Answer:
V1 = 1.721 * V2
Explanation:
To start with, we assume that both lift forces are equal, such that
L2 = L1
1 is that of the level at 10000 m, and 2 is that of the level at sea level.
Next, we try and substitute the general formula for both forces such that
C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A
On further simplification, we have
ρ1.V1² = ρ2.V2², making V1 subject of formula, we have
V1 = √(ρ2/ρ1). V2²
Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have
V1 = √(1.225/0.4135) * V2
V1 = √2.9625 * V2
V1 = 1.721 * V2
Answer:
The following query is used to display TripName. Reservationld, FirstName. LastName and TotalCost of trip by adding the trip price plus other fees and multiplying the result by the number of persons which have number of persons >4.
Query:
SELECT ReservationlD, Trip.TripName. Customer.LastName. Customer.FirstName. (TripPrice+OtherFees) 'NumPersons as TotalCost FROM Reservation, Trip, Customer WHERE NumPersons>4 AND Reservation.TriplD=Trip.TriplD AND
Customer. CustomerNum=Reservation.CustomerNum:
Explanation:
- Select clause is used to retrieve data from specified database table or relation and returns the data in the form of table.
- ReservationID. Trip.TripName. Customer.LastName. Customer.FirstName are the column name of table.
- (TripPrice+OtherFees) 'NumPersons will calculate the total cost of the Trip and stored it into TotalCost column.
- As clause is used to give new name TotalCost to resultant column.
- FROM clause specifies one or more table from where records to be retrieved. o Reservation. Trip and Customer are the table name.
- WHERE clause is used in SQL query to retrieve only those records that satisfy the specified condition
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:

It is necessary to get the Reynold's number:

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

The overall heat transfer coefficient:

Here

Substituting values:

Attached is the solution to the above question.