Bundle: Refrigeration and Air Conditioning Technology, 8th + MindTap HVAC, 4 terms (24 months) Printed Access Card
8th Edition
ISBN: 9781337190350
Author: John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher: Cengage Learning
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Chapter 19, Problem 8RQ
What causes an overload protection device to function?
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Example(1):
(Adiabatic humidification and cooling of
air). Air has to be humidified and cooled
adiabatically in a honzontal spray chamber
with recirculated water. The active part of
the chamber is Im #2m #15 m long. Under
the operating conditions, the coefficient
of heat transfer is expected to be 1300
kcal/(hr)(m2)(°C). 200 m3/min of air at 60
°C and 1 atm pressure with a humidity of
0.018 kg water/kg dry air is to be blown
through the spray chamber. Calculate the
following
(a) the temperature and hunudity of the
exit air
(b) make-up water to be supplied, windage
and blow down are neglected
(c) the expected gas-phase mass transfer
coefficient, kya
(d) the temperature and humidity of the
exit air if an identical spray chamber is
added in series with
the existing one
O
14.14. A three-stage evaporator is fed with
1.25 kg/s of a liquor which is concentrated
from 10 to 40 per cent solids by mass. The
heat transfer coefficients may be taken
as 3.1, 2.5 and 1.7 kW/m² K, respectively,
in each effect. Calculate the steam
flow at 170 kN/m² and the temperature
distribution in the three effects, if:
(a) the feed is at 294 K, and
(b) the feed is at 355 K.
Forward feed is used in each case and
the values of U are the same for the two
systems. The boiling point in the third
effect is 325 K and the liquor has no
boiling point rise.
give me solution math not explin
Chapter 19 Solutions
Bundle: Refrigeration and Air Conditioning Technology, 8th + MindTap HVAC, 4 terms (24 months) Printed Access Card
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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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