Solution Manual For
Green Building An Engineering Approach to Sustainable Construction, Edition 1 By Christian
M. Carrico
Chapters 2-15
1 Introduction to Green Building
No exercises in this chapter
2 Energy Science: Key Underlying Physics
2.1 End of Chapter Exercises
1) Concepts: On a hot day, the sweat on one‘s forehead and the condensation on one‘s cold beverage
have opposite effects. Explain.
Both processes involve latent heat. One is a condensation process (condensation on the glass), and one is
a vaporization process (perspiration). On a cold beverage, condensation may form on the outside of
the glass due to heat extraction from the atmosphere, causing the film of air near the surface to reach
the dew point temperature. Water condenses on the surface, and this causes some warming of the
glass and contents. The perspiration on the surface of the body vaporizes, extracting heat from the
skin and body volume in doing so.
2) Concepts: In Error! Reference source not found., the power consumption of the world‘s largest
laser exceeds that of the US. How does this not bring down the grid or at least dim the lights a bit
when it is operated?
The power consumption is large, but very short-duration laser pulses, meaning each shot is a relatively
small quantity of energy. The energy can be stored in capacitors and discharged at a very high rate.
3) Concepts: Design and illustrate an ideal passive to-go cup to keep your hot beverage hot. Discuss 4
features. Think about heat transfer mechanisms. How would you change it to keep your cold beverage
cold?
The pictured cup is not ideal, as can be seen from the high surface temperatures!
Maximize volume to surface area (spherical), minimize exposed surface area,
maximize thermal mass, use low conductivity material, well-sealed with a sippy
cup-type lid so no convective losses, low radiative losses by low emissivity, and
a radiant barrier to prevent radiation losses. It‘s generally the same approach for a
cold beverage, although some insulating materials work slightly better for one
application versus the other.
4) Concepts: Specify the best answer for the predominant means of heat transfer in each of these
situations:
a) Heat is transferred from a heated air dryer to one‘s hands. Convection
b) Heat is transferred from a warm front moving into New Mexico from the south. Convection
(or advection in the atmosphere)
1
, c) Standing in front of a fireplace with a roaring fire behind a glass door. Radiation
d) Burning one's hand on a hot surface. Conduction
e) A cat warming itself in a sunbeam. Radiation
5) Problem: Draw the conceptual heat engine diagram showing the flow directions of heat and work as
well as the hot and cold temperature reservoirs. Express the work output and efficiency in terms of
the hot and cold heat flows, Qh and Qc.
Wout = Qhot – Qcold
η = efficiency = Wout/Qhot
6) Problem: Convert the global average energy use of ~500 Quads to (a) an equivalent ExaJoules (1018)
per year, (b) an equivalent TWh per year, and (c) a steady-state horsepower.
Given: 5oo Quads/yr
Find: Annual energy use in EJ/yr, TWh/yr, and P (hp)
Assume: steady state
Solution:
Annual Energy Use (EJ/y) = 500E15 BTU/y × 1055 J/BTU × 1 EJ/1E18J = 527.5 EJ/y
1 kWh = 1000W×3600s = 3.6E6J
Annual Energy Use (TWh/y) = 500E15 BTU/y × 1.055 KJ/BTU × (1kWh/3600kJ) = 1.47E14kWh/y =
147,000 TWh/y
Steady State P = 527.5E18J/yr × yr/ [(365)(24)(3600)]s × 1 hp/(749J/s) = 2.2E10 hp or 22 billion horses
working for our needs continuously! Fortunately, we don‘t have to deal with that quantity of horse
apples, though we have other waste products that are a concern. Whatever way you quantify it, it‘s a
staggering rate of energy use, which has been both a blessing and a curse.
7) Problem: The first steam engine worked between the upper and lower temperatures of 10 and 100°C.
What is the maximum efficiency possible?
Efficiency = 1 – Tcold/Thot = 1 – 283K/373K = 0.24 or 24%
2
,8) Problem: Estimate the mass of the atmosphere. Assume surface P = 1 atmosphere (Earth‘s radius is
6370 km).
