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MAE 11 Homework 1: Solutions 10/5/2018

H1.1 – A gas with density 𝜌 = 0.9 lbm/ft3 weighs 4.0 lbf on the Moon, where gravitational acceleration is 𝑔 = 5.47 ft/s 2 . Determine (show all work): a. b. c. d.

weight in lbf on Mars, where 𝑔 = 12.86 ft/s 2 volume in ft 3 on Mars mass in lbm on Earth (at sea level) weight in lbf on Earth (at sea level) Given Data

Conversions

0.9 lbm/ft 3

𝜌= 𝑊𝑚𝑜𝑜𝑛 = 4.0 lbf 𝑔𝑚𝑜𝑜𝑛 = 5.47 ft/s2 𝑔𝑚𝑎𝑟𝑠 = 12.86 ft/s2

1 lbf = 32.174

lbm∙ft s2

Governing Equations 𝑊 = 𝑚𝑔

Model The density of the gas is assumed constant. This is true for a fixed amount of mass 𝑚 transported in a rigid tank of volume ∀. Analysis Part a: Determine the weight in lbf on mars 𝑊𝑚𝑎𝑟𝑠 = 𝑚𝑔𝑚𝑎𝑟𝑠 𝑚 = 𝑚𝑚𝑎𝑟𝑠 = 𝑚𝑚𝑜𝑜𝑛 = 𝑊 𝑚𝑜𝑜𝑛 /𝑔𝑚𝑜𝑜𝑛 𝑔 𝑊𝑚𝑎𝑟𝑠 = 𝑊 𝑚𝑜𝑜𝑛 𝑚𝑎𝑟𝑠 𝑔𝑚𝑜𝑜𝑛

𝑊𝑚𝑎𝑟𝑠 = (4.0 lbf) ( 𝑊𝑚𝑎𝑟𝑠 = 9.40 lbf

12.86 ft/s2 ) = 9.404 lbf 5.47 ft/s2

Part b: Volume in ft 3 on mars ∀=

𝑚

𝜌

=

𝑊𝑚𝑜𝑜𝑛 𝜌𝑔𝑚𝑜𝑜𝑛

=

∀= 26.1 ft 3

lbm∙ft 4.0 lbf [32.174 lbf∙s2 ] (0.9 lbm/ft3 )(5.47 ft/s2 )

Part c: mass in lbm on Earth (at sea level) 𝑚=

𝑊𝑚𝑜𝑜𝑛 𝑔𝑚𝑜𝑜𝑛

=

lbm∙ft 4.0 lbf [32.174 ] 5.47 fts2 lbf∙s2

𝑚 = 23.5 lbm

Part d: weight in lbf on Earth (at sea level) 𝑊𝑒𝑎𝑟𝑡ℎ = 𝑚𝑔𝑒𝑎𝑟𝑡ℎ = (23.53 lbm ) (32.174 𝑊𝑒𝑎𝑟𝑡ℎ = 23.5 lbf

1 lbf∙s2 ft ] ) [ 2 32.174 lbm∙ft s

Discussion This problem is focused on understanding mass and force and their description in English units. The force due to gravity depends on two factors: the mass and the local acceleration of gravity. In this problem the mass was fixed, and the force of gravity varied due to the location of the mass. 1

H1.2 – A U-tube water manometer is connected to a tank of gas (see text, Figure 1.7). The difference in water levels is 𝐿 = 16.5 in. The tank is located in a room where the atmospheric pressure is 14.7 psia. Assume 𝜌𝐻2𝑂 = 62.4 lbm/ft3 . Determine: a. gas pressure, psig b. gas pressure, psia c. gas pressure, kPa (absolute) Given Data

Conversions

𝐿 = 16.5 in 𝑝𝑎𝑡𝑚 = 14.7 psia lbm 𝜌𝐻2𝑂 = 62.4 3

1 lbf = 32.174 s2 1 psia = 6.89476 kPa

lbm∙ft

Diagram

ft

Model Fluid Statics Governing Equations 𝑝𝑎𝑏𝑠 = 𝑝𝑎𝑡𝑚 +𝜌𝑔𝐿 (Eq. 1.11) 𝑝𝑔𝑎𝑔𝑒 = 𝑝𝑎𝑏𝑠 −𝑝𝑎𝑡𝑚

(Eq. 1.14)

Analysis Part b: Determine tank pressure in psia 𝑝𝑎𝑏𝑠 = 𝑝𝑎𝑡𝑚 +𝜌𝐻2 𝑂 𝑔𝐿 𝑝𝑎𝑏𝑠 = 14.7

lbf in2

+(62.4

lbm ft3

) (32.174

1 lbf∙s2 1 ft 3 ft (16.5 in) [ ] ] [ ) 2 32.174 lbm∙ft 12 in s

𝑝𝑎𝑏𝑠 = 15.29 psia Part a: Determine tank pressure in psig 𝑝𝑔𝑎𝑔𝑒 = 𝑝𝑎𝑏𝑠 −𝑝𝑎𝑡𝑚 𝑝𝑔𝑎𝑔𝑒 = 15.29

lbf in2

− 14.7

lbf

in2

𝑝𝑔𝑎𝑔𝑒 = 0.596 psig Part c: Determine tank pressure in kPa (absolute) 𝑝𝑎𝑏𝑠 = (15.29 psi) [6.89476

kPa ] psi

𝑝𝑎𝑏𝑠 = 105.46 kPa Discussion: The concepts of absolute and gage pressure are explored in this problem: 𝑝𝑎𝑏𝑠 = 𝑝𝑔𝑎𝑔𝑒 + 𝑝𝑎𝑡𝑚 . Devices that measure pressure, such as the manometer, often measure pressure differences. The displacement of manometer fluid indicates the difference in pressure between the tank and the surrounding atmosphere, i.e., the gage pressure. 2

