We have defined EER by a) BTU/hr/W c) BTU/min/W b) BTU/sec/W d) BTU/hr/V

We have defined EER by
a) BTU/hr/W c) BTU/min/W
b) BTU/sec/W d) BTU/hr/V
by

1 Answer

Answer :

BTU/hr/W
by

Related questions

The relation between EER in BTU/hr/W EER in W/W is given as_________. A. EER in BTU/hr/W = 3.245 EER in W/W. B. EER in BTU/hr/W = 3.44 EER in W/W. C. EER in BTU/hr/W = 3.413 EER in W/W. D. EER in BTU/hr/W = 1.234 EER in W/W.

Description : The relation between EER in BTU/hr/W EER in W/W is given as_________. A. EER in BTU/hr/W = 3.245 EER in W/W. B. EER in BTU/hr/W = 3.44 EER in W/W. C. EER in BTU/hr/W = 3.413 EER in W/W. D. EER in BTU/hr/W = 1.234 EER in W/W.

Last Answer : EER in BTU/hr/W = 3.413 EER in W/W.

EER in BTU/hr/W = a) 3.222 * EER in W/W c) 3.413 * EER in W/W b) 3.234 * EER in W/W d) 3.400 * EER in W/W

Description : EER in BTU/hr/W = a) 3.222 * EER in W/W c) 3.413 * EER in W/W b) 3.234 * EER in W/W d) 3.400 * EER in W/W

Last Answer : 3.413 * EER in W/W

EER in BTU/HR/W = ______ X EER in W/W A. 3.410 C. 3.411 B. 3.412 D. 3.413

Description : EER in BTU/HR/W = ______ X EER in W/W A. 3.410 C. 3.411 B. 3.412 D. 3.413

Last Answer : 3.413

nergy Efficiency rating in BTU/hr/W is equal to a) 4.413 W/W b) 2.413 W/W c) 3.413 W/W

Description : Energy Efficiency rating in BTU/hr/W is equal to a) 4.413 W/W b) 2.413 W/W c) 3.413 W/W

Last Answer : 3.413w/w

Theoretical capacity of crushing rolls in tons/hr is given by (where, V = peripheral velocity, m/sec. W = width of rolls, m Dr = distance between rolls ρ = density of material to be crushed, kg/m3 here, V =πND where, N = speed of the rolls in rotation per second (rps) D = diameter of rolls, m). (A) 3.6 V.W.Dr.ρ (B) 3.6 V.W. ρ (C) 3.6 W.Dr.ρ (D) 3.6 V.W.Dr/ρ

Description : Theoretical capacity of crushing rolls in tons/hr is given by (where, V = peripheral velocity, m/sec. W = width of rolls, m Dr = distance between rolls ρ = density of material to be crushed, kg/m3 here, V =πND where, N = speed of the rolls ... V.W.Dr.ρ (B) 3.6 V.W. ρ (C) 3.6 W.Dr.ρ (D) 3.6 V.W.Dr/ρ

Last Answer : (A) 3.6 V.W.Dr.ρ

Overall heat transfer co-efficient for cooling of hydrocarbons by water is about (A) 50 -100 Kcal/hr.m2 .°C (B) 50 -100 W/m2 .°K (C) 50 -100 BTU/hr. ft. 2°F (D) 1000 - 1500 BTU/hr. ft. 2°F

Description : Overall heat transfer co-efficient for cooling of hydrocarbons by water is about (A) 50 -100 Kcal/hr.m2 .°C (B) 50 -100 W/m2 .°K (C) 50 -100 BTU/hr. ft. 2°F (D) 1000 - 1500 BTU/hr. ft. 2°F

Last Answer : (C) 50 -100 BTU/hr. ft. 2°F

What is defined as the quantity of heat needed to raise the temperature of one kilogram of water by one degrees Celsius?  a. Kilojoule  b. Btu  c. Kilocalorie  d. Latent heat

Description : What is defined as the quantity of heat needed to raise the temperature of one kilogram of water by one degrees Celsius?  a. Kilojoule  b. Btu  c. Kilocalorie  d. Latent heat

Last Answer : Kilocalorie

Find ∫ for steam at 100 psia and 600°F.If h = 1329.6 and v = 6.216  a. 1214 Btu / lb  b. 1234 Btu /lb  c. 1342 Btu / lb  d. 1324 Btu /lb formula: ∫ = h– pv/ J

