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Diesel Power Plant Problem Solving Examination NAME: Vincent Rey Olario Y. Bsme - 5 Show Complete Solutions. Solve The Following: Problem 1

The document contains 4 diesel engine problems involving calculations of fuel consumption, brake thermal efficiency, pressures, power output, and fuel tank sizing. Problem 4 involves calculating load factors, reserve capacity, plant capacity factor, plant use factor, hours off, and utilization factor for a 1 MW power plant based on hourly demand data.

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530 views5 pages

Diesel Power Plant Problem Solving Examination NAME: Vincent Rey Olario Y. Bsme - 5 Show Complete Solutions. Solve The Following: Problem 1

The document contains 4 diesel engine problems involving calculations of fuel consumption, brake thermal efficiency, pressures, power output, and fuel tank sizing. Problem 4 involves calculating load factors, reserve capacity, plant capacity factor, plant use factor, hours off, and utilization factor for a 1 MW power plant based on hourly demand data.

Uploaded by

Bensoy
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Diesel Power Plant Problem Solving Examination

NAME: Vincent Rey Olario Y.


BSME – 5
SHOW COMPLETE SOLUTIONS. SOLVE THE FOLLOWING:
PROBLEM 1.
A 2 - stroke, 4 - cylinders, 38 cm x 53 cm diesel engine is guaranteed to deliver
BP = 522 KW at 300 rpm. The fuel rate is mFB =0.26 kg/KW-hr. If the heating value of
the fuel is 44,320 KJ/kg and density of 890 kg/m3.Calculate:
a. the fuel consumption in kg/hr
b. the brake thermal efficiency
c. the brake mean effective pressure in Kpa
d. the indicated power if mechanical efficiency is 85%e. the indicated mean effective
pressure in KPa
f. the dimension of the fuel tank required for 2 weeks supply if the engine operates 10
hrs a day (assume D = 0.75H)
PROBLEM 2.

Find the volume in Liters needed for a two weeks supply of 26°API fuel oil to
operate a 750 KW engine 70 % of the time at full load, 10 % at 3/4 load and idle 20% of
the time. Fuel rate is 0.25 kg/KW-hr at full load and 0.24 kg/KW-hr at 3/4 load.
Temperature of oil is 21°C.

Solution:
T = 2 weeks (7 days) (24 hours) = 336 hrs
Mf = 336 (0.70) 0.25(750)+0.10(336)(0.75)(750)(0.24)= 48,636 kg
S = 141.5/ 131.5+26 = 0.898
S@t = 0.898-0.0007(21-15.56) =0.895
P = 895kg/m3 = 0.895 kg/L
Vf = 48,636/0.895= 54,342 Liters

PROBLEM 3.
A single acting, 4-cycle diesel engine uses 11 kg/hr of 24°API fuel when running
at 420 RPM. Engine specifications: 23 cm x 36 cm. The Prony brake used to determine
the brake power has 1 m arm and registers on the scale 130 kg gross. If the tare mass
is 12 kg, calculate the brake thermal efficiency based on the lower heating value of fuel.

Solution:
T = (P – tare)R = (130 – 12)1 = 118 kg-m (9.81 N/kg) = 1,157.58 N-m
BP = 2Πtn/ 60,000 = 51 KW
LHV = 38,105 + 139.6 1(API) = 41,455.4 KJ/kg
Ebrake=(3600BP/ mf (LHV) ) x 100% = 40.3 %

PROBLEM 4. A generation station of 1MW supplied a region which has the following
demands:
From To Demand (kW)
midnight 5 am 100
5 am 6 pm No-load
6 pm 7 pm 800
7 pm 9 pm 900
9 pm midnight 400
Neglect transmission line losses and find the following:
1. Plot the daily load curve and the load duration curve.
2. Find the load factor, the reserve capacity, plant capacity factor, plant use factor, the
hours that the plant has been off and utilization factor.

Solution:
When the transmission line losses are neglected, Pg = PL , and the demand = load
Installed capacity = 1 MW = 1000 kW and max. load = max. demand = 900 kW

Average Load = Area under the load curve (kWh) / no. of hours (h)

Average Load = (5 × 100) + (13 × 0) + (1 × 800) + (2 × 900) + (3 × 400) / 24

Average Load = 4300 kWh / 24 h = 179.16 kW


Load Factor = Average Load / Max. Load = 179.16 kW/900 kW = 0.199 = 19.19%

The Reserve Capacity = Installed Capacity − Max. Demand = 1000 − 900 = 100 kW

Plant Capacity Factor = Average Demand/Installed Capacity = 179.16 kW/1000 kW = 0.1791 = 17.91%

Plant Use Factor = Actual Energy Produced in (kWh)/Plant Capacity × no. of hours

= 4300 kWh/1000 kW × 11 h

= 0.3909 = 39.09%

Utilisation Factor = Max. Demand/installed Capacity = 900 kW/1000 kW = 0.9 = 90%P

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