Objective questions on Systems of units

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1. What is 250 degree F converted to degrees celcius?

a. 115 degree C                                 b. 121 degree C

c. 124 degree C                 d. 420 degree C

Ans - B

Explanation -

Conversion to the degrees celcius is

T ⁰C        = 5/9 (T ⁰F – 32 ⁰F)

                = (5/9)(250 – 32)

T ⁰C        = 121 ⁰ C

2. What is the Stefan-Boltzmann constant (0.1713 × 10^-8 Btu/hr-ft² – R⁴) converted from English to SI units?

a. 5.14 × 10^-10 W/m².K⁴

b. 0.95 × 10^-10 W/m².K⁴

c. 5.67 × 10^-10 W/m².K⁴

d. 7.44 × 10^-10 W/m².K⁴

Ans - C

Explanations-

The Stefan-Boltzmann constant represent the certain amount of energy (in Btu). Since it has hr-ft² -⁰R⁴  in the denominator, the energy is reported on a per hour basis and represents power. The energy is an areal value because it is reported per square basis. (For ex, on any given day, the energy that can be derived from a square meter of sunlight is greater than can be derived from a square foot of sunlight.) l/0.3048 is the conversion between feet and meters, and it is larger than 1.0 .

The Stefan-Boltzmann constant is reported on a per degree basis, and the same logic applies. One kelvin (k) is a larger (longer) interval on the temperature scale than one degree Rankine. The amount of energy per Kelvin is larger than the amount of energy must be multiplied by a number greater than 1.0 to report it on a “per kelvin” basis. This requires multiplying by 9/5. (An ncorrect conversion will result if the unit in the denominator is thought of an actual temperature. Any given numerical value of temp in degrees Rankine is larger than the same temperature in Kelvins. However, this problem does not require a temp conversions. It requires a unit conversions.

Use the following conversions factors.

            1 Btu/hr = 0.2931 W

            1 ft       = 0.3048 m

            1 degree R = 5/9 K

Performing the conversions gives

            σ  = (0.1713 × 10^-8 )(0.2931) × (1 ft / 0.3048)² (1degree R / (5/9))⁴

            σ   = 5.67 × 10^-8 W/m².K⁴

3. Approximately how many U.S. tons (2000 lbm/ton) of coal with heating value of 13000 Btu/lbm must be burn to provide as much energy as a complete nuclear conversion of 1 gm of coal? (Hint: Use Einstein’s equ. E = mc^2.)

a. 1.7 tons

b. 14 tons

c. 780 tons

d. 3300 tons

Ans – D

Explanation:

The energy produced from the nuclear conversion of any quantity of mass is

            E = mc²

The speed of light, c = 3 × 10⁸ m/s

For a mass of 1 g (0.001 kg)

            E = mc²

                            = (0.001 kg)( 3 × 10⁸ m/s)²

               = 9 × 10¹³ J

Convert to U.S. customary units with the conversion

1Btu = 1055.1 J

            E = 9 × 10¹³ J / 1055.1 J/Btu

               = 8.53 × 10¹⁰ Btu

The number of tons of 13,000 Btu/lbm coal is

            8.53 × 10¹⁰ Btu / (13000 X 2000 lbm/ton)

            = 3281 tons (3300 tons)