Power-plant-lecture-new.docx

Power Plant Lecture Chapter 4 DIESEL POWER PLANT Diesel, Rudolf Christian Karl

Diesel, Rudolf Christian Karl (1858 – 1913), German engineer, who invented the diesel engine. After studying in England, he attended the Polytechnic School in Munich, where he settled in 1893. In 1892 He patented the internal-combustion engine named for him, which employed auto-ignition of fuel. While associated with the Krupp firm in Essen, he built the first successful diesel engine, utilizing low-cost fuel. In 1913, while on a voyage to England, Diesel drowned in English Channel. Diesel Engine – is an excellent prime mover for electric power generation in capacities of 101 hp to 5070 hp. These are widely used in hotels, utility companies, municipalities and private industries.

The design of diesel electric power plant includes the following elements; the stationary diesel engine, fuel system, lubricating system, cooling system, intake and exhaust system, starting system, and the governing system.

*Advantages of the diesel engines: 1. The cost of diesel fuel is cheaper than other fuels. 2. It needs no long warming up. 3. It has no standby losses. 4. It has uniformly high efficiency of all size. 5. It has a simple plant lay out. 6. It needs no large water supply.

PERFORMANCE OF DIESEL POWER PLANT 1. Heat supplied by fuel, Q,: Qs = mfQh Where: mf = mass flow rate of fuel Qs = heating value of fuel Sample problem: (Calculating the heat supplied by Fuel) What is the heat that can be supplied by 39 kg/hr diesel fuel whose heating value is 43,912 kJ/kg? Solution:

Qs = mfQs Qs = (39/3600)43,912 kJ/s or kW

Answer:

Qs = 475.71 kW

2. Air – Fuel Ratio, A/F: 𝑨 𝑭

Where: ma = mass of air mf = mass of fuel

=

𝒎𝒂 𝒎𝒇

Sample Problem: ( Calculating the Air-Fuel Ratio) The density of air entering the engine is 1.19 kg/m3 whose volume flow rate is 0.15 m3/s. Ifthe mass flow rate of fuel is 121.38 kg/hr, what is the air-fuel ratio? Solution:

𝑨

𝒎𝒂

= 𝑭 𝒎𝒇

Where: mf = 121.38 kg/hr = 0.034 kg/s

Solving for the mass of air; ma: ma = ρaVa = 1.19(0.51) = 0.6069 kg/s then; 𝑨 𝑭

=

𝟎.𝟔𝟎𝟔𝟗 𝟎.𝟎𝟑𝟒

𝑨

Answer: 𝑭 = 18 kg air/kg fuel

3. Piston Displacement, VD: Piston displacement is the volume displaced by the piston as it moves from top dead center to bottom dead center. 𝛑𝐃²

VD = (

𝟒

) LN ncnp

Where: D = bore or diameter of the piston

NOTE: for 4-stroke engine

L = length of stroke or stroke

divide the engine speed “N”

N = engine speed

by 2.

nc = no. of cylinders np = no. of piston actions

Sample problems: (Calculating the Piston displacement) Determine the piston displacement of 35 cm x 45 cm, 4 –stroke, 1200 rpm, 8-cylinder diesel engine? Solution: πD²

Vd = ( 4 )LNncnp Vd =(

π(0.35)2 (0.45)

1200

4

2(60)

)(0.45)[

](8)

Answer: Vd = 3.46 m3/s 4. Piston Speed: Piston speed is the total distance a piston travels in a given time. Piston speed = 2 LN Where: 2L = distance by piston in one revolution N = angular speed in rpm or rps Sample problem: (calculating the piston speed of a Diesel Engine) Solution: The piston speed (V): Piston speed = 2LN = 2(0.35)(1100/60) Answer: Piston speed = 12.83 m/s 5. Indicated Power, Pind: Indicated power is the power developed by the action of piston within a cylinder, so named because it is measured by use of an indicator. Pind = PmiVD Where: Pmi = indicated mean effective pressure

@ Calculating the indicated mean effective pressure using data provided by the planimeter. Planimeter measures the area of actual P – V diagram traced by engine indicator Pmi =

𝐀𝐜𝐱𝐒𝐜 𝑳𝒄

Where: Ac = area of indicator card diagram Sc = spring scale Lc = length of indicator card diagram @ If working cylinder and crankcase are to be considered: 𝐀𝐱𝐒

Pmi = (

𝑳

𝐀𝐱𝐒

)wc – (

𝑳

)cc

NOTE: Crankcase compression is Used for scavenging.

Where: wc = working cylinder cc = crankcase Sample problem: ( Calculating the indicated power of Diesel Engine) The cylinder diameters of an eight – cylinder, single acting, four-stroke diesel engine are 750 mm and the stroke is 1125 mm. The indicated mean effective pressure in the cylinder is 586 kPawhen the engine is running at 110 rpm. Calculate the indicated power. Soulution: Pind = Pmi L A N nc np 𝛑

𝟏𝟏𝟎

= 586(1.25)[ (0.75)2][ 𝟔(𝟐𝟎)])8)(1) 𝟒

Answer

= 2,135.82 kW

6. Brake Power, Pb Brake power is the delivered power to a shaft. Brake power is always less than the indicated power for a given engine, because some of the work developed by the

cylinders is used cylinders is used to overcome the friction of running the engine. The often term of brake is shaft power.

Pb =

𝟐𝛑𝐓𝐍 𝟔𝟎

and T=Fr

NOTE:Brake power is calculated

Where:

using either prony brake or

T = brake torque

dynamometer.

N = engine rotative speed in rpm F = brake force or brake load R = brake arm or torque arm Sample problem: (Calculating the Brake power of a Diesel Engine) The flywheel of a rope brake is 1.22 m diameter and the rope is 24mm diameter. When the engine is running at 250rev/min the load on the brake is 480 N on one end of the rope and 84 N on the other end. Calculate the brake power. Solution: Pb =

𝟐𝛑𝐓𝐍 𝟔𝟎

The load (F) on the brake (F): F = 480 – 84 = 395=6 N The radius (r): 𝟏

r = 𝟐(1220 + 24) = 622 mm r = 0.622 m The torque (T): T = 396 (0.622) = 246.31 N-m The Brake power (Pb): Pb =

𝟐𝛑(𝟐𝟒𝟔.𝟑𝟏)(𝟐𝟓𝟎) 𝟔𝟎

Answer Pb = 6.448kW

= 6,448.43 W

@ Brake power in terms of brake mean effective pressure and piston displacement: Pb = PmbVD Where: Pmb = brake mean effective pressure Pb = brake power VD = piston displacement

Sample problem: (Calculating the brake mean effective pressure) A single acting, 8 cylinder, 4 stroke cycle diesel engine with a bore to stroke of 142.1-mm x 210.45-mm operates at 1200 rpm. The load on the brake arm at 1 m is 150 kg. What is the brake mean effective pressure in kPa?

