Chemical Engineering
Cork is a good insulator because- a) It is flexible
- b) It can be powered
- c) Low density
- d) It is porous
Correct answer is option 'D'. Can you explain this answer?
Cork is a good insulator because
a)
It is flexible
b)
It can be powered
c)
Low density
d)
It is porous
Telecom Tuners answered • Mar 02, 2021Even though cork is a solid material, but it’s thermal conductivity is extremely low because of air filled in it’s pores. Typical value of thermal conductivity is in the range of 0.035 to 0.043 W/m-k
A sphere, a cube and a disc all are of the same material, quality and volume, are heated to 900 K and left in air. Which of these will have the lowest rate of cooling- a) Cube
- b) Disc
- c) Sphere
- d) All will have the same rate of cooling
Correct answer is option 'C'. Can you explain this answer?
A sphere, a cube and a disc all are of the same material, quality and volume, are heated to 900 K and left in air. Which of these will have the lowest rate of cooling
a)
Cube
b)
Disc
c)
Sphere
d)
All will have the same rate of cooling
Telecom Tuners answered • Mar 02, 2021For a given volume, the sphere has a minimum surface area hence rate of heat transfer to air by convection or radiation is minimum.
What are the units of Stefan-Boltzmann constant- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
What are the units of Stefan-Boltzmann constant
a)
b)
c)
d)
Telecom Tuners answered • Mar 01, 2021By Stefan-Boltzmann law
Consider the astronaut compartments inside a space shuttle orbiting the moon. Inside these compartments air is maintained at STP conditions for comfortable breathing of astronauts. The cooling system of electronic equipment inside these compartments is based on- a) Conduction
- b) Forced convection
- c) Free convection
- d) Thermal radiation
Correct answer is option 'B'. Can you explain this answer?
Consider the astronaut compartments inside a space shuttle orbiting the moon. Inside these compartments air is maintained at STP conditions for comfortable breathing of astronauts. The cooling system of electronic equipment inside these compartments is based on
a)
Conduction
b)
Forced convection
c)
Free convection
d)
Thermal radiation
Telecom Tuners answered • Mar 01, 2021Since the equipment is in air and surface temperature is generally not high, conduction and radiation heat loss is insignificant. Also in space there is no gravity. Thus buoyancy and therefore free convection is not at all possible in the compartment air. Hence the only suitable mode for cooling is forced convection.
The minimum number of view factors that need to be known to solve a 10-surface enclosure completely, is- a)1
- b)10
- c)22
- d)45
Correct answer is option 'D'. Can you explain this answer?
The minimum number of view factors that need to be known to solve a 10-surface enclosure completely, is
a)
1
b)
10
c)
22
d)
45
Telecom Tuners answered • Mar 01, 2021For n-surface enclosure if nC2 view factors are known directly, entire enclosure can be solved. Thus for 10 surfaces, number of view factors that need to be evaluated directly will be
The wavelength for which the blackbody emissive power is maximum for a temperature of 300 K is- a)5.1μm
- b)9.7μm
- c)15.5μm
- d)38.0μm
Correct answer is option 'B'. Can you explain this answer?
The wavelength for which the blackbody emissive power is maximum for a temperature of 300 K is
a)
5.1μm
b)
9.7μm
c)
15.5μm
d)
38.0μm
Telecom Tuners answered • Mar 01, 2021By Wien’s displacement law,
∵ λmT = 2898 μm-k
λm ≃ 9.7μm
Sanidhya Chandola asked • 8 hours agoSuspended solid present in the waste water generated in blast furnace gas cooling and cleaning plant is removed by- a)biological oxygen pond.
- b)radial settling tank (thickener) using coagulant (lime & ferrous sulphate).
- c)lagoons.
- d)filtration.
Correct answer is option 'B'. Can you explain this answer?
Suspended solid present in the waste water generated in blast furnace gas cooling and cleaning plant is removed by
a)
biological oxygen pond.
b)
radial settling tank (thickener) using coagulant (lime & ferrous sulphate).
c)
lagoons.
d)
filtration.
For a given heat flow rate and for the same thickness, the temperature drop across the material will be maximum for- a) Copper
- b) Steel
- c) Glass-wool
- d) Refractory brick
Correct answer is option 'C'. Can you explain this answer?
