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CHE426: Problem set #51
1. A thermometer having a time constant of 0.2 min is placed in a temperature bath, and after
the thermometer comes to equilibrium with the bath, the temperature of the bath is increased
linearly with time at a rate of 1°/min. What is the difference between the indicated
temperature and the bath temperature (a) 0.1 min, (b) 1.0 min after the change in temperature
begins?
(c) What is the maximum deviation between indicated temperature and bath temperature, and
when does it occur?
(d) Plot the forcing function and response on the same graph. After a long enough time, by
how many minutes does the response lag the input?
(a)
(t = 0.1) =  0.2(1  e-5×0.1) =  0.0787 oF
(b)
(t = 1) =  0.2(1  e-5×1) =  0.1987 oF
(c)
Maximum deviation when t  ∞  (t) = =  0.2 oF
(d) After a long enough time, Y(t) = t  0.2, since X(t) = t
The response lags the input by 0.2 minutes
2. A mercury thermometer bulb is ½ in. long by 1/8 in. diameter. The glass envelope is very
thin. Calculate the time constant in water flowing at 10 ft/sec at a temperature of 100oF. You
can assume all resistance to heat flow is in the outside water and all heat capacity is in
mercury. The heat transfer coefficient for water can be estimated from the following
equation:
hD
= 0.1914 Re 0.618 Pr 1/ 3
k
(Note: Pr =
C p
k
)
Data:
Water: Cp = 1.0 Btu/lboF, k = 0.36 Btu/fthroF,  = 0.72 cp = 0.72×2.42 lb/ fthr
Mercury: Cp = 0.33 Btu/lboF, specific gravity = 13.6
Solution
(3.0134 10 3 )(0.33)(3600)
=
=
= 0.62 s
hA
(3981)(1.4488 10 3)
MC p
3.1 A thermometer having first-order dynamics with a time constant of 1 min is at 100oF. The
thermometer is suddenly placed in a bath at 110oF at t = 0 and left there for 1 min, after
which it is immediately returned to a bath at 100oF.
(a) Draw a sketch showing the variation of the thermometer reading with time.
(b) Calculate the thermometer reading at t = 0.5 min and at t = 2.0 min.
Solution
At t = 0.5 min  T = 103.9347
At t = 2.0 min  T = 102.32
4. A mercury thermometer, which has been on a table for some time, is registering the room
temperature, 75oF. Suddenly, it is placed in a 400oF oil bath. The following data are obtained
for the response of the thermometer.
Time, sec
Thermometer T, oF
0
75
1
107
2.5
140
5
205
8
244
10
282
15
328
30
385
Estimates the thermometer time constant.
Solution
The thermometer time constant = 10 sec
5.1 A thermometer having a time constant of 1 min is initially at 50oC. It is immersed in a
bath maintained at 100oC at t = 0. Determine the temperature reading at t =1.2 min.
Solution
y = 50 + 34.94 = 84.94oC
6.1 The level in a tank responds as a first order system with changes in the inlet flow. Given
the following level vs. time data that was gathered after the inlet flow was increased quickly
from 1.5 gal/min to 4.8 gal/min, determine the transfer function that relates the height in the
tank to the inlet flow. Be sure to use deviation variables and include units on the steady state
gain and the time constant.
Time (min)
0
0.138
0.2761
0.4141
0.5521
0.6902
0.8282
0.9663
1.1043
Level (ft)
4.8
5.3673
5.9041
6.412
6.8927
7.3475
7.7779
8.1852
8.5706
Time (min)
1.2423
1.3804
1.5184
1.6564
1.7945
1.9325
2.0705
2.2086
2.3466
Level (ft)
8.9354
9.2805
9.6071
9.9161
10.2085
10.4853
10.7471
10.9949
11.2294
Time (min)
2.4847
2.6227
2.7607
…………..
14.3558
14.4938
14.6319
14.7699
Level (ft)
11.4513
11.6612
11.8599
…………..
15.3261
15.328
15.3297
15.3313
Kp = 3.19 ft/gpm
This height occurs at t  2.5 min = . The transfer function is thus:
3.19 ft/gpm
H (s)
=
2.5 s  1
Q( s)
In this equation: H(s) = h(s) 4.8 ft and Q(s) = q-1.5 . H(s) and Q(s) are deviation variables.
7. The bypass cooling system shown is designed so that the total flow of 200 gpm of a liquid
with heat capacity of 0.8 Btu/lboF is split under the normal conditions, 20 % going around
the bypass and 80 % going through the cooler. Process inlet and outlet temperature under
these conditions are 250 and 150oF. Inlet and outlet water temperatures are 80 and 120oF.
Process side pressure drop the exchanger is 10 psi. The control valves have linear trim and
are designed to be half open at design rates with a 15 psi drop over the bypass valve and 10
psi drop over the cooler valve. Liquid density is constant at 62.3 lb/ft3.
What will the valve positions be if the total process flow is reduced to 25 percent of design
and the process outlet temperature is held at 150oF?
Set point (SP)
TC
Process bypass
TT
Water
Ti(t)
T(t)
Process
stream
Cooled stream
Solution -----------------------------------------------------------------------------------------f(x) = 0.755 and PT = 1.74 psi.
8. Express the function given the graph in the t-domain
f(t) = 2u(t - 1) + (t - 2)u(t - 2) - (t - 3)u(t - 3) - u(t -3)
9. Given f(s) = (1  2e-s + e-2s)/s2
f(t) = t  2(t  1)u(t  1) + (t  2)u(t  2)

10. Find the inverse of F (s) =
s2
s  8s  20
2
f(t) = e4tcos 2t + 3e4tsin 2t
11. Determine the transfer functions for this system at s = 0.5.
Solution
4
4
3( s  1)
C(s)/R(s) =
= 4.5 = 0.47059
4
4
1
1
3( s  1)
4.5
12. Consider the following control system with all instrumentation in electronic (4 to 20 mA)
Hot oil
o
90 F
Refrigerant
FT
o
Cooler
Heat
exchanger
50 F
Cooled oil
o
70 F
TT
TC
a) If the temperature transmitter has a range of 50-100oF, determine the value from the
temperature transmitter.
b) If the range of the orifice-differential pressure flow transmitter on the water line is 0-2000
gpm, determine the value from the flow transmitter for a water flow rate of 900 gpm.
Solution
a) If the temperature transmitter has a range of 50-100oF, determine the value from the
temperature transmitter.
 70  50 
4 + 16 
 = 10.4 mA
 100  50 
b) If the range of the orifice-differential pressure flow transmitter on the water line is 0-2000
gpm, determine the value from the flow transmitter for a water flow rate of 900 gpm.
2
 900 
4 + 16 
 = 7.24 mA
 2000 