Solution: Pb = Pmb VD

The volume displacement (VD): 𝛑

VD = [ 𝟒(0.1412)2](0.21045)[

𝟏𝟐𝟎𝟎

= 0.267 m3/s The brakepower (Pb): Pb =

𝟐𝛑𝐓𝐍 𝟔𝟎

Where; the torque(T): T = [150(0.00981)](1) T = 1.4715 kN-m

](8)(1)

𝟐(𝟔𝟎)

The brakepower (Pb): Pb =

𝟐𝛑(𝟏.𝟒𝟕𝟏𝟓)(𝟏𝟐𝟎𝟎) 𝟔𝟎

Pb = 184.91kW The mean effective pressure(Pmb): 184.91 = Pmb (0.267) Answer Pmb = 692.55 kPa 7. Friction Power, Pf: Friction power is the power dissipated in an engine through friction. Friction Power = Indicated Power – Brake Power Pf

=

Pind

-

Pb

@ Calculating friction power using Morse test method applied to multi-cylinder engines: Consider a six – cylinder engine The indicated power if all six cylinders are firing Pind(6) = Pb(6) – Pf(6) If one cylinder is cut, or five cylinder are firing Pind(5) = Pb(5) – Pf(5) Derived from two equations above, equating the friction power: Pind(6) – Pind(5) = Pind(1) = Pb(6) – Pb(5) Thus, the total indicated power for six cylinder engine is,

Pind(6) = 6(Pb(6) – Pb(5)) Note: no matter how many cylinders are firing, the friction power is constant. (Pf(6) = Pf(5)) Sample problem: ME Board Problem (Calculating the indicated power) A six cylinder, four stroke diesel engine with 76 mm bore x 89 mm stroke was run in the laboratory at 200rpm, when it was found that the engine torque was 153.5 N-m with all

cylinders firing but 123 N-m when one cylinder was out. The engine consumed 12.2 kg of fuel per hour with a heating value of 54,120 kJ/kg of air at 15.60C. Determine the indicated power. Solution: Pind(6) = 6(Pb(6) –Pb(5)) 𝟐𝛑(𝟎.𝟏𝟓𝟑𝟓)(𝟐𝟎𝟎𝟎) 𝟐𝛑(𝟎.𝟏𝟐𝟑)(𝟐𝟎𝟎𝟎)

Pind(6) =6[

𝟔𝟎

-

𝟔𝟎

]

Answer Pind(6) = 38.83 kW 8. Mechanical Efficiency, em Mechanical Efficiency is the ratio of the brake power to the indicated power. em =

𝐏𝐛

𝑷𝒊𝒏𝒅

or

em =

𝐏𝐦𝐛 𝑷𝒎𝒊

where: Pb = brake power Pind = indicated power Pmi = indicated mean effective pressure Pmb = indicated mean brake power Sample problem: (Calculating the indicated power) What is the mechanical efficiency of a 0.5 MW diesel engine if the friction power is 70 kW. Solution: The mechanical efficiency: em =

𝐏𝐛 𝑷𝒊𝒏𝒅

The indicated power: Pind = Pb + Pf

Pind = 500 + 70 kW Pind = 570 kW Substitute: em =

𝟓𝟎𝟎 𝟓𝟕𝟎

Answer: em = 0.8772 or 87.72 %

9. Electrical or Generator Efficiency, egen Electrical or Generator Efficiency is the ratio of the generator power to the brake power.

egen =

𝐏𝐠𝐞𝐧 𝑷𝒃

where: Pgen = Generator Power Pb = Brake Power Sample problem: (Calculating the generator efficiency given the brakepower) A 16-cylinder V – type diesel engine is directly coupled to ¾ MW AC generator. Calculate the generator efficiency of the engine if the brake power is 833.33 kW. Solution: egen =

𝐏𝐠𝐞𝐧 𝑷𝒃

=

𝟑/𝟒 𝟎.𝟖𝟑𝟑

Answer: egen = 0.9 or 90%

10. Thermal efficiencies,et Thermal Efficiencies is the ratio of the work done by heat engine to the heat supplied by the fuel.

a. Indicated thermal efficiency, eti Indicated thermal efficiency is the ratio indicated power to the heat supplied by the fuel.

eti =

𝐏𝐢𝐧𝐝 𝒎𝒇𝑸𝒉

Sample problem: (Calculating the indicated thermal efficiency) A 3.5 MW diesel power plant uses 3500 gallons in 24 hours period. What is the indicated thermal efficiency of the engine if the generator and mechanical efficiencies are 90% and 92% respectively? Oil is 280 API. Solution: 𝐏𝐢𝐧𝐝

eti = 𝒎𝒇𝑸𝒉 Solving for Qh: Qh = 41,130 + 139.6 (28) = 45038.80 kJ/kg Solving for mfuel: S.G.15.6 oC =

𝟏𝟒𝟏.𝟓 𝟏𝟑𝟏.𝟓+𝟐𝟖

= 0.887

ρfuel = 0.887(1000) = 887 kg/m3 = ρfuel Vfuel

mfuel

𝟑.𝟕𝟖𝟓

= 887[3500( 𝟏𝟎𝟎𝟎 )] = 11750.53 kg Mass of fuel per second: mfuel =

𝟏𝟏𝟕𝟓𝟎.𝟓𝟑 𝟐𝟒(𝟑𝟔𝟎𝟎)

= 0.136 kg/s

Then; eti =

𝐀𝐜𝐱𝐒𝐜 𝒌𝒈 𝒌𝑱 (𝟎.𝟏𝟑𝟔 )(𝟒𝟓𝟎𝟑𝟖.𝟖𝟎 ) 𝒔 𝒌𝒈

Answer: eti = 0.5714 04 57.14%

b. Brake thermal efficiency,eth Brake thermal efficiency is the ratio of the brake power to the heat supplied by the fuel.

eth =

𝐏𝐛 𝑴𝒇𝑸𝒉

Sample problem: ME Board Problem (Calculating the brake thermal efficiency) A supercharged six cylinder, four stroke cycle diesel engine of 10.48 cm bore and 12.7 cm stroke has a compression ratio of 15. When it tested on a dynamometer with a 53.34 cm arm at 2500 rpm, the scale reads 81.65 kg, 2.86 kg of fuel of 45822.20 kJ/kg heating value are burned during a 6 min. test and air metered to the cylinders at the rate of 0.182 kg/s. Find the brake thermal efficiency. Solution: etb =

𝐏𝐛 𝒎𝒇𝑸𝒉

Solving for Pb: T = 81.65(0.00981)(0.5334) T = 0.42725 N-m Pb =

𝟐𝛑𝐓𝐍 𝟔𝟎

=

𝟐𝛑(𝟎.𝟒𝟐𝟕𝟐𝟓)(𝟐𝟓𝟎𝟎) 𝟔𝟎

= 111. 854 kW

Solving for mf: mf =

𝟐.𝟖𝟔 𝟔(𝟔𝟎)

= 0.00794 kg/s

thus; etb =

𝟏𝟏𝟏.𝟖𝟓𝟒 𝟎.𝟎𝟎𝟕𝟗𝟒(𝟒𝟓,𝟖𝟖𝟐.𝟐𝟎)

= 0.307

c. Combined or Over – all thermal efficiency, etc

Combined or over – all thermal efficiency is the ratio of the electrical or generator power to heat supplied by the fuel.

etc =

𝐏𝐠𝐞𝐧 𝒎𝒇𝑸𝒉

Sample problem: (calculating the combined thermal efficiency) A 16 – cylinder V –type diesel engine is directly coupled to 1 MW AC generator. Calculate the combined thermal efficiency if the heat supplied by the fuel is 2.5 MW. Solution: etc = =

𝐏𝐠𝐞𝐧 𝑸𝒔 𝟏 𝟐.𝟓

= 0.40 or 40% 11. Engine efficiencies, et a. Indicated engine efficiency, eei Indicated engine efficiency is the ratio of the indicated thermal efficiency to the ideal thermal efficiency.

eei =

𝐞𝐭𝐢 𝒆

Sample problem: ME Board Problem(Calculating the cycle efficiency) What is the indicated engine efficiency of diesel engine if the indicated thermal efficiency is 35% and the cycle efficiency is 45%. Solution: eei =

Answer: 𝟑𝟓 𝟒𝟓

eei = 77.78%

b. Brake engine efficiency: eeb Brake engine efficiency is the ratio of the brake thermal efficienmcy to the ideal thermal efficiency.

eeb =

𝐞𝐭𝐛 𝒆

Sample problem: (Calculating engine efficiency) A 500 kW diesel has a rate of 12,000 kJ/kW-hr. The compression ratio is 16:1 cut of ratio of 2.3. Assume k = 1.32.Calculate the engine efficiency based on the output of 500kW.