H1.3 – Gas is contained in a vertical piston-cylinder device such that the piston is in equilibrium with the gas. The piston has a mass of 15 kg and diameter of 7 cm. The atmosphere exerts a pressure of 101 kPa on the top of the piston. Determine the absolute pressure of the gas (in kPa). Given Data 𝑚 = 15 kg 𝐷 = 7 cm = 0.07 m 𝑝𝑎𝑡𝑚 = 101 kPa

Model Piston-cylinder assembly in equilibrium

Diagram

Governing Equations 󰇍 = 𝑚𝒂󰇍 ∑𝑭 𝑊 = 𝑚𝑔

𝑦

Analysis Determine the absolute pressure of the gas in kPa

𝑥

Since the piston is in equilibrium the net force is zero. ∑ 𝐹𝑦 = 𝑝𝑔𝑎𝑠 𝐴−𝑝𝑎𝑡𝑚 𝐴−𝑊 = 0 𝑝𝑔𝑎𝑠 = 𝑝𝑎𝑡𝑚 +

𝑊

𝑝𝑔𝑎𝑠 = 𝑝𝑎𝑡𝑚 +

4𝑚𝑔

Where 𝐴 = piston face =

𝜋

4

𝐷2

𝐴

𝜋𝐷 2

𝑝𝑔𝑎𝑠 = 101 kPa+

4(15 kg)(9.81 m/s2 ) 𝜋(0.07 m)2

1

kPa

[ 1000 kg/m∙s2]

𝑝𝑔𝑎𝑠 = 139.24 kPa Discussion The weight of the piston compresses the gas contained within the cylinder which increases the pressure of the gas. The pressure difference between the gas and atmosphere (i.e. gage pressure) is maintained by the piston weight.

3

H1.4 – Consider a simple device or household appliance and do the following: a. Explain its basic operating principle. b. Sketch and define system (with dashed line). Identify and indicate significant energy transfers (with labeled arrows). Water-cooling systems are used to maintain the CPU temperature on high-performance computers. There are three primary components to a water-cooling system: a pump, a radiator, and a CPU block. These three components form a loop as shown below. Radiator CPU Block

Pump

System: The system consists of the coolant, tubing, radiator, pump, and CPU block. Surrounding: Everything else – fans, CPU, air, etc. Energy Transfers: Identify energy transfers between the system and surroundings. Pump – Energy is transferred into the system by work at the pump. This energy is supplied by the computer’s power supply. Considering the pump as its own system it can be shown (see Ch. 4) work input is used to raise the pressure of the coolant at the pump exit 𝑝𝑜𝑢𝑡 > 𝑝𝑖𝑛. This pressure difference is needed to overcome losses in the loop and maintain steady flow. CPU Block – Energy is transferred by heat into the system at the CPU block. Note with the given system definition, energy transferred out of the computer (surroundings) is transferred into the water cooling loop (system) so it is 𝑄󰇗𝑖𝑛 . Radiator – Energy is transferred by heat out of the system at the radiator. The radiator consists of many thin veins through which the coolant flows. Many small tubes will have more surface area than one large tube aiding the convective heat transfer. Fans are often attached to the radiator to blow cooler air over the veins. Although the fans require work to operate they were not included in the system definition. Since no energy crosses the system definition, the 𝑊󰇗 needed to operate the fans is excluded. References https://physics.stackexchange.com/questions/269904/does-fluid-speed-affect-liquid-cooling http://www.standa.lt/products/catalog/lasers_laser_accessories?item=550 4

H1.5 – Investigate energy consumption in the U.S. and two other countries (you choose). Determine the following for each country: a. total energy consumption (specify in both Btu and kWh (kilowatt-hours)) b. energy consumption per capita c. total energy consumption breakdown (%) by source (oil, natural gas, renewables, etc.) Prepare summary data table and include dates (year) of data. Be sure to cite references (use at least 2). Comment on any important/interesting findings.

Country USA Germany China Kenya Canada

Energy (QuadBtu) 97.7 13.07 119.55 0.27 14.55

Energy (TWh) 28.63 3.83 35.03 0.079 4.26

Per capita (kWh) 91.2 47.5 25.79 1.6 121.71

By Source 32% Gas, 28%, Oil 21% Coal, 11% Ren 30% Ren, 27.5% Other, 23% Lig, 17% Coal 66% Coal, 18% Oil, 5% Gas 68% Wood, 22% Oil, 10% Other 31% Oil, 28% Gas, 26% Hydro, 15% Other

References https://www.cleanenergywire.org/factsheets/germanys-energy-consumption-and-power-mix-charts https://www.iea.org/statistics/statisticssearch/ https://yearbook.enerdata.net/ https://en.wikipedia.org/wiki/World_energy_consumption https://en.wikipedia.org/wiki/List_of_countries_by_total_primary_energy_consumption_and_production https://www.eia.gov/beta/international/data/browser/#/?pa=000000001&c=ruvvvvvfvtvnvv1urvvvvfvvvvvvfvvvou20evvvvvvvvvnvvuvo&ct=0& ug=4&vs=INTL.44-2-AFG-QBTU.A&ord=SA&vo=0&v=H&start=1980&end=2014&s=INTL.44-2-DEU-QBTU.A~~INTL.44-2-KEN-QBTU.A~~INTL.44-2USA-QBTU.A~~INTL.44-2-CHN-QBTU.A~~INTL.44-2-CAN-QBTU.A

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