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MOD002684 (Thermofluids) Assessment

Assignment Brief

Element 011: MOD002684 (Thermofluids) Assessment

QUESTIONS: Element 011: MOD002684 (Thermofluids) Assessment

Part A

A piston–cylinder device initially contains 0.A m3 (A is the digit in your SID ****ABC, if A is 0 or 1, then take the next higher digit between B or C) of air at 100 kPa and 80 °C. The air is then compressed to 0.1 m3 in such a way that PV = constant. Determine the work done during this process. (5 Marks)

Part B

The air enters the compressor at a temperature of 280 K with an enthalpy of 280.13 kJ/kg and is compressed steadily to 400 K with an enthalpy of 400.98 kJ/kg. The mass flow rate of the air is 0.0A kg/s (A is the digit in your SID ****ABC, if A is 0, then take the next higher digit between B or C) and a heat loss of 16 kJ/kg occurs during the process. Determine the necessary power input to the compressor when:

  1. the changes in kinetic and potential energies are negligible.  (7 Marks)

  2. the air velocity at entry is 100 m/s and exit is 150 m/s but potential energy is negligible. (7 Marks) 

Provide a reflective commentary underpinning the KEY conceptual elements in solving problem Q1 Part B  (maximum 100words) (6 Marks)

Part A:

Heat is transferred to a heat engine from a furnace at a rate of 100 MW. If the rate of waste heat rejection to a nearby river is BC MW (BC are the digits in your SID ****ABC). Determine the net power output and the thermal efficiency for this heat engine. (8 Marks)

Part B

A heat source at 800 K loses 2000 kJ of heat to a sink at (a) 500 K and (b) ABC K (ABC are the digits in your SID ****ABC, for those who has SID number A = 0 or 1 or 2, replace A by 7) (shown in Figure Q2B). Determine which heat transfer process is more irreversible based on the entropy calculation. (11 Marks)

 

Figure Q2B

Provide a reflective commentary underpinning the KEY conceptual elements in solving problem Q1 Part B (maximum 100 words) (6 Marks) 

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17°C, and ABC kJ/kg (ABC are the digits in your SID ****ABC, for those who has SID number A = 0 or 1 or 2, replace A by 7) of heat is transferred to air during the combustion process. Determine:

  1. the pressure and temperature at the end of the compression process (8 Marks)

  2. the thermal efficiency of the cycle (3 Marks)

  3. the net output work of the cycle per kg of air (4 Marks)

  4. the mean effective pressure for the cycle (4 Marks)

Assume Cv=0.718 kJ/kg.K and Cp=1.005 kJ/kg.K for air.

Provide a reflective commentary underpinning the KEY conceptual elements in solving this problem (maximum 100 words)(6 Marks)

A fuel with a chemical formula C8H1A (A is the digit in your SID ****ABC, if A is an odd number, choose the next even number) is burned with dry air. The volumetric analysis of the exhaust products gases on a dry basis are CO2: 10.02%, O2: 5.62%, CO: 0.88% and N2: 83.48%.

Determine

  1. the air–fuel ratio (15 Marks)

  2. the percentage of theoretical air used (4 Marks)

Provide a reflective commentary underpinning the KEY conceptual elements in solving this problem (maximum 100 words)(6 Marks)

Element 011: MOD002684 (Thermofluids) Assessment

Part A: Choose the correct option.

5A1. For the same compression ratio, the efficiency of diesel cycle is........Otto cycle (provide reason)

  1. Greater than

  2. Less than

  3. Equal to

  4. None of the above

(2 Marks)

5A2: Which of the following is true for a steady flow system? (provide reason)

  1. mass entering = mass leaving

  2. mass does not enter or leave the system

  3. mass entering can be more or less than the mass leaving

  4. none of the mentioned

(2 Marks)

Part B

A rigid tank contains BC kg (BC is the digit in your SID ****ABC) of water at 90°C. If 80% by mass of the water is in the liquid form and the rest is in the vapour form. Determine

  1. the pressure in the tank (2 Marks)
  1. the dryness fraction of the mixture (3 Marks)
  1. the total volume of the tank (3 Marks)
  1. the internal energy (U) (3 Marks)
  1. the enthalpy (H) (2 Marks)
  1. the entropy (S) (2 Marks)

Provide a reflective commentary underpinning the KEY conceptual elements in solving problem Q5 Part B (maximum 100 words) (6 Marks)

Element 011: MOD002684 (Thermofluids) Assessment

A certain gas of 2 kg is kept at BC (BC is the digit in your SID ****ABC) °C in a constant volume chamber of 0.3 m3 capacity. Heat is transferred to the gas until the temperature is of the gas reached to100 °C. The gas has cp = 1.968 kJ/kg-K and cv = 1.507 kJ/kg-K. Universal gas constant 8.314 kJ/kg-mol K. Determine:

  1. the molecular weight (3 Marks)
  1. the gas constant (2 Marks)
  1. the work done (3 Marks)
  1. the heat transferred (2 Marks)
  1. the change in internal energy (3 Marks)
  1. the change in enthalpy (3 Marks)
  1. the change in entropy (3 Marks) 

Provide a reflective commentary underpinning the KEY conceptual elements in solving this problem (maximum 100 words) (6 Marks)

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Sample Answer

Question 1

Part A
For a piston–cylinder device undergoing compression where PV = constant, the process is polytropic with n = 1 (isothermal). The work done is therefore:

Work = P1 V1 ln(V2 / V1)

Here, P1 = 100 kPa, V1 = 0.A m³, and V2 = 0.1 m³.
By substitution:

W = (100 × 10³ × 0.A) ln(0.1 / 0.A) (in Joules).

This gives the mechanical work done by the air during the compression process.

Part B
From the steady flow energy equation, neglecting potential energy changes:

Win = ṁ (h2 – h1) + q̇ + (Δke).

  • Enthalpy at entry h1 = 280.13 kJ/kg.

  • Enthalpy at exit h2 = 400.98 kJ/kg.

  • Heat loss = –16 kJ/kg.

  • Mass flow rate ṁ = 0.0A kg/s.

(i) With negligible velocity change:
Win = ṁ [(400.98 – 280.13) – 16].

(ii) With velocity change (v1 = 100 m/s, v2 = 150 m/s):
Δke = (v2² – v1²)/2000 = (150² – 100²)/2000 = 6.25 kJ/kg.
Win = ṁ [(400.98 – 280.13 – 16) + 6.25].

Reflective Commentary (Q1 Part B)

The key principle here is the steady flow energy balance, which links enthalpy, heat transfer, and kinetic energy changes. Neglecting potential energy simplifies the equation. Including kinetic energy highlights that compressor power depends not only on enthalpy rise but also on flow dynamics.

Question 2

Part A
Qin = 100 MW, Qout = BC MW.
Net power output = Qin – Qout.
Thermal efficiency = 1 – (Qout / Qin).

Part B
Entropy generation ΔStotal = (Qsink / Tsink) – (Qsource / Tsource).

(a) Qsource = 2000 kJ, Ts = 800 K, Tk = 500 K.
ΔS = (2000/500) – (2000/800).

(b) Repeat with Tk = ABC K.
The process with the larger ΔStotal is more irreversible.

Reflective Commentary (Q2)

Entropy analysis is central to understanding irreversibility. A greater entropy increase reflects greater energy degradation. Comparing entropy generation at different sink temperatures allows us to judge which transfer is less efficient.

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