Given: P = 1 atm; R = 6370 km
Find: Matm
Assume: uniform pressure at the surface
Solution:
F = ma = mg = P × A = 101325 N/m2 × 4π(R)2 = 101325 N/m2 × 4π (6370E3 m)2 = 5.17E19 N
M = (5.17E19 kg·m/s2) / (9.8 m/s2) = 5.3E18 kg
9) Problem: You are using a heated 1-m diameter mass of iron at 500 °C to provide warmth in a room.
At the outset, what is the rate of heat transfer via radiation in a room-temperature environment?
Assume a perfect blackbody.
Given: D = 1 m; T = 500 °C
Find: P(t=0)
Assume: Є = 1; perfect sphere
Solution:
Area = 4πR2 = 4π(0.5)2 = 3.14159 m2
Qrad = σAϵ ((T2)4 – (T1)4) = (5.67E-8 W/m2-K4) (3.14159 m2) (1) ((773K)4 – (293K)4)
Qrad = (5.67E-8 W/m2-K4) (3.14159 m2) (3.497E11 K4)
Qrad = 62,286W
10) Problem: A heat engine is generating useful work at a rate of 50 kW. The waste heat loss rate to the
environment in the heat engine is 30kW. What is the thermal efficiency of this engine? What is the hp
rating of the engine? What is the rate of gasoline consumption (120,000 BTU/gal) in gal/h?
Given: heat engine; Wout = 50 kW; Qwaste = 30 kW
Find: P(hp); η; Qin (gal/h)
Assume: steady state
Solution:
Energy Balance Qin = Qwaste + Wout
So, Qin = Wout + Qwaste
Wout = 50 kW (1 hp/0.746 kW) = 67 hp
Qin = 80KJ/s (kBTU/1055 kJ) (gal/120kBTU) (3600 s/h) = 2.27 gal/h
3
, 11) Problem: Your power plant has an output of 2000 MW. Its flame temperature is 1227 °C and it is
exhausting to ambient conditions at 20 °C.
a) Find Carnot Efficiency.
b) For the idealized Carnot cycle heat engine, what is the minimum rate of input heat (Qhot)?
c) What is the waste heat (Qcold) being dumped into the surroundings?
d) Draw a diagram showing the energy flows in and out, and that they balance.
Given: Wout = 2000 MWe; Tflame = 1227 °C; Tamb = 20 °C
Find: For Carnot Qhot; Qcold
Assume: Carnot cycle
Solution:
Wout = 2000 MW
Carnot efficiency = 1 – Tc/Th = 1 – 293K/1500K = 0.805
Rate of heat input: Qhot = Wout/η = 2000MJ/s / 0.8 = 2484MW
Qwaste = (2484 -2000) MW = 484MW
12) Problem: Power consumed by an engine as it ramps up is given by P(t) = 100 kW + 10 kW/min × t.
Find the energy consumed in kWh in 1 hour.
Energy =
13) Problem: You successfully lift a 25 kg stone to the top of the Empire State Building (1250 ft to the
top floor). How much work did you do on the stone, assuming constant g? If you burned the calories
you ingested from drinking a Coke while accomplishing this (Coke has 182 food Calories), what is
the thermodynamic efficiency of the process (your body!)?
Given: M = 25 kg; ∆z = 1250 ft; Qin = 182 Cal
Find: Wout, η
Assume: steady state; constant g = 9.8 m/s2
Solution:
Wout = f × d = m × g × d = 25 kg × 9.8 m/s2 × 1250 ft × (m/3.28 ft) = 93368 N×m = 93.4 kJ
Qin = Energy Expended = 182,000 cal × 4.18 J/calorie = 760760 J = 760.8 kJ
η= Wout/Qin = 93.4 kJ/760.8 kJ = 0.123 = 12.3% efficient
14) Problem: An inventor has a new power plant design that is claimed to have a 2000 MW output with a
feed rate of 100 kg/s of fuel with an energy content of 25 MJ/kg. The plant is operating at hot and
cold reservoir temperatures of 1000 °C and 20 °C. Is this possible?