For a given heat flow rate and for the same thickness, the temperature drop across the material will be maximum for
a)
Copper
b)
Steel
c)
Glass-wool
d)
Refractory brick
Telecom Tuners answered • Mar 01, 2021From Fourier’s law,
I.e. T₁ - T₂ will be maximum when k is minimum. k is minimum for glass wool.
Consider a light bulb of 100 watts power. When switched on, what is the surface temperature of the tungsten filament in °C.Consider the surface to be a black surface having value of area equal to
- a) 2166
- b) 1893
- c) 800
- d) 3786
Correct answer is option 'B'. Can you explain this answer?
Consider a light bulb of 100 watts power. When switched on, what is the surface temperature of the tungsten filament in °C.
Consider the surface to be a black surface having value of area equal to
a)
2166
b)
1893
c)
800
d)
3786
Telecom Tuners answered • Mar 01, 2021By Stefan Boltzmann law
T = 2166 K (or) 1893 °C
For the composite slab shown in figure, find out when k represents thermal conductivity and L represents thickness.
- a) 110
- b) 70
- c) 60
- d) 90
Correct answer is option 'B'. Can you explain this answer?
For the composite slab shown in figure, find out when k represents thermal conductivity and L represents thickness.
a)
110
b)
70
c)
60
d)
90
Telecom Tuners answered • Mar 01, 2021The thermal circuit of the system can be represent as follows
By idea of circuits ratio of resistances is equal to ratio of temperature drops across them
150 - Ti = 2Ti - 60
210 = 3Ti
Ti = 70°C
The spectral distribution of surface irradiation is as follows
What is the total irradiation in kW/m2 ?- a)10
- b)20
- c)30
- d)40
Correct answer is option 'B'. Can you explain this answer?
The spectral distribution of surface irradiation is as follows
What is the total irradiation in kW/m2 ?
a)
10
b)
20
c)
30
d)
40
Telecom Tuners answered • Mar 01, 2021Total irradiation is given as
i.e. it is equal to area under the curve Gλ Vs λ Hence,
x (25-20) x 1000
∴ G = 20000 W/m2
I.e. G = 20 kW/m2
Sandhya Mannuru asked • 16 hours agoHot water (0.01 m3 /min) enters the tube side of a counter current shell and tube heat exchanger at 80�C and leaves at 50�C. Cold oil (0.05 m3/min) of density 800 kg/m3 and specific heat of 2 kJ/kg.K enters at 20�C. The log mean temperature difference in �C is approximately- a)32
- b)37
- c)45
- d)50
Correct answer is option 'A'. Can you explain this answer?
Hot water (0.01 m3 /min) enters the tube side of a counter current shell and tube heat exchanger at 80�C and leaves at 50�C. Cold oil (0.05 m3/min) of density 800 kg/m3 and specific heat of 2 kJ/kg.K enters at 20�C. The log mean temperature difference in �C is approximately
a)
32
b)
37
c)
45
d)
50
Solar radiation is incident on a semitransparent body at a rate of . If of this incident radiation is reflected back and is transmitted across the body, the absorptivity of the body is- a)0
- b)0.25
- c)0.30
- d)0.45
Correct answer is option 'B'. Can you explain this answer?
Solar radiation is incident on a semitransparent body at a rate of . If of this incident radiation is reflected back and is transmitted across the body, the absorptivity of the body is
a)
0
b)
0.25
c)
0.30
d)
0.45
Telecom Tuners answered • Mar 01, 2021G = 500 W/m2 Gr = 150 W/m2
Gt = 225 W/m2
By energy conservation principle,
Ga+Ge+Gt = G
Where Ga = absorbed radiation
Gr = reflected radiation
Gt = transmitted radiation
G = incident radiation
= transmissivity
α = 1-0.3 - 0.45 = 0.25
What will be the view factor for the geometry as shown in the figure (sphere within a hollow cube)?
