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Industrial Engineering Sample Problems with Solutions, Exercises of Industrial Engineering

Ateneo de Naga University (ADNU)Industrial Engineering

Includes sample problems on Industrial Engineering with solutions

Typology: Exercises

2019/2020

Available from 11/28/2022

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INDUSTRITAL PLANT ENGINEERING
1. Copra enters the dryer containing 60% water and 40% of solids and leaves with 5% water and
95% solids. Find the weight of water removed based on pound on final product and pound bone
dry material.
a. 1.375 lb/lb; 1.4475 lb/lb
b. 1.193 lb/lb
c. 1. 846 lb/lb; 2.2233 lb/lb
d. 1.472 lb/lb; .9763 lb/lb
Solution:
Consider: 1 lb of wet feed
Let x = weight of original product per lb of wet feed
Solid in wet feed = solid in dry product
0.95x = 0.40 (1)
x = 0.421
Weight of H20 removed = 1 – 0.421
= 0.579 lb of orig
lb of product
Weight of H20 removed per lb of final product
= 0.579 / 0.421
= 1.375 lb / lb of final product
Weight of H20 removed per lb of bone dry material
= 0.579 / (0.40)(1)
= 1.4475 lb/lb bone dry material
2. Copra enters the dryer containing 60% water and 40% of solids and leaves with 5% water and
95% solids. Find the weight of water removed based on pound of original product.
a. .579 lb/lb b. .795 lb/lb c. .388 lb/lb d. 1.373b/lb
Solution:
Consider: 1 lb of wet feed
Let x = weight of original product
Per lb of wet feed
Solid in wet feed = solid in dry product
0.95 x = 0.40 (1)
x= 0.579 lb / lb of original product
weight of water removed per lb of final product
= 0.579 / 0.421
= 1.373 lb/lb of final product
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INDUSTRITAL PLANT ENGINEERING

