Mindtap For Erjavec/thompson's Automotive Technology: A Systems Approach, 4 Terms Printed Access Card (mindtap Course List)
7th Edition
ISBN: 9781337794381
Author: Jack Erjavec, Rob Thompson
Publisher: Cengage Learning
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Textbook Question
Chapter 33, Problem 9SA
What types of catalyst are typically used to reduce NOx and to oxidize HC and CO?
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4.59
Using the unilateral z-transform, solve the following difference equations with the given initial
conditions.
(a) y[n]-3y[n-1] = x[n], with x[n] = 4u[n], y[− 1] = 1
(b) y[n]-5y[n-1]+6y[n-2]= x[n], with x[n] = u[n], y[-1] = 3, y[-2]= 2
Ans. (a) y[n] = -2+9(3)", n ≥ -1
(b) y[n]=+8(2)" - (3)", n ≥ -2
(30)
6. In a process design, the following process streams must be cooled or heated:
Stream
No
mCp
Temperature In Temperature Out
°C
°C
kW/°C
1
5
350
270
2
9
270
120
3
3
100
320
4
5
120
288
Use the MUMNE algorithm for heat exchanger networks with a minimum approach
temperature of 20°C.
(5)
a. Determine the temperature interval diagram.
(3)
(2)
(10)
(10)
b. Determine the cascade diagram, the pinch temperatures, and the minimum hot and
cold utilities.
c. Determine the minimum number of heat exchangers above and below the pinch.
d. Determine a valid heat exchange network above the pinch.
e. Determine a valid heat exchange network below the pinch.
Use this equation to solve it.
Chapter 33 Solutions
Mindtap For Erjavec/thompson's Automotive Technology: A Systems Approach, 4 Terms Printed Access Card (mindtap Course List)
Ch. 33 - Explain why a small PCV valve opening is adequate...Ch. 33 - At what temperature do nitrogen atoms combine with...Ch. 33 - Describe the operation of a digital EGR valve.Ch. 33 - Name the three types of emissions being controlled...Ch. 33 - The PCV system prevents from escaping to the...Ch. 33 - In a negative backpressure EGR valve, if the...Ch. 33 - Describe how a selective catalytic reduction...Ch. 33 - HC emissions may come from the tailpipe or...Ch. 33 - What types of catalyst are typically used to...Ch. 33 - Why is a PCV system critical to an engines...
Ch. 33 - What is a catalyst?Ch. 33 - Rather then rely on the AIR system for extra...Ch. 33 - Why do the EGR systems on many late-model diesel...Ch. 33 - Which of the following systems is designed to...Ch. 33 - Which of the following statements about carbon...Ch. 33 - While discussing PCV valve operation: Technician A...Ch. 33 - While discussing EGR systems with a differential...Ch. 33 - While discussing evaporative (EVAP) systems:...Ch. 33 - Technician A says that an EGR valve that is stuck...Ch. 33 - While discussing SCR systems: Technician A says...Ch. 33 - While discussing automotive emissions: Technician...Ch. 33 - Technician A says that during oxidation, as NOx...Ch. 33 - While discussing PCV systems without a PCV valve:...Ch. 33 - While diagnosing the cause of high HC and O2...Ch. 33 - While discussing electric exhaust gas...
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- 14.9. A forward feed double-effect vertical evaporator, with equal heating areas in each effect, is fed with 5 kg/s of a liquor of specific heat capacity of 4.18 kJ/kg K. and with no boiling point rise, so that 50 per cent of the feed liquor is evaporated. The overall heat transfer coefficient in the second effect is 75 per cent of that in the first effect. Steam is fed at 395 K and the boiling point in the second effect is 373 K. The feed is heated by an external heater to the boiling point in the first effect. It is decided to bleed off 0.25 kg/s of vapour from the vapour line to the second effect for use in another process. If the feed is still heated to the boiling point of the first effect by external means, what will be the change in steam consumption of the evaporator unit? For the purpose of calculation, the latent heat of the vapours and of the steam may both be taken as 2230 kJ/kgarrow_forwardExample(3): It is desired to design a double effect evaporator for concentrating a certain caustic soda solution from 12.5wt% to 40wt%. The feed at 50°C enters the first evaporator at a rate of 2500kg/h. Steam at atmospheric pressure is being used for the said purpose. The second effect is operated under 600mmHg vacuum. If the overall heat transfer coefficients of the two stages are 1952 and 1220kcal/ m2.h.°C. respectively, determine the heat transfer area of each effect. The BPR will be considered and present for the both effect 5:49arrow_forwardالعنوان ose only Q Example (7): Determine the heating surface area 개 required for the production of 2.5kg/s of 50wt% NaOH solution from 15 wt% NaOH feed solution which entering at 100 oC to a single effect evaporator. The steam is available as saturated at 451.5K and the boiling point rise (boiling point evaluation) of 50wt% solution is 35K. the overall heat transfer coefficient is 2000 w/m²K. The pressure in the vapor space of the evaporator at atmospheric pressure. The solution has a specific heat of 4.18kJ/ kg.K. The enthalpy of vaporization under these condition is 2257kJ/kg Example (6): 5:48 An evaporator is concentrating F kg/h at 311K of a 20wt% solution of NaOH to 50wt %. The saturated steam used for heating is at 399.3K. The pressure in the vapor space of the evaporator is 13.3 KPa abs. The 5:48 1 J ۲/۱ ostrarrow_forward
- Example 8: 900 Kg dry solid per hour is dried in a counter current continues dryer from 0.4 to 0.04 Kg H20/Kg wet solid moisture content. The wet solid enters the dryer at 25 °C and leaves at 55 °C. Fresh air at 25 °C and 0.01Kg vapor/Kg dry air is mixed with a part of the moist air leaving the dryer and heated to a temperature of 130 °C in a finned air heater and enters the dryer with 0.025 Kg/Kg alry air. Air leaving the dryer at 85 °C and have a humidity 0.055 Kg vaper/Kg dry air. At equilibrium the wet solid weight is 908 Kg solid per hour. *=0.0088 Calculate:- Heat loss from the dryer and the rate of fresh air. Take the specific heat of the solid and moisture are 980 and 4.18J/Kg.K respectively, A. =2500 KJ/Kg. Humid heat at 0.01 Kg vap/Kg dry=1.0238 KJ/Kg. "C. Humid heat at 0.055 Kg/Kg 1.1084 KJ/Kg. "C 5:42 Oarrow_forwardQ1: From the Figure below for (=0.2 find the following 1. Rise Time 2. Time of oscillation 3. Overshoot value 4. Maximum value 5. When 1.2 which case will be? 1.6 1.4 1.2 12 1.0 |=0.8- 0.6 0.4 0.8 0.2- 0.6 0.4 0.2 1.2 = 1.0 0 2 4 6 8 10 10 t/Tarrow_forwardPlease, I need solution in detailsarrow_forward
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