Calculating Heat Exchange for a Water Pump

In summary: The first law for a steady state device is ##\dot{Q}-\dot{W}=0## where the dot means "per unit time". In this case, the work per unit time is given by ##\dot{W}=10## watts. The heat transfer per unit time is the product of the mass flow rate, the heat capacity, and the temperature change of the water. The heat capacity is given in J/g-K, so you need to convert the mass flow rate from moles/s to g/s and the temperature difference from K to degrees C. In summary, the heat exchange required is calculated by multiplying the mass flow rate (in g/s) by the heat capacity (in J/g-K
  • #1
rachel6589
1
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Homework Statement


A pump, operating at 10 W, is used to raise the pressure of a stream of water at 2.5 mol/s from 1 bar to 2 bar. At steady state, if the water temperature should remain constant at 25oC, how much heat exchange between the pump and its surroundings is required? Note that 1 bar = 105 Pa and C liq,water = 4.18 J g-1 K -1

Homework Equations


PV=NRT

The Attempt at a Solution


I tried to set up two equations. The first one is (1)(V)=NR(373+25) and the second is (2)(V)= NR(373+25)
and I don't know find out the heat exchange since there are some unknown data( Volume? Mole?)
[/B]
 
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  • #2
rachel6589 said:

Homework Statement


A pump, operating at 10 W, is used to raise the pressure of a stream of water at 2.5 mol/s from 1 bar to 2 bar. At steady state, if the water temperature should remain constant at 25oC, how much heat exchange between the pump and its surroundings is required? Note that 1 bar = 105 Pa and C liq,water = 4.18 J g-1 K -1

Homework Equations


PV=NRT

The Attempt at a Solution


I tried to set up two equations. The first one is (1)(V)=NR(373+25) and the second is (2)(V)= NR(373+25)
and I don't know find out the heat exchange since there are some unknown data( Volume? Mole?)[/B]
The solution to this problem has nothing to do with the ideal gas law. This is a problem in application of the open system (control volume) version of the first law of thermodynamics.
 
Last edited:

1. How do you calculate heat exchange for a water pump?

To calculate heat exchange for a water pump, you need to know the temperature difference between the incoming and outgoing water, the flow rate of the water, and the specific heat capacity of water. The formula is Q = m * c * ΔT, where Q is the heat exchange in Joules, m is the mass flow rate in kilograms, c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the temperature difference in degrees Celsius.

2. What is the purpose of calculating heat exchange for a water pump?

Calculating heat exchange for a water pump allows you to determine the amount of heat energy that is transferred between the water and the pump. This is important for understanding the efficiency of the pump and how much heat may need to be removed from the system to prevent overheating.

3. How does the flow rate of water affect heat exchange for a water pump?

The flow rate of water has a direct impact on the heat exchange for a water pump. The higher the flow rate, the more water is passing through the pump, resulting in a larger amount of heat being exchanged. This is why it is important to consider the flow rate when calculating heat exchange.

4. What is the specific heat capacity of water?

The specific heat capacity of water is the amount of heat energy required to raise the temperature of one gram of water by one degree Celsius. It is a constant value of 4.184 J/g°C and is used in the formula for calculating heat exchange for a water pump.

5. How can heat exchange for a water pump impact the overall system?

The heat exchange for a water pump can have a significant impact on the overall system. If the pump is not efficient and is generating a large amount of heat, it can cause the water to overheat and potentially damage the system. On the other hand, an efficient pump with minimal heat exchange can help maintain a stable temperature and prevent any potential issues. Additionally, understanding the heat exchange can also help with energy efficiency and cost savings in the long run.

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