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LECTURE 1- 1)
Question 1 – Determine the July electric utility bill for a facility that used 112000 KWh during the month, and had a maximum power usage of 500 KW, during the peak period of time in that month. The utility has a fixed meter charge of $75 / month, and charges a flat rate of 5 cents/ KWh for energy and $12/KW for a maximum power usage during peak period in July
Ans
– Monthly Bill => Fixed Monthly + Rate of energy* Energy produced + Rate of max power * Max Power = 75+0.05*112000+12*500 = $11,675
2)
Question
2
– Determine the rate at which heat must be added in Btu/hr to a 300 cfm airstream passing through a heating coil to change its temperature from 70 F to 120 F. Assume an inlet air at a specific volume of 13.5 ft
3
/lbm and a specific heat of 0.24 Btu/lbm.F
The heat being added is sensible, as its contributing to the temp change of the airstream.
Ans
– q
s
= (Q*Cp)/V*(T
e
-T
i
) = (300 cfm * 60 m/hr *0.24)/13.5*(120-70) = 16000 Btu/hr
3)
Question 3 – Air at 1 atm and 76 F is flowing at the rate of 5000 cfm. At what rate must energy be removed, in Btu/hr, to change the temperature to 58 F assuming no dehumidification occurs? @ 76 F, density = 0.074 lbm/ft
3
C
p
= 0.34 Btu/lbmF
Ans
– q = mc
p
(T
2
-T
1
) => m = density * Q = 0.074*5000*60 = 22200 lbm/hr
q = 22200*0.24(58-76) = -95904 Btu/hr
4)
Question 4
– Determine the operative temp. for a workstation in a room near a large window where the dry bulb temp. (temp of air) and globe temp. are measured to be 75 F and 81 F respectively. The air velocity is estimated to be 30 ft/min at the station.
Ans
– Operative temp depends on the T
mrt
(Mean radiant temperature)
T
mrt
= (T
g
4
+ CV
1/2 * (T
g
– T
a
) )
1/4 = ((81+460)
4
+ 0.103x10
9
* (30)
1/2
(81-75))
1/4 = 546 R = 86 F
Operating temp = (T
mrt
+ T
a
) /2 = (86 + 75)/2 = 80.5 F
5)
Question 5
– A person breathes out CO
2
at a rate of 0.3 L/min, the concentration of CO
2
in the incoming ventilation air is 300 ppm (Parts per minute). It is desired to hold the concentration in the room below 1000 ppm. Assume that the air in the room is perfectly mixed. What is the minimum rate of air flow required to maintain the desired level. Ans
- Q
t
C
e + N = Q
t
C
s
=> N = Q
t
C
s
- Q
t
C
e
Q
t
= N/(C
s
-C
e
) = 0.3/(0.001-0.0003) = 7.1 L/s
6)
Question 6
– Carbon Dioxide is being generated in an occupied space at the rate 0.25
cfm (0.1181 m
3
/s) and outdoor air with CO
2
concentration of 200 ppm is being supplied to the space at the rate of 900 cfm (0.425 m
3
/s). What will the steady-state concertation of CO
2
be in ppm if complete mixing is assumed?
Ans
- Q
t
C
e + N = Q
t
C
s
=> N = Q
t
C
s
- Q
t
C
e
C
s
= (Q
t
C
e
+ N)/Q
t
= (900*200+0.25)/900 = 478 ppm
---------------------------------------------------------EOL------------------------------------------------------------
LECTURE 2 - 1)
Question 1 - Find the heat transfer rate required to warm 1500 cfm (ft
3
/min) of air at 60 F and 90% relative humidity to 110 F without the addition of moisture.
Ans
– Q = 1500 cfm, State 1 - T1 = 60 F,
= 90%
State 2 - T2 = 110 F
c
p
= 0.244
From psychometric chart :
- Specific volume = 13.3 ft
3
/lbm
- Enthalpy of state 1 - i
1
or h
1
= 25.1 Btu/lbm
- Enthalpy of State 2 - i
1
or h
1
= 37.4 Btu/lbm
m
a
= (1500*60)/13.3 = 6752 Q = m
a
(i
1
-i
2
) = m
a
c
p
(t
1
-t
2
) = 6752*0.244(110-60) = 82.374 Btu/hr
2)
Question 2
- 2000 cfm of air at 100 F db and 75 F wb are mixed with 1000 cfm of air at 60 F db and 50 F wb. The process is adiabatic, at a steady flow rate and at standard
sea-level pressure. Find the condition of the mixed stream using the lever rule.
State 1- DB = 60 F, WB = 50 F
State 2- DB = 100 F, WB = 75 F
From Chart: - Specific volume of state 1= 13.21 ft
3
/lbm
- Specific volume of state 2= 14.4 ft
3
/lbm
- Humidity rate of state 1
1
= 0.0054 lbm/lb
- Humidity rate of state 2
2
= 0.013 lbm/lb
m
a1
= 1000*60/13.21 = 4511.28 m
a2
= 2000*60/14.40 = 8333.33
3
=
1
+ m
a1
/ m
a2
= 0.0054 + 4511.28/8333.33 = 0.54
---------------------------------------------------------EOL------------------------------------------------------------
LECTURE 3- 1)
Question 1 – Q.5-4 What is the unit thermal resistance for an inside partition made up of 3/8 in. gypsum board on each side of 8 in. lightweight aggregate blocks with vermiculite-filled cores?
Ans
– R
gypsum
= 1/C = 1/3.1 = 0.32 (C found from table 5-1a in pg. 11 of lecture 3 slides)
R
air
= 0.68 (Table 5-2a pg. 14 lecture 3 slides)
R
vermiculite
= 1/C = 1/0.33 (C from table 5-1a pg.12 lecture 13 slides)
R = 0.68+0.32+0.30+0.32+0.68 = 5.03
2)
Question 2 -
Q.5-5 Compute the thermal resistance per unit length for a 4 in. schedule 40 steel pipe with 1½ in. of insulation. The insulation has a thermal conductivity of 0.2 Btu-in./(hr-ft
2
-F).
Ans
- R = R
steel
+ R
ins
= ln
(
r
2
r
1
)
2
π
∗
Lk
+
ln
(
r
3
r
2
)
2
π
∗
Lk
d
2
= Outer diameter = 4.5 in, r
2
= Outer Radius = 2.25 in (Table
d
1
= Inner diameter = 4.026 in, r
1
= Inner Radius = 2.013 in
r
3
= Outer Radius = r
2
+ 1.5 = 3.75 in
k
steel
= 314 Btu-in./(hr-ft
2
-F).
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