Solution: Let: eeb = brake engine efficiency eeb =

𝐞𝐭𝐛 𝒆

The brake engine efficiency,etb: etb =

𝟑𝟔𝟎𝟎

𝟏𝟐,𝟎𝟎𝟎

= 0.30 or 30%

Solving for cycle efficiency, e e = 1Answer

𝟏

(𝟐.𝟑)𝟏.𝟑𝟐−𝟏

[

𝟏𝟔(𝟏.𝟑𝟐−𝟏) 𝟏.𝟑𝟐(𝟐.𝟑−𝟏)

]

e = 0.5195 or 51.95% The engine efficiency, eeb: eeb =

𝟑𝟎 𝟓𝟏.𝟗𝟓

= .5777

eeb = 57.77% c. Combined oe over – all engine efficiency, eec Combined or over –all engine efficiency is the ratio of the combined or over- all thermal efficiency to the ideal thermal efficiency.

eec =

𝐞𝐭𝐜 𝒆

Sample problem: If the over-all thermal efficiency of a diesel engine is 33% and the diesel cycle efficiency 49%, what is the combined engine efficiency? Solution: eec = 0.33/0.49 Answer: eec = 67.35%

12. Volumetric efficiency, ev Volumetric efficiency is the ratio of the volume of air drawn into a cylinder to the piston displacement. ev =

𝐕𝐚 𝑽𝑫

where: Va =

𝐦𝐚𝐑𝐚𝐓𝐚 𝑷𝒂

VD = piston displacement Sample problem: Determine the volumetric efficiency of 35 cm x 45 cm, 4 stroke 1200 rpm, 8 cylinder diesel engine if the air drawn in the engine is 3 m3/s? Solution: The volume displacement, Vd: Vd = ( =(

πD2 4

)LNn

𝛑(𝟎.𝟑𝟓𝐱𝟎.𝟑𝟓) 𝟒

1200

)(0.45)[ 2(60)](8)

Vd = 3.46 m3/s The volumetric efficiency, ev ev =

𝐕𝐚 𝑽𝒅

Answer ev =

𝟑 𝟑.𝟒𝟔

= 0.87 or 87%

13. Specific Fuel Consumption, m Specific fuel consumption is the mass flow rate of fuel consumed per unit power developed. It is also known as specific propellant consumption.

a. Indicated specific fuel consumption, mi 𝐦𝐟(𝟑𝟔𝟎𝟎)

𝐤𝐠

𝑷𝒊𝒏𝒅

𝒌𝑾−𝒉𝒓

mi =

Sample problem: (Calculating the indicated specific fuel consumption) What is the indicated specific fuel consumption of a six cylinder, four stroke diesel engine with 76 mm bore x 89 mm stroke and indicated power of 38.83 kW if the engine consumed 12.2 kg/hr of fuel? Solution: mi =

𝟏𝟐.𝟐

𝐤𝐠

𝟑𝟖.𝟖𝟑 𝒌𝑾−𝒉𝒓

b. Brake Specific fuel consumption, mb mb =

𝐦𝐟(𝟑𝟔𝟎𝟎)

𝐤𝐠

𝑷𝒃

𝒌𝑾−𝒉𝒓

Sample problem: (Calculating the brake specific fuel consumption) A four-stroke, 8 cylinder engine with bore and stroke of 9 in and 12 in respectively and speed of 950 rpm has a brake mean effective pressure of 164 psi. What is the brake specific fuel consumption ih lb per hp-hr if the engine consumed 468.55 lbs of fuel per hour? Solution: The brake mean effective pressure consumption, mb mb =

𝐦𝐟 𝑷𝒃

Solving for Pb: 𝐏𝐦𝐛𝐋𝐀 𝐍𝐧

Pb =

𝟑𝟑𝟎𝟎𝟎

Where: π

9

A = 4 (12)2 = 0.44 ft2 Pb =

𝟏𝟕 𝟏𝟐

[𝟏𝟔𝟒(𝟏𝟒𝟒)( )(𝟎.𝟒𝟒)(

𝟗𝟓𝟎 )(𝟖) 𝟐

𝟑𝟑𝟎𝟎𝟎

= 1,201.40 Hp Then: mb =

𝟒𝟔𝟖.𝟓𝟓 𝟏𝟐𝟎𝟏.𝟒𝟎

Answer: mb = 0.39 lb/ hp-hr

c. Combined or over – all Specific consumption ,mc mc =

𝐦𝐟(𝟑𝟔𝟎𝟎)

𝐤𝐠

𝑷𝒈𝒆𝒏

𝒌𝑾−𝒉𝒓

Sample problem: (Calculating the combined specific fuel consumption) A 16-cylinder V - type diesel engine is directly coupled to 1 MW AC generator. Calculate the combined specific fuel consumption if the fuel consumed is 1 kg/s. Solution: mc =

𝟏(𝟑𝟔𝟎𝟎) 𝟏𝟎𝟎𝟎

Answer: mc = 3.6 kg/kW-hr

14.

Heat Rate, HR

Heat rate is the rate of energy charge per unit of power. To calculate the heat rate is to multiply the specific fuel consumption by the heating value of the fuel.

