Welding: Principles and Applications (MindTap Course List)
8th Edition
ISBN: 9781305494695
Author: Larry Jeffus
Publisher: Cengage Learning
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Chapter 28, Problem 19R
Referring to Figure 28-8 through Figure 28-15, which electrode has the deepest penetration?
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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
2.8
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العنوان
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Jon
14.23. A double-effect forward-feed
evaporator is required to give a product
consisting of 30 per cent crystals and a
mother liquor containing 40 per cent by
mass of dissolved solids. Heat transfer
coefficients are 2.8 and 1.7 kW/m² K in the
first and second effects respectively. Dry
saturated steam is supplied at 375 kN/m²
and the condenser operates at 13.5 kN/
m².
(a) What area of heating surface is
required in each effect assuming the
effects are identical, if the feed rate is 0.6
kg/s of liquor, containing 20 per cent by
mass of dissolved solids, and the feed
temperature is 313 K?
(b) What is the pressure above the boiling
liquid in the first effect?
The specific heat capacity may be
taken as constant at 4.18 kJ/kg K. and
the effects of boiling-point rise and of
hydrostatic head may be neglected.
Chapter 28 Solutions
Welding: Principles and Applications (MindTap Course List)
Ch. 28 - What groups have developed electrode...Ch. 28 - What types of general information about electrodes...Ch. 28 - Define tensile strength.Ch. 28 - What chemicals alloys are a. considered to be...Ch. 28 - What should CE be used for?Ch. 28 - What welding parameters should be used for a metal...Ch. 28 - What functions can the flux covering of an SMA...Ch. 28 - How does an SMA welding electrode's flux covering...Ch. 28 - What fluxing agents act as scavengers in the...Ch. 28 - How can an SMA welding electrode's flux help with...
Ch. 28 - What are the advantages of refractory-type stages?Ch. 28 - List the things that must be considered before...Ch. 28 - Why can there be more than one electrode for each...Ch. 28 - What do the following filler metal designations...Ch. 28 - Explain the parts of the AWS classified system for...Ch. 28 - Which SMA welding electrode(s) can be used to weld...Ch. 28 - Which SMA welding electrodes are commonly used to...Ch. 28 - Which SMA welding electrodes can be used with a...Ch. 28 - Referring to Figure 28-8 through Figure 28-15,...Ch. 28 - How is the E7018 molten weld pool protected?Ch. 28 - What is the purpose of the deoxidizers in ER70S-2?Ch. 28 - What alloying element used in FCA welding...Ch. 28 - What does the 15 and 16 stand for in SMA stainless...Ch. 28 - What stainless steel(s) would a. have low creep at...Ch. 28 - Referring to Table 28-5, what stainless steel...Ch. 28 - What would be a stainless steel filler metal for...Ch. 28 - What forms the basis for the AWS identification...Ch. 28 - Why must thick sections of aluminum be preheated...Ch. 28 - For what types of items would the purest aluminum...Ch. 28 - What are aluminum arc brazing electrodes used for?Ch. 28 - How do most manufacturers classify or group...
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- Example(2): Double effect evaporator is used for concentrating a certain caustic soda solution 10000kg/hr from 9wt% to 47wt%. The feed at 30°C enters the first evaporator. Backward arrangement evaporators are used. steam is available at 167.7°C and the vapor space in the second effect is 14.6Kpa. The overall heat transfer coefficients of the two effects are 8380 and 6285kcal/ W.CH ork -conce -SOLFFF and-ans.. 112.1 а DiD 3 respectively and the specific heat capacity of all caustic soda solution 3.771 KJ/Kg. °C, determine the heat transfer area of each effect معدلة 5:48 م Oarrow_forwardgive me solution math not explinarrow_forwardgive me solution math not explinarrow_forward
- use Q Strips of material 10 mm thick are dried under constant drying conditions from 28 to 13 per cent moisture in 25 ks. The equilibrium moisture content is 7 per cent. The relation between E, the ratio of the final free moisture content at time t to the initial free moisture content, and the parameter J is given by: E 1 0.64 0.49 0.38 0.295 0.22 0.14 J 0 0.1 0.2 0.3 0.5 0.6 العنوان 0.7 It may be noted that J = kt/12, where, k = constant, t = time (ks) 1 = thickness/2 of the sheet of material (mm) a. Based on the given data, plot a graph of E against J b. Determine the time taken to dry 60 mm planks from 22 to 10 per cent moisture under the same conditions assuming no loss from the edges? ina östler ۲/۱arrow_forward14.25.2.5 kg/s of a solution at 288 K containing 10 per cent of dissolved solids is fed to a forward-feed double-effect evaporator, operating at 14 kN/m² in the last effect. If the product is to consist of a liquid containing 50 per cent by mass of dissolved solids and dry saturated steam is fed to the steam coils, what PROBLEMS 1179 should be the pressure of the steam? The surface in each effect is 50 m² and the coefficients for heat transfer in the first and second effects are 2.8 and 1.7 kW/ m² K, respectively. It may be assumed that the concentrated solution exhibits a boiling-point rise of 5 deg K, that the latent heat has a constant value of 2260 kJ/kg and that the specific heat capacity of the liquid stream is constant at 3.75 kJ/kg K Oarrow_forward: +0 العنوان use only 5) A 100 kg batch of granular solids containing 30% moisture is to be dried in a tray drier to 15.5% by passing a current of air at 350 K tangentially across its surface at a velocity of 1.8 m/s. If the constant rate of drying under these conditions is 0.7 g/s m2 and the critical moisture content is 15%, calculate the approximate drying time. Assume the drying surface to be 0.03 m2 /kg dry mass. мониarrow_forward
- give me solution math not explinarrow_forward۲/۱ : +0 العنوان seoni 4) 1 Mg (dry weight) of a non-porous solid is dried under constant drying conditions with an air velocity of 0.75 m/s parallel to the drying surface. The area of drying surface is 55 m2 If initial rate of drying is 0.3 g/m2 s, how long it will take to dry a material from 0.15 to 0.025 kg water/kg dry solid? The critical moisture content is 0.125 and the equilibrium moisture is negligible. The falling rate of drying is linear in moisture content. If air velocity increases to 4 m/s, what will be the anticipated saving in drying time? 0 ostherarrow_forward14.23. A double-effect forward-feed evaporator is required to give a product consisting of 30 per cent crystals and a mother liquor containing 40 per cent by mass of dissolved solids. Heat transfer coefficients are 2.8 and 1.7 kW/m² K in the first and second effects respectively. Dry saturated steam is supplied at 375 kN/m² and the condenser operates at 13.5 kN/ m². (a) What area of heating surface is required in each effect assuming the effects are identical, if the feed rate is 0.6 kg/s of liquor, containing 20 per cent by mass of dissolved solids, and the feed temperature is 313 K? (b) What is the pressure above the boiling liquid in the first effect? The specific heat capacity may be taken as constant at 4.18 kJ/kg K. and the effects of boiling-point rise and of hydrostatic head may be neglected. O Oarrow_forward
- 5) A 100 kg batch of granular solids containing 30% moisture is to be dried in a tray drier to 15.5% by passing a current of air at 350 K tangentially across its surface at a velocity of 1.8 m/s. If the constant rate of drying under these conditions is 0.7 g/s m2 and the critical moisture content is 15%, calculate the approximate drying time. Assume the drying surface to be 0.03 m2 /kg dry mass. Oarrow_forwardSolve for v and Iarrow_forwardG = 0.350MPa, P = 900N, a=20mm, b=50mm, c=80mmarrow_forward
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