Chapter 2: Units and Dimensions (Solution)

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Problem 2.1: The viscosity of water is one centipoise (cP). Express the viscosity of water in the following units (a) Pa s (b) lb/(ft h).

Ans: (a) 10-3 Pa.s (b) 2.419 lb/(ft.h)

Explanation:

Problem 2.2: Covert 1 bar into equivalent kgf/cm2 and psi.

Ans: 1.0197 kgf/cm2, 14.5038 psi

Explanation:

Problem 2.3: The thermal conductivity of an insulating brick is 0.15 Btu/(ft h 0F). Express the thermal conductivity in SI units.

Ans: 0.2596 W/m K

Explanation:

Problem 2.4: The heat transfer coefficient of oil flowing through a pipe is 300 W/(m2 K). Determine the value of the heat transfer coefficient expressed in (a) kcal/(h m2 0C), (b) Btu/(h ft2 0F).

Ans: (a) 257.95 kcal/(h m2 0C) (b) 52.832 Btu/(h ft2 0F)

Explanation:

Problem 2.5: Thermal conductivity of pure iron is 39 Btu/(ft h oF) and that of steel containing 1 % C is 39 kcal/(m h oC). Which one is a good conductor of heat, iron or steel?

Ans: iron

Explanation:

Problem 2.6: The mass velocity of a gas through a duct is 1000 kg/m2 h. Express the velocity into lb/ft2 s.

Ans: 5.69×10-2 lb/ft2.s

Explanation:

Problem 2.7: Mass flow rate is the product of volumetric flow rate and density. If the volumetric flow rate of a liquid of specific gravity 0.78 is 100 ft3/min, find the flow rate in kg/s.

Ans: 36.81 kg/s

Explanation:

Problem 2.8: A man circling the earth in a spaceship weighed 300 N at a location where the local gravitational acceleration was 4.5 m/s2. Calculate the mass of the man and his weight on the earth where the gravitational acceleration is 9.18 m/s2

Ans: 66.67 kg, 654 N

Explanation:

Problem 2.9: A man whose weight is 600 N takes 2 min for climbing up a staircase. What is the power developed in him, if the staircase is made up of 20 stairs, each 0.18 m in height?

Ans: 18 W

Explanation:

Problem 2.10: The potential energy of a body of mass 10 kg is 1.5 kJ. What is the height of the body from the ground? If the body of mass 10 kg is moving at a velocity of 50 m/s, what is its kinetic energy?

Ans: 15.3 m, 12.5 kJ

Explanation:

Problem 2.11: Nitrogen gas is confined in a cylinder and the pressure of the gas is maintained by a weight placed on the piston. The mass of the piston and the weight together is 50 kg. The acceleration due to gravity is 9.81 m/s2 and the atmospheric pressure is 1.01325 bar. Assume a frictionless piston. Determine: (a) The force exerted by the atmosphere, the piston, and the weight on the gas if the piston is 100 mm in diameter, (b) The pressure of the gas, (c) If the gas is allowed to expand pushing up the piston and the weight by 400 mm, what is the work done by the gas in kJ and (d) What is the change in the potential energy of the piston and the weight after the expansion in part (c)?

Ans: (a) 1286.305 N (b) 1.6378 bar (c) 0.5145 kJ (d) 196.2 J

Explanation:

Problem 2.12: The volume of microbial culture is observed to increase according to the relation V (cm3) = et where t is in seconds. Find the equivalent expression where V is in inch3 and t is in hours.

Ans: V (inch3) = 0.0610e3600t

Explanation:

Problem 2.13: Stefan-Boltzmann law state that the emissive power is proportional to the fourth power of absolute temperature. State mathematically, Eb = σT4 where Eb is the emissive power in Btu/(ft2 h), σ is the Stefan-Boltzmann constant, and T is the temperature in oR. What is the value of σ in W/(m2 K4), if its value in FPS units is 0.171⨯10-8 Btu/(ft2 h oR4)?

Ans: 5.6629⨯10-8 W/(m2.K4)

Explanation:

Problem 2.14: The following empirical equation was proposed to estimate the rate of heat loss from the walls of a cylindrical pipe to the ambient air:

q=0.5\frac{{\triangle T}^{1.25}}{D^{0.25}}

Where q is the rate of heat loss in Btu/ft2 h, ∆T is the temperature difference in oF and D is the diameter in inches. a) Determine the units of the constant, b) Express the equation in SI units.

Ans: (a) \frac{Btu\;{(in)}^{0.25}}{ft^2(h){(^\circ F)}^{1.25}} (b) q=1.313\frac{\triangle T^{1.25}}{D^{0.25}}

Explanation:

Problem 2.15: The van der Waals equation of state is given as

\left(P+\frac a{V^2}\right)\left(V-b\right)=RT

Where a and b are known as the van der Waals constants. For CO2 the constants are found to be a = 0.365 N m4/mol2 and b = 4.28⨯10-5 m3/mol. If the pressure is given in atm and the molar volume in L/mol, what are the values of the constants?