A. Indicated Heat Rate, HRi: HRi =

𝐦𝐟𝐐𝐡(𝟑𝟔𝟎𝟎)

𝐤𝐉

𝑷𝒊𝒏𝒅

𝒌𝑾−𝒉𝒓

Sample problem: (Calculating the indicated heat rate) A four cylinder, 4 stroke cycle, 20 cm x 25 cm x 55 rpm diesel engine has a mean effective pressure of 150 psi. If the heat supplied by the fuel is 50 kW, what is the indicated heat rate? Solution: HRi =

𝐐𝐬𝟑𝟔𝟎𝟎 𝑷𝒊𝒏𝒅

Solving for Indicated power, Pind: Pind = Pm L A N n Pind = [150(

𝟏𝟎𝟏.𝟑𝟐𝟓 𝟏𝟒.𝟕

)](0.25)[

Pind = 148.875 kW then; HRi =

(50)3600 148.875

Answer: HRi = 1209.07

B. Brake Heat Rate, HRb: HRb =

𝐦𝐟𝐐𝐡(𝟑𝟔𝟎𝟎)

𝐤𝐉

𝑷𝒃

𝒌𝑾−𝒉𝒓

π(0.2)2 4

550

][ 2(60)](4)

Sample problem: (Calculating the brake heat rate) A 305 mm 475 mm four stroke single acting diesel engine is rated at 150 kW at 2600 rpm. Fuel consumption at rated load is 0.26 kg/kW-hr with a heating value of 43,912 kJ/kg. Calculate the brake heat rate. Solution: HRb =

mdQh(3600) 𝑃𝑏

HRb = mcQh HRb = (0.26 kg/kW-hr)(43,912 kJ/kg) Answer: HRb = 11,417.12 kJ/kW-hr

C. Combined or over –all Heat Rate, HRc: HRc =

𝐦𝐟𝐐𝐡(𝟑𝟔𝟎𝟎)

𝐤𝐉

𝑷𝒈𝒆𝒏

𝒌𝑾−𝒉𝒓

Sample problem: (Calculating the combined heat rate) The kilowatt output of a generator coupled to a diesel engine is 1.5 MW . If the mass of fuel with heating value 45,000 kJ/kg consumed by the engine is 0.04 kg/s, what is the combined vheat rate? Solution: HRc =

(0.04)(45,000)(3600)

kJ

1,500

𝑘𝑊−ℎ𝑟

Answer: kJ

HRc = 4,320 𝑘𝑊−ℎ𝑟 15. Generator Speed , N N=

𝟏𝟐𝟎𝐟 𝑷

Where: N = speed in rpm f = frequency = 60 Hz(if not given) P = no. of poles Sample problem: (Calculating the generator speed) An eight-cylinder , two cycle, single acting diesel engine rated at 1250 Hp at standard condition is to be directly coupled to a 24-pole alternator, 3 phase, 60 cycles. What is the generator speed? Solution: N=

120f 𝑛

=

120(60) 24

Answer: N = 300 rpm

16. Energy Stream ITEM 1. Useful Output (brake output) 2. Cooling Loss 3. Exhaust Loss 4. Radiation, Friction and Unaccounted Losses Total

HEATLOSS (kJ/hr) 3600 Pb

HEAT INPUT (kJ/kg) mfQf

mwCwΔTw

MfQh

mgCρgΔTg

MfQh

By difference

MfQh

(%) COOLING LOSS 3600Pb 𝑚𝑓𝑄ℎ mwCρΔTw 𝑚𝑓𝑄ℎ mgCρΔTg 𝑚𝑓𝑄ℎ 100 – ∑ 1 to 3

100%

17. Diesel Engine with closed cooling system By Energy Balance Qwj = Qw mwjCpwjΔTwj = mwCpwΔTw mwjΔTwj = mwΔTw

18. Waste Heat Recovery Boiler By energy balance Qsteam = Qgas Ms(hs-hf) = mgCpgΔtg Ms =

𝐦𝐠𝐂𝐩𝐠𝚫𝐭𝐠 𝒉𝒔−𝒉𝒇

Boiler efficiency, eboiler: eboiler =

𝐐𝐬𝐭𝐞𝐚𝐦 𝑸𝒈𝒂𝒔

=

𝐦𝐬(𝐡𝐬−𝐡𝐟) 𝒎𝒈𝑪𝒈𝜟𝒕𝒈

19. Engine operated at high altitudes 𝐏𝐚𝐜𝐭

𝐓

P = Ps (𝟏𝟎𝟏.𝟑𝟐𝟓 √𝟐𝟖𝟖.𝟔) SI units 𝐏𝐚𝐜𝐭

𝐓

P = Ps (𝟐𝟗.𝟗𝟐 √𝟓𝟐𝟎)

English units

Where: Ps = standard power or power at sea level Pact = new pressure or actual barometric pressure in in.Hg T = new temperature or actual absolute temperature in 0R

Note: 1. The decrease is pressure is approximately 1 in.Hg per 1000ft elevation. 𝒉

Pact = 29/92 - 𝟏𝟎𝟎𝟎 in.Hg 2. The decrease in temperature is approximately 3.60R per 1000ft elevation. 𝟑.𝟔 𝒉

T = 520 – 𝟏𝟎𝟎𝟎 0R Sample problem: (Calculating the maximum power delivered at a given elevation) What maximum power can be delivered by 1500 kW engine at 2800 ft elevation considering the pressure effect alone. Solution: Pact

T

P = Ps (29.92 √520) Solving for Pact h

Pactual = 29.92 - 1000 2800

= 29.92 - 1000 = 27.12 in.Hg T = 5200R (for ther’s no effect in temperature) 27.12

520

Answer: P = Ps (29.92 √520)

= 1,359.63 kW.

Chapter 11 GEOTHERMAL POWER PLANT Geothermal Power Plant – is the facility in which the electrical energy is produced from hot spring, steam vent or geyser. Geothermal Energy – is heat energy naturally occurring with the earth. It comes from two words “geo” meaning earth and “thermal” meaning concerning heat.

Magma – a molten metal within the earth which basically nickel-iron in composition whose stored energy heats the surrounding water thereby producing steam or hot water. Its temperature reaches as high as 12000C.

Well-bore product the effluent coming out from the geothermal well produced after drilling. This can be purely steam or hot water, or a mixture of both. Steam-dominated geothermal field refers to a geothermal plant with its well producing all steam as the well-bore product. Liquid-dominated geothermal field refers to a geothermal plant in which the well-bore product is practically all hot water (Pressurized). Fumarole is a crack in the earth through which geothermal substance passes. TYPES OF GEOTHERMAL PLANT 1. Dry or Superheated Geothermal Plant:

2. Separated Steam or “Single Flashed Geothermal Plant”:

3. Separated Steam/Hot-Water-Flash or “Double Flash” Geothermal Plant

4. Single Flashed Plant with Pumped Wells:

5. Binary Geothermal Plant

Performance and Design of Geothermal Power Plant 1. Mass flow rate of steam, ms : ms = x mg where: mg =mass of ground water x = quality after throlling (Process ① → ②) Solving for x: h1 = h2 = hf2 + xhfg2 2. Turbine Output: Wt = ms(h3 – h4) et Where: et = turbine efficiency 3. Generator Output: Wgen = Turbine Output x Generation Eff. Wgen = ms (h3 - h4) etegen 4. Heat Rejected in the Condenser: QR = ms (h4 – h5)

5. Over-all Plant Efficiency: ec = ec =

𝐓𝐮𝐫𝐛𝐢𝐧𝐞 𝐎𝐮𝐭𝐩𝐮𝐭 𝐦𝐠𝐡𝟏 𝐦𝐬(𝐡𝟑−𝐡𝟒)𝐞𝐭 𝐦𝐠𝐡𝟏

SOURCES OF GEOTHERMAL ENERGY 1. Hydrothermal Fluids – basically made up of hot water, steam and minerals. It is the only form of energy currently being tapped for significant commercial heat and electric energy supply. 2. Geopressurized brines – represent a special subset of hydrothermal fluids typically found at depths exceeding at 3km and is characterized as hot water existing at pressure above the normal hydraulic gradient and containing dissolved methane. 3. Hot dry rock – is a water free, impermeable rock at high temperature and practically drilling depth to extract energy, high pressure water maybe injected through one or more wells to create new to enhance existing natural fracture system with limited access to ground water flow. 4. Magma – is characterized by or practically molten rock with temperatures reaching as high as 12000C. THE GEOTHERMAL POWER PLANT IN THE PHILIPPINES 1. Tiwi Geothermal Power Plant, 330 MW (Albay) 2. Makiling – Banahaw (Mak-Ban) Geothermal Power Plant, 330 MW (Los Banos , Laguna) 3. Tongonan Geothermal Power Plant, 112.5 MW (Leyte) 4. Palimpinon Dauin Geothermal Power Plant, 112.5 MW (Negros Oriental)