Given: Wout = 2000 MWe; dM/dt = 100 kg/s; Ec = 25 MJ/kg; Thot = 1000 °C; Tcold = 20 °C
4
Green Building An Engineering Approach to Sustainable Construction, Edition 1 By Christian
M. Carrico
Chapters 2-15
1 Introduction to Green Building
No exercises in this chapter
2 Energy Science: Key Underlying Physics
2.1 End of Chapter Exercises
1) Concepts: On a hot day, the sweat on one‘s forehead and the condensation on one‘s cold beverage
have opposite effects. Explain.
Both processes involve latent heat. One is a condensation process (condensation on the glass), and one is
a vaporization process (perspiration). On a cold beverage, condensation may form on the outside of
the glass due to heat extraction from the atmosphere, causing the film of air near the surface to reach
the dew point temperature. Water condenses on the surface, and this causes some warming of the
glass and contents. The perspiration on the surface of the body vaporizes, extracting heat from the
skin and body volume in doing so.
2) Concepts: In Error! Reference source not found., the power consumption of the world‘s largest
laser exceeds that of the US. How does this not bring down the grid or at least dim the lights a bit
when it is operated?
The power consumption is large, but very short-duration laser pulses, meaning each shot is a relatively
small quantity of energy. The energy can be stored in capacitors and discharged at a very high rate.
3) Concepts: Design and illustrate an ideal passive to-go cup to keep your hot beverage hot. Discuss 4
features. Think about heat transfer mechanisms. How would you change it to keep your cold beverage
cold?
The pictured cup is not ideal, as can be seen from the high surface temperatures!
Maximize volume to surface area (spherical), minimize exposed surface area,
maximize thermal mass, use low conductivity material, well-sealed with a sippy
cup-type lid so no convective losses, low radiative losses by low emissivity, and
a radiant barrier to prevent radiation losses. It‘s generally the same approach for a
cold beverage, although some insulating materials work slightly better for one
application versus the other.
4) Concepts: Specify the best answer for the predominant means of heat transfer in each of these
situations:
a) Heat is transferred from a heated air dryer to one‘s hands. Convection
b) Heat is transferred from a warm front moving into New Mexico from the south. Convection
(or advection in the atmosphere)
1
, c) Standing in front of a fireplace with a roaring fire behind a glass door. Radiation
d) Burning one's hand on a hot surface. Conduction
e) A cat warming itself in a sunbeam. Radiation
5) Problem: Draw the conceptual heat engine diagram showing the flow directions of heat and work as
well as the hot and cold temperature reservoirs. Express the work output and efficiency in terms of
the hot and cold heat flows, Qh and Qc.
Wout = Qhot – Qcold
η = efficiency = Wout/Qhot
6) Problem: Convert the global average energy use of ~500 Quads to (a) an equivalent ExaJoules (1018)
per year, (b) an equivalent TWh per year, and (c) a steady-state horsepower.
Given: 5oo Quads/yr
Find: Annual energy use in EJ/yr, TWh/yr, and P (hp)
Assume: steady state
Solution:
Annual Energy Use (EJ/y) = 500E15 BTU/y × 1055 J/BTU × 1 EJ/1E18J = 527.5 EJ/y
1 kWh = 1000W×3600s = 3.6E6J
Annual Energy Use (TWh/y) = 500E15 BTU/y × 1.055 KJ/BTU × (1kWh/3600kJ) = 1.47E14kWh/y =
147,000 TWh/y
Steady State P = 527.5E18J/yr × yr/ [(365)(24)(3600)]s × 1 hp/(749J/s) = 2.2E10 hp or 22 billion horses
working for our needs continuously! Fortunately, we don‘t have to deal with that quantity of horse
apples, though we have other waste products that are a concern. Whatever way you quantify it, it‘s a
staggering rate of energy use, which has been both a blessing and a curse.
7) Problem: The first steam engine worked between the upper and lower temperatures of 10 and 100°C.
What is the maximum efficiency possible?
Efficiency = 1 – Tcold/Thot = 1 – 283K/373K = 0.24 or 24%
2
,8) Problem: Estimate the mass of the atmosphere. Assume surface P = 1 atmosphere (Earth‘s radius is
6370 km).