  1. Copra enters the dryer containing 60% water and 40% of solids and leaves with 5% water and 95% solids. Find the weight of water removed based on pound on final product and pound bone dry material. a. 1.375 lb/lb; 1.4475 lb/lb b. 1.193 lb/lb c. 1. 846 lb/lb; 2.2233 lb/lb d. 1.472 lb/lb; .9763 lb/lb Solution: Consider: 1 lb of wet feed Let x = weight of original product per lb of wet feed Solid in wet feed = solid in dry product 0.95x = 0.40 (1) x = 0. Weight of H 2 0 removed = 1 – 0. = 0.579 lb of orig lb of product Weight of H 2 0 removed per lb of final product = 0.579 / 0. = 1.375 lb / lb of final product Weight of H 2 0 removed per lb of bone dry material = 0.579 / (0.40)(1) = 1.4475 lb/lb bone dry material
  2. Copra enters the dryer containing 60% water and 40% of solids and leaves with 5% water and 95% solids. Find the weight of water removed based on pound of original product. a. .579 lb/lb b. .795 lb/lb c. .388 lb/lb d. 1.373b/lb Solution: Consider: 1 lb of wet feed Let x = weight of original product Per lb of wet feed Solid in wet feed = solid in dry product 0.95 x = 0.40 (1) x= 0.579 lb / lb of original product weight of water removed per lb of final product = 0.579 / 0. = 1.373 lb/lb of final product
  1. When 100 kg / min of outside air at 32 C dry bulb and 200 kg/min recirculated air at 22 C dry bulb are mixed with air conditioning system, the resulting dry bulb temperature will be, a. 25.33 C b. 46.33 C c. 35.44 C d. 26.88 C Solution: 100 kg / min x 32 C = 10.67 C 300 kg / min 200 kg / min x 22 C = 14.67 C 300 kg / min 10.67 C + 14.67 C = 25.34 C
  2. The amount of water carried by air in a cooling tower is 15 lb/min. The change in humidity ratio in outlet and inlet is .025 lb/lb. Determine the volume flow of air needed if specific volume is 13 ft^3 / lb a. 6,00 ft^3 / min b. 7500 c. 7800 d. 5000 Solution: Mw = 15 lb / min humidity ratio = ( W 2 – W 1 ) = 0.025 lb/lb Υ= 13 ft^3 / lb Volume flow rate of air = m ( υ ) Mw = ma ( W 2 – W 1 ) 15 lb / min = ma ( 0.025 lb /lb ) ma= 15 lb /min / 0.025 lb / lb ma= 600 lb /min volume flow rate of air = ( 600 lb/min ) ( 13) volume flow rate of air = 7800 ft^3 / min
  1. A metal rod 2 sq. in. in cross section and 1 foot long is heated at one end and cooled at the other. If the rate of heat input is 3 Btu/hr and the difference in the temperature of the two ends is 8 F, what is the coefficient of thermal conductivity? a. 17 Btu / ( sq. ft-hr-F) b. 27 Btu / ( sq. ft-hr-F) c. 37 Btu / ( sq. ft-hr-F) d. 47 Btu / ( sq. ft-hr-F) Solution: d= 2 in = 0.17 ft L= 1 ft Q = 3 Btu / hr ∆T= 8 F K =? Q = KA ∆T / L; K = QL / A∆T = 3 Btu / hr ( 1 ft ) л / 4 ( 0.17 ft ) 2 ( 8 F) K = 17 Btu / ft-hr-F ( Problem 9-11) A pump discharges 50 tonne of water per hour to a height of 8 meters and the overall efficiency of the pumping system is 69%
  2. Calculate the output power ( water power) of the pump. a. 1.09 kw b. 11.09 Kw c. 21.09 kw d. 31.09 kw Solution: Wp = Qhg Q= ( m /  ) hg Q= mhg Q= (50 tonne/hr ) ( 1 hr / 3600 sec) ( 1000 kg /1 tonne) (8 m )( 9.81 m/sec^2 ) Q = 13.88 kg / sec Wp = 13.88 ( 8) ( 9.81) Wp = 1090 W = 1.09 KW
  1. Calculate the input power ( Brake power ) to the pump. a. 1.58 kw b. 18.07 kw c. 30.57 kw d. 45.06 kw Solution: BP = Wp / eo BP = 1.09 / 0. BP= 1.58 Kw
  2. Calculate the energy consumed by the pump in 2 hrs. a. 11.37 MJ b. 115.70 MJ c. 220.10 MJ d. 324.06 MJ Solution: Q = [ 1.58 Kw ( 3413 Btu/ hr ) ( 2 hrs ) / 1 Kw] Q = 10785.08 Btu ( 1.055 Kj/1 Btu ) ( 1 MJ/ 1000KJ) Q = 11.37 MJ (12-14) A furnace burns natural gas with a volumetric analysis as follows; propane = 3%, Methane = 35 % , Ethane = 12 % The total fuel flow rate is 0.37 m^3 / sec and 13.5 is the actual air fuel ratio is required for complete combustion. Air is 25 C and 1 atm is supplied to the furnace by a FORCED DRAFT FAN thru a short duct of 0.25 m^2 in the area against the static pressure of 50 mm of water. The forced draft fan has an efficiency of 65 %
  3. The volume flow rate entering the furnace is? a. 1.0 m^3 / sec b. 3.0 m^3 / sec c. 5.0 m^3 / sec d. 7.0 m^3 / sec Solution: A.F. = VA / vf; VA = A.F. (vf ) = 13.05 ( 0.37 m^3 / s ) VA = 4.995 m^3 / s
  1. If the actual draft required for a furnace is 6.239 cm of water the friction losses in the stack are 15% of the theoretical draft, calculate the required stack height in meter. Assume that the flue gas have an average temperature of 149 C and molecular weight of 30. Assume air temperature of 21 C a.220 b.265 c.210 d. none of the above Solution: Hw = total draft = 6.239 + 0. = 7.34 cm water = 0.0734 ( 9.81 ) = 0.72 kPa solving for ρair ρair = P / RaTa = 101.325 / ( 0.287)( 21+273) = 1.2 kg / m^3 solving for ρfg R= 8.314 / 30 = 0. ρfg = P / Rfg Tfg = 101.325 / ( 149+273) = 0.867 kg / m^3 hw= H (ρair - ρfg ) 0.72 = H ( 1.2 ) – 0.867 / 0. H = 220 m
  2. What is the height of the chimney if the driving pressure is 30 Pa and the gas and air densities are 1 kg/m^3 and 1.5 kg /m^3 respectively? a.6.12 b.6.02 c.5.28 d.none of the above Solution: H = hw / da – dg = (30 / 100) / ( 1.5-1)(0.00981) H = 6.12 m
  1. A condensate pump at sea level take water from a surface condenser where the vacuum is 15 in of mercury. The friction and turbulence in the piping in the condenser hot well and the pump suction flange is assumed to be 6.5 ft. If the condensate pump to be installed has a required head of 9 ft, what would be the minimum height of water level in the hot well that must be maintained above the centerline of the pump to avoid cavitation? a. 2.5 ft b. 15.5 ft. c. 18 ft d. 5.5 ft. Solution: The NPSH required by the pump is 9 NPSH = P + Pd – Pv +S – hf 9.81 ( S./G) 9 = 0+S – 6. S = 15.5 ft
  2. A power plant situated at an altitude having an ambient air of 96.53 Kpa and 23.68 C. Flue gases at a rate of 5.0 kg/sec enter the stack at 200 C and leaves at 160 C. The gases gravimetric analysis are 18% CO 2 , 7% O 2 , and 75% N 2. Find the height and diameter of the stack. a. h= 75.21m and d= 1 m b. h= 54.62 m and d= 3.21 m c. h=50.21 m and d= 1m d. none of the above Solution: H = hw / a - g a = P / RTair = 96.33 KN / m^3 / (0.287)(23.88 + 273) = 1.133 Kg / m^3 MW of the gas CO 2 = 0.18 (44) = 7. C 2 = 0.07 ( 32) = 2. N 2 = 0.75 (28) = 21 / 31.16 MW g = 96.33 KN / m^3 [ ( 8.314 /31.16)( 200+160 / 2) +273] = 0.799 kg / m^3 H= 0.2 / ( 1.33 – 0.799) H = 59.88 m
  1. A DC driven pump running at 100 rpm delivers 30 liters per second of water at 40 C against a total pumping head of 27 m with a pump efficiency of 60 %. Barometer pressure is 758 mmHg abs. What pump speed and capacity would result if the pump rpm were increase to produce a pumping head of 36 m assuming no change in efficiency. a. 115.47 rpm, 34.64 L/s b. 115.47 rpm, 38.68 L/s c. 110.51 rpm, 34.46 L/s d. 110.51 rpm, 36.68 L/s Solution: New speed required N 2 / N 1 = [ H 2 / H 1 ] ½ N 2 / 100 = [ 36 /27 ] ½ N 2 = 115.47 rpm New capacity Q 2 / Q 1 = [ N 2 / N 1 ]1/ Q 2 / 30 = [ 36 /27 ] ½ Q 2 = 34.64 L /s
  2. A heat exchanger is to be designed for the following specifications; hot gas temperature = 1145 C cold gas temperature = 45 C unit surface conductance on the hot side = 230 W /sq. mtr – C unit surface conductance on the cold side = 290 W /sq. mtr – C Find the maximum thickness of metal wall between the hot gas and the cold gas so that the wall temperature does not exceed 545 C a.32 b. 20.115 c.26.325 d.30. Solution: Q /A = t 1 – to / (1/hf) + (x / k ) + ( 1/Ho Solving for Q/A Q/A = h1 ( t1 – to) = 230 ( 1145 – 545 ) = 138000 W / m^2