Hydroelectric Power Plant Hydro-Electric Power Plant is a facility at which electric energy is produced by hydroelectric generators. BASIC PARTS OF HIGH HEAD GYDRO-ELECTRIC POWER PLANTS A. MEDIUM & LOW HEAD PLANTS: RACTION TURBINE

B. HIGH HEAD IMPULSE TURBINE

DEFINITIONS Reservoir stores the water coming from the upper river or waterfalls. Head Water is the water in the reservoir. Spillway is a weir in the reservoir which discharges excess water so that the head of the plant will be maintained. Dam is the concrete structure that encloses the reservoir. Silt Sluice is a chamber which collects the mud and through which the mud is discharged. Valve is a device that opens or closes the entrance of the water into the penstock. Trash rack is a screen which prevents the leaves branches and other water contaminants to enter into the penstock. Penstock is the channel that leads the water from the reservoir to the turbine. Surge Chamber is a standpipe connected to the atmosphere and attached to the penstock so that the water will be at atmospheric pressure. Generator is a device that converts the mechanical energy of the turbine into electrical energy. Hydraulic Turbine is a device that converts the energy of water into the mechanical energy. Tail Race is channel which leads the water from the turbine to the tail water. Tail Water is the water that is discharged from the turbine. Draft Tube is a device that connects the turbine outlet to the tail water so that the turbine can be set above the tail water level. PERFORMANCE OF HYDROELECTRIC POWER PLANT

1. Gross Head, hg Gross Head is the difference between head water and tail water elevation. hg =hhw - htw where: hg= gross head hhw = head water elevation htw = tail water elevation 2. Friction Head Loss, hf Friction Head Loss is the head lost by the flow in a stream or conduit due to frictional disturbances set up by the moving fluid and its containing conduit and by intermolecular friction. a. Using Darcy’s Equation: 𝐟𝐥𝐕𝟐

hf = 𝟐𝐠𝐃

b. Using Morse Equation hf =

𝟐𝐟𝐥𝐕𝟐 𝐠𝐃

where: hf = friction head (in m) f = coefficient of friction L = total length in m g = 9.81 m/s2 D = inside diameter ( in m) Note: Friction head loss is usually expressed as a percentage of the gross head.

3. Net head or Effective head, h: Net Head or Effective head is the difference between the gross head and the friction head loss. h = hg -hf 4. Penstock efficiency,ep: Penstock efficiency is the ratio of the net head to the gross head. 𝐡

ep = 𝐡𝐠 5. Volume flow rate of water, Q: The volume flow rate of water is the product of the velocity and the cross-sectional area. Q = AV 6. Water Power, Pw Water power is the power generated from an elevated water supply by the use of hydraulic turbines. Pw= γQh Where: γ = specific weight of water = 9.81 kN/m3 7. Turbine efficiency, et: Turbine efficiency is the ratio of the turbine power output to the water power output. et = 𝐏𝐭

𝐓𝐮𝐫𝐛𝐢𝐧𝐞 𝐏𝐨𝐰𝐞𝐫 𝐖𝐀𝐭𝐞𝐫 𝐏𝐨𝐰𝐞𝐫

et = 𝐏𝐰 or Pt = γQhet

8. Electrical or Generator Efficiency, egen: Electrical or Generator Efficiency is the ratio of the generator output to the turbine power output. eG = egen =

𝐏𝐠𝐞𝐧 𝐏𝐭

𝐆𝐞𝐧𝐞𝐫𝐚𝐭𝐨𝐫 𝐎𝐮𝐭𝐩𝐮𝐭 𝐓𝐮𝐫𝐛𝐢𝐧𝐞 𝐎𝐮𝐭𝐩𝐮𝐭

or Pgen = Pt egen = γQhetegen

9. Generator Speed, N N=

𝟏𝟐𝟎𝐟 𝐏

Where: N = angular frequency, rpm f = frequency (usually 60 hertz) P = no. of poles (even number) 10. Hydraulic Efficiency, eh: Hydraulic efficiency is the ratio of the utilized head to the net head. eh =

𝐡𝐰 𝐡

where: hw = utilized head h = net head 11. Head of Impulse Turbine (Pelton) Impulse Turbine is a power-generation prime mover in which fluid under pressure enters a stationary nozzle where its pressure (potential) energy is converted to velocity (kinetic) energy and absorbed by the rotor.

12. Head of Reaction Turbine (Francis and Kaplan) Reaction Turbine is a power – generation prime mover utilizing the steady flow principle of fluid acceleration where nozzles are mounted on the moving element.

13. Peripheral Coefficient, Φ Peripheral coefficient is the ratio of the peripheral velocity (Vp) to the velocity of the jet (Vj). Φ= Φ=

𝐕𝐩 𝐕𝐣 𝛑𝐃𝐍 √𝟐𝐠𝐡

Where: D = diameter of runner N = angular speed H = net head 14. Specific Speed of Hydraulic Turbine, Ns Specific speed is a number used to predict the performance of the hydraulic turbines. a. In English units: Ns =

where:

𝐍√𝐇𝐏

N = angular speed, rpm

𝐡𝟓/𝟒

h = net head, ft b. In SI units: Ns =

where:

𝟎.𝟐𝟔𝟐𝟑𝐍√𝐤𝐖 𝐡𝟓/𝟒

N = angular speed, rpm h = net head, m

15. Total efficiency, etotal: etotal = ehemev where: eh = hydraulic efficiency em = mechanical efficiency ev = volumetric efficiency 16. Turbine type recommendation based on head Net Head Up tp 70 ft 70 ft to 110 ft 110 ft to 800 ft 800 ft to 1300 ft 1300 ft and above

Type of Turbine Propeller Type Propeller or Francis Type Francis Type Francis Type or Impulse Type Impulse Type

TYPES OF HYDRAULIC TURBINES: 1. Impulse Turbine (also known as Tangential Wheel or Pelton Wheel) * A turbine that utilizes kinetic energy of high velocity jet which acts upon a small part of the circumference at an instant. * Movements of the water is tangential * Suited for (very) high head plants (150 m and above) & low volume of water * No exact value for critical head, hence heads are given in range * Impulse turbine has no draft tube * Typical turbine efficiencies are in the range of 80% to 90% * Higher efficiencies are associated with turbines having two or more runners 2. Reaction Turbines (Francis Type or Propeller Type) *A turbine which develops power from the combined action of pressure and velocity of the water that completely fills the runner and the water passage * Movement of water for reaction turbines can be radial for Francis Type and axial for Propeller /Kaplan type * Reaction turbine has draft tube which keeps the turbine up to 5 m (15 ft) above the tailwater * Reaction turbines conversion efficiency is usually higher than that of impulse turbine * Reaction turbines conversion efficiency is about 85% to 95% Types of Reaction Turbines A. Francis Type – for medium head. It is named after JAMES B. FRANCIS of California, who developed such turbine type for which conversion efficiency in range of 90-94 %.

