Control Systems Engineering
7th Edition
ISBN: 9781118170519
Author: Norman S. Nise
Publisher: WILEY
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Chapter 7, Problem 15P
To determine
The type of the given system.
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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.
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
O
give me solution math not explin
Chapter 7 Solutions
Control Systems Engineering
Ch. 7 - Prob. 1RQCh. 7 - A position control, tracking with a constant...Ch. 7 - Name the test inputs used to evaluate steady-state...Ch. 7 - Prob. 4RQCh. 7 - Increasing system gain has what effect upon the...Ch. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - The forward transfer function of a control system...
Ch. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Is the forward-path actuating signal the system...Ch. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Name two methods for calculating the steady-state...Ch. 7 - Prob. 1PCh. 7 - Figure P7.2 shows the ramp input r(t) and the...Ch. 7 - Prob. 3PCh. 7 - Prob. 4PCh. 7 - Prob. 5PCh. 7 - Prob. 6PCh. 7 - Prob. 7PCh. 7 - Prob. 8PCh. 7 - A system has Kp = 4. What steady-state error can...Ch. 7 - Prob. 10PCh. 7 - Prob. 11PCh. 7 - Prob. 12PCh. 7 - For the system shown in Figure P7.4. [Section:...Ch. 7 - Prob. 14PCh. 7 - 1515. Find the system type for the system of...Ch. 7 - Prob. 16PCh. 7 - Prob. 17PCh. 7 - Prob. 18PCh. 7 - Prob. 19PCh. 7 - Given the system of Figure P7.8, design the value...Ch. 7 - Prob. 21PCh. 7 - Prob. 22PCh. 7 - Prob. 23PCh. 7 - Prob. 24PCh. 7 - Prob. 25PCh. 7 - Prob. 26PCh. 7 - Prob. 27PCh. 7 - Prob. 28PCh. 7 - Prob. 29PCh. 7 - Prob. 30PCh. 7 - Prob. 31PCh. 7 - Prob. 32PCh. 7 - Given the system in Figure P7.9, find the...Ch. 7 - Repeat Problem 33 for the system shown in Figure...Ch. 7 - Prob. 36PCh. 7 - Prob. 37PCh. 7 - Prob. 38PCh. 7 - Design the values of K1and K2in the system of...Ch. 7 - Prob. 41PCh. 7 - For each system shown in Figure P7.17, find the...Ch. 7 - For each system shown in Figure P7.18, find the...Ch. 7 - Prob. 44PCh. 7 - 45. For the system shown in Figure P7.20,...Ch. 7 - Prob. 47PCh. 7 - Prob. 48PCh. 7 - Prob. 49PCh. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - Prob. 53PCh. 7 - Prob. 54PCh. 7 - Prob. 55PCh. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - Prob. 62PCh. 7 - Prob. 63PCh. 7 - Prob. 64PCh. 7 - Prob. 65PCh. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Prob. 68P
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- give me solution math not explinarrow_forwarduse 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_forwardgive 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_forward
- 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. O Oarrow_forward5) 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_forward
- G = 0.350MPa, P = 900N, a=20mm, b=50mm, c=80mmarrow_forward[20 Points] Use Method of Sections to draw the shear-force and bending-moment diagrams for the simply supported beam shown. Determine the maximum bending moment that occurs in the span. 1.5 kN/m 4 m 2 m Carrow_forward14.9. 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/kg Ад Oarrow_forward
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