the temperature difference between the inlet and outlet of an air_cooled engine is 30.0K. The engine generates 7.0kWof waste power that the air extracts from the engine. calculate the rate of flow of of air in kg/s needed to extract this power. shc of air at constant pressure is 1.01×10^3JKg^-1K^-1

To calculate the rate of flow of air needed to extract the waste power from the engine, we can use the formula:

Power (P) = mass flow rate (m) x specific heat capacity (c) x temperature difference (ΔT)

Given:
Power (P) = 7.0 kW = 7,000 W
Temperature difference (ΔT) = 30.0 K
Specific heat capacity (c) = 1.01 x 10^3 J/kg·K

Now, let's rearrange the formula to solve for the mass flow rate:
m = P / (c x ΔT)

Substituting the given values:
m = 7,000 W / (1.01 x 10^3 J/kg·K x 30.0 K)

Calculating:
m = 7,000 W / (3.03 x 10^4 J/kg)
m ≈ 0.230 kg/s

Therefore, the rate of flow of air needed to extract the waste power is approximately 0.230 kg/s.

To calculate the rate of flow of air needed to extract the given waste power from the engine, we can use the formula:

Rate of Flow = Waste Power / (Specific Heat Capacity × Temperature Difference)

Now let's substitute the given values into the formula:

Waste Power = 7.0 kW
Temperature Difference = 30.0 K
Specific Heat Capacity (SHC) = 1.01 × 10^3 J/(kg·K)

First, let's convert the waste power from kilowatts to watts, as the specific heat capacity is given in joules:

Waste Power = 7.0 kW × 1000 = 7000 W

Now, we can substitute the values into the formula:

Rate of Flow = 7000 W / (1.01 × 10^3 J/(kg·K) × 30.0 K)

Simplifying, we get:

Rate of Flow = 7000 W / (30300 J/(kg·K))

To calculate this, we need to remember that power is energy divided by time, and the energy is given by the specific heat capacity times the mass times the temperature difference:

Rate of Flow = (7000 W) / ((1.01 × 10^3 J/(kg·K)) × (30.0 K))

Let's assume the mass of air needed to extract the power is 'm':

Rate of Flow = (Specific Heat Capacity × m × Temperature Difference) / time

Now, we rearrange the equation to solve for 'm':

m = (Rate of Flow × time) / (Specific Heat Capacity × Temperature Difference)

Since we want to calculate the rate of flow of air in kg/s, let's choose the time to be 1 second:

m = (Rate of Flow × 1 s) / (Specific Heat Capacity × Temperature Difference)

Now, substitute the known values and solve for 'm':

m = (7000 W × 1 s) / ((1.01 × 10^3 J/(kg·K)) × 30.0 K)

m = 7000 / (1.01 × 10^3 × 30.0) kg

Calculating this expression gives us the rate of flow of air needed in kg/s to extract the given waste power from the engine.

air flow= 30K*7kW/(1.01e3J/(KgK))

air flow= 30 kW/(1.10kj)*Kg
but kiloWatts= kiloJoule /sec so
air flow= 30Kg/1.01sec= ... check my work.