1. The heat generated by fuel is 2500 𝑘�𝑊�. The water jacket loss is 30%. If the
allowance temperature rise of the coolant is 8°𝐶�, determine the volume flow rate
in gallons per minute of circulating water required.
Given:
𝐸𝐶 = 2500�𝑘𝑊
𝑊𝑎𝑡𝑒𝑟�𝐽𝑎𝑐𝑘𝑒𝑡�𝐿𝑜𝑠𝑠 = 30%
∆𝑇 = 8�℃
Required:
𝑉𝑐𝑤 �𝑖𝑛
𝑔𝑎𝑙𝑙𝑜𝑛𝑠
𝑚𝑖𝑛𝑢𝑡𝑒
Solution:
𝑄𝑙𝑜𝑠𝑠 = 𝐸𝐶(𝑊𝑎𝑡𝑒𝑟�𝐽𝑎𝑐𝑘𝑒𝑡�𝐿𝑜𝑠𝑠)
𝑄𝑙𝑜𝑠𝑠 = (2500�𝑘𝑊)(30%)
𝑄𝑙𝑜𝑠𝑠 = 750�𝑘𝑊
𝑄𝑙𝑜𝑠𝑠 = 𝑚𝑐𝑤 𝐶𝑝 ∆𝑇
750�𝑘𝑊 = 𝑚𝑐𝑤 (4.187
𝑚𝑐𝑤 = 22.39
𝑘𝑔
𝑠
𝜌𝑐𝑤 =
𝑉𝑐𝑤 =
𝑉𝑐𝑤 =
𝑘𝐽
) (8�𝐾)
𝑘𝑔𝐾
𝑚𝑐𝑤
𝑉𝑐𝑤
𝑚𝑐𝑤
𝜌𝑐𝑤
𝑘𝑔
22.39 𝑠
𝑘𝑔
𝑚3
𝑚3 3.283 �𝑓𝑡 3 7.48�𝑔𝑎𝑙𝑙𝑜𝑛𝑠
60�𝑠
𝑉𝑐𝑤 = 0.0224
×
×
×
3
3
𝑠
1�𝑚
1�𝑓𝑡
1�𝑚𝑖𝑛
𝑔𝑎𝑙𝑙𝑜𝑛𝑠
𝑉𝑐𝑤 = 354.603
𝑚𝑖𝑛𝑢𝑡𝑒
1000
2. A 4−𝑠�𝑡�𝑟�𝑜�𝑘�𝑒�, 8 −𝑐�𝑦�𝑙�𝑖�𝑛�𝑑�𝑒�𝑟�, I.C.E with bore and stroke of 9 − 𝑖�𝑛� and 12 −
𝑖�𝑛�, respectively and speed of 950 𝑟�𝑝�𝑚� has a break mean effective pressure of
164 𝑝�𝑠�𝑖�. The brake specific fuel consumption is 0.39 𝑙�𝑏�/𝐻�𝑝� − ℎ𝑟� and the fuel
heating value is 18,500 𝐵�𝑇�𝑈�/𝑙�𝑏�. Find the engine’s thermal efficiency.