Ans: a=3.602\frac{atm.L^2}{{mol}^2} , b=4.28\times10^{-2}\frac L{mol}

Explanation:

Problem 2.16: The rate of a gas phase reaction (A↔B) taking place at 298 K and 1 atm is given by

-\frac{dC_A}{dt}=k_1p_A^{2.2}-k_2p_B^{0.5}

With k1 = 17.3 and k2 = 0.2, where CA is the concentration in kmol/m3, t is the time in minutes, and pA, pB are partial pressure in atm. (a) Find the units of k1 and k2 (b) If the rate is to be written in the form of

-\frac{dp_A}{dt}=k_1^`C_A^{2.2}-k_2^`C_B^{0.5}

What are the values of k1’ and k2’?

Ans: (a) k_1=\frac{kmol}{m^3.min.\left(atm\right)^{2.2}};\;\;k_2=\frac{kmol}{m^3.min.\left(atm\right)^{0.5}} ) (b) k_1^`=4.795\times10^5;\;\;k_2^`=24.185

Explanation:

Problem 2.17: The pressure drops for a fluid flowing through a packed bed of solids is given by the relation:

\triangle P=3.61\left[\frac{\mu^{0.15}H\rho^{0.85}v^{1.85}}{D_P^{1.15}}\right]

Where ∆P is a pressure drop in Pa, 𝛍 is fluid viscosity in poise, H is the bed depth in meters, ρ is the fluid density in kg/m3, v is the fluid velocity in m/s and DP is packing diameter in meters. Modify the relation such that the pressure drop is still in Pa, but other terms are in FPS units.

Ans: \triangle P=7.58\left[\frac{\mu^{0.15}H\rho^{0.85}v^{1.85}}{D_P^{1.15}}\right]

Explanation:

Problem 2.18: The Antoine equation predicts the effect of temperature on vapor pressure by the relation:

\ln P^s=A-\frac B{T-C}

Where A, B, and C are Antoine constants. For water, the values of the constants when the vapor pressure is in kPa and temperature is in K are found to be A = 16.26205, B = 3799.887, and C = 46.854. What are these values when (a) Vapor pressure is in mm Hg and temperature in oC, (b) vapor pressure is in psia and temperature in oF?

Ans: (a) A=18.2773, B=3799.887, C=-226.296 (b) A=14.331, B=6839.7966 C=-375.0628

Explanation:

Problem 2.19: The heat capacity of sulphuric acid in Btu/(lb-mol oR) is given by

C_P=23.06+2.071\times10^{-2}T

Where T is in oR. What is the equivalent expression if heat capacity is given in (a) kJ/(kmol K) and T is in K (b) kcal/(kmol oC) and T in oC.

Ans: (a) CP=96.55+0.1562T (b) CP=33.24+0.0373T

Explanation:

Problem 2.20: The following equation relates the heat capacity of CO gas with the temperature

C_P=26.586+7.582\times10^{-3}T-1.12\times10^{-6}T^2

Where CP is expressed in kJ/(kmol K) and T in K. Obtain equations which will express (a) The heat capacity in cal/mol K, temperature in K (b) The heat capacity in cal/mol oC, temperature in oC (c) The heat capacity in J/kg K, temperature in K (d) The heat capacity in Btu/lb-mol oF, temperature in oF.

Ans: (a) CP=6.35+1.811×10-3T-2.675×10-7T2 (b) CP=6.825+1.665×10-3T-2.675×10-7T2 (c) CP=9.495×102+0.271T-4.0×10-5T2 (d) CP=6.795+0.93×10-3T-8.256×10-8T2

Explanation:

Problem 2.21: The heat capacity of CO2 gas is given by

C_P=0.1978+1.059\times10^{-4}T-2.395\times10^{-8}T^2

Where CP is in Btu/(lb oF) and T is in oF. Change the equation into the form in which CP is given in kJ/(kmol K) and T is in K.

Ans: CP=26.5386+4.24×10-2T-1.4295×10-5T2

Explanation:

Problem 2.22: For fluids in turbulent motion through tubes, the heat transfer coefficient is given as

h=a\left(\frac{C_PG^{0.8}}{D^{0.2}}\right)

The numerical value of the constant a = 10.1, when h is measured in Btu/(ft2 h oF), CP is the specific heat of the fluid given in Btu/(lb oF), G is the mass velocity lb/(ft2 s) and D is in ft. Determine the value of a (a) when h is measured in W/(m2 K), CP is given in kJ/(kg K), G is in kg/(m2 s) and (b) when h is in kcal/(m2 h oC), CP is in kcal/(kg oC), G is in kg/(m2 s) and D in m.

Ans: (a) 3.038 (b) 10.94

Explanation:

Problem 2.23: The diffusivity of a gas pair A-B is given by

D_{AB}=\frac{1.8583\times10^{-3}T^{3/2}\left(\frac1{M_A}+\frac1{M_B}\right)^{1/2}}{P\sigma^2\mathrm\Omega}

Where DAB is in cm2/s, T is the absolute temperature in K, MA and MB are molecular weights of species A and B, respectively, P is the pressure in atm, σ is the Lennard-Jones parameter in angstrom units, and is a dimensionless parameter. If the equation is written as

D_{AB}=\frac{CT^{3/2}\left(\frac1{M_A}+\frac1{M_B}\right)^{1/2}}{P\sigma^2\mathrm\Omega}

Where DAB is in m2/s, T is in K, P is the pressure in Pa, σ is the L-J parameter in nanometres, what is the value of C?

Ans: 1.8829⨯10-4 J

Explanation:


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