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
What will be the view factor for the geometry as shown in the figure (sphere within a hollow cube)?
a)
b)
c)
d)
Telecom Tuners answered • Mar 01, 2021
1 is a spherical surface
∴ F11 = 0
By summation rule,
F12 + F11 = 1
∴ F12 = 1
By Reciprocity theorem,
A1F12 = A2F21
Consider monochromatic emissive power (Eλ) Vs Wavelength(λ) of a black and grey surface both at same temperature. The ratio AB/AC is given as
- a)Monochromatic emissivity (λ) of grey surface
- b)Total emissivity (ε) of grey surface
- c)Total emissive power of grey surface
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Consider monochromatic emissive power (Eλ) Vs Wavelength(λ) of a black and grey surface both at same temperature. The ratio AB/AC is given as
a)
Monochromatic emissivity (λ) of grey surface
b)
Total emissivity (ε) of grey surface
c)
Total emissive power of grey surface
d)
None of the above
Telecom Tuners answered • Mar 01, 2021AB & AC are monochromatic emissive power of grey and black surfaces for same wavelength. Hence, ratio is
which is monochromatic emissivity of grey surface.
Find out the heat flux (in W/m²) between two fluids separated by a planar wall as shown in the figure.
- a)200
- b)200
Correct answer is between '200,200'. Can you explain this answer?
Find out the heat flux (in W/m²) between two fluids separated by a planar wall as shown in the figure.
a)
200
b)
200
Telecom Tuners answered • Mar 01, 2021Consider the thermal circuit for the system
Where nodes H, 1, 2 & C represent hot fluid, left surface of wall, right surface of wall and cold fluid respectively. Therefore,
Solar radiation of falls on a grey opaque surface at steady state. The surface has a temperature of and emissivity of 0.8. Find radiosity from the surface?Correct answer is 'A'. Can you explain this answer?
Solar radiation of falls on a grey opaque surface at steady state. The surface has a temperature of and emissivity of 0.8. Find radiosity from the surface?
Telecom Tuners answered • Mar 01, 2021
J = E+Gr
E = εσT4
E = 0.8 x 5.67 x 10-8 x (50 + 273)4
E = 493.72 W/m2
Given G = 1200 W/m2
But as surface is at steady state
α = ε = 0.8
∴ ρ = 1 -α = 0.2
∴ Gr = ρG
Gr = 0.2 x 1200 = 240 W/m2
Therefore
J = E+Gr
J = 493.72 + 240
J = 733.72 W/m2
Consider H gas, both at temperature of 30°C. The thermal conductivity of H is ________ than that of .- a) greater
- b) equal
- c) lower
- d) depends upon pressure
Correct answer is option 'A'. Can you explain this answer?
Consider H gas, both at temperature of 30°C. The thermal conductivity of H is ________ than that of .
a)
greater
b)
equal
c)
lower
d)
depends upon pressure
Telecom Tuners answered • Mar 01, 2021The thermal conductivity of a gaseous medium is dominantly based on diffusion rate, which depends on root mean square velocity.The root mean square velocity for molecules is given by
where M is the molecular mass of the gas & T is the temperature of the gas and ̅ is universal gas constant. Thus diffusion rate and therefore thermal conductivity is lower for gases with higher molecular masses because
Therefore H₂ gas has more thermal conductivity than O₂.
Two radiating surfaces A1= 6m2 and A2=4m2 have shape factor Then shape factor will be- a)0.18
- b)0.15
- c)0.12
- d)0.10
Correct answer is option 'B'. Can you explain this answer?
Two radiating surfaces A1= 6m2 and A2=4m2 have shape factor Then shape factor will be
a)
0.18
b)
0.15
c)
0.12
d)
0.10
Telecom Tuners answered • Mar 01, 2021By Reciprocity theorem
A1F12 = A2F21
= 0.15
Consider a hemispherical furnace. The view factor of it’s roof (hemisphere) with respect to itself is __________.- a)0
- b)0.25
- c)0.5
- d)1.0
Correct answer is option 'C'. Can you explain this answer?
Consider a hemispherical furnace. The view factor of it’s roof (hemisphere) with respect to itself is __________.
a)
0
b)
0.25
c)
0.5
d)
1.0
Telecom Tuners answered • Mar 01, 2021
By reciprocity theorem
A1F12 = A2F12
1 is a planar surface, i.e F11 = 0
∴ F12 = 1
Hence F21 = 0.5
F21 + F22 = 0.5
F22 + 1-0.5 = 0.5
With increase in pressure, the thermal conductivity of an ideal gas __________.- a) increases
- b) remains constant
- c) decrease
- d) remains constant
Correct answer is option 'B'. Can you explain this answer?