Given:
4 − 𝑠𝑡𝑟𝑜𝑘𝑒
8 − 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
𝐷 × 𝐿 = 9�𝑖𝑛 × 12�𝑖𝑛
𝑁 = 950�𝑟𝑝𝑚
𝑃𝑏 = 164�𝑝𝑠𝑖
𝐵𝑆𝐹𝐶 = 0.39
𝑙𝑏
ℎ𝑝 − ℎ𝑟
𝐻𝐻𝑉 = 18500
𝐵𝑇𝑈
𝑙𝑏𝑚
Required:
𝑒𝑏
Solution:
𝑒𝑏 =
𝐵𝑃
× 100%
𝐸𝐶
𝐵𝑃 = 𝑃𝑏 𝐿𝐴𝑁
𝑙𝑏𝑓
𝜋
𝑟𝑒𝑣
2𝜋
1�𝑚𝑖𝑛 1
𝐵𝑃 = (164 2 ) (12�𝑖𝑛) ( (9�𝑖𝑛)2 ) (950
×
×
× )
𝑖𝑛
4
𝑚𝑖𝑛 1�𝑟𝑒𝑣
60�𝑠
2
𝑖𝑛 − 𝑙𝑏𝑓 1�𝑓𝑡
60�𝑠
𝐵𝑃 = 6227621.681
×
×
𝑠
12�𝑖𝑛 1�𝑚𝑖𝑛
𝑓𝑡 − 𝑙𝑏𝑓
1�ℎ𝑝
𝐵𝑃 = 31138108.41
×
𝑓𝑡 − 𝑙𝑏𝑓
𝑚𝑖𝑛
33000 𝑚𝑖𝑛
0.746�𝑘𝑊
𝐵𝑃 = 943.579�ℎ𝑝 ×
1�ℎ𝑝
𝐵𝑃 = 703.910�𝑘𝑊
𝐵𝑆𝐹𝐶 =
𝑚𝑓
𝐵𝑃
𝑚𝑓 = 𝐵𝑆𝐹𝐶 × 𝐵𝑃
𝑙𝑏𝑚
𝑚𝑓 = 0.39
× 943.579�ℎ𝑝
ℎ𝑝 − ℎ𝑟
𝑙𝑏𝑚
1�𝑘𝑔
𝑚𝑓 = 367.995
×
ℎ𝑟 2.205�𝑙𝑏𝑚
𝑘𝑔
𝑚𝑓 = 166.891
ℎ𝑟
𝐸𝐶 = 𝑚𝑓 𝐻𝐻𝑉
𝑘𝐽
1
𝑘𝑔
𝐵𝑇𝑈
𝑘𝑔
𝐸𝐶 = (166.891 ) (18500
×
)
ℎ𝑟
𝑙𝑏𝑚 0.42987 𝐵𝑇𝑈
𝑙𝑏𝑚
𝐸𝐶 = 7182388.142
𝑘𝐽
1�ℎ𝑟
1�𝑚𝑖𝑛
×
×
ℎ𝑟 60�𝑚𝑖𝑛
60�𝑠
𝐸𝐶 = 1995.107�𝑘𝑊
𝐵𝑃
𝐸𝐶
703.91�𝑘𝑊
𝑒𝑏 =
× 100%
1995.107�𝑘𝑊
𝑒𝑏 = 35.282%
𝑒𝑏 =
3. The engine thermal efficiency of 1 − 𝑀�𝑊� Diesel electric plant is 36%. For a
generator efficiency of 89%, determine the heat generated by the fuel in unit of
horsepower.
Given:
𝐸𝑃 = 1�𝑀𝑊
𝑒𝑏 = 36%
𝜂𝑔 = 89%
Required:
𝐸𝐶
Solution:
𝜂𝑔 =
𝐸𝑃
𝐵𝑃
𝐸𝑃
𝜂𝑔
1�𝑀𝑊
𝐵𝑃 =
89%
𝐵𝑃 =
1000�𝑘𝑊
1�𝑀𝑊
𝐵𝑃 = 1123.595�𝑘𝑊
𝐵𝑃 = 1.123�𝑀𝑊 ×
𝑒𝑏 =
𝐵𝑃
𝐸𝐶
𝐵𝑃
𝑒𝑏
1123.595�𝑘𝑊
𝐸𝐶 =
36%
𝐸𝐶 =
𝐸𝐶 = 3121.098�𝑘𝑊 ×
𝐸𝐶 = 4183.777�ℎ𝑝
1�ℎ𝑝
0.746�𝑘𝑊
4. A diesel electric plant of 700-kW capacity has an engine thermal efficiency of
30%. For a 2,500 𝑘�𝑊� heat generation by the fuel used, determine the generator
efficiency.