Description : Find ∫ for steam at 100 psia and 600°F.If h = 1329.6 and v = 6.216  a. 1214 Btu / lb  b. 1234 Btu /lb  c. 1342 Btu / lb  d. 1324 Btu /lb formula: ∫ = h– pv/ J

Last Answer : 1214Btu / lb

Minimum EER of 5-star rating is a. 3.50 c) 3.99 b. 4.00 d) 4.50

Description : Minimum EER of 5-star rating is a. 3.50 c) 3.99 b. 4.00 d) 4.50

Last Answer : 4.50

Higher the number of stars, ________ will be the efficiency or EER a) Lesser c) Greater b) Likewise e) average

Description : Higher the number of stars, ________ will be the efficiency or EER a) Lesser c) Greater b) Likewise e) average

Last Answer : Greater

Full form of EER is _____ A. Energy Electric Ratio C. Energy Efficiency Ratio B. Energy Effective Ratio D. Energy Electron Ratio

Description : Full form of EER is _____ A. Energy Electric Ratio C. Energy Efficiency Ratio B. Energy Effective Ratio D. Energy Electron Ratio

Last Answer : Energy Efficiency Ratio

Higher the number stars, EER or efficiency will be. A. Greater C. Lesser B. Moderate D. None of the above

Description : Higher the number stars, EER or efficiency will be. A. Greater C. Lesser B. Moderate D. None of the above

Last Answer : Greater

___________ is the minimum EER for 5-star in India in 2019. A) 4.50 C) 4.00 B) 7.10 D) 4.49

Description : ___________ is the minimum EER for 5-star in India in 2019. A) 4.50 C) 4.00 B) 7.10 D) 4.49

Last Answer : 4.50

The ratio of the total amount of heat that equipment can remove from the indoors to travel and amount of energy consumed by the equipment is known as A) Energy performance ratio (EPR). C) Energy efficiency RATIO (EER). B) Energy consumption ratio (ECR). D) None of the above

Description : . The ratio of the total amount of heat that equipment can remove from the indoors to travel and amount of energy consumed by the equipment is known as A) Energy performance ratio (EPR). C) Energy efficiency RATIO (EER). B) Energy consumption ratio (ECR). D) None of the above

Last Answer : Energy efficiency RATIO (EER).

Which one is the correct relation between energy efficiency ratio (EER) and coefficient of performance (COP)?  A. EER = 2.34 COP  B. EER = 3.24 COP  C. EER = 3.42 COP  D. EER = 4.23 COP

Description : Which one is the correct relation between energy efficiency ratio (EER) and coefficient of performance (COP)?  A. EER = 2.34 COP  B. EER = 3.24 COP  C. EER = 3.42 COP  D. EER = 4.23 COP

Last Answer : EER = 3.42 COP

Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Description : Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Last Answer : -364 BTU/lbm

The capacity of a classifier in 'tons of solid/hr' is given by (where, A = cross-sectional area in m2 , V = rising velocity of fluid in m/sec, S = percentage of solids in the suspension by volume, ρ = density of solids in kg/m3 ) (A) 3.6 AVS.ρ (B) 3.6 A.V.ρ (C) 3.6 A.S. ρ (D) 3.6 AVS/ρ

Description : The capacity of a classifier in 'tons of solid/hr' is given by (where, A = cross-sectional area in m2 , V = rising velocity of fluid in m/sec, S = percentage of solids in the suspension by volume, ρ = density of solids in kg/m3 ) (A) 3.6 AVS.ρ (B) 3.6 A.V.ρ (C) 3.6 A.S. ρ (D) 3.6 AVS/ρ

Last Answer : (A) 3.6 AVS.ρ

1 BTU/hr.ft.°F is equal to __________ kcal/hr. m.°C. (A) 1.49 (B) 1 (C) 4.88 (D) None of these

Description : 1 BTU/hr.ft.°F is equal to __________ kcal/hr. m.°C. (A) 1.49 (B) 1 (C) 4.88 (D) None of these

Last Answer : (A) 1.49

The unit of heat transfer co-efficient is (A) BTU/hr. ft 2°F (B) BTU/hr. °F. ft (C) BTU/hr. °F (D) BTU/hr. ft

Description : The unit of heat transfer co-efficient is (A) BTU/hr. ft 2°F (B) BTU/hr. °F. ft (C) BTU/hr. °F (D) BTU/hr. ft

Last Answer : (A) BTU/hr. ft 2°F

A BTU/hr.ft. 2 °F is equal to (A) 1 kcal/hr. m2°C (B) 4.88 kcal/hr. m. 2°C (C) 1 kcal/hr. m2 .°K (D) None of these

Description : A BTU/hr.ft. 2 °F is equal to (A) 1 kcal/hr. m2°C (B) 4.88 kcal/hr. m. 2°C (C) 1 kcal/hr. m2 .°K (D) None of these

Last Answer : (B) 4.88 kcal/hr. m. 2°C

Pick out the correct statement. (A) 1 kcal/hr.m.°C is equal to 1 BTU/hr. ft.°F (B) In steady state heat conduction, the only property of the substance which determines the temperature distribution, is the thermal conductivity (C) In unsteady state heat conduction, heat flows in the direction of temperature rise (D) In heat transfer by forced convection, Grashoff number is very important

Description : Pick out the correct statement. (A) 1 kcal/hr.m.°C is equal to 1 BTU/hr. ft.°F (B) In steady state heat conduction, the only property of the substance which determines the ... in the direction of temperature rise (D) In heat transfer by forced convection, Grashoff number is very important

Last Answer : (B) In steady state heat conduction, the only property of the substance which determines the temperature distribution, is the thermal conductivity

One ton of refrigeration capacity is equivalent to the heat removal rate of (A) 50 kcal/hr (B) 200 BTU/hr (C) 200 BTU/minute

Description : One ton of refrigeration capacity is equivalent to the heat removal rate of (A) 50 kcal/hr (B) 200 BTU/hr (C) 200 BTU/minute

Last Answer : (C) 200 BTU/minute

One ton of refrigeration is not equivalent to the heat removal rate of (A) 200 BTU/minute (B) 50 kcal/minute (C) 50 kJ/sec (D) 3.5 KW

Description : One ton of refrigeration is not equivalent to the heat removal rate of (A) 200 BTU/minute (B) 50 kcal/minute (C) 50 kJ/sec (D) 3.5 KW

Last Answer : (C) 50 kJ/sec

In the lathe it is the peripheral speed of the work piece cutting speed of tool is expressed inm/min, V is a.= 𝜋𝐷𝑁/1000 b.= 𝜋𝐷/1000 c.= 𝐷𝑁/1000 d.=𝜋𝐷𝑁/10

Description : In the lathe it is the peripheral speed of the work piece cutting speed of tool is expressed inm/min, V is a.= 𝜋𝐷𝑁/1000 b.= 𝜋𝐷/1000 c.= 𝐷𝑁/1000 d.=𝜋𝐷𝑁/10

Last Answer : a.= 𝜋𝐷𝑁/1000

Calculate the recoverable waste heat (Q, in kCal/hour) from flue gases using the followingparameters: V (flow rate of the substance) 2000 m3/hr r (density of the flue gas): 0.9 kg/m3 Cp (specific heat of the substance): 0.20 kCal/kg oC DT (temperature difference): 120 oC h (recovery factor): 50% a. 21600 b. 43200 c. 25600 d. 34000

Description : Calculate the recoverable waste heat (Q, in kCal/hour) from flue gases using the followingparameters: V (flow rate of the substance) 2000 m3/hr r (density of the flue gas): 0.9 kg/m3 Cp (specific heat ... (temperature difference): 120 oC h (recovery factor): 50% a. 21600 b. 43200 c. 25600 d. 34000

Last Answer : 21600

Water is flowing through a 1 foot diameter pipe at the rate of 10ft/sec. What is the volume flow rate of water in ft³/sec?  a. 7.85  b. 6.85  c. 8.85  d. 5.85 V = Aν

Description : Water is flowing through a 1 foot diameter pipe at the rate of 10ft/sec. What is the volume flow rate of water in ft³/sec?  a. 7.85  b. 6.85  c. 8.85  d. 5.85 V = Aν

Last Answer : 7.85

What Is The Meaning Of Btu In Air Conditioners?

Description : What Is The Meaning Of Btu In Air Conditioners?

Last Answer : Btu - British Thermal Unit (Btu) is the international measure of energy. A Btu is the amount of heat needed to raise 1 (one) pound of water by 1(one) degree Fahrenheit. In HVAC industry, Btu's ... heat a conditioning unit can remove from a room per hours. One BTU per hour is equal to 0293 watts.

What Is Btu?

Description : What Is Btu?

Last Answer : The British thermal unit (Btu or BTU) is a traditional unit of heat; it is defined as the amount of heat required to raise the temperature of one pound of water by one degree Fahrenheit. ... Heat is ... to be equivalent to energy, for which the metric unit is the joule; one BTU is about 1055 joules.

78 India‘s total primary energy consumption is a) 24.3 BTU b) 19.01 BTU c) 120 BTU d) 30.1 BTU

Description : 78 India‘s total primary energy consumption is a) 24.3 BTU b) 19.01 BTU c) 120 BTU d) 30.1 BTU

Last Answer : 30.1 BTU

The mechanical equivalent of heat is  a. joule  b. calorie  c. Btu  d. Specific heat

Description : The mechanical equivalent of heat is  a. joule  b. calorie  c. Btu  d. Specific heat

Last Answer : joule

A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. T2 = 2.620ºR, ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

Description : A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. ... , ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

Last Answer : T2=1.620ºR, ∫H = 122.83 Btu

Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Description : Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Last Answer : 15.7 BTU

Calculate the entropy of steam at 60psiawith a quality of 0.8  A. 0.4274 BTU/lbm-˚R  B. 0.7303 BTU/lbm-˚R  C. 1.1577 BTU/lbm-˚R  D. 1.2172 BTU/lbm-˚R Formula: fromthe steamtable at 60 psia: sƒ = 0.4274 BTU/lbm-˚R sƒg = 1.2172 BTU/lbm-˚R) s = sƒ + x sƒg where x = is the quality

Description : Calculate the entropy of steam at 60psiawith a quality of 0.8  A. 0.4274 BTU/lbm-˚R  B. 0.7303 BTU/lbm-˚R  C. 1.1577 BTU/lbm-˚R  D. 1.2172 BTU/lbm-˚R Formula: fromthe steamtable at 60 psia: sƒ = 0.4274 BTU/lbm-˚R sƒg = 1.2172 BTU/lbm-˚R) s = sƒ + x sƒg where x = is the quality

Last Answer : 1.1577 BTU/lbm-˚R

_________ is the quantity of heat required to raise the temperature of one pound of water 1 F˚ from 63 ˚F to 64 ˚F.  a. one Joule  b. one calorie  c. one watt  d. one BTU

Description : _________ is the quantity of heat required to raise the temperature of one pound of water 1 F˚ from 63 ˚F to 64 ˚F.  a. one Joule  b. one calorie  c. one watt  d. one BTU

Last Answer : one BTU

1 British thermal unit (BTU) is equivalent to how many joules?  A. 1016  B. 1043  C. 1023  D. 1054

Description : 1 British thermal unit (BTU) is equivalent to how many joules?  A. 1016  B. 1043  C. 1023  D. 1054

Last Answer : 1054

The amount of heat required to raise the temperature of 1 pound of water by 1°F is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Description : The amount of heat required to raise the temperature of 1 pound of water by 1°F is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Last Answer : BTU

The amount of heat required to raise the temperature of 1kg of water through 1 °C is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Description : The amount of heat required to raise the temperature of 1kg of water through 1 °C is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Last Answer : Kilocalorie

What is the SI unit of energy?  A. Newton  B. Btu  C. Calorie  D. Joule

Description : What is the SI unit of energy?  A. Newton  B. Btu  C. Calorie  D. Joule

Last Answer : Joule

The normal glucose tolerance curve reaches peak is (A) 15 min (B) 1 hr (C) 2 hrs (D) 2 ½ hrs

Description : The normal glucose tolerance curve reaches peak is (A) 15 min (B) 1 hr (C) 2 hrs (D) 2 ½ hrs

Last Answer : B

One tonrefrigeration corresponds to (a) 50 kcal/min (b) 50 kcal/kr (c) 80 kcal/min (d) 80 kcal/hr

Description : One tonrefrigeration corresponds to (a) 50 kcal/min (b) 50 kcal/kr (c) 80 kcal/min (d) 80 kcal/hr

Last Answer : Ans: a

What is the Time Scale for Meso Climate? [a] 1-10 hr [b] 1-10 min [c] 1-10 seconds [d] 1-10 days [e] None of the above

Description : What is the Time Scale for Meso Climate? [a] 1-10 hr [b] 1-10 min [c] 1-10 seconds [d] 1-10 days [e] None of the above

Last Answer : [d] 1-10 days

A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time and distance to PSR is: a. 1hr 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Description : A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time ... 1hr 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Last Answer : d. 1hr 16 min, 616nm

A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time and distance to PSR is: a. 1hr 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Description : A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time ... 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Last Answer : d. 1hr 16 min, 616nm

A train M speeding with 60km/hr crooses another train N, running in the same direction in 1 min. If the lengths of the trains M and N be 50m and 100m respectively. What is the speed of train N? A) 15 km/hr B) 61km/hr C) 51km/hr D) 50k/hr

Description : A train M speeding with 60km/hr crooses another train N, running in the same direction in 1 min. If the lengths of the trains M and N be 50m and 100m respectively. What is the speed of train N? A) 15 km/hr B) 61km/hr C) 51km/hr D) 50k/hr

Last Answer : ANSWER: C Explanation:  Let the speed of be train N be ‘x’km/hr  Speed of M relative to N = (60 – x)km/hr  = (60-x) * 5/18]m/s = (300 – 5x / 18)m/s  150 / (300 – 5x / 18) =60  2700/300-5x=60  2700=18000-300x 300x=15300  x= 153/3  x=51km/hr  Therefore the speed of the train is 51 km/hr

Dhanvanth can swim at 15 km/hr in stagnant water. In a river with 3 km/hr current, he swims to a certain distance and comes back within 100 min. What is the distance between the two points? A) 11km B) 13km C) 19km D) 12km

Description : Dhanvanth can swim at 15 km/hr in stagnant water. In a river with 3 km/hr current, he swims to a certain distance and comes back within 100 min. What is the distance between the two points? A) 11km B) 13km C) 19km D) 12km

Last Answer : ANSWER: D Explanation: Speed in still water (a) = 15 km/hr Speed in current( b) = 3 km/hr Upstream speed = a - b = 15 - 3 = 12 km/hr Downstream speed = a + b = 15 + 3 = 18 km/hr Let the ... 36 = 5 / 3 15x = 36 * 5 x = 36 * 5 / 15 = 12 km Distance between the two points is 12km

How do you convert 36 cm per sec to m per min?

Description : How do you convert 36 cm per sec to m per min?

Last Answer : To convert cm to m divide by 100To convert s to min divide by 60→ 36 cm/s = 36 × (1 cm)/(1 s)= 36 × (1 ÷ 100 m )/(1 ÷ 60 min)= 36 × (1/100 m ÷ 1/60 min)= 36 × 60/100 m/min= 21.6 m/min

How do you convert 36 cm per sec to m per min?

Description : How do you convert 36 cm per sec to m per min?

Last Answer : To convert cm to m divide by 100To convert s to min divide by 60→ 36 cm/s = 36 × (1 cm)/(1 s)= 36 × (1 ÷ 100 m )/(1 ÷ 60 min)= 36 × (1/100 m ÷ 1/60 min)= 36 × 60/100 m/min= 21.6 m/min

The earth completes one rotation on its axis in : (1) 23 hrs. 56 min. 4.9 sec (2) 23 hrs. 10 min. 2 sec (3) 23 hrs. 30 min. (4) 24 hrs.

Description : The earth completes one rotation on its axis in : (1) 23 hrs. 56 min. 4.9 sec (2) 23 hrs. 10 min. 2 sec (3) 23 hrs. 30 min. (4) 24 hrs.

Last Answer : (1) 23 hrs. 56 min. 4.9 sec Explanation: The Earth rotates around its axis once in about 24 hours with respect to the sun and once every 23 hours 56 minutes and 4 seconds with respect to the stars. The Earth rotates from the west towards east. Seasons are caused by Earth's rotation around its axis.