Note: the movement is radial B. Propeller & Kaplan Type Reaction Turbines for (very) low head. Propeller (Fixed Blade) – a type of reaction turbine with reduced number of fixed blades. The flow is inward flow axial. Suited for low head plant and has usual conversion efficiency is 80% Kaplan (Adjustable Blade – a type of reaction turbine with a movable blades. The flow is inward flow axial. Suited for low headed and large volume flow of water and usual conversion efficiency of more or less 93%. Note: the movement is axial

Chapter 13 Non-Conventional Power Plant Solar Power Plant is the conversion of the energy of the sun’s radiation to useful work.

TYPES OF SOLAR COLLECTORS ① Flat plate ② Concentrating ③ Focusing

SOLAR ENERGY RECEIVED AT THE EARTH’S SURFACE Es = q S ( 1 – I ) A Where: Qs = solar energy without atmospheric interference i = atmospheric interference A = surface area of solar collector Qsun = Qw + PE + Qloss

II. Wind power Plant Windmills are any various mechanisms, such as mill, pump or electric generator, operated by the force of wind against vanes or sails radiating about a horizontal shafts.

PERFORMANCE OF WIND POWER PLANT 1. Pump Power, Wp Wp =

γQh ep

Where: γ = specific weight of water γ = 9.81 kN/m3 Q = volume flow rate H = net head ep = pump efficiency

2. Kinetic Energy, KEair 1

KEair = 2maVa2 Where: ma = mass flow rate of air Va = average velocity of air

3. Volume of air, Va Va =

maRaTa Pa

If wind velocity is given: Volume = Area x Velocity Va =

πD2 4

x velocity

4. Aerodynamic Efficiency, ea: Pump Power

ea = Kinetic Energy of Air OTHER NON-CONVENTIONAL POWER SOURCES ① Tidal Power ② Thermoionic Converter ③ Fuel Cell ④ Low Thermal Head Plant ⑤ Magneto Hydrodynamic Plant

Sample problem: 1. A solar collecting panel 20 m2 in area receives solar energy at the rate of 750 W/m2. It is estimated that 35% of the incident energy is lost to surrounding; Water enters the panel at steady flow rate of 0.05 kg/s at 150C. Calculate the temperature of water leaving the collecting panel. Change in elevation of water is 2 meters. Solution: By Energy Balance: Qsun = Qw + Qloss + P.E. Where: W

Qsun = 750 m2 (20m2) = 15,000 Watts Qw = mCwΔtw = (0.05 kg/s)(4.187 kJ/kg.K)(to-15)(1000 W/kW) = 209.35 (to-15) Qloss = 0.35(15,000) = 5,250 W P.E. = mgh = (0.05kg/s)(9.81m/s2)(2m) = 0.981 W Substituting to the above formula: 15,000 = 209.35 (to – 15) + 5250 + 0.981 Thus; Answer t = 61.56 0C

Chapter 14 NUCLEAR POWER PLANT NUCLEAR POWER PLANT is a power plant in which nuclear energy is converted into heat to be used in producing steam for turbines, which in turn drive generators that produce electric power. TYPICAL NUCLEAR POWER PLANT

Fuel Core is a radioactive material, U235 and U238 which is source of energy. Moderator is a device that slows down the neutrons to thermal energy, made of Carbon and Beryllium. Control Rods are boron coated steel rods used to control the reactor. Reflector is a device made of lead or carbon which surrounds the core to bounds back any leakage of neutrons. Thermal Shield is a device that prevents escape of radiation from reactor vessel. Reactor Drum is a device that encloses the fuel core and components.

Biological Shield is a concrete or lead which absorbs any leakage of radiation and protects operators from exposure to radioactivity. Control Cubicle is a device that contains the meters that show the operating quantities in the reactor. Containment Vessel is a concrete the prevent the spread of radiation in case of major explosion. Coolant is a substance that absorbs the heat from the fuel core then releases the heat to the water in the steam generator. Coolant Pump is a device (pump) that circulates the coolant. Turbine generator is a device that generates the electric power. Condenser is a device that converts steam coming from the turbine into liquid. Feedwater Pump a device that delivers the feedwater to the steam generator.

TYPES OF NUCLEAR POWER REACTORS 1. PRESSURIZED WATER NUCLEAR POWER REACTOR A pressurized water reactor uses water under pressure as both the coolant and moderator in the reactor. The water pressure must be sufficiently high so that we can have water at 550 -6600F without boiling in the core. The fission of the fuel produces heat that is carried away by circulating water under pressure. 2. FAST-BREEDER NUCLEAR POWER REACTOR A fast - breeder reactor depends upon the fission of fast neutrons rather than thermal neutrons. Therefore, no moderator or moderating material cane be used. Further the term “breeder” implies that the reactor produces more fuel due to the absorption of neutrons than is burned up. This is possible since on the average 2.5 neutrons are produced due to its fission of a uranium atom and only one of this is needed to keep the chain reaction going. 3. GAS – COOLED NUCLEAR POWER REACTOR Gas cooled nuclear power reactor uses carbon dioxide as coolant.

Advantages of Gas Cooled Power Reactor 1. High temperature is possible at low pressure since boiling is not a problem. 2. Plant thermal efficiencies increasing cycle temperatures Steam can be produced in heat exchangers heated by gas. 3. Gases can be non hazardous. Disadvantages of Gas Cooled Power Reactor 1. The power required to pump the gas compared to that for pumping liquid. 2. Large flow rates and high velocities are required. 3. Care must be taken not to use gases that are corrosive to moderating materials at the high temperatures employed.

4. LIQUID – COOLED NUCLEAR POWER REACTOR There are two types of liquid – cooled reactor: the water-cooled reactor and the liquid metal – cooled reactor. Advantages of water-cooled reactor 1. 2. 3. 4.

Low cost of coolant Coolant is also a moderator Pumping power is small and design of pumps is relatively simple Low viscosity

Disadvantages of water-cooled reactor 1. High-pressures are required to obtain high temperatures. 2. Detrimental to large conversion ratios because hydrogen in the water is a good absorber. 3. Problem of corrosion of materials in contact with the water. Advantages of liquid metal-cooled reactor 1. 2. 3. 4.

High thermal conductivity Can operate atmospheric pressure Good heat transfer characteristic Stable at high temperatures

Disadvantages of liquid metal-cooled reactor 1. High Cost 2. Danger of reaction with air or water makes for difficulty in handling.

5. BOILING WATER NUCLEAR POWER REACTOR Boiling water nuclear power reactor is the simplest nuclear reactor. The feedwater from the power turbine goes directly into the reactor and picks up the heat from the fuel core, thus the feedwater also serves as the coolant.

