Solutions

ES 3003 Homework #1

1. Problem 1-34

2. Problem 1-41

3. Problem 1-55

4. Problem 1-56
 Chapter 1 Introduction and Basic Concepts
1-33
SOLUTION During an analysis, a relation with inconsistent units is obtained. A correction is to be found, and the probable
cause of the error is to be determined.

Analysis The two terms on the right-hand side of the equation

E = 25 kJ + 7 kJ/kg

do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by
mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally
homogeneous; that is, every term in the equation will have the same unit.

Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage.

1-34
Solution A resistance heater is used to heat water to desired temperature. The amount of electric energy used in kWh and
kJ are to be determined.

Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of electric
energy used in 2 hours becomes

Total energy = (Energy per unit time)(Time interval)

= (4 kW)(2 h)
= 8 kWh

Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ,

Total energy = (8 kWh)(3600 kJ/kWh)

= 28,800 kJ

Discussion Note kW is a unit for power whereas kWh is a unit for energy.

1-35
Solution A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is
to be obtained for the filling time.

Assumptions Gasoline is an incompressible substance and the flow rate is constant.

Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the
unit of time is ‘seconds’. Therefore, the independent quantities should be arranged such that we end up with the unit of
seconds. Putting the given information into perspective, we have

t [s] ↔ V [L], and V [L/s}

It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate.
Therefore, the desired relation is
V
t=
V
Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities.

1-14
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 Chapter 1 Introduction and Basic Concepts
1-40E
Solution We are to estimate the rate of heat transfer into a room and the cost of running an air conditioner for one
hour.

Assumptions 1 The rate of heat transfer is constant. 2 The indoor and outdoor temperatures do not change significantly
during the hour of operation.

Analysis
(a) In one hour, the air conditioner supplies 5,000 Btu of cooling, but runs only 60% of the time. Since the indoor and
outdoor temperatures remain constant during the hour of operation, the average rate of heat transfer into the room is the
same as the average rate of cooling supplied by the air conditioner. Thus,
0.60 ( 5000 Btu ) ⎛ 1 kW ⎞
Q = = 3, 000 Btu/h ⎜ ⎟ = 0.879 kW
1h ⎝ 3412.14 Btu/h ⎠

(b) Energy efficiency ratio is defined as the amount of heat removed from the cooled space in Btu for 1 Wh (watt-
hour) of electricity consumed. Thus, for every Wh of electricity, this particular air conditioner removes 9.0 Btu from the
room. To remove 3,000 Btu in one hour, the air conditioner therefore consumes 3,000/9.0 = 333.33 Wh = 0.33333 kWh of
electricity. At a cost of 7.5 cents per kWh, it costs only 2.50 cents to run the air conditioner for one hour.

Discussion Notice the unity conversion ratio in the above calculation. We also needed to use some common sense and
dimensional reasoning to come up with the appropriate calculations. While this may seem very cheap, if this air conditioner
is run at these conditions continuously for one month, the electricity will cost ($0.025/h) (24 h/day) (30 day/mo) = $18/mo.

1-41
Solution We are to calculate the volume flow rate and mass flow rate of water.

Assumptions 1 The volume flow rate, temperature, and density of water are constant over the measured time.

Properties The density of water at 20oC is ρ = 998.0 kg/m3.

Analysis The volume flow rate is equal to the volume per unit time, i.e.,
V 2.0 L ⎛ 60 s ⎞
V = = = 42.105 L/min ≅ 42.1 Lpm
Δt 2.85 s ⎜⎝ 1 min ⎟⎠

where we give our final answer to 3 significant digits, but retain 5 digits to avoid round-off error in the second part of the
problem. Since density is mass per unit volume, mass flow rate is equal to volume flow rate times density. Thus,

⎛ 1 min ⎞ ⎛ 1 m ⎞
( )
3
m = ρV = 998.0 kg/m3 ( 42.105 L/min ) ⎜ ⎟ ⎜ ⎟ = 0.700 kg/s
⎝ 60 s ⎠ ⎝ 1000 L ⎠

Discussion We used one unity conversion ratio in the first calculation, and two in the second. If we were interested
only in the mass flow rate, we could have eliminated the intermediate calculation by solving for mass flow rate directly,
i.e.,
2.0 L ⎛ 1 m3 ⎞
m = ρ
V
Δt
(
= 998.0 kg/m3 ) ⎜ ⎟ = 0.700 kg/s .
2.85 s ⎝ 1000 L ⎠

1-17
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
 Chapter 1 Introduction and Basic Concepts

Review Problems

1-54
SOLUTION The flow of air through a wind turbine is considered. Based on unit considerations, a proportionality relation
is to be obtained for the mass flow rate of air through the blades.

