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AS101- Midterm Exercise No.1 - Solution

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University of Cebu-METC
Alumnos, Mambaling, Cebu City, 6000
2nd Semester School Year 2020-2021
AS101 (Thermodynamics 1) – Energy, Work and Heat
Midterm – Exercise No.1 - Solution
1. Air is compressed in a cylinder such that the volume changes from 120 to 20 𝑖𝑛3 . The initial
pressure is 60 psia and the temperature is held constant at 100℉. Calculate the work.
Solution:
𝑣2
𝑣2
π‘Š = 𝑃1 𝑉1 ∫ 𝑃 𝑑𝑉 = ∫
2
2
𝑑𝑉
𝑣2
= 𝑃1 𝑉1 𝐼𝑛
𝑉
𝑣2
𝑙𝑏
𝑖𝑛2
𝑓𝑑 3
20
= (60 2 ) (144 2 ) (120 𝑖𝑛3 ) (
= −πŸπŸŽπŸ•πŸ“ 𝒇𝒕 − 𝒍𝒃𝒇
) 𝐼𝑛
3
𝑖𝑛
𝑓𝑑
1728 𝑖𝑛
120
2. A 40 m x 20 m swimming pool is filled with water to a depth of 2.8 m. Determine the heat required
to raise the temperature of the water in the pool from 12℃ to 29℃.
Solution:
𝑉 = π΄π‘Ÿπ‘’π‘Ž π‘₯ π·π‘’π‘π‘‘β„Ž
= (40 π‘š)(20 π‘š)(2.8 π‘š)
𝑉 = 2240 π‘š3
π‘š = πœŒπ‘‰ = (1000
π‘˜π‘”
) (2240 π‘š3 ) = 2.240 π‘₯ 106 π‘˜π‘”
π‘š3
π‘˜π½
) π‘₯(29℃ − 12℃)
π‘˜π‘” − 𝐾
𝑸 = πŸπŸ“πŸ—. πŸπŸ• 𝒙 πŸπŸŽπŸ” π’Œπ‘±
𝑄 = π‘šπ‘π‘ βˆ†π‘‡ = (2.240 π‘₯ 106 π‘˜π‘”) (4.18
3. A car engine with a power output of 85 hp has a thermal efficiency of 26 percent. Find the fuel
consumption rate of this car if the fuel has a heating value of 20,400 Btu/π‘™π‘π‘š
Solution:
πœ‚π‘‘β„Ž =
π‘Š
𝑄𝐴
0.26 =
85
𝑄𝐴
𝑄𝐴 = 326.923 β„Žπ‘ π‘₯
𝐡𝑑𝑒
β„Žπ‘Ÿ
1 β„Žπ‘
2545
𝑄𝐴 = 832,019.2 𝐡𝑑𝑒/β„Žπ‘Ÿ
𝑄𝐴 = π‘šπ‘“ π‘₯ (𝐻𝐻𝑉)
𝐡𝑑𝑒
𝐡𝑑𝑒
832,019.2
= π‘šπ‘“ (20 400
)
β„Žπ‘Ÿ
π‘™π‘π‘š
π’π’ƒπ’Ž
π’Žπ’‡ = πŸ’πŸŽ. πŸ•πŸ–
𝒉𝒓
4. Water is heated in an insulated, constant diameter tube by a 18-kw electric resistance heater. If
the water enters steadily at 20 ℃ and leaves at 100 ℃, find the mass flow rate of water.
Solution:
𝑄 = π‘šπ‘π‘ βˆ†π‘‡
π‘˜π½
π‘˜π½
18
= π‘š (4.18
) (100 ℃ − 20 ℃)
𝑠
π‘˜π‘” − 𝐢
π’Ž = 𝟎. πŸŽπŸ“πŸ’ π’Œπ’ˆ/𝒔
5. A freshwater lake with an area of 2400 π‘š2 and an average depth of 5 m is located in a
mountainous region which is 350 m above a valley floor. What is the potential energy stored in
this lake of water?
Solution:
π‘šπ‘Žπ‘ π‘  = π΄π‘Ÿπ‘’π‘Ž π‘₯ π‘‘π‘’π‘π‘‘β„Ž π‘₯ 𝐷𝑒𝑛𝑠𝑖𝑑𝑦 π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ
π‘˜π‘”
π‘š = (2400 π‘š2 )(5 π‘š) (1000 3 )
π‘š
π‘š = 12 π‘₯ 106 π‘˜π‘”
𝑃𝐸 = π‘šπ‘”π‘§
= (12 π‘₯ 106 π‘˜π‘”) (9.81
π‘š
) (350 π‘š)
𝑠2
𝑷𝑬 = πŸ’. 𝟏𝟐𝟎𝟐 𝒙 𝟏𝟎𝟏𝟎 𝑱 = πŸ’πŸπŸπŸŽπŸ 𝑴𝑱
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