Thermodynamics Miscellaneous
- A solar collector receiving solar radiation at the rate of 0.6 kW/m2 transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to run a heat engine which rejects heat at 313 K. If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be
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10. Given; Receiving solar radiation at the rate of 0.6 kW/m2
Internal energy of fluid after absorbing solar radiation = 0.6 × 1 kW / m2 = 0.3 kW / m2 2 ηengine = 1 - 315 = 0.1 350 0.1 = W Q1 ∴ Q1 = 2.5 = 25 k W 0.1
Let A be minimum area of collector
∴ Q1 = 0.3 × A or 25 kW = 0.3 kW/m2or A = 25 = 83.33 m2 0.3
Correct Option: A
10. Given; Receiving solar radiation at the rate of 0.6 kW/m2
Internal energy of fluid after absorbing solar radiation = 0.6 × 1 kW / m2 = 0.3 kW / m2 2 ηengine = 1 - 315 = 0.1 350 0.1 = W Q1 ∴ Q1 = 2.5 = 25 k W 0.1
Let A be minimum area of collector
∴ Q1 = 0.3 × A or 25 kW = 0.3 kW/m2or A = 25 = 83.33 m2 0.3
- For an ideal gas with constant values of specific heats, for calculation of the specific enthalpy,
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it is sufficient to know only the temperature
Correct Option: A
it is sufficient to know only the temperature
- A mixture of ideal gases has the following composition by mass :
N2 O2 CO2 60% 30% 10%
If the universal gas constant is 8314 J/kmolK, the characteristic gas constant of the mixture (in J/kgK) is _______.
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Gas constant of mixture,
Rm = Universal gas constant Average molar mass Rm = Universal gas constant Average molar mass Average molar mass = 100 = 30.233 kg/kmol 60 + 30 + 10 28 32 44 Rm = 8314 = 274.996 J/kg -K 30.233
Correct Option: A
Gas constant of mixture,
Rm = Universal gas constant Average molar mass Rm = Universal gas constant Average molar mass Average molar mass = 100 = 30.233 kg/kmol 60 + 30 + 10 28 32 44 Rm = 8314 = 274.996 J/kg -K 30.233
- The internal energy of an ideal gas is a function of
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temperature only
Correct Option: D
temperature only
- A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m3. It expands quasistatically at constant temperature to a final volume of 0.030 m3. The work output (in kJ) during this process will be
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W = P1V1 ln V2 = 8.317 kJ V1 Correct Option: A
W = P1V1 ln V2 = 8.317 kJ V1