiency = (power gained by the fluid)/(power supplied by the shaft)

Power gained by the fluid = Q(ΔP) + m(((v2)^2) - ((v1)^2))/2

Where Q = volumetric flow rate = 10 ft3/s = 0.283 m3/s

ΔP = 1.2 psia = 8273.709 Pa

m = mass flow rate = density × volumetric flow rate

Density = 62.4 lbm/ft3 = 999.52 kg/m3

m = 999.52 × 0.283 = 282.86 kg/s

v2 = 5ft/s = 1.524 m/s

v1 = 2ft/s = 0.61 m/s

Q(ΔP) = 0.283 × 8273.709 = 2341.46 W

Power from change in kinetic energy = m(((v2)^2) - ((v1)^2))/2 = 275.92 W

Power gained by the fluid = 2341.46 + 275.92 = 2617.38 W

Power supplied by shaft = 6hp = 6 × 746 = 4476 W

Efficiency = 2617.38/4476 = 0.58475 = 0.585 to 3s.f

QED!

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The human body gets its energy via the combustion of blood sugar (glucose). if all of the chemical bond energy in 10 g of glucos
Gnesinka [82]

Answer:

speed by mass attain is 55.86 m/s

Explanation:

given data

glucose = 10 g

mass = 100 kg

to find out

speed by mass attain

solution

we know glucose have 180 g molecular weight and

that 1 g glucose produce energy = 2816/180 × 10³ J

so here 10 g of glucose produce energy =  1.56 × 10^{5} J

so here energy release = 1/2 × mv²

1.56 × 10^{5}  = 1/2 × (100)v²

v² = 3.12 × 10³

and v = 55.86 m/s

so speed by mass attain is 55.86 m/s

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3 years ago
2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?
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2 years ago
Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure
iris [78.8K]

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3  = √9 = 3

P5/P6 = P7/P8  = √9 =3  

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860  

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238  

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158  

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg  

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg  

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =  

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))  

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg  

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33  

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg  

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg  

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =  

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg  

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg  

Calculate the net work done

Wnet = (Wturb)out  - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg  

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

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3 years ago
Is Tesla French, American, German, or Russian?
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3 years ago
Read 2 more answers
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
2 years ago
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