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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:
	 = 0.1914	(Note: Pr = )
Data: 
Water: Cp = 1.0 Btu/lboF, k = 0.36 Btu/fthroF, = 0.72 cp = 0.72×2.42 lb/ fthr
Mercury: Cp = 0.33 Btu/lboF, specific gravity = 13.6
Solution
 = = = 0.62 s
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
	0
	1
	2.5
	5
	8
	10
	15
	30
	Thermometer T, oF
	75
	107
	140
	205
	244
	282
	328
	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)
	Level (ft)
	Time (min)
	Level (ft)
	Time (min)
	Level (ft)
	0 
	4.8
	1.2423 
	8.9354
	2.4847
	11.4513
	0.138 
	5.3673
	1.3804
	9.2805
	2.6227
	11.6612
	0.2761 
	5.9041
	1.5184
	9.6071
	2.7607
	11.8599
	0.4141 
	6.412
	1.6564
	9.9161
	…………..
	…………..
	0.5521 
	6.8927
	1.7945
	10.2085
	14.3558
	15.3261
	0.6902 
	7.3475
	1.9325
	10.4853
	14.4938
	15.328
	0.8282 
	7.7779
	2.0705
	10.7471
	14.6319
	15.3297
	0.9663 
	8.1852
	2.2086
	10.9949
	14.7699
	15.3313
	1.1043 
	8.5706
	2.3466
	11.2294
	
	
Kp = 3.19 ft/gpm
This height occurs at t 2.5 min = . The transfer function is thus:
	 = 
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/lboF 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?
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(s) = 
		
 	 f(t) = e4tcos 2t + 3e4tsin 2t
11. Determine the transfer functions for this system at s = 0.5.	 
Solution
C(s)/R(s) = = = 0.47059	
12. Consider the following control system with all instrumentation in electronic (4 to 20 mA)
 
	 
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. 
4 + 16 = 10.4 mA
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.
4 + 16 = 7.24 mA