then; 138000 = 1145 – 45 (1/230)+(x/115)+(1/290) X= 20.115 mm

  1. If one has 10 pounds of ice at -18 F and wishes to change it to 10 pounds of steam at 312 F how much heat is required? a. 13643 b.15000 c.12354 d. none of the above Solution: QT = mc∆t + mLf + mc∆t + mHv + mc∆t = 10 ( 0.50)( 32+18) + 10 ( 144) + 10 (1)(212-32)+ 10 ( 970.3) + 10 ( 0.45)( 312-212) QT = 13643 BTU
  2. A single stage air compressor handles 0.454 m^3 / s of atmospheric pressure, 27 C air and delivers it to a receiver at 652.75 Kpa. Its volumetric efficiency is .72.. its compression efficiency on isothermal basis is 0.85 and its mechanical efficiency is 0.90. if it rotates at 350 rpm, what powering kw is required to drive it? a. 95 b. 112 c. 120 d. 100 Solution: Wi = isothermal power = P 1 V 1 ln ( P 2 / P 1 ) = 101.3 ( 0.454 ) ln ( 652.75 / 101.3) = 85.685 KW drive power = 85.685 / 0.85 ( 0.90 ) = 112 KW
  3. A rotary compressor receives 8 m^3 / min of gas ( R=0.410 Kj / kg-K ) at 108 Kpa, 27 C and delivers it at 650 Kpa. Find the work if compression is polytrophic with PV1.3^ = C a. -32.02 Kw b. 33.02 Kw c. 50.01 Kw d. -50.01 Kw Solution: Wksf = n / n-1 [ P 2 V 2 – P 1 V 1

