Answer:
a) Q = 1436 kW
b) P ≈ 776 kW
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