With increase in pressure, the thermal conductivity of an ideal gas __________.
a)
increases
b)
remains constant
c)
decrease
d)
remains constant
Telecom Tuners answered • Mar 01, 2021For an ideal gas, thermal conductivity is given as
Where ⊽ is the mean travel velocity of molecules Cv, is the specific heat of gas at constant volume, L is the mean free path of molecules
and is the density of the gas.
and is the density of the gas.As pressure increases, density of gas increases but based on experiments it is observed that the mean free path diminishes in the same proportions. Thus the product
remains constant.
remains constant.Also ⊽ depends upon temperature only hence if temperature is constant, change in pressure does not bring any change in ⊽ Also for an ideal gas Cv is independent of pressure.
Shubham Kumar asked • 16 hours agoIf average heat transfer co-efficient is ha and the local coefficient at the end of the plate is hl then in case of heat transfer to a fluid flowing over a flat plate, heated over its entire length- a)ha = hl
- b)ha = 2hl
- c)ha = 0.5 hl
- d)ha = 0.75 hl
Correct answer is option 'B'. Can you explain this answer?
If average heat transfer co-efficient is ha and the local coefficient at the end of the plate is hl then in case of heat transfer to a fluid flowing over a flat plate, heated over its entire length
a)
ha = hl
b)
ha = 2hl
c)
ha = 0.5 hl
d)
ha = 0.75 hl
Consider a composite slab as shown in the figure.
If the left most and right most surfaces are maintained at temperatures of 100°C & 40°C respectively, find out the interface temperature in 
in °CCorrect answer is 'Range: 66 to 67'. Can you explain this answer?
Consider a composite slab as shown in the figure.
If the left most and right most surfaces are maintained at temperatures of 100°C & 40°C respectively, find out the interface temperature in in °C
in °C Telecom Tuners answered • Mar 01, 2021The slab can be represented by the following thermal circuit.
Considering the width of slab to plane of paper to be equal to W,
R₂ and R₃ are arranged in parallel and can be shown by single equivalent resistance Rₑ
By idea of series circuit, that ratio of temperature drops across resistances is equal to ratio of respective thermal resistances.
I.e.,
Which of the following events represents thermal radiation?- a) Microwave heating
- b) Nuclear reaction in a sun
- c) Light bulb
- d) None of the above
Correct answer is option 'C'. Can you explain this answer?
Which of the following events represents thermal radiation?
a)
Microwave heating
b)
Nuclear reaction in a sun
c)
Light bulb
d)
None of the above
Telecom Tuners answered • Mar 01, 2021In microwave ovens, the radiation emitted is a result of excitation of crystals like magnetrons and klystrons and in microwave wavelength band which is not in the range of thermal radiation. Also during nuclear reactions like those inside sun, -rays are produced which is actually not considered as thermal radiation. Once these -rays get absorbed by the matter in the sun then temperature rises and thermal radiation is emitted. Thus nuclear reaction does not directly emit thermal radiation.
But during lighting of bulb energy is emitted mainly in form of light and IR due to temperature of tungsten filament. Thus this event is termed as thermal radiation.
Which of the following is used to measure the temperature inside a furnace?- a) Gas thermometer
- b) Optical pyrometer
- c) Alcohol thermometer
- d) Mercury thermometer
Correct answer is option 'B'. Can you explain this answer?
Which of the following is used to measure the temperature inside a furnace?
a)
Gas thermometer
b)
Optical pyrometer
c)
Alcohol thermometer
d)
Mercury thermometer
Telecom Tuners answered • Mar 01, 2021In a furnace heat transfer is a result of thermal radiation due to high temperature values and contribution of conduction and convection to the phenomenon is insignificant. Hence optical pyrometers are best suited devices to measure temperature inside the furnace. These devices detect wavelength corresponding to maximum monochromatic emissive power in the infrared radiation and using Wien’s law, i... more
Karan Gohil asked • 21 hours agoThe thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP are- a)(COP)R = (COP)HP = 0.6
- b)(COP)R = 2.5; (COP)HP = 1.5
- c)(COP)R = 1.5; (COP)HP = 2.5
- d)(COP)R = (COP)HP = 2.5
Correct answer is option 'C'. Can you explain this answer?