Advantages of boiling water reactor 1. Moderator and coolant are the same (water). This is the same as in a pressurized-water reactor. 2. The steam formed goes directly to the turbine, thus eliminating the extra heat exchange present in the pressurized water plant. 3. The pumping power is much less than that for pressurized water plants because of lower operating pressures to produce the same temperature and elimination of the extra loop. 4. The stresses in the pressure vessel are lower because of lower operating pressures, thus decreasing costs. Disadvantages of boiling water reactor 1. Presence of the steam in the reactor will result in more neutron leakage and will require a more highly enriched fuel. (For example: enrichment of U235) 2. There will be some depositing radioactive elements in the pump, turbine, and condensing. This can become a hazard under normal maintenance procedures. 3. The leakage of the less dense steam is more apt to occur than in a pressurized water reactor

Chapter 17 VARIABLE LOAD PROBLEMS DAILY LOAD CURVE

1. RESERVE OVER PEAK RESERVE OVER PEAK = PLANT CAPACITY – PEAK LOAD 2. AVERAGE LOAD AVERAGE LOAD =

𝐄𝐍𝐄𝐑𝐆𝐘 𝐏𝐑𝐎𝐃𝐔𝐂𝐄𝐃 (𝐢𝐧 𝐤𝐖−𝐡𝐫𝐬) 𝐍𝐔𝐌𝐁𝐄𝐑 𝐎𝐅 𝐇𝐎𝐔𝐑𝐒

3. LOAD FACTOR LOAD FACTOR =

𝐀𝐕𝐄𝐑𝐀𝐆𝐄 𝐋𝐎𝐀𝐃 𝐏𝐄𝐀𝐊 𝐋𝐎𝐀𝐃

4. CAPACITY FACTOR CAPACITY FACTOR =

𝐀𝐂𝐓𝐔𝐀𝐋 𝐄𝐍𝐄𝐑𝐆𝐘 𝐏𝐑𝐎𝐃𝐔𝐂𝐄𝐃 𝐌𝐀𝐗𝐈𝐌𝐔𝐌 𝐄𝐍𝐄𝐑𝐆𝐘 𝐏𝐑𝐎𝐃𝐔𝐂𝐄𝐃 𝐎𝐍 𝐓𝐇𝐄 𝐒𝐀𝐌𝐄 𝐏𝐄𝐑𝐈𝐎𝐃

5. ANNUAL CAPACITY FACTOR ANNUAL CAPACITY FACTOR =

𝐀𝐍𝐍𝐔𝐀𝐋 𝐄𝐍𝐄𝐑𝐆𝐘 𝐏𝐑𝐎𝐃𝐔𝐂𝐓𝐈𝐎𝐍 (𝐢𝐧 𝐤𝐖−𝐡𝐫𝐬) 𝐤𝐖 𝐏𝐋𝐀𝐍𝐓 𝐂𝐀𝐏𝐀𝐂𝐈𝐓𝐘 𝐗 𝟖𝟕𝟔𝟎 𝐡𝐫𝐬

6. USE FACTOR 𝐀𝐍𝐍𝐔𝐀𝐋 𝐤𝐖−𝐡𝐫𝐬

USE FACTOR = 𝐤𝐖 𝐏𝐋𝐀𝐍𝐓 𝐂𝐀𝐏𝐀𝐂𝐈𝐓𝐔 𝐗 𝐍𝐎.𝐎𝐅 𝐇𝐎𝐔𝐑𝐒 𝐎𝐏𝐄𝐑𝐀𝐓𝐈𝐎𝐍

7. DEMAND FACTOR DEMAND FACTOR =

𝐀𝐂𝐓𝐔𝐀𝐋 𝐌𝐀𝐗.𝐃𝐄𝐌𝐀𝐍𝐃 𝐂𝐎𝐍𝐍𝐄𝐂𝐓𝐄𝐃 𝐋𝐎𝐀𝐃

8. DIVERSITY FACTOR 𝐒𝐔𝐌 𝐎𝐅 𝐈𝐍𝐃𝐈𝐕𝐈𝐃𝐔𝐀𝐋 𝐌𝐀𝐗.𝐃𝐄𝐌𝐀𝐍𝐃

DIVERSITY FACTOR = 𝐌𝐀𝐗𝐈𝐌𝐔𝐌 𝐒𝐈𝐌𝐔𝐋𝐓𝐀𝐍𝐄𝐎𝐔𝐒 𝐃𝐄𝐌𝐀𝐍𝐃𝐒 9. UTILIZATION FACTOR UTILIXATION FACTOR =

𝐌𝐀𝐗𝐈𝐌𝐔𝐌 𝐃𝐄𝐌𝐀𝐍𝐃 𝐎𝐅 𝐓𝐇𝐄 𝐒𝐘𝐒𝐓𝐄𝐌 𝐑𝐀𝐓𝐄𝐃 𝐂𝐀𝐏𝐀𝐂𝐈𝐓𝐘 𝐎𝐅 𝐓𝐇𝐄 𝐒𝐘𝐒𝐓𝐄𝐌

10. OPERATION FACTOR 𝐃𝐔𝐑𝐀𝐓𝐈𝐎𝐍 𝐎𝐅 𝐀𝐂𝐓𝐔𝐀𝐋 𝐒𝐄𝐑𝐕𝐈𝐂𝐄

OPERATION FACTOR = 𝐓𝐎𝐓𝐀𝐋 𝐃𝐔𝐑𝐀𝐓𝐈𝐎𝐍 𝐎𝐅 𝐓𝐇𝐄 𝐂𝐎𝐍𝐒𝐈𝐃𝐄𝐑𝐄𝐃 𝐏𝐄𝐑𝐈𝐎𝐃 11. PLANT FACTOR PLANT FACTOR =

𝐀𝐕𝐄𝐑𝐀𝐆𝐄 𝐋𝐎𝐀𝐃 𝐄𝐐𝐔𝐈𝐏𝐌𝐄𝐍𝐓 𝐑𝐀𝐓𝐈𝐍𝐆 𝐒𝐔𝐏𝐏𝐋𝐘𝐈𝐍𝐆 𝐓𝐇𝐄 𝐋𝐎𝐀𝐃

DEFINITION Load Curve refers to the graph use to represent the relationship between the demanded load and time sequence. Monthly Load Curve is the average of the daily load curves over in a one month period that is used in a establishing the rates. Annual Load Curve is the average of the daily load curves over a period of one year that is used in determinating the annual load factor. Load Duration Curve is obtained from the same data as the daily load curves for one year or a period of 8,760 hours.

BASE LOAD AND PEAK LOAD POWER PLANTS Base Load Power Plant include steam, hydro-electric and geothermal power plants. Peak Load Power Plant include diesel – electric and gas turbine power plants.

