Automotive Technology
7th Edition
ISBN: 9781337794213
Author: ERJAVEC, Jack.
Publisher: Cengage,
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Chapter 3, Problem 1SA
Describe Newton’s first law of motion and give an application of this law in automotive theory.
Expert Solution & Answer
To determine
Newton’s first law of motion with an application in terms of automotive theory.
Answer to Problem 1SA
Newton’s first law of motion or Newton’s law of inertia: “Every object in its state of rest or uniform motion(Acceleration is zero) in a straight line unless it is compelled to change that state by forces impressed on it”.
Explanation of Solution
When an object is in rest or motion, that object will always be in rest or motion until an external force work on that object.
Example: Assume a car is in rest, the car will be in rest until the force applied. Now the car is moving, and this car will be moving until the force applied on it.
Conclusion:
Newton’s first law of motion is working every condition when any object is in rest or trying to rest.
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Students have asked these similar questions
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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Can you solve the question by finding initial angular acceleration of D then ueing it to calculate angular velocity of D.
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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 3 Solutions
Automotive Technology
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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
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