The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. Then the coefficient of performance as a refrigerator (COP)R and the co-efficient of performance as a heat pump (COP)HP are
a)
(COP)R = (COP)HP = 0.6
b)
(COP)R = 2.5; (COP)HP = 1.5
c)
(COP)R = 1.5; (COP)HP = 2.5
d)
(COP)R = (COP)HP = 2.5
Consider two infinitely long blackbody concentric cylinders with a diameter ratio . The shape factor of the inner surface of outer cylinder with respect to itself will be
- a)0
- b)1/3
- c)2/3
- d)1
Correct answer is option 'C'. Can you explain this answer?
Consider two infinitely long blackbody concentric cylinders with a diameter ratio . The shape factor of the inner surface of outer cylinder with respect to itself will be
a)
0
b)
1/3
c)
2/3
d)
1
Telecom Tuners answered • Feb 28, 2021
By Energy conservation
F12 + F22 =1
∴ F22 = 1-F21 ….①
By Reciprocity theorem
A1F12 = A2F21
πD1L x F12 = πD2L x F21
If the temperature of a solid surface changes from then its emissive power will increases in the ratio of- a)3
- b)9
- c)27
- d)81
Correct answer is option 'D'. Can you explain this answer?
If the temperature of a solid surface changes from then its emissive power will increases in the ratio of
a)
3
b)
9
c)
27
d)
81
Telecom Tuners answered • Feb 28, 2021By Stefan Boltzmann law
E ∝ T4
For an opaque plane surface at steady state the irradiation, radiosity and emissive power are respectively 20, 12 and . What is the emissivity of the surface?- a)0.9
- b)0.4
- c)0.8
- d)1.0
Correct answer is option 'A'. Can you explain this answer?
For an opaque plane surface at steady state the irradiation, radiosity and emissive power are respectively 20, 12 and . What is the emissivity of the surface?
a)
0.9
b)
0.4
c)
0.8
d)
1.0
Telecom Tuners answered • Feb 28, 2021
J = pG + εEb
For an opaque body
∵ α + ρ = 1
ρ = 1- α
∴ J = (1-α)G + εEb
By kirchoff’s law, α = ε
∴ J = (1-α)G + εEb
12 = (1- ε)20 + 10
ε = 0.9
Navya Rajulapati asked • yesterdayChoose the wrong statement with respect to Nusselt number and convective heat transfer coefficient- a)Nusselt number represents the ratio of temperature gradient at the surface, to an overall or reference temperature gradient
- b)Nusselt number represents the dimensionless slope of the temperature distribution curve at the surface
- c)The convective coefficient can be evaluated from a knowledge of fluid temperature distribution in the neighbourhood of the surface
- d)For a given Nusselt number, the convective coefficient is inversely proportional to thermal conductivity of the fluid.
Correct answer is option 'D'. Can you explain this answer?
Choose the wrong statement with respect to Nusselt number and convective heat transfer coefficient
a)
Nusselt number represents the ratio of temperature gradient at the surface, to an overall or reference temperature gradient
b)
Nusselt number represents the dimensionless slope of the temperature distribution curve at the surface
c)
The convective coefficient can be evaluated from a knowledge of fluid temperature distribution in the neighbourhood of the surface
d)
For a given Nusselt number, the convective coefficient is inversely proportional to thermal conductivity of the fluid.
What is net radiation heat exchange per square meter per unit time for two very large plates at temperature values of 800 K and 500 K? Emissivities of hot and cold plates are 0.8 and 0.6 respectively. Stefan-Boltzmann constant is - 5.67x10-8W/m2K4- a)1026 kW/m2
- b)10.26 kW/m2
- c)102.6 kW/m2
- d)1.026 kW/m2
Correct answer is option 'B'. Can you explain this answer?
What is net radiation heat exchange per square meter per unit time for two very large plates at temperature values of 800 K and 500 K? Emissivities of hot and cold plates are 0.8 and 0.6 respectively. Stefan-Boltzmann constant is - 5.67x10-8W/m2K4
a)
1026 kW/m2
b)
10.26 kW/m2
c)
102.6 kW/m2
d)
1.026 kW/m2
Telecom Tuners answered • Feb 28, 2021For two very large parallel plates, the net radiation heat exchange (using thermal circuit) is given by,
= 10.26 kW/m2
Sandeep Agrawal asked • yesterdayPick out the wrong statement.- a)Orifice baffles are never used in a shell and tube heat exchanger.