Q = 333.33 ft 3 / sec δd = 14.7 lb / in^2 ( 144 in^2 / ft 2 53.3 ft-lb / lb R ( 530 R ) = 0,075 lb / ft^3 HT = Vp + Sp V = Q /A = 333.33 Ft^3 /sec / ( 3 x 4 )ft = 27.78 ft / sec P = δ h δwhw = δaha ( 62.4 lb / ft^3 ) ( 3 in / 12 ) = 0.075 lb / ft^3 ha= 208 ft of air Vp = V^2 / 2g = ( 27.78 )^2 / ( 2 x 32.174 ) = 12 ft of air HT = Vp + Sp = 12 ft + 208 ft = 220 ft of air TAP = Q δa HT / 550 = 333.33 ft^3 / s ( 0.075 lb / ft^3 )( 220 ft of air) 550 TAP = 9.99 = 10 Hp

  1. Compute the humidity ratio of air at 65 % relative humidity and 34 C when the barometric pressure is 101.325 Kpa. a. 0.022 b. 0.023 c. 0.024 d. 0. Solution: T = 34 C P = 101.325 kPa Psat @ 34 C = 5. W = 0.622 Pv / Pt – Pv Pv= RHPsat = 0.65 (5.32) = 3.458 Kpa W= 0.622 ( 3.458 Kpa / 101.325 – 3.458 Kpa RH= 0.
  2. What is the enthalpy of air – vapor mixture in problem 1 ( KJ / Kg) a. 90.61 b. 92.45 c. 90.50 d. 91. Solution: hg @ 34 C = 2563.5 Kj / kg

h= Cpt + WHg h= 1 Kj / kg K ( 34 C ) + ( 0.022) 2563.5 Kj /kg h= 90.4 KJ /Kg

  1. What is the specific volume cum / kg dry air of an air vapor mixture at 30 C and a relative humidity of 40 % at 101.3 Kpa pressure? a. 0.873 b. 0.90 c. 0.864 d. 0. Solution: Ϋ = 1 / ρ Ρ = P / RT R = 8.314 / 29 Kj / kgm K = 0.287 Kj / kgm K Ρ= 101.325 Kn / m^3 / ( 8.314 Kn m / Kg K) ( 30+273) = 1.165 kg / m^3 Ϋ= 1 / 1.165 kg / m^3 = 0.864 m^3 / kg
  2. Determine the quantity of heat removed from 14 m^2 per min of air when cooled from 37 C dry bulb and 21 C wet bulb temperature to 15 C. what are the initial and final relative humidities? a. 15.75 & 6.04 b. 16.30 & 3.42 c. 14.76 & 6.90 d. 15.3 & -6. Solution: m= 14 m^3 / min / 0.889 m^3 / kg m= 15.75 kg / min From psychometric chart @ td1 = 37 c and twl = 21 C ф 1 = 23 % ф 2 = 85 % h1 = 61 Kj/ kg h2= 38 kj / kg V1= 0.889 m^3 / kg Q= m ( h2-h1) Q= 15.75 ( 38-61)

Tmix =? wb mm Dbm = mr Dbr + mo Dbo (1) Dbm = (26 C) ( 2/3) + 35 ( 1/3) Dbm= 29 C mmWbm = mrWbr + moWbo (1)9Wbm) = (19 C ) ( 2/3 ) + 24 ( 1/3) Wbm= 20.7 C 29 C db; 20.7 Wb

  1. In a space the sensible heat load is 13.5 kw and the latent heat load is 3.4 kw. Outside air is at 32 C and 50 % relative humidity. The space is to be maintained at 25 C dry bulb end 18 C wet bulb temperature. All outside air is supplied with reheater to satisfy the space conditions. The conditioned air leaves the supply fan at 17 C. Determine (a) the refrigeration load, (b) the capacity of the supply fan (c), the heat supplied in the reheater. a. 37kj/g b. 39kj/kg c. 32kj/kg d. 40kj/kg Solution: Qs = ( 1.0062) ( m) ( t 4 – t 3 ) 135 = ( 1.0062) ( m)( 25 – 17) m= 1.68 kg /sec Pt 4 @ t 4 = 25 C & tw 4 = 18 C W 4 = 0.0101 kg / kg Pt 3 @ QL = ( 2500 ) ( m) ( W 4 –W 3 ) 3.4 = ( 2500)(m)( 0.0101 – W 3 ) W = 0.0093 kg / kg @ t 3 = 17 C & W 3 = 0. V 3 = 0.835^3 / kg; h 3 = 40.5 Kj / Kg Pt 2 @ W 2 = W 3 = 0.0093 kg / kg and saturated h 2 = 37 KJ / kg
  1. An air- conditioned theater is to be maintained at 26.7 C db temperature and 50 % relative humidity. The calculated total sensible heat load in the theater is 126240kcal / hr and latent heat load is 82920kcal / hr. the air mixture at 28.9 C db and 22.2 C wb temperature is cooled to 17. C db and 15 C wb temperature by chilled water cooling coils and delivered as supplied air to the theater. Calculate the tons of refrigeration required. a. 125 b. 120 c. 129 d. 130 Solution: Ein = Eout Q + m4h4 = m3h Q= m ( h3-h4) = 67476 kg / hr ( 19.9 -14.3 ) Q= 376482.6 Kcal / hr Q= 376482.6 kcal / hr x ( 4.187 Kj/ 1 Kcal ) ( 1 Btu / 1.055 KJ )( TR / 12000 Btu / Hr) Q= 124.51 = 125 TOR
  2. Calculate the horsepower of the boiler generating 4000lb of dry steam per hr, with a factor of evaporation of 1.06. The latent heat of steam at 212 F is 970 Btu / lb a. 132 boiler hp b. 128 boiler hp c. 123 boiler hp d. 135 boiler hp Solution: Ws=m = 4000 lbs dry steam / hr F.E. = 1. T = 212 F ∆H = 970 Btu / lb F.E. = ∆H / 970.3______ eq. 1 Dev. Boiler hp = ws ∆H/ 33480_____ eq 2 Equate 1 to 2 F.E. (970.3) = bo.hp (33480)/ W Bo. Hp = F.E. (970.3) Ws / 33480 Bo. Hp = 122.88 = 123 hp
  1. A refrigeration plant is rated at 20 ton capacity. How many pounds of air will it cool 90 to 70 F at constant pressure. a. 50000 lb / hr b. 70000 lb /hr c. 60000 lb / hr d. 40000 lb/hr Solution: Q = 20 + R T1= 90 F T2 = 70 F m=? Q= mCp ( T1 – T2 ) m= Q / Cp∆T m= ( 20 Tr x 1200 Btu /hr / 1 Tr) 0.24Btu/ lb-F( 90-70) F m= 50000 lb
  2. A refrigeration plant is rated at 20 ton capacity. What is the approximate engine horsepower required to operate the plant? Assume COP = 4 a. 23.6 hp b. 25.3 hp c. 24.3 hp d. 22.3 hp Solution: Q= 20 TR COP = 4 We =? COP = Qe / We We= Qe / COP = ( 20TR x 3.52 kw / 1 Tr ) / 4 We = 17.6 kw ( 1 hp / 0.746 kw ) We= 23.59 hp
  3. A boiler installed where the atmospheric pressure is 752 mm Hg has a pressure of 12 kg per sq. cm. What is the absolute pressure in MPa/ a. 1.772 Mpa b. 1.727 Mpa c. 1.279 Mpa d. 1.327 Mpa Solution: Pb = 12 kg / cm^2 Pabs = Pb + Pg Pabs = 12 kg / cm^2 + ( 101.325 Kpaa / 1.032 kg / cm^2 )

Pabs = 1279.52 Kpa ( 1 Mpa / 1000 Kpa) Pabs = 1.279 MPaa

  1. ME Board April 1999 A counter flow heat exchanger is designed to heat fuel oil fro 45 C to 100 C while the heating fluid enters at 150 C and leaves at 115 C. Calculate the arithmetic mean temperature difference. a. 40 C b. 50 C c. 60 C d. 70 C Solution: AMTD= (∆t)max + (∆t)min 2 Where: (∆t)max= 115- = 70 C (∆t)min= 150- = 50 C AMTD= 70+ 2 AMTD= 60 C
  2. ME Board October 1999 The stroke of the piston of an air compressor is 250 mm and the clearance volume is equal to 6% of the stroke volume. The pressure of the air at the beginning of the compression is 0.98 bar and it is discharge at 3.8 bar. Assuming compression to follow the law PVn^ = C , where n= 1.25, calculate the distance moved by the piston from the beginning of its pressure stroke before the discharge valves opens and express this is a percentage of the stroke. a. 70.15% b. 75.25% c. 82.75% d. 78.25% Solution: V= V 1 – V 2 Where: V 1 = c + 250 = ( 0.06 )( 250 ) + 250 = 265