Given:
𝐸𝑃 = 700�𝑘𝑊
𝑒𝑏 = 30%
𝐸𝐶 = 2500�𝑘𝑊
Required:
𝜂𝑔
Solution:
𝜂𝑔 =
𝐸𝑃
× 100%
𝐵𝑃
𝑒𝑏 =
𝐵𝑃
𝐸𝐶
𝐵𝑃 = 𝑒𝑏 𝐸𝐶
𝐵𝑃 = (30%)(2500�𝑘𝑊)
𝐵𝑃 = 750�𝑘𝑊
700�𝑘𝑊
× 100%
750�𝑘𝑊
𝜂𝑔 = 93.333%
𝜂𝑔 =
5. An engine having a thermal efficiency of 35% uses 25°𝐴�𝑃�𝐼� fuel at 40 𝑘�𝑔�/ℎ𝑟�.
Determine the rating of the engine in kW.
Given:
𝑒𝑏 = 35%
°𝐴𝑃𝐼 = 25
𝑚𝑓 = 40
𝑘𝑔
ℎ𝑟
Required:
𝐵𝑃
Solution:
𝑒𝑏 =
𝐵𝑃
𝐸𝐶
𝐸𝐶 = 𝑚𝑓 𝐻𝐻𝑉
𝐻𝐻𝑉 = 17680 + 60(°𝐴𝑃𝐼)
𝐻𝐻𝑉 = 17680 + 60(25)
𝑘𝐽
1
𝐵𝑇𝑈
𝑘𝑔
𝐻𝐻𝑉 = 19180
×
𝑙𝑏𝑚 0.42987 𝐵𝑇𝑈
𝑙𝑏𝑚
𝑘𝐽
𝐻𝐻𝑉 = 44618.140
𝑘𝑔
𝐸𝐶 = 𝑚𝑓 𝐻𝐻𝑉
𝑘𝑔
1�ℎ𝑟
1�𝑚𝑖𝑛
𝑘𝐽
𝐸𝐶 = (40
×
×
) (44618.140 )
ℎ𝑟 60�𝑚𝑖𝑛
60�𝑠
𝑘𝑔
𝐸𝐶 = 495.757�𝑘𝑊
𝑒𝑏 =
𝐵𝑃
𝐸𝐶
𝐵𝑃 = 𝑒𝑏 𝐸𝐶
𝐵𝑃 = (35%)(495.757�𝑘𝑊)
𝐵𝑃 = 173.514�𝑘𝑊
6. A six cylinder, 4-stroke Gasoline engine power plant of 10.5 cm by 12.7 cm
dimension has a compression ratio of 15. When it is tested on a dynamometer
with 54 cm dynamometer arm at 2,500 rpm, the scale reads 82 kg. 3 kg of fuel
with a 45, 820 kJ/kg heating value are burned during a 6 minute testing.
Considering that friction is present in the system and is determined to be 10% of
the brake torque. Determine the following: (a) brake thermal efficiency, (b) brake
specific fuel consumption, (c) friction power, (d) indicated Power, (e) indicated
thermal efficiency, (f) indicated specific fuel consumption, and (g) mechanical
engine efficiency.
