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E234  Thermodynamics
Assignment – Week 2
1. My weight is kg =(lbs)x0.453
My height in m = 2.54x102x(inches)
2. Convert the formula for water density to be for T in degrees Kelvin.
ρ = 1008 – TC/2 [kg/m3]
ρ = 1008 – TC/2 [kg/m3]
We need to express degrees Celsius in degrees Kelvin
TC = TK – 273.15
and substitute into formula
ρ = 1008 – TC/2 = 1008 – (TK – 273.15)/2 = 1144.6 – TK/2
3. How much mass is there approximately in 1 L of mercury (Hg)? Atmospheric air?
What is the specific volume of mercury? of air
Check tables at end of book
Mercury:
density = 13 560 kg/m3 1 L has a mass of 13.56 kg,
v =1/density = 7.4x105 m3/kg
Water:
density = 1000 kg/m3 1 L has a mass of 1.0 kg,
v = 1/density = 0.001 m3/kg
4. You exercise on a stairmaster. Each step is 25 cm high. You do 120 steps/minute.
You weigh xx kg (see 1. above). How much energy will you expand in one hour?
What is your average power? Results are to be given in cal/minute and SI units.
You work against gravity; Each step requires an energy E = mgh. Assume a mass
of 80 kg. 120 steps per minute is 2 steps per second. You power is thus
P = mghx2 = 80x10x.25x2 = 400 W
Energy in one hour = Pxtime = 400x3600 = 1.44x106 J
Energy calorie per hour = 1.44x106/(60x4.180) = 5740 cal/min
To lose one pound of fat, you would not to exercise for
t = 3500/5.74 = 609 minutes = 10 hours!
5. A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg
resting on the stops, as shown below. With an outside atmospheric pressure of
100 kPa, what should the water pressure be to lift the piston? Fall 2008 The force acting down on the piston comes from gravitation and the outside
atmospheric pressure acting over the top surface.
Force balance: F↑ = F↓ = PA = mpg + P0A
Now solve for P (divide by 1000 to convert to kPa for 2nd term)
P = Po + mpgA = 100 kPa + 100 × 9.806650.01 × 1000 kPa
= 100 kPa + 98.07 kPa = 198 kPa
6. The difference in height between the columns of a manometer is 200 mm with a
fluid of density 900 kg/m3. What is the pressure difference? What is the height
difference if the same pressure difference is measured using mercury, density
13600 kg/ m3, as manometer fluid?
ΔP = ρ1gh1 = 900 kg/m3 × 9.807 m/s2 × 0.2 m = 1765.26 Pa = 1.77 kPa
hHg = ΔP/ (ρhg g) = (ρ1 gh1) / (ρhg g) = 90013600 × 0.2
= 0.0132 m = 13.2 mm
7. Cengel and Boles 6th edition problem
272
Assumptions 1 The wind is
blowing steadily at a constant
uniform velocity. 2 The efficiency
Wind
of the wind turbine is independent
of the wind speed.
12
m/s
Properties The density of air is
3
given to be = 1.25 kg/m .
Analysis Kinetic energy is the
only form of mechanical energy
the wind possesses, and it can be
converted to work entirely.
Therefore, the power potential of
the wind is its kinetic energy,
which is V2/2 per unit mass, and mV 2 / 2 for a given mass flow rate:
emech ke Wind
turbine V 2 (12 m/s)2 1 kJ/kg 0.072 kJ/kg 2 2
2
2 1000 m /s m VA V D 2
4 (1.25 kg/m3 )(12 m/s) (50 m)2
29,450 kg/s
4 Wmax Emech memech (29,450 kg/s)(0.072 kJ/kg) 2121 kW The actual electric power generation is determined by multiplying the power
generation potential by the efficiency, Welect wind turbineWmax (0.30)(2121 kW) 636 kW Fall 2008 50 m Therefore, 636 kW of actual power can be generated by this wind turbine at the
stated conditions.
Discussion The power generation of a wind turbine is proportional to the cube of
the wind velocity, and thus the power generation will change strongly with the
wind conditions.
278
Assumptions 1 The pump operates
steadily. 2 The elevations of the
reservoirs remain constant. 3 The
2
changes in kinetic energy are
negligible.
Reservoi
Properties We take the density of
3
r
water to be = 1000 kg/m .
45 m
Pump
Analysis The elevation of water and
thus its potential energy changes
1
during pumping, but it experiences
no changes in its velocity and
Reservoir
pressure. Therefore, the change in
the total mechanical energy of water
is equal to the change in its potential
energy, which is gz per unit mass, and mgz for a given mass flow rate.
That is, E mech me mech mpe mgz Vgz 1N
(1000 kg/m 3 )(0.03 m 3 /s)(9.81 m/s 2 )(45 m) 1 kg m/s 2 1 kW 1000 N m/s 13.2 kW Then the mechanical power lost because of frictional effects becomes Wfrict W pump, in E mech 20 13.2 kW 6.8 kW Discussion The 6.8 kW of power is used to overcome the friction in the piping
system. The effect of frictional losses in a pump is always to convert mechanical
energy to an equivalent amount of thermal energy, which results in a slight rise in
fluid temperature. Note that this pumping process could be accomplished by a
13.2 kW pump (rather than 20 kW) if there were no frictional losses in the
system. In this ideal case, the pump would function as a turbine when the water is
allowed to flow from the upper reservoir to the lower reservoir and extract 13.2
kW of power from the water. Fall 2008 ...
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 Spring '09
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