Chapter 19 Miscellaneous ME Board Problems SUPPLIMENTARY PROBLEMS

1. The over-all efficiency of an electric system, coal pile to lights, is 11%. What is the fraction of a kg of coal containing 30,250 kJ/kg must be consumed to light a 100 W lamp for 45 minutes? a. 0.78 kg c. 0.081 kg b. 0.095 kg d. 0.065 kg Solution: kJ equivalent of 1 to 100 W lamp lighted for 45 minutes J s

= (100 )(45 min)(

60 s min

)

= 270,000 J = 270 kJ Required head input from fuel for 11% efficiency: =

270 0.11

= 2454.545 kJ

Fraction of 30,250

=

kJ kg

coal to supply 2454.545 kJ

2545.545 kJ 30,250 kJ/kg

Thus; Answer = 0.081 kg 2. A power plant serves a factory having two 22 kW motors and ten 3.7 kW motors. Assume the efficiency of motors 80%, of transmission line 95%, of generator 92%. What should be the rated capacity of the engine? a. 115.85 kW c. 245.56 kW b. 200.75 kW d. 457.35 kW Solution: Max. total kW output: = 2(22) + 10(3.7) = 81 kW

The kW capacity of generator to take care of this load at 80% motor efficiency and 95% transmission efficiency: =

81 (0.80)(0.95)

= 106. 58 kW The rated capacity of engine at 92% generator efficiency: =

106.58 0.92

Thus, Answer = 115.85 kW

3. An electric motor converted 700 W of electrical input into work at 58.186 kg-

m/s. The speed was 1750 rpm. Find the kg-m of driven torque. a. 0.183 kg-m c. 0.318 kg-m b. 0.381 kg-m d. 0.138 kg-m Solution: kW output of motor kg−m

= (58.186 = 570.80

s N−m

m

)(9.81s2)

s

= 0.571 kW From: P=

2πTN 60

0.571 =

Answer:

2πT(1750) 60

T = 0.003116 kN – m = 3.116 N – m T = 0.318 kg – m

4. In a charge over from the steam to the electric heating, a unit which had been considering 4.5 kg dry and saturated steam at 2.109 kg/cm2 absolute each 15 min is to be electrical heated from 220 V circuit. How many 12 ohms resistor in

parallel would be needed to supply the same heat? Latent heat of vaporization at 2.109 kg/cm2 is 2,198.535 kJ/kg. a. 3 c. 4 b. 2 d. 5 Solution: Heat given off by 4.5 kg/min dry and sat. steam at 2.109 kg/cm2 = hfg (4.5) =

2198.535(4.5) 15

= 659.56 kJ/min Heat given off in 1 hr. = 659.56 kJ/ min (60min/hr) = 39,573.6 kJ/hr If the same heat is to come from an electric source. = 39,573.6 = 10.993 kW = 10, 993 W Total resistance required: P=

V2 R

10,993 =

(22)2 R

R = 4.403 ohms No. of Resistors required: 12

n = 4.403 = 2.725 say 3 Actual Total Resistance: Answer:

=12/3

= 4

5. An electric heater is to heat 11 kg of oil per minute from 4.5 0C to 65.50C. Specific heat of the oil is 2.1 kJ/kg0C. How many watts should this heater consume? a. 24.385 kW c. 28.385 kW b. 23. 485 kW d. 32. 854 kW Solution Q = m Cp ∆T = 11 (2.1)(65.5 – 4.5) kJ

1 min

= 1409.1 min ( 60 s ) Thus; Q = 23.485 kW 6. The flow of a river is 4.25 m3/s at a site where a 22.86 m hydrostatic head can be created by the erection of a dam. What is the potential capacity of a hydroelectric power plant if installed at this site? Consider that the hydraulic efficiency of energy conversion can be 80%, and that the electric efficiency can be 90%. a. 866.522 kW c. 566.822 kW b. 686.225 kW d. 656.522 kW Solution KW that can be develop from the water flow: = y Q h eh = 9.81 (4.25) (22.86)(0.80) = 762.47 kW KW capacity at 90% electric efficiency: = 762.47 (0.90) Thus; = 686.225 kW

7. A hoist is to raise a 11147.727 N mine cage at the rate of 4.57 m/s. Mechanical efficiency of the hoist is 92%. What is the kW required to drive at this speed? a. 73.522 kW c. 55.375 kW b. 73. 355 kW d. 53. 753 kW Solution Work done by the hoist: = (11,147.727) (4.57) = 50945.114 = 50.945 kW

N.m s

The kW required at 92% efficiency of the hoist: =

50.945 0.92

Thus; = 55.375 kW

8. A turbo generator rotating mass has a moment of inertia of 555 hyls/m2. It is delivering 2500 kW at 1800 rpm. The load then suddenly increased to 2550 kW, the developed steam remaining unchanged. What is the resulting speed in rpm after 10 seconds? a. 177.82 rpm c. 128.77 rpm b. 182.77 rpm d. 217.87 rpm Solution Kinetic Energy of Rotating mass: 1

= 2 I ω2 1

1800

= 2 (555) [

60

(2π)]2

= 9,859,734.797 kg-m = 96723.998 kN-m = 96723.998 kJ Change in Load = 2550 – 2500 = 50 kW Total Change of Energy in 10 s: = 50 kJ (10 s) = 500 kJ Net KE of rotating mass = 95723.998 – 500 = 96223.998 kJ Let, ω2 = angular velocity after 10 seconds. ω22 =

2(96223.988) 555

= 18.62 rad/s Thus, ω22 = 18.62 rad/s ω2 = 177.82 rpm

9. Forty percent if the electrical input to a motor driven pump is converted into a hydraulic jet 0.013 m in diameter for the purpose of washing down ashes. Find the jet velocity in m/s. The motor has a 3-phase, 220 V, 7.5 amp rating. Power factor 85%. a. 52.38 m/s c. 38.53 m/s b. 58.32 m/s d. 35.83 m/s Solution Watts input to motor = √3 E I cos ө = √3 (220) (7.5) (0.85) = 2429.20 W = 2.43 W Energy Converted to jet: = 0.40 (2.43) = 0972 kW = 972 W Then; V2

972 = (ρ A V) 2(9.81) π

V2

972 = [1000x4 (0.013)2V] 2(9.81) Thus; V = 52.38 m/s 10. The difference in tension between the sides of a belt running over a 0.762 m diameter pulley is 22.727 kg. Pulley speed is 500 rpm. What kW is transmitted? a. 4,45 kW

c. 3.45 kW

b. 3.45 kW

d. 6.44 kW

Solution: P=

2πTN 60

Solving for torque: T = net tension x radius of pulley 0.762

= 22.727 (

2

= 8.659 kg-m

) = 84.94N-m

Then; P=

2π(84.94)(500) 60

= 4447.70 W Thus; Answer

P = 4.45 kW

11. What is the power of a steam jet 0.015m in diameter moving at 761.963 m/s? Steam condition, 1.406 kg/cm2 dry and saturated (v = 0.19 m3/kg) a. 220.27 kW c. 208.72 kW b. 202.78 kW d. 280.27 kW Solution: The mass of the steam leaving the nozzle AV

m=

v1

where: A = area of jet V = velocity of jet v1 = specific volume π

m=4

(0.015)2(761.963) 0.19

= 0.719 kg/s Kinetic Energy of the jet: mV2

KE =

2g

=

0.719(761.963)2 2(9.81)

= 21,276.376 = 208.72 kW

kg− m 0.00981 kN s

(

1 kg

)

12. A pump is lifting water through 5.486 m to fill a 566.254 m 3 tank. The over-all energy efficiency is 80%. Calculate the length of time that 7.46 kW applied to the pump will require to complete the job. a. 85.11 min c. 75.11 min b. 81.51 min d. 71.51 min Solution: Total mass of water pumped to tank at 1000 kg/m3: = 566.254 (1000) = 566254 kg Total work input to pump at 80% efficiency: =

(566254 kg)(5.486) 0.80

= 3,883,086.805 kg – m = 38093.08 kN- m Work output of 7.46 kW: = 7.46

kN−m 60 s s

(min)

kN−m

= 447.60

min

Time required to fill the tank: = =

total work input to pump rate of doing work 38093.08 kN−m kN−m min

447.60

Thus; Answer

= 85.11 min.