Question

2. Air at a temperature of 20 ºC passes through a pipe with a constant velocity of 40 m/s. The pipe goes through a heat exchanger in which hot gases outside the pipe heat up the air to 820 ºC. It then enters a turbine with a velocity of 40 m/s and expands till the temperature falls to 620 ºC. The air stream loses 4.3 kW heat in the turbine. If the air flow rate is 2.5 kg/s, calculate (a) How much heat is transferred to the air in the heat exchanger. (b) The power output of the turbine.

Answer 1

a) Q = 1436 kW

b) P ≈ 776 kW

Step-by-step explanation:

Let's begin by listing out the given parameters:

T1 = 20 °C, u = 40 m/s, T2 = 820 °C, P = 4.3 kW, m = 2.5 kg/s, T3 = 510 °C, V1 = 40 m/s,

V2 = 40 m/s, V3 = 55 m/s, ṁ = 2.5 kg/s

To solve the question, we make this assumption that the size of the pipe is constant

a) No change in velocity implies that heat added is isochoric

Q = m * C * ΔT

Cv of air at 300 K(≈20 °C) = 0.718

Q = 2.5 * 0.718 * (820 − 20)

Q = 1436 kW

b) P = ṁ * Cp * ΔT + ṁ * (V2² - V3²) ÷ 2000] - Ql

V2² - V3² = 55² - 40² = 1425

ΔT = T2 - T3 = 820 - 510 = 310 °C

Cp of air at 300 K(≠20 °C) = 1.005 kJ/kgK

Ql = 4.3 kW

P = 2.5 * (1.005 * 310) + 2.5 * (1425 ÷ 2000) - 4.3

P = 778.875 + 1.78125 - 4.3 = 776.35625

P ≈ 776 kW