Find the pressure due to a column of mercury 74.0 cm high.  a. 91.80 x 10^3 N/m^2  b. 73.56 x 10^2 N/m^2  c. 9.86 x 10^4 N/m^2  d. 87.2 x 10^4 N/m^2

Find the pressure due to a column of mercury 74.0 cm high.  a. 91.80 x 10^3 N/m^2  b. 73.56 x 10^2 N/m^2  c. 9.86 x 10^4 N/m^2  d. 87.2 x 10^4 N/m^2
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Answer :

9.86 x 10^4 N/m^2
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A 0.064 kg of octane vapor (MW = 114) is mixed with0.91 kg of air (MW = 29.0) in the manifold of an Engine. The total pressure in the manifold is 86.1 kPa, and a temperature is 290 K. assume octane behaves ideally. What is the partial pressure of the air in the mixture in KPa?  a. 46.8  b. 48.6  c. 84.6  d. 64.8

Description : A 0.064 kg of octane vapor (MW = 114) is mixed with0.91 kg of air (MW = 29.0) in the manifold of an Engine. The total pressure in the manifold is 86.1 kPa, and a temperature is 290 K. assume octane behaves ideally. What is ... of the air in the mixture in KPa?  a. 46.8  b. 48.6  c. 84.6  d. 64.8

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Last Answer : Both a & b

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Last Answer : (B) CO2 : 7.42, CO : 1.86, O2 : 4.64, N2 :86.02

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Last Answer : 97.6%,2713 kJ/kg

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Last Answer : Answer: D

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Last Answer : 33.9 ft

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Last Answer : 1.3 x 108 J

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Last Answer : 48.75 N

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Last Answer : 4474.23 N

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

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Last Answer : 12 kJ

The volume of a gas under standard atmospheric pressure & 76 cmHg is 200m³. What is the volume when pressure is 80 cmHg if the temperature is unchanged?  a) 180 in³  b) 170 in³  c) 160 in³  d) 190 in³ Formula: P2V2 = P1V1

Description : The volume of a gas under standard atmospheric pressure & 76 cmHg is 200m³. What is the volume when pressure is 80 cmHg if the temperature is unchanged?  a) 180 in³  b) 170 in³  c) 160 in³  d) 190 in³ Formula: P2V2 = P1V1

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Last Answer : 111 psi

A 600kg hammer of a pile driver is lilted 2m the pilling head. What is the change of potential energy? If the hammer is realest. What will be its velocity and the instant if it sticks the pilling?  a. 10,772 N-m and 5.26m/s  b. 13,200 N-m and 5.26m/s  c. 11,772 N-m and 6.26m/s  d. 11,77 2N-m and5.26m/s ∫PE = mgo(∫Z)/gc

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Last Answer : 11,772 N-m and 6.26m/s

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Last Answer : 7.9 °C

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Last Answer : Answer: C

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Last Answer : 14.223 psi

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Last Answer : 9 Newton’s

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Last Answer : 2474.23 N

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Last Answer : N/m^2

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Last Answer : Answer : a

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

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Description : What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

Last Answer : 1000 kgf/m3

Give a reading as the length of some liquid column: water, alcohol, etc.  a. Banometer  b. Nanometer  c. Thermometer  d. Manometer

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Last Answer : Manometer

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Last Answer : 24.9 ft^3

The SI unit of pressure  a. Pa  b. N  c. J  d. None of the above

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Last Answer : Answer : c

If value of n is infinitely large in a polytropic process pV” = C, then the process is known as constant  (a) volume  (b) pressure  (c) temperature  (d) enthalpy  (e) entropy

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In a non-flow reversible process for which p = (- 3V+ 15) x 105N/m2,V changes from 1 m to 2 m3. The work done will be about  (a) 100 xlOO5 joules  (b) lxlO5 joules  (c) 10 xlO5 joules  (d) 10 xlO5 kilo joules  (e) 10xl04ki\ojoules.

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Last Answer : Answer : c

A certain fluid is flowing in a 0.5m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.  a. 267 kg/s  b. 378 kg/s  c. 375 kg/s  d. 456.5 kg/s m = Aν/V

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Last Answer : 375 kg/s

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

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Last Answer : 13.33 ft^3/lbm

What mass of nitrogen is contained in a10 ft3 vessel at a pressure of 840atm and 820°R? Make a computation by using ideal gas equation.  a. 194lb  b. 214lb  c. 394 lb  d. 413lb formula: m=pV /RT

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Last Answer : 394 lb

The heat supplied to the gaS at constant volume is (where m = Mass of gas, cv = Specific heat at constant volume, cp = Specific heat at constant pressure, T2 – T1 = Rise in temperature, and R = Gas constant)  A. mR(T2 – T1)  B. mcv(T2 – T1)  C. mcp(T2 – T1)  D. mcp(T2 + T1)

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A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Description : A gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04 m3 to 0.10 m3 at a constant pressure of 200 kPa. Find the work done on the system.  a. 5 kJ  b. 15 kJ  c. 10 kJ  d. 12 kJ

Last Answer : 12 kJ

Minimum work in compressor is possible when the value of adiabatic index n is equal to  (a) 0.75  (b) 1  (c) 1.27  (d) 1.35  (e) 2.

Description : Minimum work in compressor is possible when the value of adiabatic index n is equal to  (a) 0.75  (b) 1  (c) 1.27  (d) 1.35  (e) 2.

Last Answer : Answer : b

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

Water flow to a terminal 3 mm diameter and has an average speed of 2 m/s. What is the rate of flow in cubic meter/mm?  a. 0.0001m³/min  b. 0.076 m³/min  c. 0.085 m³/min  d. 0.097 m³/min

Description : Water flow to a terminal 3 mm diameter and has an average speed of 2 m/s. What is the rate of flow in cubic meter/mm?  a. 0.0001m³/min  b. 0.076 m³/min  c. 0.085 m³/min  d. 0.097 m³/min

Last Answer : 0.085 m³/min

In the above problem, compute for the mass. (Formula: m = p1V1 / RT1)  a. 0.2148 lb  b. 0.2134 lb  c. 0.1248 lb  d. None of the above

Description : In the above problem, compute for the mass. (Formula: m = p1V1 / RT1)  a. 0.2148 lb  b. 0.2134 lb  c. 0.1248 lb  d. None of the above

Last Answer : 0.2148 lb

A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C. 2.54 kg  D. 3.16 kg Formula: from the steam table at 600kPa vƒ = 0.001101 m^3 / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

Description : A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C ... / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

Last Answer : 3.16 kg