The unit of gcis (A) lbm/lbf. ft/sec2 (B) lbf/lbm. ft/sec2 (C) ft/sec2 (D) lbm/lbf. sec2/ft

The unit of gcis
(A) lbm/lbf. ft/sec2
(B) lbf/lbm. ft/sec2
(C) ft/sec2
(D) lbm/lbf. sec2/ft
by

1 Answer

Answer :

(A) lbm/lbf. ft/sec2
by

Related questions

3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Description : 3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Last Answer : 24.9 ft^3

Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Description : Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Last Answer : 3300 lbf-ft/lbm

If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Description : If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Last Answer : 13.33 ft^3/lbm

The unit of velocity head is (A) ft-lb/sec (B) ft-lb/ft 3 (C) ft-lbf/lb (D) ft-lbf/sec

Description : The unit of velocity head is (A) ft-lb/sec (B) ft-lb/ft 3 (C) ft-lbf/lb (D) ft-lbf/sec

Last Answer : (C) ft-lbf/lb

One horsepower is equal to (A) 550 lbf.ft/second (B) 550 kgf.m/second (C) Both (A) and (B) (D) 550 lbf.ft./h

Description : One horsepower is equal to (A) 550 lbf.ft/second (B) 550 kgf.m/second (C) Both (A) and (B) (D) 550 lbf.ft./h

Last Answer : (A) 550 lbf.ft/second

The weight of an object is 50lb. What is its mass at standard condition?  a. 50 lbm  b. 60 lbm  c. 70 lbm  d. 80 lbm formula: m = Fgk /g

Description : The weight of an object is 50lb. What is its mass at standard condition?  a. 50 lbm  b. 60 lbm  c. 70 lbm  d. 80 lbm formula: m = Fgk /g

Last Answer : 50 lbm

What is the weight of a66-kgm man at standard condition? (Formula: Fg= mg / k)  a. 66 kgf  b. 66 kgm  c. 66 lbm  d. 66 gf

Description : What is the weight of a66-kgm man at standard condition? (Formula: Fg= mg / k)  a. 66 kgf  b. 66 kgm  c. 66 lbm  d. 66 gf

Last Answer : 66 kgf

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

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

A particle starts from rest and moves in a straight line whose equation of motion is given by S = 2t3 - t2 - 1. The acceleration of the particle after one sec will be a.6 m/sec2 b.4 m/sec2 c.12 m/sec2 d.10 m/sec2 e.8 m/sec2

Description : A particle starts from rest and moves in a straight line whose equation of motion is given by S = 2t3 - t2 - 1. The acceleration of the particle after one sec will be a.6 m/sec2 b.4 m/sec2 c.12 m/sec2 d.10 m/sec2 e.8 m/sec2

Last Answer : d. 10 m/sec2

The angular acceleration of a flywheel decreases uniformly from 8 rad/sec2 to 2 rad sec2 in 6 sec at which time its angular velocity is 42 rad/sec. The initial angular velocity was a.12 rad/sec b.24 rad/sec c.30 rad/sec d.3 rad/sec e.6 rad/sec

Description : The angular acceleration of a flywheel decreases uniformly from 8 rad/sec2 to 2 rad sec2 in 6 sec at which time its angular velocity is 42 rad/sec. The initial angular velocity was a.12 rad/sec b.24 rad/sec c.30 rad/sec d.3 rad/sec e.6 rad/sec

Last Answer : a. 12 rad/sec

A man pulls a cart with 100 N force from rest with an acceleration of 10 m/sec2 for 10 seconds. The work done by him is a.500 KJ b.50.96 KJ c.101.92 KJ d.None of the above e.Tapered bearing

Description : A man pulls a cart with 100 N force from rest with an acceleration of 10 m/sec2 for 10 seconds. The work done by him is a.500 KJ b.50.96 KJ c.101.92 KJ d.None of the above e.Tapered bearing

Last Answer : a. 500 KJ

The number of revolutions through which a pulley will rotate from rest if its angular acceleration is increased uniformly from zero to 12 rad per sec2 during 4 sec and then uniformly decreased to 0 rad per sec2 during the next 3 sec would be a.46.4 b.26.74 c.11.6 d.3.2 e.17.7

Description : The number of revolutions through which a pulley will rotate from rest if its angular acceleration is increased uniformly from zero to 12 rad per sec2 during 4 sec and then uniformly decreased to 0 rad per sec2 during the next 3 sec would be a.46.4 b.26.74 c.11.6 d.3.2 e.17.7

Last Answer : b. 26.74

A couple applied to a flywheel of mass 5 kg and of radius of gyration 20 cm produces an angular acceleration of 4 radians/sec2. The torque applied in dyne cm is a.8 x 104 b.8 x 106 c.8 x 107 d.8 x 108 e.8 x 1010

Description : A couple applied to a flywheel of mass 5 kg and of radius of gyration 20 cm produces an angular acceleration of 4 radians/sec2. The torque applied in dyne cm is a.8 x 104 b.8 x 106 c.8 x 107 d.8 x 108 e.8 x 1010

Last Answer : b. 8 x 106

The force which produces an acceleration of 1 m/sec2 in a mass of 1 kg is called a.100 ergs b.one Newton c.one watt d.One erg e.one joule

Description : The force which produces an acceleration of 1 m/sec2 in a mass of 1 kg is called a.100 ergs b.one Newton c.one watt d.One erg e.one joule

Last Answer : b. one Newton

A body whose true weight is 14 kg appears to weigh 15 kg when weighed with the help of a spring balance in a moving lift. The acceleration of the lift at the time was a.1 m/sec2 b.0.7 m/sec2 c.0.5 m/sec2 d.1.4 m/sec2 e.0.35 m/sec2

Description : A body whose true weight is 14 kg appears to weigh 15 kg when weighed with the help of a spring balance in a moving lift. The acceleration of the lift at the time was a.1 m/sec2 b.0.7 m/sec2 c.0.5 m/sec2 d.1.4 m/sec2 e.0.35 m/sec2

Last Answer : b. 0.7 m/sec2

The value of acceleration due to gravity is 980 cm/sec2. Its value in meter per (hour)2 will be a.1.27 x 108 b.127 c.1270 d.1.27 x 107 e.1270000

Description : The value of acceleration due to gravity is 980 cm/sec2. Its value in meter per (hour)2 will be a.1.27 x 108 b.127 c.1270 d.1.27 x 107 e.1270000

Last Answer : a. 1.27 x 108

In which case power required would be more a.In lifting up a weight of 100 kg at a velocity of 3 m/sec but accelerating upwards at 0.5 m/sec b.In lowering a weight of 100 kg along an a inclined plane at an angle of 45? with c.In lifting up a weight of 100 kg at a constant speed of 3 m/sec d.In lowering the weight at a constant speed of 3 m/sec e.In lowering the weight at a velocity of 3 m/sec and acceleration of 1 m/sec2

Description : In which case power required would be more a.In lifting up a weight of 100 kg at a velocity of 3 m/sec but accelerating upwards at 0.5 m/sec b.In lowering a weight of 100 kg along an a inclined plane ... speed of 3 m/sec e.In lowering the weight at a velocity of 3 m/sec and acceleration of 1 m/sec2

Last Answer : a. In lifting up a weight of 100 kg at a velocity of 3 m/sec but accelerating upwards at 0.5 m/sec

What is the total acceleration of a point on the rim of a flywheel 4 mts in diameter 0.6 see after it has started from rest if it accelarates uniformly from rest to 2000 rpm in 20 secs. a.68.9 mts/sec2 b.102.6 mts/sec2 c.78.94 mts/sec2 d.107 dynes e.33.2 mts/sec2

Description : What is the total acceleration of a point on the rim of a flywheel 4 mts in diameter 0.6 see after it has started from rest if it accelarates uniformly from rest to 2000 rpm in 20 secs. a.68.9 mts/sec2 b.102.6 mts/sec2 c.78.94 mts/sec2 d.107 dynes e.33.2 mts/sec2

Last Answer : c. 78.94 mts/sec2

A jet of water of cross-sectional area 20 sq. cm impinges on a plate at an angle of 60? with a velocity of 10 m/sec. Neglecting the friction between the jet and plate (density of water 1000/9.81 kgm-1 sec2 per cubic meter), the force exerted by the jet on the plate would be in the range a.15 - 20 kg b.20 - 25 kg c.5 - 10 kg d.25 - 30 kg e.10 - 15 kg

Description : A jet of water of cross-sectional area 20 sq. cm impinges on a plate at an angle of 60? with a velocity of 10 m/sec. Neglecting the friction between the jet and plate (density of water 1000/9.81 kgm-1 sec2 per cubic meter), the ... a.15 - 20 kg b.20 - 25 kg c.5 - 10 kg d.25 - 30 kg e.10 - 15 kg

Last Answer : a. 15 - 20 kg

A wheel accelerates from rest with an acceleration of 6 rad/sec2. How many revolutions are experienced in 4 seconds? a.12.36 b.25.6 c.9.82 d.107 dynes e.7.64

Description : A wheel accelerates from rest with an acceleration of 6 rad/sec2. How many revolutions are experienced in 4 seconds? a.12.36 b.25.6 c.9.82 d.107 dynes e.7.64

Last Answer : e. 7.64

A crate weighing 40 kg rests on a cart moving with an acceleration W. The coefficient of friction between the cart and the crate is a.2.45 m/sec2 b.4.00 m/sec2 c.0.25 m/sec2 d.4.16 m/sec2 e.1.25 m/sec2

Description : A crate weighing 40 kg rests on a cart moving with an acceleration W. The coefficient of friction between the cart and the crate is a.2.45 m/sec2 b.4.00 m/sec2 c.0.25 m/sec2 d.4.16 m/sec2 e.1.25 m/sec2

Last Answer : a. 2.45 m/sec2

Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Description : Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Last Answer : 18 kg

A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

Description : A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

Last Answer : 625 fpm

The SI unit of Planck’s constant is a) Joule – Sec-1 b) Joule – Sec-2 c) Joule – Sec d) Joule – Sec2

Description : The SI unit of Planck’s constant is a) Joule – Sec-1 b) Joule – Sec-2 c) Joule – Sec d) Joule – Sec2

Last Answer : c) Joule – Sec

Due to scan geometry of a satellite sensor: (A) The off-nadir resolution is degraded (B) The ground distance swept by the senor, IFOV is proportional to sec2 of scan measured from the nadir (C) The details towards the edge of the scan get compressed (D) All of these

Description : Due to scan geometry of a satellite sensor: (A) The off-nadir resolution is degraded (B) The ground distance swept by the senor, IFOV is proportional to sec2 of scan measured from the nadir (C) The details towards the edge of the scan get compressed (D) All of these

Last Answer : Answer: Option D

The maximum comfortable retardation applied to moving vehicles, is (A) 3.42 m/sec2 (B) 4.42 m/sec2 (C) 5.56 m/sec2 (D) 7.80 m/sec2

Description : The maximum comfortable retardation applied to moving vehicles, is (A) 3.42 m/sec2 (B) 4.42 m/sec2 (C) 5.56 m/sec2 (D) 7.80 m/sec2

Last Answer : Answer: Option A

The total length of a valley formed by two gradients - 3% and + 2% curve between the two tangent points to provide a rate of change of centrifugal acceleration 0.6 m/sec2 , for a design speed 100 kmph, is (A) 16.0 m (B) 42.3 m (C) 84.6 m (D) None of these

Description : The total length of a valley formed by two gradients - 3% and + 2% curve between the two tangent points to provide a rate of change of centrifugal acceleration 0.6 m/sec2 , for a design speed 100 kmph, is (A) 16.0 m (B) 42.3 m (C) 84.6 m (D) None of these

Last Answer : Answer: Option C

Over taking time required for a vehicle with design speed 50 km ph and overtaking acceleration 1.25 m/sec2 to overtake a vehicle moving at a speed 30 km ph, is (A) 5.0 secs (B) 6.12 secs (C) 225.48 secs (D) 30 secs

Description : Over taking time required for a vehicle with design speed 50 km ph and overtaking acceleration 1.25 m/sec2 to overtake a vehicle moving at a speed 30 km ph, is (A) 5.0 secs (B) 6.12 secs (C) 225.48 secs (D) 30 secs

Last Answer : Answer: Option B

An automobile weighing 2800 lb and travelling at 30 miles/hr, hits a depression in the road which has a radius of curvature of 50 ft. What is the total force to which springs are subjected. a.8230 lb b.6170 lb c.2800 ib d.107 dynes e.3280 lb

Description : An automobile weighing 2800 lb and travelling at 30 miles/hr, hits a depression in the road which has a radius of curvature of 50 ft. What is the total force to which springs are subjected. a.8230 lb b.6170 lb c.2800 ib d.107 dynes e.3280 lb

Last Answer : b. 6170 lb

A bullet is fired vertically upwards with a velocity of 2200 ft/s. Bullet ascends to a height of a.56,900 ft b.107 dynes c.75,000 ft d.48.300 ft e.65,800 ft

Description : A bullet is fired vertically upwards with a velocity of 2200 ft/s. Bullet ascends to a height of a.56,900 ft b.107 dynes c.75,000 ft d.48.300 ft e.65,800 ft

Last Answer : c. 75,000 ft

If the above tape weights 0.001 kg/ft, sag is a.2.73 ft b.3.13 ft c.2.90 ft d.3.03 ft e.107 dynes

Description : If the above tape weights 0.001 kg/ft, sag is a.2.73 ft b.3.13 ft c.2.90 ft d.3.03 ft e.107 dynes

Last Answer : c. 2.90 ft

The sag in a tape 100 ft long held between two supports 99.8 ft aprat at the same level is a.3.32 ft b.0.98 ft c.2.73 ft d.1.87 ft e.107 dynes

Description : The sag in a tape 100 ft long held between two supports 99.8 ft aprat at the same level is a.3.32 ft b.0.98 ft c.2.73 ft d.1.87 ft e.107 dynes

Last Answer : c. 2.73 ft

If span is 150 mts and sag is 10 ft of a cable carrying a load of 3200 kg per horizontal metre then tension of ends is a.104,000 kg b.107 dynes c.1200,000 kg d.932,000 kg e.584,000 kg

Description : If span is 150 mts and sag is 10 ft of a cable carrying a load of 3200 kg per horizontal metre then tension of ends is a.104,000 kg b.107 dynes c.1200,000 kg d.932,000 kg e.584,000 kg

Last Answer : d. 932,000 kg

A stone dropped from a baloon that was ascending at the rate of 30 ft/sec, reaches ground after 10 secs. The height of the baloon where stone was dropped was a.3420 ft b.107 dynes c.4520 ft d.2310 ft e.1310 ft

Description : A stone dropped from a baloon that was ascending at the rate of 30 ft/sec, reaches ground after 10 secs. The height of the baloon where stone was dropped was a.3420 ft b.107 dynes c.4520 ft d.2310 ft e.1310 ft

Last Answer : e. 1310 ft

A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Description : A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Last Answer : 1620°R

The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Description : The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Last Answer : 111 psi

Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Description : Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Last Answer : -420Btu

A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

Description : A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

Last Answer : 33.9 ft

A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Description : A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Last Answer : 33.9 ft

Calculate: a. Mass flow rate in lb/hr. b. The velocity at section 2 in fps  a. 800,000lb/hr;625ft/s  b. 900,000lb/hr;625 ft/s  c. 888,000lb/hr;269 ft/s  d. 700,000lb/hr;269 ft/s m = A1V!/V1

Description : Calculate: a. Mass flow rate in lb/hr. b. The velocity at section 2 in fps  a. 800,000lb/hr;625ft/s  b. 900,000lb/hr;625 ft/s  c. 888,000lb/hr;269 ft/s  d. 700,000lb/hr;269 ft/s m = A1V!/V1

Last Answer : 900,000 lb/hr;625 ft/s

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

A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Description : A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Last Answer : 1620 °R

Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Description : Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Last Answer : 209 psia

What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Description : What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Last Answer : 108 hp

A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Description : A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Last Answer : 42.41 ft^3

Two 12 ft. sections of aluminum siding are placed end to end on the outside wall of house. How large a gap should be left between the pieces to prevent buckling if the temperature can change by 55˚C?  a. 0.21 m  b. 0.18 m  c. 0.31 in  d. 0.18 in

Description : Two 12 ft. sections of aluminum siding are placed end to end on the outside wall of house. How large a gap should be left between the pieces to prevent buckling if the temperature can change by 55˚C?  a. 0.21 m  b. 0.18 m  c. 0.31 in  d. 0.18 in

Last Answer : 0.18 in

Carnot cycle is  (a) a reversible cycle (ft) an irreversible cycle  (c) a semi-reversible cycle  (d) a quasi static cycle  (e) an adiabatic irreversible cycle.

Description : Carnot cycle is  (a) a reversible cycle (ft) an irreversible cycle  (c) a semi-reversible cycle  (d) a quasi static cycle  (e) an adiabatic irreversible cycle.

Last Answer : Answer : a

Solids and liquids have  (a) one value of specific heat (ft) two values of specific heat  (c) three values of specific heat  (d) no value of specific heat  (e) one value under some conditions and two values under other conditions.

Description : Solids and liquids have  (a) one value of specific heat (ft) two values of specific heat  (c) three values of specific heat  (d) no value of specific heat  (e) one value under some conditions and two values under other conditions.

Last Answer : Answer : a