- b)Pressure drop on the shell side of a heat exchanger depends upon tube pitch also.
- c)In a horizontal tube evaporator, surface blanketing by air is avoided.
- d)Split ring type and pull through type floating heads are two commonly used floating heads is heat exchangers.
Correct answer is option 'A'. Can you explain this answer?
Pick out the wrong statement.
a)
Orifice baffles are never used in a shell and tube heat exchanger.
b)
Pressure drop on the shell side of a heat exchanger depends upon tube pitch also.
c)
In a horizontal tube evaporator, surface blanketing by air is avoided.
d)
Split ring type and pull through type floating heads are two commonly used floating heads is heat exchangers.
For the radiation between two infinite parallel planes of emissivity respectively, which one of the following is the expression for emissivity factor or equivalent emissivity?- a)ε1ε2
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
For the radiation between two infinite parallel planes of emissivity respectively, which one of the following is the expression for emissivity factor or equivalent emissivity?
a)
ε1ε2
b)
c)
d)
Telecom Tuners answered • Feb 28, 2021The equivalent thermal resistance between two gray bodies is given by
Rth = 1-ε1/A1ε1+1/A1F12+1-ε2/A2ε2
Now for infinite parallel panes, F12 =1
A1 = A2 = A
For unit area, A = 1 m2
Rth = 1/ε1+1/ε2-1
Hence equivalent emissivity
Εe = 1/Rth
A metal storage tank with planar surfaces of external area contains water at and is located inside a large enclosure with walls at 65oC . The tank is initially painted black on the outside and its emissivity is approximated to 0.95. How much percentage reduction in heat loss would occur if the outside surface of tank is coated with aluminium paint (instead of black) of emissivity 0.55? Take radiation constantσb = 5.67 x 10-8 W/m2K4- a)58
- b)42
- c)29
- d)21
Correct answer is option 'A'. Can you explain this answer?
A metal storage tank with planar surfaces of external area contains water at and is located inside a large enclosure with walls at 65oC . The tank is initially painted black on the outside and its emissivity is approximated to 0.95. How much percentage reduction in heat loss would occur if the outside surface of tank is coated with aluminium paint (instead of black) of emissivity 0.55? Take radiation constant
σb = 5.67 x 10-8 W/m2K4
a)
58
b)
42
c)
29
d)
21
Telecom Tuners answered • Feb 28, 2021Consider outer surface of metal tank to be 1 & inner surface of enclosure to be surface 2.
By concept of thermal circuits
As the surfaces of 1 are planar, F11 = 0
∴ F21 = 1
Also ,
A2 >> A1
∴ (Q1➝2)net = σA1ε1(T14 - T24)
∴ Q ∝ ε1
Q0.55/Q0.95 = 0.55/0.95 = 0.4211
Thus heat exchange rate decreases by 57.89 %, by painting the tank with aluminium paint.
Radha Ganji asked • yesterdayAir initially at 101. 3 kPa and 40oC and with a relative humidity of 50%, is cooled at constant pressure to 30oC. The cooled air has a- a)higher dew point.
- b)higher absolute (specific) humidity.
- c)higher relative humidity
- d)higher wet bulb temperature.
Correct answer is option 'C'. Can you explain this answer?
Air initially at 101. 3 kPa and 40oC and with a relative humidity of 50%, is cooled at constant pressure to 30oC. The cooled air has a
a)
higher dew point.
b)
higher absolute (specific) humidity.
c)
higher relative humidity
d)
higher wet bulb temperature.
Sajan Kumar asked • 2 days agoPhenolic water generated in coke ovens & by-product plant attached to an integrated steel plant containing phenol in concentration of less than 100 mg/litre can be removed by- a)chlorination
- b)treating in biological oxygen pond
- c)chemical coagulation
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Phenolic water generated in coke ovens & by-product plant attached to an integrated steel plant containing phenol in concentration of less than 100 mg/litre can be removed by
a)
chlorination
b)
treating in biological oxygen pond
c)
chemical coagulation
d)
none of these
Shamaim Malik asked • 3 days agoIn the equation Q = UAΔt; Δt is- a)geometric mean temperature difference.