Given:
6�𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟𝑠
4 − 𝑠𝑡𝑟𝑜𝑘𝑒
𝐷 × 𝐿 = 10.5�𝑐𝑚 × 12.7�𝑐𝑚
𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛�𝑅𝑎𝑡𝑖𝑜 = 15
𝐿𝑎𝑟𝑚 = 54�𝑐𝑚
𝑁 = 2500�𝑟𝑝𝑚
𝐹 = 82�𝑘𝑔𝑓
𝑚𝑓 = 3�𝑘𝑔
𝐻𝐻𝑉 = 45820
𝑘𝐽
𝑘𝑔
𝑡 = 6�𝑚𝑖𝑛𝑠
𝐹𝑙𝑜𝑠𝑠 = 10%𝑇𝑏𝑟𝑎𝑘𝑒
Required:
𝑒𝑏
𝐵𝑆𝐹𝐶
𝐹𝑃
𝐼𝑃
𝑒𝐼
𝐼𝑆𝐹𝐶
𝜂𝑚
Solution:
𝑒𝑏 =
𝐵𝑃
× 100%
𝐸𝐶
𝐵𝑃 =
𝑇𝑁
9.549 × 103
𝑇𝑏 = 𝐹𝐿𝑎𝑟𝑚
9.81�𝑁
1�𝑚
) (54�𝑐𝑚 ×
)
1�𝑘𝑔𝑓
100�𝑐𝑚
𝑇𝑏 = 434.386�𝑁𝑚
𝑇𝑏 = (82�𝑘𝑔𝑓 ×
𝑇𝑏 𝑁
9.549 × 103
(434.386�𝑁𝑚)(2500�𝑟𝑝𝑚)
𝐵𝑃 =
9.549 × 103
𝐵𝑃 = 113.725�𝑘𝑊
𝐵𝑃 =
𝐸𝐶 = 𝑚𝑓 𝐻𝐻𝑉
3�𝑘𝑔 1�𝑚𝑖𝑛
×
6�𝑚𝑖𝑛
60�𝑠
𝑘𝑔
𝑚𝑓 = 0.00833
𝑠
𝑚𝑓 =
𝐸𝐶 = 𝑚𝑓 𝐻𝐻𝑉
𝑘𝑔
𝑘𝐽
) (45820 )
𝑠
𝑘𝑔
𝐸𝐶 = 381.833�𝑘𝑊
𝐸𝐶 = (0.00833
𝐵𝑃
× 100%
𝐸𝐶
113.725�𝑘𝑊
𝑒𝑏 =
× 100%
381.833�𝑘𝑊
𝑒𝑏 = 29.784%
𝑒𝑏 =
𝐵𝑆𝐹𝐶 =
𝐵𝑆𝐹𝐶 =
𝑚𝑓
𝐵𝑃
𝑘𝑔
0.00833 𝑠
113.725�𝑘𝑊
𝑘𝑔
𝐵𝑆𝐹𝐶 = 0.263
𝑘𝑊ℎ
×
60�𝑠
60�𝑚𝑖𝑛
×
1�𝑚𝑖𝑛 1�ℎ𝑜𝑢𝑟
𝑇𝑓 = 10%𝑇𝑏
𝑇𝑓 = (10%)(434.386�𝑁𝑚)
𝑇𝑓 = 43.438�𝑁𝑚
𝑇𝑓 𝑁
9.549 × 106
(43.438�𝑁𝑚)(2500�𝑟𝑝𝑚)
𝐹𝑃 =
9.549 × 103
𝐹𝑃 = 11.372�𝑘𝑊
𝐹𝑃 =
𝐼𝑃 − 𝐵𝑃 = 𝐹𝑃
𝐼𝑃 = 𝐹𝑃 + 𝐵𝑃
𝐼𝑃 = 11.372�𝑘𝑊 + 113.725�𝑘𝑊
𝐼𝑃 = 125.097�𝑘𝑊
𝐼𝑃
× 100%
𝐸𝐶
125.097�𝑘𝑊
𝑒𝐼 =
× 100%
381.833�𝑘𝑊
𝑒𝐼 = 32.762%
𝑒𝐼 =
𝐼𝑆𝐹𝐶 =
𝐼𝑆𝐹𝐶 =
𝑚𝑓
𝐼𝑃
𝑘𝑔
0.00833 𝑠
125.097�𝑘𝑊
𝑘𝑔
𝐼𝑆𝐹𝐶 = 0.239
𝑘𝑊ℎ
×
60�𝑠
60�𝑚𝑖𝑛
×
1�𝑚𝑖𝑛 1�ℎ𝑜𝑢𝑟
𝐵𝑃
× 100%
𝐼𝑃
113.725�𝑘𝑊
𝜂𝑚 =
× 100%
125.097�𝑘𝑊
�𝜂𝑚 = 90.909%
𝜂𝑚 =
Diagram: