Answer:
Explanation:
From the question we are told that:
Water flow Rate
Initial Temperature
Final Temperature
Let
Specific heat of water
And
Generally the equation for Heat transfer rate of water is mathematically given by
Heat transfer rate to water= mass flow rate* specific heat* change in temperature
Therefore
Answer:
Speed of aircraft ; (V_1) = 83.9 m/s
Explanation:
The height at which aircraft is flying = 3000 m
The differential pressure = 3200 N/m²
From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3
Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.
Thus, let's apply the Bernoulli equation :
P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2
Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.
We'll obtain ;
P1/ρg + (V_1)²/2g = P2/ρg
Let's make V_1 the subject;
(V_1)² = 2(P1 - P2)/ρ
(V_1) = √(2(P1 - P2)/ρ)
P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question
Thus,
(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