- b)arithmetic mean temperature difference.
- c)logarithmic mean temperature difference.
- d)the difference of average bulk temperatures of hot and cold fluids.
Correct answer is option 'C'. Can you explain this answer?
In the equation Q = UAΔt; Δt is
a)
geometric mean temperature difference.
b)
arithmetic mean temperature difference.
c)
logarithmic mean temperature difference.
d)
the difference of average bulk temperatures of hot and cold fluids.
Manish Swaraj asked • 3 days agoRelative humidity is the ratio of the- a)partial pressure of the vapour to the vapour pressure of the liquid at room temperature.
- b)partial pressure of the vapour to the vapour pressure of the liquid at gas temperature.
- c)actual humidity to saturation humidity.
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Relative humidity is the ratio of the
a)
partial pressure of the vapour to the vapour pressure of the liquid at room temperature.
b)
partial pressure of the vapour to the vapour pressure of the liquid at gas temperature.
c)
actual humidity to saturation humidity.
d)
none of these
Radha Ganji asked • 3 days agoA process stream of dilute aqueous solution flowing at the rate of10 Kg.s-1 is to be heated. Steam condensate at 95�C is available for heating purpose, also at a rate of 10 Kg.s-1. A 1 - 1 shell and tube heat exchanger is available. The best arrangement is- a)counter flow with process stream on shell side.
- b)counter flow with process stream on tube side.
- c)parallel flow with process stream on shell side.
- d)parallel flow with process stream on tube side.
Correct answer is option 'A'. Can you explain this answer?
A process stream of dilute aqueous solution flowing at the rate of10 Kg.s-1 is to be heated. Steam condensate at 95�C is available for heating purpose, also at a rate of 10 Kg.s-1. A 1 - 1 shell and tube heat exchanger is available. The best arrangement is
a)
counter flow with process stream on shell side.
b)
counter flow with process stream on tube side.
c)
parallel flow with process stream on shell side.
d)
parallel flow with process stream on tube side.
Radha Ganji asked • 3 days agoOverall efficiency of the distillation column is- a)the ratio of number of ideal plates to actual plates.
- b)the ratio of number of actual plates to ideal plates.
- c)same as the Murphree efficiency.
- d)always more than the point efficiency.
Correct answer is option 'A'. Can you explain this answer?
Overall efficiency of the distillation column is
a)
the ratio of number of ideal plates to actual plates.
b)
the ratio of number of actual plates to ideal plates.
c)
same as the Murphree efficiency.
d)
always more than the point efficiency.
Anshul Rasyotra asked • 4 days ago200 kg of solids (on dry basis) is subjected to a drying process for a period of 5000 seconds. The drying occurs in the constant rate period with the drying rate as, Nc = 0.5 x 10-3 kg/m2.s. The initial moisture content of the solid is 0.2 kg moisture/kg dry solid. The interfacial area available for drying is 4 m2/1000 kg of dry solid. The moisture content at the end of the drying period is (in kg moisture/kg dry solid)- a)0.5
- b)0.05
- c)0.1
- d)0.15
Correct answer is option 'C'. Can you explain this answer?
200 kg of solids (on dry basis) is subjected to a drying process for a period of 5000 seconds. The drying occurs in the constant rate period with the drying rate as, Nc = 0.5 x 10-3 kg/m2.s. The initial moisture content of the solid is 0.2 kg moisture/kg dry solid. The interfacial area available for drying is 4 m2/1000 kg of dry solid. The moisture content at the end of the drying period is (in kg moisture/kg dry solid)
a)
0.5
b)
0.05
c)
0.1
d)
0.15
Shafi Shaik asked • 4 days agoCompound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is- a)25%
- b)50%
- c)70%
- d)90%
Correct answer is option 'B'. Can you explain this answer?
Compound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is
a)
25%
b)
50%
c)
70%
d)
90%
Sana Shakoor asked • 4 days agoWhich is the best and the most effective method for the removal of organic contaminant present in the polluted water in very small quantity (say < 200 mg/litre)?- a)Lagooning
- b)Activated carbon adsorption
- c)Biological oxidation pond
- d)Chemical coagulation
Correct answer is option 'B'. Can you explain this answer?
