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4 years ago

Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of

Engineering

1 answer:

Savatey [412]4 years ago

8 0

Answer:

option C

option A

Explanation:

Enthalpy gained by air= 1023-290

= 733 kJ/kg

Rate of energy gain= mass flow rate × Enthalpy gained by air

= 0.1 × 733

= 73.3 kJ/s

rate of heat transfer between compressor and air= 77kW

Heat loss by air to surroundings= 77-73.3

=3.7kW

Enthalpy lost by steam in turbine= 1407.6-1236.4

= 171.2 Btu/lb

Rate of energy transfer to turbine= Enthalpy lost by steam× mass flow rate

= 171.2×5

= 856 Btu/s

Net rate of energy transfer to turbine=rate of Energy transfer to turbine- rate of heat transfer to turbine

= 856-40

= 816 Btu/s

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An insulated piston–cylinder device initially contains 1 m3 of air at 120 kPa and 17°C. Air is now heated for 15 min by a 200-W

irakobra [83]

Answer:

∆S1 = 0.5166kJ/K

∆S2 = 0.51826kJ/K

Explanation:

Check attachment for solution

4 0

3 years ago

Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca

Andru [333]

Answer:

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

Explanation:

a)

Identify the unknown: 

The charge and energy stored if the capacitors are connected in series

List the Knowns:

Capacitance of the first capacitor: $C_{1}$ = 2цF = 2 x 10-6 F

Capacitance of the second capacitor $C_{2}$ = 7.4цF = 7.4 x 10-6 F

Voltage of battery: V = 9 V

Set Up the Problem:

Capacitance of a series combination:

$\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }$ +............

$\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\$

Capacitance of a series combination is given by:

$C_{s}=\frac{Q}{V}$

Then the charge stored in the series combination is:

$Q=C_{s} V$

Energy stored in the series combination is:

$U_{c}=\frac{1}{2} V^{2} C_{s}$

Solve the Problem: 

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

b)

Identify the unknown:

The charge and energy stored if the capacitors are connected in parallel

Set Up the Problem: 

Capacitance of a parallel combination:

$C_{p} =C_{1} +C_{2} +C_{3}$

$C_{p} =2+7.4=9.4*10^-6F$

Capacitance of a parallel combination is given by

$C_{p} =\frac{Q}{V}$

Then the charge stored in the parallel combination is

$Q=C_{p} V$

Energy stored in the parallel combination is:

$U_{c}=\frac{1}{2} V^2C_{p}$

Solve the Problem:

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

5 0

3 years ago

Read 2 more answers

Assuming that the following three variables have already been declared, which variable will store a Boolean value after these st

Yuki888 [10]

Answer:

Explanation:

Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.

5 0

3 years ago

What are the laws that apply to one vehicle towing another?

Sergeu [11.5K]

The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.

5 0

3 years ago

. In order to prevent injury from inflating air bags, it is recommended that vehicle occupants sit at least __________ inches aw

scZoUnD [109]

10 inches is the answer

8 0

3 years ago

Read 2 more answers