Given: P = 1 atm; R = 6370 km
Find: Matm
Assume: uniform pressure at the surface
Solution:
F = ma = mg = P × A = 101325 N/m2 × 4π(R)2 = 101325 N/m2 × 4π (6370E3 m)2 = 5.17E19 N
M = (5.17E19 kg·m/s2) / (9.8 m/s2) = 5.3E18 kg
9) Problem: You are using a heated 1-m diameter mass of iron at 500 °C to provide warmth in a room.
At the outset, what is the rate of heat transfer via radiation in a room-temperature environment?
Assume a perfect blackbody.
Given: D = 1 m; T = 500 °C
Find: P(t=0)
Assume: Є = 1; perfect sphere
Solution:
Area = 4πR2 = 4π(0.5)2 = 3.14159 m2
Qrad = σAϵ ((T2)4 – (T1)4) = (5.67E-8 W/m2-K4) (3.14159 m2) (1) ((773K)4 – (293K)4)
Qrad = (5.67E-8 W/m2-K4) (3.14159 m2) (3.497E11 K4)
Qrad = 62,286W
10) Problem: A heat engine is generating useful work at a rate of 50 kW. The waste heat loss rate to the
environment in the heat engine is 30kW. What is the thermal efficiency of this engine? What is the hp
rating of the engine? What is the rate of gasoline consumption (120,000 BTU/gal) in gal/h?
Given: heat engine; Wout = 50 kW; Qwaste = 30 kW
Find: P(hp); η; Qin (gal/h)
Assume: steady state
Solution:
Energy Balance Qin = Qwaste + Wout
So, Qin = Wout + Qwaste
Wout = 50 kW (1 hp/0.746 kW) = 67 hp
Qin = 80KJ/s (kBTU/1055 kJ) (gal/120kBTU) (3600 s/h) = 2.27 gal/h
3
, 11) Problem: Your power plant has an output of 2000 MW. Its flame temperature is 1227 °C and it is
exhausting to ambient conditions at 20 °C.
a) Find Carnot Efficiency.
b) For the idealized Carnot cycle heat engine, what is the minimum rate of input heat (Qhot)?
c) What is the waste heat (Qcold) being dumped into the surroundings?
d) Draw a diagram showing the energy flows in and out, and that they balance.
Given: Wout = 2000 MWe; Tflame = 1227 °C; Tamb = 20 °C
Find: For Carnot Qhot; Qcold
Assume: Carnot cycle
Solution:
Wout = 2000 MW
Carnot efficiency = 1 – Tc/Th = 1 – 293K/1500K = 0.805
Rate of heat input: Qhot = Wout/η = 2000MJ/s / 0.8 = 2484MW
Qwaste = (2484 -2000) MW = 484MW
12) Problem: Power consumed by an engine as it ramps up is given by P(t) = 100 kW + 10 kW/min × t.
Find the energy consumed in kWh in 1 hour.
Energy =
13) Problem: You successfully lift a 25 kg stone to the top of the Empire State Building (1250 ft to the
top floor). How much work did you do on the stone, assuming constant g? If you burned the calories
you ingested from drinking a Coke while accomplishing this (Coke has 182 food Calories), what is
the thermodynamic efficiency of the process (your body!)?
Given: M = 25 kg; ∆z = 1250 ft; Qin = 182 Cal
Find: Wout, η
Assume: steady state; constant g = 9.8 m/s2
Solution:
Wout = f × d = m × g × d = 25 kg × 9.8 m/s2 × 1250 ft × (m/3.28 ft) = 93368 N×m = 93.4 kJ
Qin = Energy Expended = 182,000 cal × 4.18 J/calorie = 760760 J = 760.8 kJ
η= Wout/Qin = 93.4 kJ/760.8 kJ = 0.123 = 12.3% efficient
14) Problem: An inventor has a new power plant design that is claimed to have a 2000 MW output with a
feed rate of 100 kg/s of fuel with an energy content of 25 MJ/kg. The plant is operating at hot and
cold reservoir temperatures of 1000 °C and 20 °C. Is this possible?
Given: Wout = 2000 MWe; dM/dt = 100 kg/s; Ec = 25 MJ/kg; Thot = 1000 °C; Tcold = 20 °C
4