Which is the best and the most effective method for the removal of organic contaminant present in the polluted water in very small quantity (say < 200 mg/litre)?
a)
Lagooning
b)
Activated carbon adsorption
c)
Biological oxidation pond
d)
Chemical coagulation
CHAMELI DEVI asked • 2 weeks agoAn oil cooler in a high performance engine has an outside surface area 0.12 m2 and a surface temperature of 65 degree Celsius. At any intermediate time air moves over the surface of the cooler at a temperature of 30 degree Celsius and gives rise to a surface coefficient equal to 45.4 W/ m 2 K. Find out the heat transfer rate?- a)238.43 W
- b)190.68 W
- c)543.67 W
- d)675.98 W
Correct answer is option 'B'. Can you explain this answer?
An oil cooler in a high performance engine has an outside surface area 0.12 m2 and a surface temperature of 65 degree Celsius. At any intermediate time air moves over the surface of the cooler at a temperature of 30 degree Celsius and gives rise to a surface coefficient equal to 45.4 W/ m 2 K. Find out the heat transfer rate?
a)
238.43 W
b)
190.68 W
c)
543.67 W
d)
675.98 W
Krish Jat asked • 2 weeks agoA forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)?
Related: Evaporators (Part - 1)
SHUBHAM SINGH asked • Apr 11, 2021For an air(A)-water vapour(B) mixture, the partial pressure of the air is 5 pa and the total pressure of the system is 15 pa. The absolute humidity in mass of air/mass of water vapour is- a)0.805
- b)0.311
- c)0.705
- d)0.411
Correct answer is option 'A'. Can you explain this answer?
For an air(A)-water vapour(B) mixture, the partial pressure of the air is 5 pa and the total pressure of the system is 15 pa. The absolute humidity in mass of air/mass of water vapour is
a)
0.805
b)
0.311
c)
0.705
d)
0.411
ADITYA LAMBA asked • Mar 30, 2021Of the radiant energy 350W/m2 incident upon a surface 250W/m2 is absorbed, 60W/m2 is reflected and the remainder is transmitted through the surface. Workout the value for absorptivity for the surface material- a)0.113
- b)0.114
- c)0.115
- d)0.116
Correct answer is option 'C'. Can you explain this answer?
Of the radiant energy 350W/m2 incident upon a surface 250W/m2 is absorbed, 60W/m2 is reflected and the remainder is transmitted through the surface. Workout the value for absorptivity for the surface material
a)
0.113
b)
0.114
c)
0.115
d)
0.116
KANISHK DESHWAL asked • Mar 26, 2021A gas mixture containing 21% O2 and 79% N2 flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2 is 0.04m/s and N2 is 0.035m/s.Q. What will be the value of mass average velocity (m/s)?- a)0.981
- b)0.891
- c)0.985
- d)0.857
Correct answer is option 'A'. Can you explain this answer?
A gas mixture containing 21% O2 and 79% N2 flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2 is 0.04m/s and N2 is 0.035m/s.
Q. What will be the value of mass average velocity (m/s)?
a)
0.981
b)
0.891
c)
0.985
d)
0.857
TAMANNA RAVINDER asked • Mar 26, 2021For a component A of a mixture, concentration as a function of x is given:
CA=5e-10x (x is in cm and CA in mol/liter)
Calculate the value of diffusion velocity (m/s)of the component A at the point x=0, if diffusivity of A in the mixture is 2.567*10-5m2/s. - a)0.2567
- b)2.567
- c)0.0025
- d)3.541
Correct answer is option 'A'. Can you explain this answer?
For a component A of a mixture, concentration as a function of x is given:
CA=5e-10x (x is in cm and CA in mol/liter)
Calculate the value of diffusion velocity (m/s)of the component A at the point x=0, if diffusivity of A in the mixture is 2.567*10-5m2/s.
CA=5e-10x (x is in cm and CA in mol/liter)
Calculate the value of diffusion velocity (m/s)of the component A at the point x=0, if diffusivity of A in the mixture is 2.567*10-5m2/s.
a)
0.2567
b)
2.567
c)
0.0025
d)
3.541
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