Tion A. Classwork

A heat exchanger takes compressed liquid water and converts it into steam with a temperature and pressure of 20 Mpa and 480 C; r

belka [17]

Answer:

See explaination

Explanation:

Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.

Please kindly check attachment for the step by step solution of the given problem.

6 0

3 years ago

Viscous effects are negligible outside of the hydrodynamic boundary layer. (3 points) a. True b. False

Valentin [98]

Answer:

I would say false but I am not for sure

8 0

3 years ago

1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperatur

Anit [1.1K]

Answer:

Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, $T_L = 36^0 F = 275.372 K$

Outside temperature, $T_H = 100^0F = 310.928 K$

rate of insulation, $Q = 100 Btu/h$

To get the minimum electrical power required, use the relation below:

$\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt$

V = 5 V

Power = IV

$W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A$

If the cooler is supposed to work for 4 hours, t = 4 hours

$I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr$

Minimum battery size needed = 3.03 Amp-hr

6 0

3 years ago

4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and

hodyreva [135]

Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out

6 0

3 years ago

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

Dafna11 [192]

Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

4 0

3 years ago