Assumptions Wind approaches the turbine blades with a uniform velocity.

Analysis The mass flow rate depends on the air density, average wind velocity, and the cross-sectional area which depends
on hose diameter. Also, the unit of mass flow rate m is kg/s. Therefore, the independent quantities should be arranged
such that we end up with the proper unit. Putting the given information into perspective, we have

m [kg/s] is a function of ρ [kg/m3], D [m], and V [m/s}

It is obvious that the only way to end up with the unit “kg/s” for mass flow rate is to multiply the quantities ρ and V with
the square of D. Therefore, the desired proportionality relation is

m is proportional to ρ D 2V
or,
m = CρD 2V
 =
Where the constant of proportionality is C =π/4 so that m ρ (πD 2 / 4 )V

Discussion Note that the dimensionless constants of proportionality cannot be determined with this approach.

1-55
Solution The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at
different locations are to be determined.
Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into
the relation
⎛ ⎞
W = mg = (80 kg)(9.807 − 3.32 × 10−6 z ) ⎜
1N
2 ⎟ ( where z is in units of m/s )2

⎝ 1 kg ⋅ m/s ⎠

Sea level: (z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 N

Denver: (z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 N
Mt. Ev.: (z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×9.778 = 782.2 N

Discussion We report 4 significant digits since the values are so close to each other. The percentage difference in
weight from sea level to Mt. Everest is only about -0.3%, which is negligible for most engineering calculations.

1-22
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
 Chapter 1 Introduction and Basic Concepts
1-56
Solution A man is considering buying a 12-oz steak for $3.15, or a 320-g steak for $2.80. The steak that is a better
buy is to be determined.
Assumptions The steaks are of identical quality.
Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. We choose 1
kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be

⎛ $3.15 ⎞ ⎛ 16 oz ⎞ ⎛ 1 lbm ⎞
12 ounce steak: Unit Cost = ⎜ ⎟⎜ ⎟ ⎜⎜ ⎟⎟ = $9.26/kg
⎝ 12 oz ⎠ ⎝ 1 lbm ⎠ ⎝ 0.45359 kg ⎠
320 gram steak:
⎛ $2.80 ⎞ ⎛ 1000 g ⎞
Unit Cost = ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = $8.75/kg
⎝ 320 g ⎠ ⎝ 1 kg ⎠
Therefore, the steak at the international market is a better buy.

Discussion Notice the unity conversion factors in the above equations.

1-57
Solution The thrust developed by the jet engine of a Boeing 777 is given to be 85,000 pounds. This thrust is to be
expressed in N and kgf.
Analysis Noting that 1 lbf = 4.448 N and 1 kgf = 9.81 N,
the thrust developed is expressed in two other units as

Thrust in N:
⎛ 4.448 N ⎞
Thrust = (85,000 lbf )⎜ ⎟ = 3.78 × 10 N
5
⎝ 1 lbf ⎠
Thrust in kgf:
⎛ 1 kgf ⎞
Thrust = (37.8 × 10 5 N )⎜ ⎟ = 3.85 × 10 kgf
4
⎝ 9.81 N ⎠
Discussion Because the gravitational acceleration on earth is close to 10 m/s2, it turns out that the two force units N and
kgf differ by nearly a factor of 10. This can lead to confusion, and we recommend that you do not use the unit kgf.

1-58
SOLUTION A relation for the air drag exerted on a car is to be obtained in terms of on the drag coefficient, the air density,
the car velocity, and the frontal area of the car.
Analysis The drag force depends on a dimensionless drag coefficient, the air density, the car velocity, and the frontal area.
Also, the unit of force F is newton N, which is equivalent to kg⋅m/s2. Therefore, the independent quantities should be
arranged such that we end up with the unit kg⋅m/s2 for the drag force. Putting the given information into perspective, we
have
FD [ kg⋅m/s2] ↔ CDrag [], Afront [m2], ρ [kg/m3], and V [m/s]
It is obvious that the only way to end up with the unit “kg⋅m/s2” for drag force is to multiply mass with the square of the
velocity and the fontal area, with the drag coefficient serving as the constant of proportionality. Therefore, the desired
relation is

FD = C Drag ρAfrontV 2
Discussion Note that this approach is not sensitive to dimensionless quantities, and thus a strong reasoning is required.

1-23
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.