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max2010maxim [7]
3 years ago
13

Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature

of 300 K, and a turbine inlet temperature of 1000 K. The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4. Determine the required mass flow rate of air for a net power output of 70 MW, assuming both the compressor and the turbine have an isentropic efficiency of: a. 100 percent b. 85 percent.
Engineering
1 answer:
7nadin3 [17]3 years ago
4 0

Answer:

A) m' = 351.49 kg/s

B) m'= 1036.91 kg/s

Explanation:

We are given;

Pressure Ratio;r_p = 12

Inlet temperature of compressor;T1 = 300 K

Inlet temperature of turbine;T3 = 1000 K

cp = 1.005 kJ/kg·K

k = 1.4

Net power output; W' = 70 MW = 70000 KW

A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;

m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))

At, 100% efficiency, plugging in the relevant values, we have;

m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))

m' = 70000/199.1508

m' = 351.49 kg/s

B) At 85% efficiency, the formula will now be;

m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))

Where η is efficiency = 0.85

Thus;

m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))

m' = 70000/(1.005*(432.09129 - 364.9189)

m'= 1036.91 kg/s

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\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

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Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

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